Free math worksheets

- Find the critical times in the stated interval. - For an absolute maximum on a closed interval, compare the critical-point value with both endpoint values. - Evaluate the original height function only after identifying the candidates.

1. Differentiate: \(h'(t)=-0.3t^2+0.4t+1.5\). 2. Solving \(h'(t)=0\) gives \(t=3\) and \(t=-\frac53\). Only \(t=3\) lies in the interval \([0,5]\). 3. For the absolute maximum on the closed interval, compare the endpoint and critical-point values: \(h(0)=0\), \(h(3)=3.6\), and \(h(5)=0\). 4. Therefore, the absolute maximum height is \(3.6\,\text{km}\), reached at \(t=3\,\text{h}\).

The model reaches its absolute maximum height of \(3.6\,\text{km}\) at \(t=3\,\text{h}\).

- How do you form profit from revenue and cost? - What does a negative profit value mean? - Find the critical numbers of the profit function. - Verify that the candidate gives a maximum. - Compare with the endpoints of the allowed production interval.

1. Form the profit function: \(P(x) = 25x - (x^3 - 6x^2 + 10x + 50) = -x^3 + 6x^2 + 15x - 50\). 2. Evaluate \(P(2) = -8 + 24 + 30 - 50 = -4\). Because the units are thousands of dollars, this is a loss of \(\$4{,}000\). 3. Differentiate: \(P'(x) = -3x^2 + 12x + 15\). Solving \(P'(x) = 0\) gives \(x = -1\) and \(x = 5\). Only \(x = 5\) lies in the production interval. 4. Use \(P''(x) = -6x + 12\). Since \(P''(5) = -18 < 0\), \(x = 5\) gives a local maximum. 5. Check the endpoints: \(P(0) = -50\), \(P(5) = 50\), and \(P(7) = 6\). Therefore, the absolute maximum occurs at \(x = 5\).

a) \(P(2) = -4\), so the company loses \(\$4{,}000\). b) The maximum profit occurs when the company produces \(5\) tons.

- The quantity to optimize is already given as a function of the parameter. - Use the full closed interval when building the candidate list. - Compare the critical number with both endpoint values.

1. Differentiate: \(A'(a)=a^2-2\). 2. The only critical number in \([0,2]\) is \(a=\sqrt2\). 3. Since \(A''(a)=2a>0\) at \(a=\sqrt2\), this is a local minimum. 4. Compare the closed-interval candidates: \(A(0)=\frac83\), \(A(2)=\frac43\), and \(A(\sqrt2)=\frac{8-4\sqrt2}{3}\). 5. Therefore, the absolute minimum occurs at \(a=\sqrt2\), with value \(\frac{8-4\sqrt2}{3}\approx0.78\).

The absolute minimum occurs at \(a=\sqrt2\), and the minimum value is \(\frac{8-4\sqrt2}{3}\approx0.78\).

- Differentiate the given one-variable volume model. - Use the sign of the derivative on both sides of the critical number. - Compare the critical number with the volume's limiting behavior at both ends of the domain.

1. Differentiate: \(V'(a)=\frac{\pi(2-a)}{a^3}\). 2. The only critical number for \(a>1\) is \(a=2\). 3. Since \(V'(a)>0\) for \(1<a<2\) and \(V'(a)<0\) for \(a>2\), the volume increases and then decreases. 4. Also, \(V(a)\to0\) as \(a\to1^+\) and as \(a\to\infty\). Therefore, \(a=2\) gives the absolute maximum. 5. \(V(2)=\frac{\pi}{4}\).

The absolute maximum occurs at \(a=2\), and the maximum volume is \(\frac{\pi}{4}\).

- Substitute the given time into the function. - Use the product and chain rules to differentiate the time-dependent term. - Find where the derivative changes from positive to negative. - Exponential decay eventually dominates polynomial growth.

1. Substitute \(t=0\): \(k(0)=0.8\,\text{mg/L}\). 2. Differentiate using the product and chain rules: \(k'(t)=(12-6t)e^{-0.5t}\). Since the exponential factor is positive, the derivative is zero at \(t=2\), positive before \(2\), and negative after \(2\). Therefore, the maximum occurs after \(2\) hours. 3. The maximum concentration is \(k(2)=0.8+24e^{-1}\approx9.63\,\text{mg/L}\). 4. After \(12\) hours, \(k(12)=0.8+144e^{-6}\approx1.16\,\text{mg/L}\). 5. Since \(te^{-0.5t}\to0\), \(\lim_{t\to\infty}k(t)=0.8\,\text{mg/L}\). In the model, the concentration approaches a baseline level of \(0.8\,\text{mg/L}\).

a) \(0.8\,\text{mg/L}\). b) After \(2\) hours; maximum concentration \(0.8+\frac{24}{e}\approx9.63\,\text{mg/L}\). c) Approximately \(1.16\,\text{mg/L}\). d) \(\lim_{t\to\infty}k(t)=0.8\,\text{mg/L}\); the model approaches that baseline concentration.

- Determine where the path changes from rising to falling. - Check the derivative's sign on both sides of the critical number. - Substitute the maximizing horizontal distance into the height function.

1. Differentiate: \(h'(x)=-0.1x+0.8\). 2. Solve \(h'(x)=0\): \(-0.1x+0.8=0\), so \(x=8\,\text{m}\). 3. For \(0\le x<8\), \(h'(x)>0\), and for \(8<x\le17\), \(h'(x)<0\). Thus the modeled path rises up to \(x=8\) and falls afterward, so this point is the absolute maximum on \([0,17]\). 4. Evaluate \(h(8)=-0.05(8^2)+0.8(8)+1.5=4.7\,\text{m}\).

a) \(8\,\text{m}\) b) \(4.7\,\text{m}\) c) \(h'(x)\) changes from positive to negative at \(x=8\), so the path rises before that point and falls after it; therefore the maximum is absolute.

- Identify the function that represents marginal cost before optimizing. - Use the derivative of marginal cost to locate where its behavior changes. - Check the sign on both sides of the critical production level to decide whether the minimum is global on the feasible domain.

1. Differentiate to obtain the marginal cost: \(K'(x)=\frac{1}{100}x^2-x+40\). 2. Differentiate the marginal-cost function: \(K''(x)=\frac{1}{50}x-1\). 3. Solve \(K''(x)=0\): \(\frac{1}{50}x-1=0\), so \(x=50\). 4. For \(0\le x<50\), \(K''(x)<0\), and for \(x>50\), \(K''(x)>0\). Thus marginal cost decreases and then increases, so \(x=50\) gives its absolute minimum on \(x\ge0\). 5. Evaluate \(K'(50)=\frac{1}{100}\cdot50^2-50+40=15\).

The marginal cost is absolutely minimized at \(x=50\) units, and the minimum marginal cost is \(\$15\) per additional unit.

- Keep the launch angle within the stated first-quadrant interval. - Use the sign of the derivative on both sides of the only critical angle. - Convert the final exact angle from radians to degrees.

1. Differentiate: \(d'(\alpha)=120(\cos^2\alpha-\sin^2\alpha)=120\cos2\alpha\). 2. The only critical number in \(0<\alpha<\frac{\pi}{2}\) is \(\alpha=\frac{\pi}{4}\). 3. For \(0<\alpha<\frac{\pi}{4}\), \(\cos2\alpha>0\), so \(d'>0\). For \(\frac{\pi}{4}<\alpha<\frac{\pi}{2}\), \(\cos2\alpha<0\), so \(d'<0\). 4. Thus the range increases and then decreases across the entire feasible interval, so \(\alpha=\frac{\pi}{4}\) gives the absolute maximum. 5. \(\frac{\pi}{4}=45^\circ\).

\(\alpha=\frac{\pi}{4}=45^\circ\)

- Relate the bottom width and the two side-wall heights to the total strip width. - Write the rectangular cross-sectional area in terms of \(h\) alone. - Maximize the resulting quadratic function. - Check that the answer is physically possible.

1. Let \(b\) be the bottom width. The strip-width constraint is \(2h+b=24\), so \(b=24-2h\). 2. The cross-sectional area is \(A(h)=h(24-2h)=24h-2h^2\), for \(0\le h\le 12\). 3. Differentiate: \(A'(h)=24-4h\). The critical number is \(h=6\,\text{cm}\). Since \(A''(h)=-4<0\), the area function is concave down, so this critical number gives the absolute maximum.

The side walls should be \(6\,\text{cm}\) high.

- Use the demand model to identify the feasible sales interval. - Form profit from revenue and cost before differentiating. - Verify the critical sales quantity by checking the profit derivative on both sides over the entire feasible interval.

1. Revenue is \(R(x)=xp(x)=-0.4x^2+84x\), so profit is \(P(x)=R(x)-C(x)=-0.4x^2+72x-400\). 2. Differentiate: \(P'(x)=-0.8x+72\). The only interior critical number is \(x=90\). 3. On \(0<x<90\), \(P'(x)>0\), and on \(90<x<210\), \(P'(x)<0\). Therefore profit increases and then decreases over the full feasible interval, so \(x=90\) gives the absolute maximum. 4. The corresponding price is \(p(90)=48\), or \(\$48\) per unit. 5. The maximum profit is \(P(90)=2840\), or \(\$2840\).

The absolute maximum profit occurs at \(x=90\) units. The corresponding price is \(\$48\) per unit, and the maximum profit is \(\$2840\).

- Use the fixed wire length to find the complete allowable range of radii. - Express sector area using radius and arc length before eliminating the arc length. - Include both degenerate boundary cases when verifying the absolute maximum.

1. The boundary constraint is \(2r+b=20\), so \(b=20-2r\). Feasibility gives \(0\le r\le10\). 2. The sector area is \(A=\frac12rb\), so \(A(r)=10r-r^2\). 3. Differentiate: \(A'(r)=10-2r\). The only interior critical point is \(r=5\,\text{cm}\). 4. The derivative is positive on \((0,5)\) and negative on \((5,10)\), while \(A(0)=A(10)=0\). Therefore \(r=5\) gives the absolute maximum. 5. Then \(b=10\,\text{cm}\). Since \(b=r\varphi\), \(10=5\varphi\), so \(\varphi=2\) radians.

The absolute maximum-area sector has radius \(r=5\,\text{cm}\) and central angle \(\varphi=2\) radians.

- Solve the intersection equation before optimizing. - Differentiate the supplied area model only with respect to the parameter. - Use both derivative signs and boundary behavior to justify a global maximum.

1. Set the functions equal: \(a^2=\frac{a^2}{(6-a)^2}x^2\). Since \(a>0\), \(x^2=(6-a)^2\), so the intersections are \((-(6-a),a^2)\) and \((6-a,a^2)\). 2. Differentiate the supplied area model: \(A'(a)=16a-4a^2=4a(4-a)\). The only critical number in \((0,6)\) is \(a=4\). 3. \(A'(a)>0\) for \(0<a<4\) and \(A'(a)<0\) for \(4<a<6\), so \(A\) increases and then decreases. 4. Also \(A(a)\to0\) as \(a\to0^+\) or \(a\to6^-\). Therefore, \(a=4\) gives the absolute maximum. 5. \(A(4)=\frac{128}{3}\) square units.

a) \((-(6-a),a^2)\) and \((6-a,a^2)\) b) The absolute maximum occurs at \(a=4\), with area \(\frac{128}{3}\) square units.

- What is the distance from a point \((x, y)\) to the x-axis? - Since the graph stays above the x-axis, which values of \(f\) give the smallest distance? - Find and classify the critical points of \(f\). - Give complete ordered pairs in your answer.

1. Because \(f(x) > 0\), the distance from a point \((x, f(x))\) to the x-axis is \(f(x)\). Minimize \(f\). 2. Differentiate: \(f'(x) = 4x^3 - 4x = 4x(x - 1)(x + 1)\). The critical numbers are \(x = -1\), \(x = 0\), and \(x = 1\). 3. Use \(f''(x) = 12x^2 - 4\). The points at \(x = -1\) and \(x = 1\) are local minima, while the point at \(x = 0\) is a local maximum. 4. Since \(f(x) \to \infty\) as \(x \to \pm\infty\), the local minima are absolute minima. Evaluate \(f(-1) = 4\) and \(f(1) = 4\). 5. Therefore, the closest points are \((-1, 4)\) and \((1, 4)\).

\((-1, 4)\) and \((1, 4)\)

- When every function value is negative, is the point closest to the x-axis at a maximum or a minimum? - Examine every critical number; not every horizontal tangent is an extremum. - Use the end behavior to decide whether the local extremum is absolute.

1. Since \(f(x) < 0\), the point closest to the x-axis occurs where \(f\) has its greatest value. 2. Differentiate: \(f'(x) = -x^3 + 3x^2 = -x^2(x - 3)\). The critical numbers are \(x = 0\) and \(x = 3\). 3. At \(x = 0\), the derivative does not change sign, so this critical point is not an extremum. At \(x = 3\), the derivative changes from positive to negative, so \(f\) has a local maximum. 4. Because \(f(x) \to -\infty\) as \(x \to \pm\infty\), this local maximum is also the absolute maximum. 5. Evaluate \(f(3) = -\frac{1}{4}(3^4) + 3^3 - 10 = -\frac{13}{4} = -3.25\). The closest point is \((3, -3.25)\).

\((3, -3.25)\)

- Marginal cost is the first derivative of total cost. - Which derivative locates an extremum of marginal cost? - An inflection point marks a change in concavity. - Compare how the slope behaves on each side of the inflection-point x-coordinate.

1. The marginal-cost function is \(K'(x)=x^2-12x+45\). Its derivative is \(K''(x)=2x-12\). 2. Set \(K''(x)=0\): \(2x-12=0\), so \(x=6\). Since \(K'''(x)=2>0\), the marginal cost has a minimum at \(x=6\). 3. Evaluate the total cost: \(K(6)=72-216+270+120=246\). The inflection point is \((6, 246)\). 4. Before \(x=6\), the marginal cost is decreasing, so the total-cost graph is concave down. After \(x=6\), marginal cost is increasing, so the total-cost graph is concave up. Thus, \(x=6\) is the transition from decreasing marginal cost to increasing marginal cost.

a) \(x=6\) units b) \((6, 246)\), representing a total cost of \(\$246\). At this production level, marginal cost changes from decreasing to increasing.

- Profit equals revenue minus cost. - In this model, what quantity does \(x = 1\) represent? - Use derivatives to locate the maximum of the profit function. - Compare interior critical points with the endpoints.

1. Form the profit function: \(P(x) = 25x - (x^3 - 9x^2 + 40x + 10) = -x^3 + 9x^2 - 15x - 10\). 2. Producing \(100\) components means \(x = 1\). Then \(P(1) = -1 + 9 - 15 - 10 = -17\), so the company has a loss of \(\$17{,}000\). 3. Differentiate: \(P'(x) = -3x^2 + 18x - 15\). Solving \(P'(x) = 0\) gives \(x = 1\) and \(x = 5\). 4. Use \(P''(x) = -6x + 18\). Since \(P''(1) = 12 > 0\), \(x = 1\) is a local minimum. Since \(P''(5) = -12 < 0\), \(x = 5\) is a local maximum. 5. Compare the candidate and endpoint values: \(P(0) = -10\), \(P(5) = 15\), and \(P(8) = -66\). Thus, the absolute maximum is \(15\) thousand dollars at \(x = 5\).

a) The company loses \(\$17{,}000\) when it produces \(100\) components. b) The maximum profit is \(\$15{,}000\), achieved by producing \(500\) components.

- How does the sign of the first derivative determine whether a function is increasing? - Use the discriminant to decide whether the derivative can equal zero. - Express revenue in the same units as cost. - Profit is revenue minus cost, and its maximum must be checked on the entire interval.

1. Differentiate the cost function: \(K'(x) = \frac{1}{10}x^2 - x + 4\). 2. The discriminant of \(\frac{1}{10}x^2 - x + 4 = 0\) is \((-1)^2 - 4(\frac{1}{10})(4) = -\frac{3}{5}\). Because the quadratic has no real zeros and a positive leading coefficient, \(K'(x) > 0\) for every \(x\). Thus, \(K\) is strictly increasing. 3. Revenue, in thousands of dollars, is \(R(x) = 6.4x\). Therefore, profit is \(P(x) = R(x) - K(x) = -\frac{1}{30}x^3 + 0.5x^2 + 2.4x - 20\). 4. Differentiate: \(P'(x) = -0.1x^2 + x + 2.4\). Solving \(P'(x) = 0\) gives \(x = -2\) and \(x = 12\). Only \(x = 12\) lies in the domain. 5. Since \(P''(x) = -0.2x + 1\) and \(P''(12) = -1.4 < 0\), \(x = 12\) gives a local maximum. 6. Compare values: \(P(0) = -20\), \(P(12) = 23.2\), and \(P(20) = -\frac{116}{3}\). Therefore, the maximum profit is \(23.2\) thousand dollars, or \(\$23{,}200\), at \(x = 12\).

a) \(K'(x) = 0.1x^2 - x + 4\) is positive for all \(x\), so \(K\) is strictly increasing. b) The maximum profit is \(\$23{,}200\), achieved by producing \(12\) tons.

- Factor each function before comparing the graphs. - Analyze the sign of the difference on the interior of the interval. - Treat the supplied area model as the objective function. - Use the derivative and the behavior near both ends of the parameter interval to establish an absolute maximum.

1. Factor \(h_a(x)=x(a-x)\), so its zeros are \(x=0\) and \(x=a\). Also \(p_a(x)=\frac{x}{a}\left(1-\frac{x}{a}\right)\), so its zeros are again \(0\) and \(a\). 2. The difference is \(d_a(x)=\frac{1-a^2}{a^2}x(a-x)\). For \(0<a<1\) and \(0<x<a\), every factor is positive, so \(d_a(x)>0\). 3. Differentiate the supplied area model: \(A'(a)=\frac16(1-3a^2)\). The only critical number in \((0,1)\) is \(a=\frac1{\sqrt3}\). 4. Since \(A'(a)>0\) for \(0<a<\frac1{\sqrt3}\) and \(A'(a)<0\) for \(\frac1{\sqrt3}<a<1\), the area increases and then decreases. Also \(A(a)\to0\) at both ends of the interval, so the maximum is absolute.

a) Both functions have zeros \(x=0\) and \(x=a\). b) \(d_a(x)=\frac{1-a^2}{a^2}x(a-x)>0\) for \(0<x<a\). c) The area is maximized when \(a=\frac1{\sqrt3}=\frac{\sqrt3}{3}\).

- Differentiate the supplied one-variable model. - Keep only critical values in the stated interval. - An absolute-extremum decision on a closed interval requires endpoint comparison.

1. Differentiate: \(A'(c)=2c^2-1\). 2. The critical value in \([0,1]\) is \(c=\frac1{\sqrt2}\). 3. Since \(A''(c)=4c>0\) at this value, it is a local minimum. 4. Compare the closed-interval candidates: \(A(0)=\frac23\), \(A(1)=\frac13\), and \(A\left(\frac1{\sqrt2}\right)=\frac{2-\sqrt2}{3}\). 5. Therefore, the absolute minimum occurs at \(c=\frac1{\sqrt2}\), with value \(\frac{2-\sqrt2}{3}\approx0.195\).

The absolute minimum occurs at \(c=\frac1{\sqrt2}=\frac{\sqrt2}{2}\), and the minimum value is \(\frac{2-\sqrt2}{3}\approx0.195\).

- Differentiate the score as a quotient. - Find where the derivative changes sign. - Compare the critical number with the behavior near both ends of the domain.

1. Differentiate: \(P'(k)=\frac{1-\ln k}{k^2}\). 2. The only critical number for \(k>1\) satisfies \(1-\ln k=0\), so \(k=e\). 3. Since \(P'(k)>0\) for \(1<k<e\) and \(P'(k)<0\) for \(k>e\), \(P\) increases and then decreases. 4. Also, \(P(k)\to0\) as \(k\to1^+\) and as \(k\to\infty\). Therefore, the maximum is absolute and occurs at \(k=e\). 5. The maximum score is \(P(e)=\frac1e\).

The absolute maximum occurs at \(k=e\), with maximum score \(\frac1e\).

- First relate the total amount of fencing to the three fenced sides of the rectangle. - Rewrite the area so that it depends on only one variable. - Determine the smallest and largest possible values of the perpendicular side length. - Use the derivative to find an interior critical point. - Compare the critical-point value with the values at both endpoints.

1. Let \(x\) be the side parallel to the wall and \(y\) be each side perpendicular to the wall. The fencing constraint is \(x+2y=60\), so \(x=60-2y\). Therefore, \(A(y)=xy=(60-2y)y=60y-2y^2\). The side lengths must be nonnegative, so \(0\le y\le 30\). Thus the domain is \([0, 30]\). 2. Differentiate: \(A'(y)=60-4y\). Set the derivative equal to zero: \(60-4y=0\Rightarrow y=15\). Then \(x=60-2(15)=30\). Also, \(A''(y)=-4<0\), so the critical point is a local maximum. 3. The area is \(A(15)=30\cdot 15=450\,\text{m}^2\). At the endpoints, \(A(0)=0\) and \(A(30)=0\). Because \(A\) is continuous on \([0, 30]\) and the critical-point value is greater than both endpoint values, \(450\,\text{m}^2\) is the absolute maximum.

1. \(A(y)=60y-2y^2\), with domain \([0, 30]\). 2. The perpendicular sides should each be \(15\,\text{m}\), and the side parallel to the wall should be \(30\,\text{m}\). 3. The maximum area is \(450\,\text{m}^2\). Since \(A(0)=A(30)=0\), the interior maximum is the absolute maximum on the domain.

- Recall the perimeter formula for a square and the volume formula for a rectangular prism. - Use the given sum to express the height in terms of the base side length. - Apply the power rule when differentiating the volume function. - Check that the resulting dimensions are positive and compare with the endpoint cases.

1. The square base has perimeter \(4x\), so the constraint is \(h+4x=108\). Thus \(h=108-4x\). Since \(V=x^2h\), \(V(x)=x^2(108-4x)=108x^2-4x^3\). The physically meaningful domain is \(0\le x\le 27\). 2. Differentiate: \(V'(x)=216x-12x^2=12x(18-x)\). The interior critical point is \(x=18\). The second derivative is \(V''(x)=216-24x\), and \(V''(18)=-216<0\), so this point gives a local maximum. The corresponding height is \(h=108-4(18)=36\,\text{cm}\). The endpoint volumes are both \(0\), so this local maximum is also the absolute maximum on the domain. 3. \(V(18)=18^2\cdot 36=11{,}664\,\text{cm}^3\).

1. Using \(h=108-4x\) and \(V=x^2h\) gives \(V(x)=108x^2-4x^3\). 2. The volume is maximized when \(x=18\,\text{cm}\) and \(h=36\,\text{cm}\). 3. The maximum volume is \(11{,}664\,\text{cm}^3\).

- Express both the fee and membership count using the same change variable. - Determine which changes keep both quantities nonnegative. - Revenue is the product of price per member and number of members. - Use the shape of the revenue function to justify a global maximum.

1. Let \(x\) be the change in the monthly fee, in dollars. Then the fee is \(p(x)=45+x\), and the number of members is \(n(x)=800-10x\). 2. Requiring a nonnegative fee and membership count gives the feasible interval \(-45\le x\le80\). 3. Revenue is \(R(x)=(45+x)(800-10x)=-10x^2+350x+36{,}000\). 4. Differentiate: \(R'(x)=-20x+350\). The critical number is \(x=17.5\), which lies in the feasible interval. 5. Because \(R\) is a concave-down quadratic, this critical number gives the absolute maximum revenue on the feasible interval. 6. The optimal fee is \(\$62.50\), the membership is \(625\), and the maximum monthly revenue is \(\$39{,}062.50\).

The revenue is maximized with a monthly fee of \(\$62.50\). The center then has \(625\) members and monthly revenue of \(\$39{,}062.50\).

- Use the product constraint to eliminate one variable. - Keep the positivity condition when defining the one-variable domain. - A local classification is not enough by itself; compare with behavior near both ends of the positive domain.

1. Let the first number be \(x>0\) and the second be \(y>0\). The constraint \(xy=36\) gives \(y=\frac{36}{x}\). 2. The quantity to minimize is \(S=x+9y\), so \(S(x)=x+\frac{324}{x}\), for \(x>0\). 3. Differentiate: \(S'(x)=1-\frac{324}{x^2}\). The only positive critical number is \(x=18\). 4. Since \(S''(x)=\frac{648}{x^3}>0\) for \(x>0\), and \(S(x)\to\infty\) as \(x\to0^+\) or \(x\to\infty\), \(x=18\) gives the absolute minimum. 5. Then \(y=\frac{36}{18}=2\).

The absolute minimum occurs for first number \(18\) and second number \(2\).

- Write an equation for the sum of the two numbers. - Express the product as a function of one variable. - Determine the allowable interval for the first number. - Compare all critical points and endpoints.

1. Let the first number be \(x\) and the second number be \(y\), where \(x \ge 0\) and \(y \ge 0\). 2. Since \(x + y = 15\), write \(y = 15 - x\). 3. The product is \(P(x) = x^2(15 - x) = 15x^2 - x^3\), with \(0 \le x \le 15\). 4. Differentiate: \(P'(x) = 30x - 3x^2 = 3x(10 - x)\). The critical numbers are \(x = 0\) and \(x = 10\). 5. Evaluate the candidates: \(P(0) = 0\), \(P(10) = 500\), and \(P(15) = 0\). Therefore, the absolute maximum occurs at \(x = 10\). 6. The second number is \(y = 15 - 10 = 5\).

The first number is \(10\), and the second number is \(5\).

- Express the rectangle's area using the coordinates of \(P\). - Write an area function that depends only on \(x\). - How do you find the largest value of a function on a closed interval? - Remember to check the endpoints as well as the critical numbers.

1. The rectangle has width \(x\) and height \(f(x)=(x-6)^2\), so its area is \(A(x)=x(x-6)^2=x^3-12x^2+36x\). 2. Differentiate: \(A'(x)=3x^2-24x+36=3(x-2)(x-6)\). 3. The critical numbers in \([0,6]\) are \(x=2\) and \(x=6\). 4. Evaluate the candidates: \(A(0)=0\), \(A(2)=32\), and \(A(6)=0\). Therefore, the absolute maximum occurs at \(x=2\). 5. The corresponding height is \(f(2)=(2-6)^2=16\), so \(P=(2,16)\).

\(P=(2,16)\)

- Use the volume formula for a cylinder to express the height in terms of the radius. - Write the total surface area of a closed cylinder and substitute for the height. - Find a critical number of the one-variable surface-area function. - Verify that the critical number gives a minimum. - Use the critical-point equation to compare the height with the radius.

1. The volume constraint is \(\pi r^2h=15\), so \(h=\frac{15}{\pi r^2}\), with \(r>0\). 2. A closed cylinder has surface area \(S=2\pi r^2+2\pi rh\). Substituting for \(h\) gives \(S(r)=2\pi r^2+2\pi r\left(\frac{15}{\pi r^2}\right)=2\pi r^2+\frac{30}{r}\). 3. Differentiate: \(S'(r)=4\pi r-\frac{30}{r^2}\). Set the derivative equal to zero: \(4\pi r-\frac{30}{r^2}=0\Rightarrow 4\pi r^3=30\Rightarrow r=\sqrt[3]{\frac{7.5}{\pi}}\approx 1.337\,\text{m}\). 4. Since \(S''(r)=4\pi+\frac{60}{r^3}>0\) for \(r>0\), this critical number is a local minimum. Also, \(S(r)\to\infty\) as \(r\to 0^+\) or as \(r\to\infty\), so it is the absolute minimum. 5. The height is \(h=\frac{15}{\pi r^2}\approx 2.673\,\text{m}\). 6. At the critical number, \(4\pi r^3=30\), so \(15=2\pi r^3\). Therefore, \(h=\frac{15}{\pi r^2}=\frac{2\pi r^3}{\pi r^2}=2r\). Thus the height equals the diameter.

a) \(S(r)=2\pi r^2+\frac{30}{r}\), for \(r>0\). b) \(r=\sqrt[3]{\frac{7.5}{\pi}}\approx 1.337\,\text{m}\) and \(h\approx 2.673\,\text{m}\). c) At the minimum, \(h=2r\), so the height equals the diameter.

- Recall the distance formula for two points in the coordinate plane. - Why can you minimize the square of the distance instead of the distance itself? - Keep the domain restriction \(x\ge -5\) in mind. - Verify that the critical number gives a minimum.

1. The squared distance from \(Q\) to \(P\) is \(D(x)=(x-3)^2+(\sqrt{x+5})^2=x^2-5x+14\). 2. Minimizing the distance is equivalent to minimizing \(D(x)\), because the square-root function is increasing. 3. Differentiate: \(D'(x)=2x-5\). Setting \(D'(x)=0\) gives \(x=\frac{5}{2}\). 4. Since \(D(x)\) is an upward-opening quadratic, this critical number gives its global minimum on \([-5,\infty)\). 5. The minimum distance is \(\sqrt{D\left(\frac{5}{2}\right)}=\sqrt{\frac{31}{4}}=\frac{\sqrt{31}}{2}\approx 2.78\).

The distance is minimized at \(x=\frac{5}{2}\), and the minimum distance is \(\frac{\sqrt{31}}{2}\approx 2.78\).

- Identify which dimensions determine the rectangle’s area and which segments make up the outside perimeter. - Use the perimeter constraint to express one dimension in terms of the other. - Remember that the semicircle is attached to only one short side; the opposite short side remains part of the perimeter. - Maximize the resulting one-variable area function.

1. The area of the rectangle is \(A=bh\). Because the semicircle has diameter \(b\), its arc length is \(\frac{\pi b}{2}\). The perimeter constraint is \(b+2h+\frac{\pi b}{2}=200\). 2. Solve for \(h\): \(h=100-\left(\frac{1}{2}+\frac{\pi}{4}\right)b\). Therefore, \(A(b)=100b-\left(\frac{1}{2}+\frac{\pi}{4}\right)b^2\). 3. Differentiate: \(A'(b)=100-\left(1+\frac{\pi}{2}\right)b\). Setting the derivative equal to zero gives \(b=\frac{100}{1+\frac{\pi}{2}}=\frac{200}{2+\pi}\approx 38.90\,\text{m}\). 4. Substitute into the constraint: \(h=100-\frac{200}{2+\pi}\cdot\frac{2+\pi}{4}=50\,\text{m}\). Since \(A''(b)=-\left(1+\frac{\pi}{2}\right)<0\), the area function is concave down, so these dimensions give the absolute maximum on the physical domain.

The width should be \(b=\frac{200}{2+\pi}\approx 38.90\,\text{m}\), and the length should be \(h=50\,\text{m}\).

- Use the volume relationship among the three dimensions of a rectangular prism. - Write the surface area of a closed rectangular prism in terms of its edge lengths. - Use the length condition and the volume constraint to reduce the problem to one variable. - Find and verify the minimum of the resulting surface-area function.

1. Let the width be \(b\), the length be \(2b\), and the height be \(h\). The volume constraint is \(2b^2h=72\), so \(h=\frac{36}{b^2}\). 2. The surface area is \(S=2(2b\cdot b)+2(2b\cdot h)+2(b\cdot h)=4b^2+6bh\). Substituting for \(h\) gives \(S(b)=4b^2+\frac{216}{b}\), for \(b>0\). 3. Differentiate: \(S'(b)=8b-\frac{216}{b^2}\). Setting \(S'(b)=0\) gives \(8b^3=216\Rightarrow b^3=27\Rightarrow b=3\,\text{ft}\). 4. Since \(S''(b)=8+\frac{432}{b^3}>0\) for \(b>0\), the critical number is a local minimum. Because \(S(b)\to\infty\) as \(b\to 0^+\) or as \(b\to\infty\), it is the absolute minimum. 5. The length is \(2b=6\,\text{ft}\), and the height is \(h=\frac{36}{3^2}=4\,\text{ft}\).

The width is \(3\,\text{ft}\), the length is \(6\,\text{ft}\), and the height is \(4\,\text{ft}\).

- Write the area of the right triangle as one-half times base times height. - How is the height related to \(f(u)\)? - Use the quotient rule to differentiate the area function. - What sign change in the derivative identifies a maximum? - Check that the critical number is in the domain.

1. The triangle has base \(u\) and height \(f(u)\), so \(A(u)=\frac{1}{2}u f(u)=\frac{5\ln(u)}{u}\). 2. Differentiate using the quotient rule: \(A'(u)=\frac{5-5\ln(u)}{u^2}\). 3. Set the derivative equal to zero: \(5-5\ln(u)=0\), so \(\ln(u)=1\) and \(u=e\). 4. For \(1\le u<e\), \(A'(u)>0\), and for \(u>e\), \(A'(u)<0\). Therefore, \(A\) increases and then decreases at \(u=e\). 5. Also, \(A(1)=0\) and \(A(u)\to 0\) as \(u\to\infty\), so \(u=e\) gives the global maximum.

The triangle has maximum area when \(u=e\).

- Express the rectangle's side lengths using the coordinates of \(Q\). - Use both the product rule and the chain rule to differentiate the area function. - The exponential factor is never zero, so which factor determines the critical number? - Verify the maximum using a derivative sign change or the second derivative.

1. The rectangle has width \(u\) and height \(g(u)\), so \(A(u)=12u e^{-0.25u^2}\). 2. Differentiate using the product and chain rules: \(A'(u)=(12-6u^2)e^{-0.25u^2}\). 3. Since the exponential factor is positive, \(A'(u)=0\) when \(12-6u^2=0\). Because \(u\ge 0\), \(u=\sqrt{2}\). 4. The derivative is positive before \(\sqrt{2}\) and negative after \(\sqrt{2}\), so this critical number gives the maximum. 5. The maximum area is \(A(\sqrt{2})=12\sqrt{2}e^{-1/2}\approx 10.29\).

The area is maximized at \(u=\sqrt{2}\), and the maximum area is \(12\sqrt{2}e^{-1/2}\approx 10.29\).

- A maximum can occur where the first derivative changes from positive to negative. - Factor the derivative before solving for critical times. - Keep the exponential factor when differentiating. - Exponential decay eventually dominates a polynomial factor.

1. Substitute \(t=0\): \(f(0)=15\,^{\circ}\text{C}\). 2. Differentiate: \(f'(t)=(0.4t-0.02t^2)e^{-0.1t}=t(0.4-0.02t)e^{-0.1t}\). 3. For \(t>0\), the derivative is zero at \(t=20\), positive before \(20\), and negative after \(20\). Therefore, the maximum occurs after \(20\) seconds. 4. The maximum temperature is \(f(20)=15+80e^{-2}\approx25.83\,^{\circ}\text{C}\). 5. After \(100\) seconds, \(f(100)=15+2000e^{-10}\approx15.09\,^{\circ}\text{C}\). 6. Since \(t^2e^{-0.1t}\to0\), \(\lim_{t\to\infty}f(t)=15\,^{\circ}\text{C}\). In the model, the component temperature approaches \(15\,^{\circ}\text{C}\).

a) \(15\,^{\circ}\text{C}\). b) After \(20\) seconds; approximately \(25.83\,^{\circ}\text{C}\). c) Approximately \(15.09\,^{\circ}\text{C}\). d) \(\lim_{t\to\infty}f(t)=15\,^{\circ}\text{C}\); the modeled temperature approaches \(15\,^{\circ}\text{C}\).

- Distinguish the fill height from the total can height. - Reduce the surface-area model to the positive radius domain. - Use convexity and end behavior to upgrade the critical number from a local to an absolute minimum.

1. Since \(800\,\text{mL}=800\,\text{cm}^3\), the fill height is \(h_{\text{fill}}=\frac{800}{\pi r^2}\), and the total height is \(h=\frac{800}{\pi r^2}+2\), with \(r>0\). 2. The surface area is \(S(r)=2\pi r^2+\frac{1600}{r}+4\pi r\). 3. Differentiate: \(S'(r)=4\pi r-\frac{1600}{r^2}+4\pi\). The critical-number equation is \(\pi r^3+\pi r^2-400=0\), whose positive solution is \(r\approx4.719\,\text{cm}\). 4. Since \(S''(r)=4\pi+\frac{3200}{r^3}>0\) for every \(r>0\), \(S'\) is strictly increasing, so there is only one critical number. Also \(S(r)\to\infty\) as \(r\to0^+\) or \(r\to\infty\). Therefore, this critical number gives the absolute minimum. 5. The corresponding total height is \(h\approx13.437\,\text{cm}\).

a) \(S(r)=2\pi r^2+\frac{1600}{r}+4\pi r\), for \(r>0\) b) \(r\approx4.719\,\text{cm}\) and \(h\approx13.437\,\text{cm}\)

- Express the water depth and total height separately before writing the material area. - Keep the base side length positive. - Convexity plus end behavior can establish an absolute minimum on the whole positive domain.

1. Write the clearance as \(0.2\,\text{m}\). If \(h_w\) is the water depth, \(a^2h_w=2\), so \(h_w=\frac2{a^2}\) and \(H=\frac2{a^2}+0.2\), with \(a>0\). 2. The open-top material area is \(A(a)=a^2+4aH=a^2+\frac8a+0.8a\). 3. Differentiate: \(A'(a)=2a-\frac8{a^2}+0.8\). The critical-number equation is \(a^3+0.4a^2-4=0\), whose positive solution is \(a\approx1.465\,\text{m}\). 4. Since \(A''(a)=2+\frac{16}{a^3}>0\) for \(a>0\), the area function is strictly convex. Also \(A(a)\to\infty\) as \(a\to0^+\) or \(a\to\infty\). Thus the critical number gives the absolute minimum. 5. The corresponding total height is \(H\approx1.132\,\text{m}\).

The absolute minimum occurs at \(a\approx1.465\,\text{m}\) and \(H\approx1.132\,\text{m}\).

- Find the horizontal base length and vertical height of the triangle. - Use \(A=\frac{1}{2}bh\). - Express the point coordinates and height in terms of \(k\). - Differentiate the area function to locate its largest value. - Check the derivative sign and the behavior near the ends of the domain.

1. For \(k=1\), \(f(1)=\frac{1}{5}\). The base \(P_1Q_1\) has length \(2\), and the height from \(R\) is \(1-\frac{1}{5}=\frac{4}{5}\). Thus the area is \(\frac{1}{2}\cdot 2\cdot\frac{4}{5}=\frac{4}{5}\). 2. In general, the base has length \(2k\), and the height is \(1-f(k)=1-\frac{k^2}{k^2+4}=\frac{4}{k^2+4}\). 3. Therefore, \(A(k)=\frac{1}{2}(2k)\left(\frac{4}{k^2+4}\right)=\frac{4k}{k^2+4}\). 4. Differentiate: \(A'(k)=\frac{16-4k^2}{(k^2+4)^2}\). 5. For \(k>0\), the only critical number is \(k=2\). The derivative is positive for \(0<k<2\) and negative for \(k>2\), so this is a maximum. 6. Since \(A(k)\to 0\) as \(k\to 0^+\) and as \(k\to\infty\), the maximum is global. Its value is \(A(2)=1\).

a) The area for \(k=1\) is \(\frac{4}{5}=0.8\), and \(A(k)=\frac{4k}{k^2+4}\). b) The maximum occurs at \(k=2\), and the maximum area is \(1\).

- Express the triangle's width in terms of its height \(k\). - Use the triangle area formula. - Use the parabola's symmetry to find the full base length. - Confirm that the maximizing value lies in \(0<k<12\).

1. The intersections satisfy \(k=12-x^2\), so \(x=\pm\sqrt{12-k}\). 2. The horizontal base \(PQ\) has length \(2\sqrt{12-k}\), and the height from the origin to \(y=k\) is \(k\). 3. Therefore, \(A(k)=\frac{1}{2}\left(2\sqrt{12-k}\right)k=k\sqrt{12-k}\), for \(0<k<12\). 4. Differentiate: \(A'(k)=\frac{24-3k}{2\sqrt{12-k}}\). 5. The derivative is zero at \(k=8\), positive before \(8\), and negative after \(8\). Also, \(A(k)\to 0\) as \(k\to 0^+\) and as \(k\to 12^-\), so \(k=8\) gives the global maximum. 6. The maximum area is \(A(8)=8\sqrt{4}=16\).

The area is maximized when \(k=8\), and the maximum area is \(16\) square units.

- Find the equation of the line through \(B\) and \(C\). - Express the horizontal length \(DE\) in terms of \(k\). - The height of triangle \(DEM\) relative to base \(DE\) is the distance from \(M\) to the line \(y=k\). - Consider whether the x-coordinate of \(M\) affects that distance.

1. The line through \(B=(12, 0)\) and \(C=(0, 6)\) has slope \(-\frac{1}{2}\), so the hypotenuse has equation \(y=-\frac{1}{2}x+6\). 2. Since \(D\) lies on the y-axis, \(D=(0, k)\). At point \(E\), \(k=-\frac{1}{2}x+6\), so \(x=12-2k\) and \(E=(12-2k, k)\). 3. The horizontal base \(DE\) has length \(12-2k\). The perpendicular distance from \(M\) to the line \(y=k\) is \(k\). Thus \(A(k)=\frac{1}{2}(12-2k)k=6k-k^2\), for \(0<k<6\). 4. Differentiate: \(A'(k)=6-2k\). Setting the derivative equal to zero gives \(k=3\). Since \(A''(k)=-2<0\), the function is concave down and \(k=3\) gives the absolute maximum.

Segment \(DE\) should be drawn at \(k=3\).

- Use similar triangles in a cross section through the cone’s axis. - Express the cylinder height in terms of its radius. - Maximize the resulting one-variable cylinder-volume function. - Divide the maximum cylinder volume by the cone volume.

1. Similar triangles in an axial cross section give \(\frac{h}{H}=\frac{R-r}{R}\), so \(h=H\left(1-\frac{r}{R}\right)\). 2. The cylinder volume is \(V(r)=\pi r^2H\left(1-\frac{r}{R}\right)=\pi H\left(r^2-\frac{r^3}{R}\right)\), for \(0\le r\le R\). 3. Differentiate: \(V'(r)=\pi H\left(2r-\frac{3r^2}{R}\right)=\pi Hr\left(2-\frac{3r}{R}\right)\). The interior critical number is \(r=\frac{2R}{3}\). Since the volume is \(0\) at \(r=0\) and \(r=R\), this critical number gives the absolute maximum. 4. The corresponding height is \(h=H\left(1-\frac{2}{3}\right)=\frac{H}{3}\). 5. The maximum cylinder volume is \(V_{\text{cyl}}=\pi\left(\frac{2R}{3}\right)^2\left(\frac{H}{3}\right)=\frac{4}{27}\pi R^2H\). The cone volume is \(V_{\text{cone}}=\frac{1}{3}\pi R^2H\), so \(\frac{V_{\text{cyl}}}{V_{\text{cone}}}=\frac{4}{9}\).

a) \(r=\frac{2R}{3}\) and \(h=\frac{H}{3}\). b) The cylinder occupies at most \(\frac{4}{9}\), or about \(44.4\%\), of the cone’s volume.

- Express the box’s length, width, and height in terms of the corner-square size \(x\). - Determine the physically meaningful domain for \(x\). - Differentiate the volume function and discard critical numbers outside the domain. - Compare the remaining critical-number value with the endpoint values.

1. The box dimensions are \(48-2x\), \(30-2x\), and \(x\). Thus \(V(x)=x(48-2x)(30-2x)\). The physical domain is \(0\le x\le 15\). 2. Expand: \(V(x)=4x^3-156x^2+1440x\). Differentiate: \(V'(x)=12x^2-312x+1440=12(x^2-26x+120)\). The critical numbers are \(x=6\) and \(x=20\). Only \(x=6\) lies in the physical domain. 3. The endpoint volumes are \(V(0)=V(15)=0\), while \(V(6)=6(36)(18)=3888\,\text{cm}^3\). Therefore, \(x=6\,\text{cm}\) gives the absolute maximum.

The squares should have side length \(x=6\,\text{cm}\). The maximum volume is \(3888\,\text{cm}^3\).

- At equilibrium, the population does not change from one year to the next. - Write natural growth as a function of the current population. - Find the maximum of that quadratic function. - Compare the fixed harvest with the largest possible natural increase.

1. At equilibrium, \(B_{n+1} = B_n\), so the harvest must equal the natural annual growth: \(A = G(B) = 0.12B\left(1 - \frac{B}{12500}\right)\). 2. Differentiate: \(G'(B) = 0.12 - \frac{0.24}{12500}B\). 3. Set \(G'(B) = 0\): \(B = 6250\). Because \(G\) is a downward-opening quadratic, this value gives the maximum growth. 4. Evaluate \(G(6250) = 0.12(6250)\left(1 - \frac{6250}{12500}\right) = 375\). 5. Therefore, the greatest sustainable fixed harvest in this model is \(375\) trees per year, with equilibrium population \(6250\).

The maximum fixed harvest is \(375\) trees per year. At that harvest level, the equilibrium population is \(6250\) trees.

- Profit equals revenue minus cost. - Use the first derivative to find candidates for maximum profit. - Use the second derivative or a sign chart to verify the maximum. - A loss occurs where the profit function is negative. - Approximate the break-even points numerically.

1. Revenue is \(R(x)=500x\), so profit is \(P(x)=R(x)-C(x)=-x^3+30x^2-1000\). 2. Differentiate: \(P'(x)=-3x^2+60x=-3x(x-20)\). The critical numbers for \(x\ge 0\) are \(x=0\) and \(x=20\). 3. Since \(P''(20)=-60<0\), and \(P\) increases on \((0,20)\) and decreases for \(x>20\), the global maximum occurs at \(x=20\). 4. The maximum profit is \(P(20)=-8000+12{,}000-1000=3000\), or \(\$3000\). 5. Numerically, the nonnegative break-even points are \(x\approx 6.53\) and \(x\approx 28.79\). Thus the continuous model gives a loss for \(0\le x<6.53\) and for \(x>28.79\). 6. For whole-number production levels, the company loses money at \(x=0,1,\ldots,6\) and at \(x\ge 29\).

a) \(P(x)=-x^3+30x^2-1000\) b) The maximum occurs at \(x=20\) units, with a profit of \(\$3000\). c) In the continuous model, profit is negative for \(0\le x<6.53\) and \(x>28.79\). For whole units, there is a loss at \(0\) through \(6\) units and at \(29\) or more units.

- Use the nonnegative production domain when interpreting derivative roots. - A derivative sign chart over the full feasible domain can establish an absolute maximum directly. - Evaluate the profit function at the requested production levels only after the model is formed.

1. Revenue is \(600x\), so \(G(x)=-0.05x^3+6x^2+135x-4700\). 2. Differentiate: \(G'(x)=-0.15x^2+12x+135=-0.15(x+10)(x-90)\). On the feasible domain \(x\ge0\), \(G'>0\) for \(0\le x<90\) and \(G'<0\) for \(x>90\). 3. Therefore, \(G\) increases and then decreases on the entire feasible domain, so the absolute maximum occurs at \(x=90\). 4. \(G(90)=\$19{,}600\). 5. \(G(10)=-2800\), so producing \(10\,\text{kg}\) gives a loss of \(\$2800\).

a) \(G(x)=-0.05x^3+6x^2+135x-4700\) b) \(90\,\text{kg}\); absolute maximum profit \(\$19{,}600\) c) Loss of \(\$2800\)

- First distinguish total cost from the marginal-cost function you must optimize. - Determine where marginal cost changes from decreasing to increasing. - For the last part, combine the sign information for the first two derivatives with the nonnegative production domain.

1. The marginal-cost function is \(K'(x)=0.6x^2-24x+300\). 2. Differentiate again: \(K''(x)=1.2x-24\). The only critical number of the marginal-cost function is \(x=20\). 3. For \(0\le x<20\), \(K''(x)<0\), and for \(x>20\), \(K''(x)>0\). Thus marginal cost decreases and then increases, so its absolute minimum occurs at \(x=20\). The minimum value is \(K'(20)=60\), or \(\$60\) per additional unit. 4. Since \(K'(x)=0.6(x-20)^2+60>0\), total cost is increasing for every feasible production level. 5. Total cost increases at a decreasing rate where \(K''(x)<0\), which on the feasible domain is \(0\le x<20\).

a) \(K'(x)=0.6x^2-24x+300\) b) At \(x=20\) units; the absolute minimum marginal cost is \(\$60\) per additional unit. c) \(0\le x<20\).

- Fixed cost is the cost when production is zero. - Marginal cost is the first derivative of total cost. - Minimize the marginal-cost function. - Show that the first derivative is always positive.

1. The fixed cost is the cost at zero production: \(K(0)=\$60\). 2. The marginal-cost function is \(K'(x)=x^2-8x+21\). Its derivative is \(K''(x)=2x-8\), which is zero at \(x=4\). Since \(K'''(x)=2>0\), marginal cost is minimized at \(x=4\). 3. Complete the square: \(K'(x)=x^2-8x+21=(x-4)^2+5>0\) for every real \(x\). Therefore, \(K\) is strictly increasing on \([0, \infty)\).

a) \(\$60\) b) \(x=4\) production units c) \(K'(x)=(x-4)^2+5>0\), so \(K\) is strictly increasing for all \(x\ge0\).

- No fertilizer corresponds to \(x=0\). - Find the absolute maximum on the stated interval. - Marginal yield is the first derivative. - Maximize the first derivative to find the greatest marginal increase.

1. With no fertilizer, \(x=0\), so \(f(0)=45\) bushels per acre. 2. The derivative is \(f'(x)=-0.00015x^2+0.012x+0.3\). Solving \(f'(x)=0\) gives \(x=-20\) and \(x=100\), so the relevant critical number is \(x=100\). Since \(f''(100)=-0.018<0\), this is a local maximum. Checking the endpoints confirms that the maximum modeled yield is \(f(100)=85\) bushels per acre. 3. The marginal increase in yield is \(f'(x)\). It is greatest where \(f''(x)=-0.0003x+0.012=0\), which gives \(x=40\) pounds per acre.

a) \(45\) bushels per acre b) \(100\) pounds per acre; \(85\) bushels per acre c) \(40\) pounds per acre

- Identify the fixed surface area and the volume to be maximized. - Write the surface area of a closed rectangular box with a square base. - Use the surface-area constraint to eliminate the height. - Find and verify the maximum of the one-variable volume function.

1. Let \(a\) be the side length of the square base and \(h\) be the height. The volume is \(V=a^2h\), and the surface-area constraint is \(2a^2+4ah=150\). 2. Solve for \(h\): \(h=\frac{150-2a^2}{4a}=\frac{37.5}{a}-\frac{a}{2}\). Positive dimensions require \(0<a<\sqrt{75}\). 3. Substitute into the volume: \(V(a)=37.5a-\frac{1}{2}a^3\). Differentiate: \(V'(a)=37.5-1.5a^2\). The positive critical number satisfies \(a^2=25\), so \(a=5\,\text{cm}\). 4. Since \(V''(a)=-3a<0\) for \(a>0\), the critical number gives a local maximum. The volume approaches \(0\) at both ends of the physical domain, so it is the absolute maximum. 5. The height is \(h=\frac{150-2(5)^2}{4(5)}=5\,\text{cm}\).

The base side length is \(5\,\text{cm}\), and the height is \(5\,\text{cm}\). The maximizing box is a cube.

- Write the area as a function of \(x\). - Use the derivative to find critical numbers inside the interval. - Evaluate the area at every critical number and at both endpoints. - An absolute maximum can occur at more than one input.

1. The area function is \(A(x)=x((x-2)^2+1)=x^3-4x^2+5x\). 2. Differentiate: \(A'(x)=3x^2-8x+5=(3x-5)(x-1)\). 3. The critical numbers are \(x=1\) and \(x=\frac{5}{3}\). 4. Evaluate the area at the critical numbers and endpoints: \(A(0)=0\), \(A(1)=2\), \(A\left(\frac{5}{3}\right)=\frac{50}{27}\), and \(A(2)=2\). 5. The largest value is \(2\), attained at both \(x=1\) and \(x=2\).

The maximum area is \(2\) square units, attained at \(x=1\) and \(x=2\).

- Use the fixed volume to eliminate one dimension from the material-area expression. - Check the entire positive domain, not only the critical number. - Once the critical-number relation is known, use it to simplify the ratio of the dimensions.

1. The material area is \(A=a^2+4ah\), and the volume constraint is \(V=a^2h\). Thus \(h=\frac{V}{a^2}\). 2. Substitute into the area formula: \(A(a)=a^2+\frac{4V}{a}\), for \(a>0\). 3. Differentiate: \(A'(a)=2a-\frac{4V}{a^2}\). Setting \(A'(a)=0\) gives \(a^3=2V\). 4. Since \(A''(a)=2+\frac{8V}{a^3}>0\) for \(a>0\), \(A\) is strictly convex. Also, \(A(a)\to\infty\) as \(a\to0^+\) and as \(a\to\infty\). Therefore the critical number gives the absolute minimum. 5. Using \(a^3=2V\), \(h=\frac{V}{a^2}=\frac{a}{2}\). Hence \(a:h=2:1\).

The material is absolutely minimized when \(a=2h\), so the optimal ratio is \(a:h=2:1\).

- Write a formula for the total enclosed area. - Account for all three perpendicular fence segments in the fencing constraint. - Use the constraint to eliminate one variable. - Find the vertex of the resulting area function and verify that it is a maximum.

1. The area is \(A=xy\), and the fencing constraint is \(3x+y=120\). Thus \(y=120-3x\). 2. Substitute into the area: \(A(x)=x(120-3x)=120x-3x^2\), for \(0\le x\le 40\). 3. Differentiate: \(A'(x)=120-6x\). The critical number is \(x=20\,\text{m}\). Since \(A''(x)=-6<0\), the area function is concave down, so this point gives the absolute maximum. 4. The corresponding length is \(y=120-3(20)=60\,\text{m}\).

The maximum area occurs when \(x=20\,\text{m}\) and \(y=60\,\text{m}\).

- Distinguish the printed dimensions from the full poster dimensions. - Use the fixed printed area to express one printed dimension in terms of the other. - After finding a critical number, check the behavior of the area over the entire positive domain.

1. Let \(x\) and \(y\) be the width and height of the printed region. Then \(xy=384\), while the full poster dimensions are \(x+4\) and \(y+6\). 2. The total area is \(A=(x+4)(y+6)\). Using \(y=\frac{384}{x}\), \(A(x)=6x+\frac{1536}{x}+408\), for \(x>0\). 3. Differentiate: \(A'(x)=6-\frac{1536}{x^2}\). The only positive critical number satisfies \(x^2=256\), so \(x=16\,\text{cm}\). 4. Since \(A''(x)=\frac{3072}{x^3}>0\) for \(x>0\), \(A\) is strictly convex. Also, \(A(x)\to\infty\) as \(x\to0^+\) and as \(x\to\infty\). Therefore the critical number gives the absolute minimum. 5. The printed height is \(y=\frac{384}{16}=24\,\text{cm}\), so the overall dimensions are \(20\,\text{cm}\) by \(30\,\text{cm}\).

The absolute minimum paper area occurs for a poster \(20\,\text{cm}\) wide and \(30\,\text{cm}\) high.

- Separate the exterior perimeter from the shared interior edge. - Use the perimeter condition to reduce the model to one variable. - Determine the feasible interval from the requirement that both dimensions be positive. - The shape of the resulting quadratic can establish a global maximum.

1. The area is \(A=bh+\frac{\sqrt3}{4}b^2\). The shared edge is internal, so the outside perimeter is \(3b+2h=300\), giving \(h=150-\frac32b\). 2. Physical dimensions require \(b>0\) and \(h>0\), so \(0<b<100\). 3. Substitute: \(A(b)=150b-\frac{6-\sqrt3}{4}b^2\). 4. Differentiate: \(A'(b)=150-\frac{6-\sqrt3}{2}b\). The critical number is \(b=\frac{300}{6-\sqrt3}\approx70.29\,\text{cm}\). 5. Since \(A\) is a concave-down quadratic on the feasible interval, this critical number gives the absolute maximum. 6. The corresponding height is \(h=150-\frac32\left(\frac{300}{6-\sqrt3}\right)=\frac{450-150\sqrt3}{6-\sqrt3}\approx44.56\,\text{cm}\).

The area is maximized when \(b=\frac{300}{6-\sqrt3}\approx70.29\,\text{cm}\) and \(h=\frac{450-150\sqrt3}{6-\sqrt3}\approx44.56\,\text{cm}\).

- Use the equation of the diagonal cut to relate \(x\) and \(y\). - Write the rectangle’s area as a function of \(x\). - Find the critical number within \([6, 16]\). - Compare the critical-number area with both endpoint areas.

1. Along the cut edge, \(y=-0.5x+13\). The area of a rectangle whose upper-right corner lies on this edge is \(A(x)=x(-0.5x+13)=-0.5x^2+13x\), for \(6\le x\le 16\). 2. Differentiate: \(A'(x)=-x+13\). The critical number is \(x=13\). Since \(A''(x)=-1<0\), it gives a local maximum. 3. The corresponding height is \(y=-0.5(13)+13=6.5\). For \(0\le x\le6\), the full sheet height \(10\,\text{in}\) is available, so the area is \(10x\le60\). On the cut-edge interval, the endpoint areas are \(A(6)=60\) and \(A(16)=80\), while \(A(13)=84.5\). Therefore, the absolute maximum over the entire remaining sheet occurs at \(x=13\,\text{in}\) and \(y=6.5\,\text{in}\).

The maximizing side lengths are \(x=13\,\text{in}\) and \(y=6.5\,\text{in}\).

- Account for the different material rates before eliminating a variable. - Use the fixed volume to reduce the cost to one positive variable. - After locating a critical number, check the behavior of the cost over the entire positive domain.

1. Let the side-wall material cost \(c>0\) dollars per square foot. The base and top have total area \(2x^2\) at cost rate \(2c\), while the four walls have total area \(4xh\) at cost rate \(c\). Thus \(C=4cx^2+4cxh\). Since \(c>0\), minimize \(F=4x^2+4xh\). 2. The volume constraint \(x^2h=54\) gives \(h=\frac{54}{x^2}\), so \(F(x)=4x^2+\frac{216}{x}\) for \(x>0\). 3. Differentiate: \(F'(x)=8x-\frac{216}{x^2}\). The only positive critical number satisfies \(8x^3=216\), so \(x=3\,\text{ft}\). 4. Since \(F''(x)=8+\frac{432}{x^3}>0\) for \(x>0\), \(F\) is strictly convex. Also, \(F(x)\to\infty\) as \(x\to0^+\) and as \(x\to\infty\). Therefore \(x=3\) gives the absolute minimum. 5. The corresponding height is \(h=\frac{54}{3^2}=6\,\text{ft}\).

The absolute minimum material cost occurs when \(x=3\,\text{ft}\) and \(h=6\,\text{ft}\).

- Use the side relationships in an isosceles right triangle to write the perimeter constraint. - Convert each absorption rate to a transmission rate. - Weight each region’s area by its transmission rate. - Substitute the perimeter constraint and maximize the resulting quadratic function.

1. If the triangle’s hypotenuse is \(x\), each leg has length \(\frac{x}{\sqrt{2}}\). The perimeter constraint is \(x+2y+\sqrt{2}x=8\), so \(y=4-\frac{1+\sqrt{2}}{2}x\). 2. The rectangle area is \(xy\). The triangle area is \(\frac{1}{2}\left(\frac{x}{\sqrt{2}}\right)^2=\frac{x^2}{4}\). The transmitted-light objective is proportional to \(L=0.9xy+0.6\left(\frac{x^2}{4}\right)\). 3. Substitute for \(y\): \(L(x)=3.6x-(0.3+0.45\sqrt{2})x^2\). Differentiate: \(L'(x)=3.6-(0.6+0.9\sqrt{2})x\). The critical number is \(x=\frac{3.6}{0.6+0.9\sqrt{2}}\approx 1.922\,\text{m}\). 4. Because the quadratic coefficient is negative, this critical number gives the absolute maximum. The corresponding rectangle height is \(y=4-\frac{1+\sqrt{2}}{2}(1.922263\ldots)\approx 1.680\,\text{m}\).

The transmitted light is maximized when \(x\approx 1.922\,\text{m}\) and \(y\approx 1.680\,\text{m}\).

- Use the volume constraint to eliminate the height. - Apply each material rate only to the surfaces made from that material. - Verify the minimizing critical point against the full positive domain.

1. The volume constraint \(x^2h=6\) gives \(h=\frac{6}{x^2}\), with \(x>0\). 2. The base costs \(12x^2\), and the four side walls cost \(8(4xh)=32xh\). Thus \(C(x)=12x^2+\frac{192}{x}\). 3. Differentiate: \(C'(x)=24x-\frac{192}{x^2}\). The only positive critical point satisfies \(24x^3=192\), so \(x=2\,\text{m}\). 4. Since \(C''(x)=24+\frac{384}{x^3}>0\) for \(x>0\), \(C\) is strictly convex. Also, \(C(x)\to\infty\) as \(x\to0^+\) and as \(x\to\infty\). Therefore \(x=2\) gives the absolute minimum. 5. The corresponding height is \(h=\frac{6}{2^2}=1.5\,\text{m}\).

The absolute minimum material cost occurs when \(x=2\,\text{m}\) and \(h=1.5\,\text{m}\).

- Use similar triangles in an axial cross section to relate \(r\) and \(h\). - The lateral area of a cylinder is circumference times height. - Maximize the resulting quadratic function of \(h\). - Use the cone’s slant height to find its lateral surface area.

1. Similar triangles give \(\frac{r}{R}=\frac{H-h}{H}\), so \(r=R\left(1-\frac{h}{H}\right)\). Therefore, \(M(h)=2\pi rh=2\pi R\left(h-\frac{h^2}{H}\right)\), for \(0\le h\le H\). 2. Differentiate: \(M'(h)=2\pi R\left(1-\frac{2h}{H}\right)\). The critical point is \(h=\frac{H}{2}\). Since \(M''(h)=-\frac{4\pi R}{H}<0\), this point gives the absolute maximum. 3. For \(R=5\) and \(H=12\), the cone’s slant height is \(s=\sqrt{5^2+12^2}=13\,\text{cm}\), so its lateral area is \(M_{\text{cone}}=\pi Rs=65\pi\,\text{cm}^2\). At \(h=6\,\text{cm}\), the cylinder radius is \(r=2.5\,\text{cm}\), and its lateral area is \(M_{\text{cyl}}=2\pi(2.5)(6)=30\pi\,\text{cm}^2\). Thus the fraction is \(\frac{M_{\text{cyl}}}{M_{\text{cone}}}=\frac{6}{13}\approx 46.2\%\).

1. \(M(h)=2\pi R\left(h-\frac{h^2}{H}\right)\). 2. The maximum occurs at \(h=\frac{H}{2}\). 3. The fraction is \(\frac{6}{13}\), or about \(46.2\%\).

- Use an axial cross section to relate the cylinder dimensions to the hemisphere radius. - Reduce the volume to one variable over its full feasible interval. - Check the derivative on both sides of the interior critical point and compare with the boundary cases.

1. In an axial cross section, \(r^2+h^2=100\), so \(r^2=100-h^2\), with \(0\le h\le10\). 2. The cylinder volume is \(V(h)=\pi(100-h^2)h=100\pi h-\pi h^3\). 3. Differentiate: \(V'(h)=100\pi-3\pi h^2\). The interior critical point is \(h=\frac{10}{\sqrt3}\). 4. On \(0<h<\frac{10}{\sqrt3}\), \(V'(h)>0\); on \(\frac{10}{\sqrt3}<h<10\), \(V'(h)<0\). Also \(V(0)=V(10)=0\). Thus this critical point gives the absolute maximum. 5. The corresponding radius is \(r=\sqrt{100-\frac{100}{3}}=10\sqrt{\frac23}\approx8.16\,\text{cm}\), while \(h\approx5.77\,\text{cm}\).

The maximum-volume cylinder has \(h=\frac{10}{\sqrt3}\approx5.77\,\text{cm}\) and \(r=10\sqrt{\frac23}\approx8.16\,\text{cm}\).

- Use similar triangles to relate the rectangle dimensions. - Determine the full range of possible rectangle widths before optimizing. - Check how the area changes on both sides of its interior critical point.

1. Similar triangles show that the horizontal width available at height \(b\) is \(a=12\left(1-\frac{b}{9}\right)\), so \(b=9-\frac34a\). Feasibility gives \(0\le a\le12\). 2. The area is \(A(a)=a\left(9-\frac34a\right)=9a-\frac34a^2\). 3. Differentiate: \(A'(a)=9-\frac32a\). The only interior critical point is \(a=6\,\text{cm}\). 4. Since \(A'(a)>0\) for \(0<a<6\) and \(A'(a)<0\) for \(6<a<12\), the area increases and then decreases on the full feasible interval. Therefore \(a=6\) gives the absolute maximum. 5. The corresponding height is \(b=9-\frac34(6)=4.5\,\text{cm}\).

The absolute maximum area occurs when \(a=6\,\text{cm}\) and \(b=4.5\,\text{cm}\).

- Use the axial cross-section to relate the radius, height, and fixed slant height. - State the physically meaningful interval before optimizing. - Compare the interior critical point with the boundary cases of the feasible interval.

1. The cone volume is \(V=\frac13\pi r^2h\). The cross-section gives \(r^2+h^2=144\), so \(r^2=144-h^2\), with \(0\le h\le12\). 2. Substitute: \(V(h)=\frac{\pi}{3}(144h-h^3)\). 3. Differentiate: \(V'(h)=\frac{\pi}{3}(144-3h^2)\). The positive interior critical point satisfies \(h^2=48\), so \(h=4\sqrt3\approx6.93\,\text{cm}\). 4. Since \(V''(h)=-2\pi h<0\) for \(h>0\), the critical point is a local maximum. It is the only interior critical point, and \(V(0)=V(12)=0\), so it is the absolute maximum on the feasible interval. 5. The corresponding radius is \(r=\sqrt{144-48}=4\sqrt6\approx9.80\,\text{cm}\).

The maximum-volume cone has \(r=4\sqrt6\approx9.80\,\text{cm}\) and \(h=4\sqrt3\approx6.93\,\text{cm}\).

- Express both rectangle dimensions using the same input. - Check the behavior of the perimeter near both ends of the positive domain. - For the final part, decide whether the area remains bounded as the input grows.

1. The side lengths are \(x\) and \(f(x)\), so \(p(x)=2(x+f(x))=4x+\frac{16}{x}\), for \(x>0\). 2. Differentiate: \(p'(x)=4-\frac{16}{x^2}\). The only critical number in the feasible domain is \(x=2\). 3. Since \(p''(x)=\frac{32}{x^3}>0\) for \(x>0\), and \(p(x)\to\infty\) as \(x\to0^+\) or \(x\to\infty\), \(x=2\) gives the absolute minimum perimeter. 4. The side lengths are \(2\) and \(f(2)=6\), so the minimum perimeter is \(16\). 5. The area is \(A(x)=xf(x)=x^2+8\). Since \(A(x)\to\infty\) as \(x\to\infty\), no maximum area exists.

a) Side lengths \(2\) and \(6\) b) Minimum perimeter \(16\) c) No maximum area exists because \(A(x)=x^2+8\) is unbounded above on \(x>0\).

- Set the equations of the line and the curve equal to find the intersections. - Recall the distance formula between two points. - To minimize a square root, you can minimize its radicand. - Use the condition \(m>0\) when solving for the slope.

1. At an intersection, \(mx=\frac{9}{x}\), so \(x^2=\frac{9}{m}\). Therefore, \(x=\pm\frac{3}{\sqrt{m}}\). 2. Since \(y=mx\), the points are \(P_1=\left(\frac{3}{\sqrt{m}},3\sqrt{m}\right)\) and \(P_2=\left(-\frac{3}{\sqrt{m}},-3\sqrt{m}\right)\). 3. To minimize \(d(m)\), minimize \(h(m)=m+\frac{1}{m}\) for \(m>0\). 4. Differentiate: \(h'(m)=1-\frac{1}{m^2}\). The only critical number in the domain is \(m=1\). 5. Since \(h''(m)=\frac{2}{m^3}>0\), and \(h(m)\to\infty\) as \(m\to 0^+\) or \(m\to\infty\), this gives the global minimum. 6. When \(m=1\), the points are \(P_1=(3,3)\) and \(P_2=(-3,-3)\), and \(d(1)=\sqrt{72}=6\sqrt{2}\approx 8.49\).

a) \(P_1=\left(\frac{3}{\sqrt{m}},3\sqrt{m}\right)\) and \(P_2=\left(-\frac{3}{\sqrt{m}},-3\sqrt{m}\right)\) b) \(m=1\) c) \(P_1=(3,3)\), \(P_2=(-3,-3)\), and the minimum distance is \(6\sqrt{2}\approx 8.49\).

- Use the demand restriction to identify the feasible sales interval. - Express revenue as quantity times price before differentiating. - Verify the critical quantity by checking the direction of change across the whole feasible interval.

1. Revenue is quantity times price: \(R(x)=x(360-1.5x)=360x-1.5x^2\), for \(0\le x\le240\). 2. Differentiate: \(R'(x)=360-3x\). The only interior critical number is \(x=120\). 3. For \(0<x<120\), \(R'(x)>0\), and for \(120<x<240\), \(R'(x)<0\). Thus revenue increases and then decreases on the full feasible interval, so \(x=120\) gives the absolute maximum. 4. The maximum revenue is \(R(120)=21{,}600\), or \(\$21{,}600\). 5. The corresponding price is \(p(120)=180\), or \(\$180\) per backpack.

1. \(R(x)=360x-1.5x^2\) 2. \(120\) backpacks 3. The absolute maximum revenue is \(\$21{,}600\), at a price of \(\$180\) per backpack.

- Average cost is total cost divided by quantity. - Set the derivative of the average-cost function equal to zero. - Multiply by \(x^2\) to eliminate fractions. - Use the second derivative and end behavior to verify the minimum.

1. The average-cost function is \(C_{\text{avg}}(x)=\frac{C(x)}{x}=x^2-12x+60+\frac{256}{x}\). 2. Differentiate: \(C_{\text{avg}}'(x)=2x-12-\frac{256}{x^2}\). 3. Setting \(C_{\text{avg}}'(x)=0\) and multiplying by \(x^2\) gives \(x^3-6x^2-128=0\). 4. Factor: \(x^3-6x^2-128=(x-8)(x^2+2x+16)\). The quadratic factor has no real zeros, so the only critical number is \(x=8\). 5. Since \(C_{\text{avg}}''(x)=2+\frac{512}{x^3}>0\) for \(x>0\), and \(C_{\text{avg}}(x)\to\infty\) as \(x\to 0^+\) or \(x\to\infty\), this is the global minimum. 6. The minimum average cost is \(C_{\text{avg}}(8)=60\), or \(\$60\) per unit.

The average cost is minimized at \(x=8\) units, and the minimum average cost is \(\$60\) per unit.

- Form profit by subtracting total cost from revenue. - Use the given break-even value to help factor the profit function. - For the maximum, examine the derivative over the entire nonnegative production domain, not just near one critical point.

1. Revenue is \(R(x)=52x\), so \(P(x)=R(x)-C(x)=-x^3+15x^2-48x-10\). 2. Using the given zero, \(P(x)=-(x-5)(x^2-10x-2)\). The nonnegative zeros are \(x=5\) and \(x=5+3\sqrt3\approx10.20\), so profit is positive for \(5<x<5+3\sqrt3\). 3. Differentiate: \(P'(x)=-3x^2+30x-48=-3(x-2)(x-8)\). Thus \(P\) decreases on \((0,2)\), increases on \((2,8)\), and decreases on \((8,\infty)\). 4. Since \(P(0)=-10\) and the function decreases after \(x=8\), the local maximum at \(x=8\) is the absolute maximum on \(x\ge0\). 5. The maximum profit is \(P(8)=54\), or \(\$54\).

a) \(P(x)=-x^3+15x^2-48x-10\). Profit is positive for \(5<x<5+3\sqrt3\), approximately \(5<x<10.20\). b) The absolute maximum profit is \(\$54\) at \(x=8\) lamps in the continuous model.

- Start by finding the full first-quadrant domain of the square-root function. - Express the rectangle area using the coordinates of its upper-right corner. - Use the derivative and the boundary cases to verify an absolute maximum.

1. The radicand must be nonnegative: \(27-3x\ge0\). In the first quadrant, the feasible interval is \(0\le x\le9\). 2. The rectangle has width \(x\) and height \(f(x)\), so \(A(x)=x\sqrt{27-3x}\). 3. Differentiate: \(A'(x)=\frac{54-9x}{2\sqrt{27-3x}}\) for \(0<x<9\). The only interior critical number is \(x=6\). 4. The derivative is positive on \((0,6)\) and negative on \((6,9)\), while \(A(0)=A(9)=0\). Therefore \(x=6\) gives the absolute maximum. 5. The corresponding height is \(f(6)=3\), so \(P=(6,3)\), and the maximum area is \(18\) square units.

a) \([0,9]\) b) \(P=(6,3)\), and the absolute maximum area is \(18\) square units.

- The points have the same \(x\)-coordinate, so their distance is vertical. - Subtract the two function values to create a distance function. - Find where the derivative of the distance function is zero. - Compare all critical points and interval endpoints.

1. On \([-4,-1]\), \(f(a)\ge g(a)\), so the vertical distance is \(d(a)=f(a)-g(a)=-\frac{1}{3}a^3-2a^2-3a+3\). 2. Differentiate: \(d'(a)=-a^2-4a-3=-(a+1)(a+3)\). 3. The critical numbers in the interval are \(a=-3\) and \(a=-1\). 4. Evaluate the distance at all candidates: \(d(-4)=\frac{13}{3}\), \(d(-3)=3\), and \(d(-1)=\frac{13}{3}\). 5. Therefore, the minimum occurs at \(a=-3\), and the minimum distance is \(3\).

The segment is shortest when \(a=-3\), and its minimum length is \(3\).

- Use symmetry to express the full width in terms of one positive x-coordinate. - Determine how far the upper corner can move before the rectangle becomes degenerate. - Check the derivative across that complete feasible interval and include the boundary cases.

1. If the upper-right corner is \((x,f(x))\), symmetry gives rectangle width \(2x\) and height \(f(x)\). Feasibility requires \(0\le x\le\sqrt{12}\). 2. The area is \(A(x)=2x(-0.5x^2+6)=-x^3+12x\). 3. Differentiate: \(A'(x)=-3x^2+12\). The positive interior critical number is \(x=2\). 4. For \(0<x<2\), \(A'(x)>0\), and for \(2<x<\sqrt{12}\), \(A'(x)<0\). Also the area is \(0\) at both boundary cases. Therefore \(x=2\) gives the absolute maximum. 5. The width is \(4\,\text{m}\), the height is \(f(2)=4\,\text{m}\), and the maximum area is \(16\,\text{m}^2\).

The absolute maximum-area rectangle is \(4\,\text{m}\) wide and \(4\,\text{m}\) high, with area \(16\,\text{m}^2\).

- Marginal cost is the derivative of total cost. - Minimize a function by setting its derivative equal to zero and checking the result. - Average cost is total cost divided by quantity. - The equation in part b) can be solved by testing likely positive values.

1. The marginal-cost function is \(C'(x)=0.003x^2-0.12x+5\). 2. Differentiate marginal cost: \(C''(x)=0.006x-0.12\). Setting this equal to zero gives \(x=20\). Since \(C'''(x)=0.006>0\), marginal cost is minimized there. 3. The average-cost function is \(C_{\text{avg}}(x)=0.001x^2-0.06x+5+\frac{100}{x}\). 4. Differentiate: \(C_{\text{avg}}'(x)=0.002x-0.06-\frac{100}{x^2}\). 5. Setting \(C_{\text{avg}}'(x)=0\) gives \(0.002x^3-0.06x^2-100=0\), whose positive solution is \(x=50\). 6. Since \(C_{\text{avg}}''(x)=0.002+\frac{200}{x^3}>0\), and \(C_{\text{avg}}(x)\to\infty\) at both ends of its domain, this is the global minimum. 7. At \(x=50\), \(C'(50)=0.003(50)^2-0.12(50)+5=6.5\), and \(C_{\text{avg}}(50)=0.001(50)^2-0.06(50)+5+\frac{100}{50}=6.5\). Thus marginal cost equals average cost.

a) Marginal cost is minimized at \(x=20\) units. b) Average cost is minimized at \(x=50\) units. c) \(C'(50)=C_{\text{avg}}(50)=\$6.50\) per unit.

- Use \(V=x^2h\) to express the height in terms of \(x\). - Separate the cost of the base and lid from the cost of the side walls. - Substitute the volume constraint before differentiating. - Verify the critical number using the second derivative and end behavior.

1. The volume constraint is \(x^2h=16\), so \(h=\frac{16}{x^2}\). 2. The base and lid have total area \(2x^2\) and cost \(3(2x^2)=6x^2\). The four side walls have total area \(4xh\) and cost \(2(4xh)=8xh\). 3. Substitute \(h=\frac{16}{x^2}\): \(C(x)=6x^2+8x\left(\frac{16}{x^2}\right)=6x^2+\frac{128}{x}\). 4. Differentiate: \(C'(x)=12x-\frac{128}{x^2}\). Setting \(C'(x)=0\) gives \(12x^3=128\), so \(x=\sqrt[3]{\frac{32}{3}}\approx 2.20\,\text{ft}\). 5. Since \(C''(x)=12+\frac{256}{x^3}>0\), and \(C(x)\to\infty\) as \(x\to 0^+\) or \(x\to\infty\), this is the global minimum. 6. The corresponding height is \(h=\frac{16}{x^2}=\frac{16}{(32/3)^{2/3}}\approx 3.30\,\text{ft}\).

1. \(C(x)=6x^2+\frac{128}{x}\) 2. \(x=\sqrt[3]{\frac{32}{3}}\approx 2.20\,\text{ft}\) 3. \(h=\frac{16}{(32/3)^{2/3}}\approx 3.30\,\text{ft}\)

- Use the mailing constraint to express the tube length in terms of its radius. - Determine which radii keep the tube length nonnegative. - Verify the maximizing critical point over the entire feasible radius interval.

1. The constraint \(l+2\pi r=120\) gives \(l=120-2\pi r\). Feasibility requires \(0\le r\le\frac{60}{\pi}\). 2. The volume is \(V(r)=\pi r^2(120-2\pi r)=120\pi r^2-2\pi^2r^3\). 3. Differentiate: \(V'(r)=240\pi r-6\pi^2r^2=6\pi r(40-\pi r)\). The positive interior critical number is \(r=\frac{40}{\pi}\). 4. On \(0<r<\frac{40}{\pi}\), \(V'(r)>0\), and on \(\frac{40}{\pi}<r<\frac{60}{\pi}\), \(V'(r)<0\). Also \(V(0)=V(\frac{60}{\pi})=0\). Thus the critical number gives the absolute maximum. 5. The corresponding length is \(l=40\,\text{cm}\), and \(V_{\max}=\frac{64000}{\pi}\approx20{,}371.8\,\text{cm}^3\).

a) \(V(r)=120\pi r^2-2\pi^2r^3\), for \(0\le r\le\frac{60}{\pi}\). b) \(r=\frac{40}{\pi}\approx12.73\,\text{cm}\) and \(l=40\,\text{cm}\). c) \(V_{\max}\approx20{,}371.8\,\text{cm}^3\).

- Write an equation for the total length of the eight metal-edged segments. - Substitute the resulting expression for \(h\) into the volume formula. - Use the minimum-height condition to determine the feasible interval for \(a\). - Check whether the unconstrained critical number lies in that interval; otherwise inspect the boundary.

1. The edging constraint is \(4a+4h=360\), so \(h=90-a\). Therefore, \(V(a)=a^2h=a^2(90-a)=90a^2-a^3\). 2. The requirement \(h\ge 40\) gives \(90-a\ge 40\Rightarrow 0<a\le 50\). 3. Differentiate: \(V'(a)=180a-3a^2=3a(60-a)\). This derivative is positive throughout \((0, 50]\), because the unconstrained interior maximum would occur at \(a=60\), outside the allowed domain. Therefore, \(V\) is increasing on the entire feasible interval. 4. The constrained maximum occurs at the right endpoint: \(a=50\,\text{cm}\). Then \(h=90-50=40\,\text{cm}\), and \(V_{\max}=50^2(40)=100{,}000\,\text{cm}^3\).

a) \(V(a)=90a^2-a^3\). b) \(a=50\,\text{cm}\) and \(h=40\,\text{cm}\). c) \(V_{\max}=100{,}000\,\text{cm}^3\).

- Use the area formula for a right triangle. - The height is the function value \(f(x)\). - Apply both the product rule and the chain rule. - Compare the critical number with the interval endpoints.

1. The triangle has base \(x\) and height \(f(x)\), so \(A(x)=\frac{1}{2}x(4-x)e^{0.5x}=(2x-0.5x^2)e^{0.5x}\). 2. Differentiate using the product and chain rules: \(A'(x)=(2-0.25x^2)e^{0.5x}\). 3. The exponential factor is positive, so \(A'(x)=0\) when \(2-0.25x^2=0\). Thus \(x^2=8\). 4. In \([0,4]\), the relevant critical number is \(x=\sqrt{8}=2\sqrt{2}\). 5. The derivative is positive before \(2\sqrt{2}\) and negative after it. Also, \(A(0)=A(4)=0\), so the absolute maximum occurs at \(x=2\sqrt{2}\approx 2.83\).

\(x=2\sqrt{2}\approx 2.83\)

- Express both the area and the outside perimeter in terms of the radius and rectangular height. - Use the perimeter restriction to find the full feasible range of radii. - Check the sign of the area derivative across that entire interval.

1. The rectangle has width \(2r\). The area is \(A=2rh+\frac12\pi r^2\), and the outside perimeter condition is \(2h+(2+\pi)r=6\). 2. Hence \(h=3-r\left(1+\frac\pi2\right)\). Feasibility requires \(0\le r\le\frac{6}{2+\pi}\). 3. Substitute: \(A(r)=6r-r^2\left(2+\frac\pi2\right)\). 4. Differentiate: \(A'(r)=6-(4+\pi)r\). The only interior critical number is \(r=\frac{6}{4+\pi}\). 5. Since \(A'(r)>0\) before this value and \(A'(r)<0\) after it throughout the feasible interval, the critical number gives the absolute maximum. 6. The corresponding rectangular height is \(h=\frac{6}{4+\pi}\), so \(r=h\approx0.840\,\text{m}\).

The absolute maximum area occurs when \(r=h=\frac{6}{4+\pi}\approx0.840\,\text{m}\).

- Translate the material-price information into one budget equation. - Use the budget equation to determine the possible base sizes and eliminate the height. - Check how the volume changes throughout the full feasible interval.

1. Measured in equivalent square meters of side-wall material, the budget constraint is \(3a^2+4ah=9\). Thus \(h=\frac{9-3a^2}{4a}\), and feasibility gives \(0<a<\sqrt3\). 2. The volume is \(V(a)=a^2h=\frac{9a-3a^3}{4}\). 3. Differentiate: \(V'(a)=\frac94(1-a^2)\). The only positive critical number is \(a=1\,\text{m}\). 4. The derivative is positive on \((0,1)\) and negative on \((1,\sqrt3)\). Also \(V(a)\to0\) at both ends of the feasible interval. Therefore \(a=1\) gives the absolute maximum. 5. The height is \(h=\frac{9-3}{4}=1.5\,\text{m}\), and the maximum volume is \(1.5\,\text{m}^3\).

a) \(a=1\,\text{m}\) and \(h=1.5\,\text{m}\). b) \(V_{\max}=1.5\,\text{m}^3\).

- Let one sheet dimension be the tube circumference and use the perimeter constraint for the other. - Convert circumference to radius before forming the volume. - Include both degenerate boundary cases when verifying the absolute maximum.

1. Let \(x\) be the side that becomes the base circumference and \(h\) the cylinder height. The sheet-perimeter constraint \(2x+2h=120\) gives \(h=60-x\), so \(0\le x\le60\). Since \(x=2\pi r\), \(r=\frac{x}{2\pi}\). 2. The volume is \(V(x)=\pi\left(\frac{x}{2\pi}\right)^2(60-x)=\frac{60x^2-x^3}{4\pi}\). 3. Differentiate: \(V'(x)=\frac{3x(40-x)}{4\pi}\). The interior critical number is \(x=40\,\text{cm}\). 4. The derivative is positive on \((0,40)\) and negative on \((40,60)\), and \(V(0)=V(60)=0\). Therefore \(x=40\) gives the absolute maximum. 5. The other sheet dimension is \(h=20\,\text{cm}\).

The sheet should measure \(40\,\text{cm}\times20\,\text{cm}\), with the \(40\,\text{cm}\) side forming the circumference.

- Use the net dimensions to express the base length and width in terms of \(h\). - Determine the physical domain for the box height. - Write and differentiate the volume function. - Discard any critical number outside the domain and verify the maximum.

1. Let \(l\) and \(w\) be the base dimensions. The net constraints are \(2l+2h=80\Rightarrow l=40-h\) and \(w+2h=50\Rightarrow w=50-2h\). 2. The volume is \(V(h)=h(40-h)(50-2h)=2h^3-130h^2+2000h\). Positive dimensions require \(0<h<25\). 3. Differentiate: \(V'(h)=6h^2-260h+2000\). Solving \(V'(h)=0\) gives \(h=10\) or \(h=\frac{100}{3}\). Only \(h=10\) lies in the physical domain. 4. Since \(V''(10)=12(10)-260=-140<0\), the critical number gives a maximum. The volume approaches \(0\) at both ends of the physical domain, so it is the absolute maximum. 5. The maximum volume is \(V(10)=10(30)(30)=9000\,\text{cm}^3\).

The box height should be \(h=10\,\text{cm}\), and the maximum volume is \(9000\,\text{cm}^3\).

- Use the volume condition to eliminate the cylindrical height. - Determine which radii keep that height nonnegative. - Verify the critical radius over this feasible domain rather than only classifying it locally.

1. The exterior area is \(S=3\pi r^2+2\pi rh\). The volume constraint is \(\pi r^2h+\frac23\pi r^3=1.5\), so \(h=\frac{1.5}{\pi r^2}-\frac23r\). 2. Feasibility requires \(r>0\) and \(h\ge0\), giving \(0<r\le\sqrt[3]{\frac{2.25}{\pi}}\). 3. Substitute into the area: \(S(r)=\frac53\pi r^2+\frac3r\). 4. Differentiate: \(S'(r)=\frac{10}{3}\pi r-\frac{3}{r^2}\). The critical number satisfies \(r^3=\frac{9}{10\pi}\), so \(r\approx0.659\,\text{m}\), which is feasible. 5. Since \(S''(r)=\frac{10}{3}\pi+\frac{6}{r^3}>0\), \(S\) is strictly convex. Also \(S(r)\to\infty\) as \(r\to0^+\). Hence the feasible critical number gives the absolute minimum. 6. Using the critical-number relation in the volume constraint gives \(h=r\approx0.659\,\text{m}\).

The absolute minimum-material dimensions are \(r=\sqrt[3]{\frac{9}{10\pi}}\approx0.659\,\text{m}\) and \(h\approx0.659\,\text{m}\).

- Write the fixed-cost constraint using the two material rates. - Solve the cost constraint for the height. - Substitute into the volume and find the critical number. - Check the physically feasible radius interval before concluding that the maximum is absolute.

1. Let \(k\) be the side-material cost per unit area and \(C\) be the fixed total cost. The cost constraint is \(C=2k\pi r^2+2k\pi rh\). Solving for \(h\) gives \(h=\frac{C}{2k\pi r}-r\). 2. The volume is \(V(r)=\pi r^2h=\frac{C}{2k}r-\pi r^3\). 3. Differentiate: \(V'(r)=\frac{C}{2k}-3\pi r^2\). At the critical number, \(C=6k\pi r^2\). Substitute this into the original cost constraint: \(6k\pi r^2=2k\pi r^2+2k\pi rh\). Therefore, \(h=2r\). 4. The feasibility condition \(h\ge 0\) gives \(0<r\le\sqrt{\frac{C}{2k\pi}}\). On this interval, \(V''(r)=-6\pi r<0\), so \(V\) is strictly concave down. Also, \(V(r)\to0\) as \(r\to0^+\), and \(V=0\) at the other boundary where \(h=0\). Therefore the unique critical number gives the absolute maximum.

The absolute maximum occurs when \(h=2r\), so the optimal ratio is \(h:r=2:1\).

- Identify the quantity to maximize and write its formula. - Use the fixed surface area to express one variable in terms of the other. - Find the critical number of the one-variable volume function. - Check that the dimensions are physically meaningful.

1. The volume is \(V=\pi r^2h\). The fixed surface-area condition is \(2\pi r^2+2\pi rh=600\). 2. Solve the constraint for \(h\): \(h=\frac{300}{\pi r}-r\). 3. Substitute into the volume: \(V(r)=\pi r^2\left(\frac{300}{\pi r}-r\right)=300r-\pi r^3\). 4. Differentiate and set the derivative equal to zero: \(V'(r)=300-3\pi r^2=0\). Thus \(r=\sqrt{\frac{100}{\pi}}\approx 5.64\,\text{cm}\). 5. Since \(V''(r)=-6\pi r<0\) for \(r>0\), the critical number gives a maximum. Also, the volume approaches \(0\) at both ends of the physical domain, so this is the absolute maximum. 6. The height is \(h=\frac{300}{\pi r}-r=2r\approx 11.28\,\text{cm}\). 7. The maximum volume is \(V=\pi r^2h\approx 1128.38\,\text{cm}^3\).

The can should have radius \(r\approx 5.64\,\text{cm}\) and height \(h\approx 11.28\,\text{cm}\). Its maximum volume is \(V\approx 1128.38\,\text{cm}^3\).

- Because the tank is open, include only the base and the curved side. - Build the different material prices directly into the cost function. - Use the volume equation to eliminate \(h\). - Find where the derivative of the cost function is zero.

1. The material cost is \(K(r, h)=15\pi r^2+10(2\pi rh)\). The volume constraint is \(\pi r^2h=10\). 2. Solve for \(h\): \(h=\frac{10}{\pi r^2}\). 3. Substitute into the cost function: \(K(r)=15\pi r^2+20\pi r\left(\frac{10}{\pi r^2}\right)=15\pi r^2+\frac{200}{r}\). 4. Differentiate and set the derivative equal to zero: \(K'(r)=30\pi r-\frac{200}{r^2}=0\). Therefore, \(30\pi r^3=200\), so \(r=\sqrt[3]{\frac{20}{3\pi}}\approx 1.29\,\text{m}\). 5. Because \(K''(r)=30\pi+\frac{400}{r^3}>0\) for \(r>0\), and \(K(r)\to\infty\) as \(r\to 0^+\) or \(r\to\infty\), this critical number gives the absolute minimum. 6. The corresponding height is \(h=\frac{10}{\pi r^2}\approx 1.93\,\text{m}\). 7. The minimum cost is \(K\approx \$233.45\).

The minimum cost occurs at \(r\approx 1.29\,\text{m}\) and \(h\approx 1.93\,\text{m}\). The minimum material cost is \(K\approx \$233.45\).

- Use the fixed volume to eliminate the cylindrical height from the weighted cost. - Include the condition that the cylindrical height cannot be negative. - Verify the critical radius over the full feasible domain before using its equation to simplify the ratio.

1. The fixed volume is \(V=\pi r^2h+\frac23\pi r^3\). If cylindrical-side material costs one unit per square meter, the weighted cost is \(C=2\pi rh+4\pi r^2\). 2. Solve for the cylindrical height: \(h=\frac{V}{\pi r^2}-\frac23r\). Feasibility requires \(r>0\) and \(h\ge0\), so \(0<r\le\sqrt[3]{\frac{3V}{2\pi}}\). 3. Substitute: \(C(r)=\frac{2V}{r}+\frac83\pi r^2\). 4. Differentiate: \(C'(r)=-\frac{2V}{r^2}+\frac{16}{3}\pi r\). The critical-number equation gives \(V=\frac83\pi r^3\). 5. Since \(C''(r)=\frac{4V}{r^3}+\frac{16}{3}\pi>0\), \(C\) is strictly convex, and \(C(r)\to\infty\) as \(r\to0^+\). The feasible critical number therefore gives the absolute minimum. 6. Substituting \(V=\frac83\pi r^3\) into the height formula gives \(h=2r\), so \(h:r=2:1\).

The absolute minimum cost occurs when \(h=2r\), so the optimal ratio is \(h:r=2:1\).

- Use the symmetry of the region to express the rectangle's width from one positive x-coordinate. - Express the height using the parabola. - Write the area as a function of that x-coordinate. - Determine the allowable interval and use the derivative to find the absolute maximum.

1. By symmetry, let the upper-right vertex be \((x,12-x^2)\), where \(0<x<\sqrt{12}\). The rectangle has width \(2x\) and height \(12-x^2\). 2. Its area is \(A(x)=2x(12-x^2)=24x-2x^3\). 3. Differentiate: \(A'(x)=24-6x^2\). Setting the derivative equal to zero gives \(x=2\). 4. Since \(A''(2)=-24<0\), this critical number gives a local maximum. The area approaches \(0\) at both ends of the domain, so it is the global maximum. 5. The width is \(2x=4\), and the height is \(12-x^2=8\).

The side lengths are \(4\) and \(8\).

- Use a vertical cross section to relate prism height to its square-base side length. - State the complete interval of possible base side lengths before optimizing. - Use the derivative sign and the degenerate boundary cases to establish an absolute maximum.

1. In a vertical cross section, similarity gives \(\frac{h-y}{h}=\frac{x}{a}\), so \(y=h\left(1-\frac{x}{a}\right)\). 2. Feasibility gives \(0\le x\le a\). The prism volume is \(V(x)=x^2y=hx^2-\frac{h}{a}x^3\). 3. Differentiate: \(V'(x)=hx\left(2-\frac{3x}{a}\right)\). Thus \(V'(x)>0\) for \(0<x<\frac{2a}{3}\) and \(V'(x)<0\) for \(\frac{2a}{3}<x<a\). 4. Since \(V(0)=V(a)=0\), the absolute maximum occurs at \(x=\frac{2a}{3}\). Then \(y=\frac h3\). 5. The maximum prism volume is \(\frac{4}{27}a^2h\), while the pyramid volume is \(\frac13a^2h\). Their ratio is \(\frac49\).

1. \(y=h\left(1-\frac{x}{a}\right)\) 2. \(x=\frac{2a}{3}\) and \(y=\frac h3\) 3. The maximum-volume prism occupies \(\frac49\) of the pyramid's volume.

- Use the axial cross section shown to relate the height, half the base diagonal, and the fixed lateral edge. - Express the base area in terms of \(h\). - Determine the possible interval for the height. - Differentiate the resulting volume function and check the boundary cases.

1. Let \(a\) be the base edge length. The volume is \(V=\frac{1}{3}a^2h\). 2. A right triangle formed by the height, half the base diagonal, and a lateral edge gives \(9^2=h^2+\left(\frac{a\sqrt{2}}{2}\right)^2\). Therefore, \(a^2=2(81-h^2)\). 3. Substitute into the volume: \(V(h)=\frac{2}{3}(81h-h^3)\), where \(0<h<9\). 4. Differentiate: \(V'(h)=\frac{2}{3}(81-3h^2)\). The positive critical point satisfies \(h^2=27\), so \(h=3\sqrt{3}\,\text{cm}\approx 5.20\,\text{cm}\). 5. Since \(V''(h)=-4h<0\) for \(h>0\), and the volume is \(0\) at the endpoints of the physical domain, this point gives the absolute maximum. 6. The maximum volume is \(V(3\sqrt{3})=108\sqrt{3}\,\text{cm}^3\approx 187.06\,\text{cm}^3\).

The maximizing height is \(h=3\sqrt{3}\,\text{cm}\approx 5.20\,\text{cm}\). The maximum volume is \(108\sqrt{3}\,\text{cm}^3\approx 187.06\,\text{cm}^3\).

- Write the cylinder volume formula. - Use an axial cross section to relate \(r\), \(h\), and the sphere radius. - Eliminate one variable from the volume formula. - Include the endpoints of the physical domain in your maximum check.

1. The cylinder volume is \(V=\pi r^2h\). In an axial cross section of the sphere, \(r^2+\left(\frac{h}{2}\right)^2=6^2\). Thus \(r^2=36-\frac{h^2}{4}\). 2. Substitute into the volume: \(V(h)=\pi\left(36-\frac{h^2}{4}\right)h=36\pi h-\frac{\pi}{4}h^3\), where \(0\le h\le12\). 3. Differentiate: \(V'(h)=36\pi-\frac{3\pi}{4}h^2\). Setting the derivative equal to zero gives \(h^2=48\), so \(h=4\sqrt{3}\,\text{m}\approx 6.93\,\text{m}\). 4. Since \(V''(h)=-\frac{3\pi}{2}h<0\) at the positive critical number, and the endpoint volumes are \(0\), this is the absolute maximum. 5. The radius is \(r=\sqrt{36-\frac{48}{4}}=2\sqrt{6}\,\text{m}\approx 4.90\,\text{m}\). 6. The maximum volume is \(V=\pi(24)(4\sqrt{3})=96\sqrt{3}\pi\,\text{m}^3\approx 522.37\,\text{m}^3\).

The cylinder has radius \(r\approx 4.90\,\text{m}\) and height \(h\approx 6.93\,\text{m}\). Its maximum volume is \(V\approx 522.37\,\text{m}^3\).

- Write the volume in terms of \(a\) and \(h\). - Use the fixed space diagonal to eliminate one variable. - Differentiate the one-variable volume function. - Compare the resulting values of \(a\) and \(h\).

1. The volume is \(V=a^2h\). The space-diagonal constraint is \(D^2=2a^2+h^2\). Therefore, \(a^2=\frac{D^2-h^2}{2}\). 2. Substitute into the volume: \(V(h)=\frac{1}{2}(D^2h-h^3)\), where \(0<h<D\). 3. Differentiate: \(V'(h)=\frac{1}{2}(D^2-3h^2)\). The positive critical number is \(h=\frac{D}{\sqrt{3}}\). 4. Then \(a^2=\frac{D^2-D^2/3}{2}=\frac{D^2}{3}\), so \(a=\frac{D}{\sqrt{3}}\). The volume is zero at both endpoints of the physical domain, so the interior critical number gives the maximum. 5. Therefore, \(\frac{h}{a}=1\).

The optimal ratio is \(\frac{h}{a}=1\). The maximum-volume box is a cube.

- Use the Pythagorean theorem to relate \(r\), \(h\), and \(s\). - Use the fixed volume to eliminate one variable. - Minimize \(s^2\) rather than \(s\). - Express the critical-number equation as a ratio of \(h\) and \(r\).

1. The slant height satisfies \(s^2=r^2+h^2\). Since \(s>0\), minimize \(s^2\). 2. The fixed-volume condition is \(V=\frac{1}{3}\pi r^2h\), so \(r^2=\frac{3V}{\pi h}\). 3. Substitute into the squared slant height: \(f(h)=\frac{3V}{\pi h}+h^2\). 4. Differentiate and set the derivative equal to zero: \(f'(h)=-\frac{3V}{\pi h^2}+2h=0\). Therefore, \(2h^3=\frac{3V}{\pi}\). 5. Using \(V=\frac{1}{3}\pi r^2h\), this becomes \(2h^3=r^2h\). Hence \(r^2=2h^2\), and \(h:r=1:\sqrt{2}\). The objective approaches infinity as \(h\to0^+\) or \(h\to\infty\), so the critical point gives the absolute minimum.

The slant height is minimized when \(h:r=1:\sqrt{2}\), or \(r=\sqrt{2}h\).

- Revenue equals quantity times price. - Use the product and chain rules to differentiate the exponential revenue function. - The exponential factor never equals zero. - Include units and round monetary values to two decimal places.

1. Revenue is \(R(x)=x p(x)=300x e^{-0.005x}\). 2. Differentiate using the product and chain rules: \(R'(x)=300e^{-0.005x}(1-0.005x)\). 3. The exponential factor is positive, so \(R'(x)=0\) when \(1-0.005x=0\). Thus \(x=200\). 4. For \(0\le x<200\), \(R'(x)>0\), and for \(x>200\), \(R'(x)<0\). Thus revenue increases and then decreases on its entire domain, so \(x=200\) gives the absolute maximum. 5. The corresponding price is \(p(200)=300e^{-1}\approx \$110.36\). 6. The maximum revenue is \(R(200)=60{,}000e^{-1}\approx \$22{,}072.77\).

The absolute maximum revenue occurs at \(x=200\) units. The corresponding price is approximately \(\$110.36\) per unit, and the maximum revenue is approximately \(\$22{,}072.77\).

- Use the price-demand function to identify the feasible sales interval. - Form profit from revenue and cost before differentiating. - Check the sign of the profit derivative on both sides of the feasible critical point.

1. Revenue is \(R(x)=xp(x)=300x-10x^2\), so profit is \(P(x)=R(x)-C(x)=-x^3+5x^2+200x-500\). 2. Differentiate: \(P'(x)=-3x^2+10x+200=-3(x-10)\left(x+\frac{20}{3}\right)\). The only critical point in \([0,30]\) is \(x=10\). 3. On \(0<x<10\), \(P'(x)>0\); on \(10<x<30\), \(P'(x)<0\). Thus profit increases and then decreases across the full feasible interval, so \(x=10\) gives the absolute maximum. 4. The maximum profit is \(P(10)=1000\), or \(\$1000\).

The absolute maximum profit occurs when \(10\) bags are produced and sold, and the maximum profit is \(\$1000\).

- Represent a general point on the parabola as \((x, x^2)\). - Minimize the squared distance to avoid a square root. - Try small integer values when solving the cubic equation. - Use the second derivative to verify the minimum.

1. A point on the graph is \((x, x^2)\). Minimize the squared distance \(D(x)=(x-3)^2+x^4=x^4+x^2-6x+9\). 2. Differentiate: \(D'(x)=4x^3+2x-6\). The equation \(D'(x)=0\) has the solution \(x=1\). 3. Since \(D''(x)=12x^2+2>0\) for every real \(x\), \(D\) is strictly convex. Therefore, its only critical point is the absolute minimum. 4. The corresponding y-coordinate is \(f(1)=1\). Thus the closest point is \((1, 1)\).

The closest point is \(P=(1, 1)\).

- Write a squared-distance function for \((x, f(x))\) and \(S\). - Use the parabola's symmetry to check your results. - Find and classify every critical number. - Compare with the distance from the vertex.

1. A point on the graph is \((x, 0.5x^2-1.5)\). Its squared distance from \(S\) is \(D(x)=x^2+(0.5x^2-2.5)^2=0.25x^4-1.5x^2+6.25\). 2. Differentiate: \(D'(x)=x^3-3x=x(x^2-3)\). The critical numbers are \(x=0\) and \(x=\pm\sqrt{3}\). 3. The second derivative is \(D''(x)=3x^2-3\). Thus \(x=0\) gives a local maximum, while \(x=\pm\sqrt{3}\) give local minima. Since \(D(x)\to\infty\) as \(|x|\to\infty\), these are the absolute minima. 4. At \(x=\pm\sqrt{3}\), \(f(x)=0\). The minimum distance is \(d=\sqrt{(\sqrt{3})^2+(0-1)^2}=2\).

The closest points are \(P_1=(-\sqrt{3}, 0)\) and \(P_2=(\sqrt{3}, 0)\). The minimum distance is \(2\) units.

- Use the given relationship between length and width. - Write the total surface area of a closed rectangular box. - Use the fixed volume to express \(h\) in terms of \(b\). - Find the minimum of the resulting surface-area function.

1. Let \(l=3b\). The surface area is \(O=2(lb+lh+bh)=6b^2+8bh\). The volume constraint is \(lbh=3b^2h=36\). 2. Solve for \(h\): \(h=\frac{12}{b^2}\). Substitute into the surface area: \(O(b)=6b^2+\frac{96}{b}\). 3. Differentiate: \(O'(b)=12b-\frac{96}{b^2}\). Setting the derivative equal to zero gives \(12b^3=96\), so \(b=2\,\text{ft}\). 4. Then \(l=3b=6\,\text{ft}\) and \(h=\frac{12}{2^2}=3\,\text{ft}\). 5. Since \(O''(b)=12+\frac{192}{b^3}>0\) for \(b>0\), and \(O(b)\to\infty\) at both ends of the positive domain, these dimensions give the absolute minimum.

The container dimensions are \(l=6\,\text{ft}\), \(b=2\,\text{ft}\), and \(h=3\,\text{ft}\).

- Substitute the fixed values before optimizing. - Rewrite \(\frac{1}{\cos^2(\alpha)}\) using \(\tan(\alpha)\). - Let \(t=\tan(\alpha)\) to obtain a quadratic objective function. - Use the vertex or derivative of that quadratic.

1. For \(\alpha=45^\circ\), \(y(30)=30\tan(45^\circ)-\frac{10(30)^2}{2(20)^2\cos^2(45^\circ)}\). Using \(\tan(45^\circ)=1\) and \(\cos^2(45^\circ)=0.5\), \(y(30)=30-22.5=7.5\,\text{m}\). 2. At the wall, the height as a function of angle is \(h(\alpha)=30\tan(\alpha)-\frac{11.25}{\cos^2(\alpha)}\). Let \(t=\tan(\alpha)\). Since \(\sec^2(\alpha)=1+t^2\), \(h(t)=30t-11.25(1+t^2)\), where \(t\ge0\). 3. Differentiate: \(h'(t)=30-22.5t\). The critical point is \(t=\frac{4}{3}\). Because the quadratic opens downward, this point gives the absolute maximum. 4. Thus \(\alpha=\arctan\left(\frac{4}{3}\right)\approx53.13^\circ\). The maximum height is \(h_{\max}=30\left(\frac{4}{3}\right)-11.25\left(1+\frac{16}{9}\right)=8.75\,\text{m}\).

a) The capsule hits the wall at a height of \(7.5\,\text{m}\). b) The maximum impact height is \(8.75\,\text{m}\), reached at \(\alpha\approx53.13^\circ\).

- Simplify the objective with a double-angle identity before differentiating. - Ignore positive constant factors when locating the maximum. - Use the product and chain rules. - Check the endpoint behavior.

1. Multiply the two expressions: \(P(\alpha)=\frac{v^4}{2g^2}\sin(2\alpha)\sin^2(\alpha)\). Using \(\sin(2\alpha)=2\sin(\alpha)\cos(\alpha)\), \(P(\alpha)=\frac{v^4}{g^2}\sin^3(\alpha)\cos(\alpha)\). 2. The constant factor does not affect the maximizing angle, so maximize \(f(\alpha)=\sin^3(\alpha)\cos(\alpha)\). 3. Differentiate: \(f'(\alpha)=3\sin^2(\alpha)\cos^2(\alpha)-\sin^4(\alpha)\) \(=\sin^2(\alpha)\left(3\cos^2(\alpha)-\sin^2(\alpha)\right)\). 4. The interior critical point satisfies \(3\cos^2(\alpha)=\sin^2(\alpha)\), so \(\tan^2(\alpha)=3\). In the given interval, \(\alpha=60^\circ\). 5. The product is zero at \(0^\circ\) and approaches zero as \(\alpha\to90^\circ-\), so \(60^\circ\) gives the absolute maximum.

The product of range and maximum height is maximized at \(\alpha=60^\circ\).

- Treat marginal cost, average cost, and profit as three different functions to optimize. - Respect the positive production domain in every part. - For each result, verify what happens on the entire domain rather than only applying a local test.

1. Marginal cost is \(C'(x)=3x^2-18x+30\). Its derivative is \(C''(x)=6x-18\), which changes from negative to positive at \(x=3\). Thus marginal cost has its absolute minimum at \(x=3\) on \(x>0\). 2. Average cost is \(C_{\text{avg}}(x)=\frac{C(x)}{x}=x^2-9x+30+\frac{25}{x}\). 3. Its derivative is \(C_{\text{avg}}'(x)=2x-9-\frac{25}{x^2}\). Setting it equal to zero gives \((x-5)(2x^2+x+5)=0\), so the only positive critical point is \(x=5\). 4. Since \(C_{\text{avg}}''(x)=2+\frac{50}{x^3}>0\), average cost is strictly convex; moreover it tends to \(\infty\) as \(x\to0^+\) and as \(x\to\infty\). Therefore \(x=5\) gives the absolute minimum. 5. Revenue is \(R(x)=51x\), so profit is \(P(x)=-x^3+9x^2+21x-25\). Its derivative is \(P'(x)=-3(x-7)(x+1)\). 6. For \(0<x<7\), \(P'(x)>0\), and for \(x>7\), \(P'(x)<0\). Thus profit has its absolute maximum at \(x=7\).

a) \(x=3\) units b) \(x=5\) units c) \(x=7\) units

- Separate the expression into one factor involving \(T_1\) and one involving \(T_2\). - Maximize each one-variable factor on its interval. - Why does maximizing both nonnegative factors maximize their product? - Include the interval endpoints in each comparison.

1. Write \(A(T_1, T_2) = F(T_1)G(T_2)\), where \(F(T_1) = 3T_1^2 - T_1^3\) and \(G(T_2) = 12T_2 - T_2^2\). Both factors are nonnegative on the stated intervals, so the product is maximized when both factors are maximized. 2. Differentiate \(F\): \(F'(T_1) = 6T_1 - 3T_1^2 = 3T_1(2 - T_1)\). The candidates are \(T_1 = 0\), \(T_1 = 2\), and the endpoint \(T_1 = 3\). Since \(F(0) = F(3) = 0\) and \(F(2) = 4\), the maximum occurs at \(T_1 = 2\). 3. Differentiate \(G\): \(G'(T_2) = 12 - 2T_2\). The candidates are \(T_2 = 6\) and the endpoints. Since \(G(0) = G(12) = 0\) and \(G(6) = 36\), the maximum occurs at \(T_2 = 6\). 4. Therefore, the yield is maximized when \(T_1 = 2\) and \(T_2 = 6\).

\(T_1 = 2\) and \(T_2 = 6\)

- Use the coastline schematic to express both the land distance and the underwater distance in terms of one variable. - Use the right triangle for the underwater segment. - Build a total-cost function using the two construction rates. - Compare the critical point with both endpoints of the coastal interval.

1. Let \(x\) be the distance in miles from \(A\) to \(P\) along the coast. The land distance is \(x\), and the underwater distance is \(\sqrt{(15-x)^2+8^2}\). 2. The cost function is \(K(x)=30{,}000x+50{,}000\sqrt{(15-x)^2+64}\), where \(0\le x\le15\). 3. Differentiate: \(K'(x)=30{,}000-\frac{50{,}000(15-x)}{\sqrt{(15-x)^2+64}}\). Setting the derivative equal to zero gives \(3\sqrt{(15-x)^2+64}=5(15-x)\). Squaring and simplifying yields \(15-x=6\), so \(x=9\). 4. Compare the critical point and endpoints: \(K(0)=\$850{,}000\), \(K(15)=\$850{,}000\), and \(K(9)=30{,}000(9)+50{,}000(10)=\$770{,}000\). Therefore, the critical point gives the minimum.

The pipeline should enter the water \(9\) miles east of station \(A\). The minimum total cost is \(\$770{,}000\).

- Use time equals distance divided by speed. - Express the forest and road distances using one variable. - Differentiate the total-time function. - Convert the fractional part of an hour to minutes.

1. Let \(y\) be the distance from \(C\) to \(P\). The forest distance is \(\sqrt{4^2+y^2}\), and the road distance is \(10-y\). 2. The travel-time function is \(T(y)=\frac{\sqrt{16+y^2}}{3}+\frac{10-y}{5}\), where \(0\le y\le10\). 3. Differentiate: \(T'(y)=\frac{y}{3\sqrt{16+y^2}}-\frac{1}{5}\). Setting the derivative equal to zero gives \(5y=3\sqrt{16+y^2}\). Squaring yields \(25y^2=9(16+y^2)\), so \(y=3\). 4. Compare the critical point and endpoints: \(T(0)=\frac{10}{3}\), \(T(10)=\frac{\sqrt{116}}{3}\), and \(T(3)=\frac{5}{3}+\frac{7}{5}=\frac{46}{15}\) hours. Therefore, the minimum occurs at \(y=3\) miles. 5. Since \(\frac{46}{15}=3+\frac{1}{15}\), the travel time is \(3\) hours and \(4\) minutes.

The hiker should reach the road \(3\) miles from \(C\) toward \(T\). The minimum travel time is \(3\) hours \(4\) minutes.

- How is the annual increase \(G(x)\) related to the reproduction model \(f(x)\)? - What derivative condition holds at an interior maximum? - How can the population value for the following year be written as an equation involving \(f\)? - Use the two conditions to create a system for \(a\) and \(b\).

1. Write the annual-increase function: \(G(x)=f(x)-x=ax^2+(b-1)x\). 2. Since \(f(400)=500\), \(160{,}000a+400b=500\). 3. The increase is maximized at \(x=400\), so \(G'(400)=0\). Because \(G'(x)=2ax+b-1\), this gives \(800a+b=1\). 4. From the second equation, \(b=1-800a\). Substitute into the first equation: \(160{,}000a+400(1-800a)=500\). 5. Simplify: \(-160{,}000a=100\), so \(a=-\frac{1}{1600}=-0.000625\). 6. Then \(b=1-800\left(-\frac{1}{1600}\right)=\frac{3}{2}=1.5\). 7. Since \(G''(x)=2a=-\frac{1}{800}<0\), the critical point is a maximum, as required.

\(a=-\frac{1}{1600}=-0.000625\) and \(b=\frac{3}{2}=1.5\)

- Use the relationship between a circle's radius and circumference. - Express the volume as a function of the height. - Keep the height restriction in the domain. - Compare the interior critical point with both endpoints.

1. The volume is \(V=\pi r^2h\). The constraint is \(h+2\pi r=180\), so \(r=\frac{180-h}{2\pi}\). 2. Substitute into the volume: \(V(h)=\frac{(180-h)^2h}{4\pi}\), where \(20\le h\le100\). 3. Differentiate: \(V'(h)=\frac{3h^2-720h+32{,}400}{4\pi}\). The critical numbers are \(h=60\) and \(h=180\). Only \(h=60\) lies in the allowed interval. 4. Compare the critical point and endpoints: \(V(20)=\frac{128{,}000}{\pi}\approx40{,}743.67\,\text{cm}^3\), \(V(60)=\frac{216{,}000}{\pi}\approx68{,}754.94\,\text{cm}^3\), and \(V(100)=\frac{160{,}000}{\pi}\approx50{,}929.58\,\text{cm}^3\). Thus the absolute maximum occurs at \(h=60\,\text{cm}\). 5. The radius is \(r=\frac{180-60}{2\pi}=\frac{60}{\pi}\,\text{cm}\approx19.10\,\text{cm}\).

a) The maximizing dimensions are \(h=60\,\text{cm}\) and \(r=\frac{60}{\pi}\,\text{cm}\approx19.10\,\text{cm}\). b) The maximum volume is \(\frac{216{,}000}{\pi}\,\text{cm}^3\approx68{,}754.94\,\text{cm}^3\).

- A rectangular box has four edges of each dimension. - Use \(l=1.5w\) in the edge-sum constraint. - Use the height restriction to determine the domain for \(w\). - Compare the critical point with the boundary value and boundary behavior.

1. The volume is \(V=wlh=1.5w^2h\). The edge-length constraint is \(4(w+l+h)=240\). Using \(l=1.5w\), \(2.5w+h=60\), so \(h=60-2.5w\). 2. The volume function is \(V(w)=1.5w^2(60-2.5w)=90w^2-3.75w^3\). The height restriction \(h\le30\) gives \(w\ge12\), while \(h>0\) gives \(w<24\). Thus \(12\le w<24\). 3. Differentiate: \(V'(w)=180w-11.25w^2\). The interior critical point is \(w=16\). 4. Compare the critical point and boundary behavior: \(V(12)=6480\,\text{in}^3\), \(V(16)=7680\,\text{in}^3\), and \(V(w)\to0\) as \(w\to24^-\). Therefore, the absolute maximum occurs at \(w=16\,\text{in}\). 5. The remaining dimensions are \(l=1.5(16)=24\,\text{in}\) and \(h=60-2.5(16)=20\,\text{in}\).

a) The box should have width \(16\,\text{in}\), length \(24\,\text{in}\), and height \(20\,\text{in}\). b) The maximum volume is \(7680\,\text{in}^3\).

- Set the two functions equal and factor the resulting expression. - Treat the supplied area formula as the objective function. - Use the sign of its derivative on the two sides of the interior critical value. - Evaluate the area only after locating the maximizing parameter.

1. Set the functions equal: \(4k^2-4x^2=k^3-kx^2\). Rearranging gives \((4-k)(k^2-x^2)=0\). Since \(0<k<4\), the intersections occur at \(x=\pm k\). 2. Differentiate the supplied area model: \(A'(k)=16k^2-\frac{16}{3}k^3=16k^2(1-\frac{k}{3})\). The only interior critical number is \(k=3\). 3. The derivative is positive for \(0<k<3\) and negative for \(3<k<4\), so \(A\) increases and then decreases. Thus \(k=3\) gives the absolute maximum. 4. \(A(3)=36\) square units.

a) The graphs intersect at \(x=\pm k\). b) The area is maximized at \(k=3\), and the maximum area is \(36\) square units. c) \(A\) increases on \((0,3)\) and decreases on \((3,4)\), so the maximum is absolute.

- Set the two parabola equations equal and simplify before solving for the intersections. - Differentiate the supplied one-variable area function. - Determine the sign of the derivative on both sides of its positive zero. - Use that sign change to justify an absolute minimum.

1. Set the equations equal. Factoring the positive quantity \(k+\frac1{k^2}\) gives \(x^2=4\), so \(x=\pm2\). 2. Differentiate the supplied area function: \(A'(k)=\frac{32}{3}(1-\frac{2}{k^3})\). The only positive critical number is \(k=\sqrt[3]{2}\). 3. \(A'(k)<0\) for \(0<k<\sqrt[3]{2}\) and \(A'(k)>0\) for \(k>\sqrt[3]{2}\), so this critical value gives the absolute minimum. 4. \(A(\sqrt[3]{2})=16\sqrt[3]{2}\approx20.16\) square units.

a) \(x=\pm2\) b) \(k=\sqrt[3]{2}\); minimum area \(16\sqrt[3]{2}\approx20.16\) square units. c) The area decreases before this value and increases after it, so the minimum is absolute.

- Factor the intersection equation before solving it. - Treat the supplied area expression as the objective function. - Its derivative factors into terms whose signs are easy to track for \(a>0\). - Use the derivative sign change to establish the global behavior.

1. Solve \(ax^2=(a+1)x\). The intersection x-coordinates are \(x=0\) and \(x=\frac{a+1}{a}\). 2. Differentiate the supplied area model: \(A'(a)=\frac{(a+1)^2(a-2)}{6a^3}\). The only positive critical number is \(a=2\). 3. Since \(A'(a)<0\) on \(0,2\) and \(A'(a)>0\) on \(2,\infty\), the area decreases and then increases. Thus \(a=2\) gives the absolute minimum. 4. \(A(2)=\frac98=1.125\) square units.

a) \(x=0\) and \(x=\frac{a+1}{a}\) b) The area is minimized at \(a=2\); the minimum area is \(\frac98=1.125\) square units. c) The derivative changes from negative to positive at \(a=2\), so the minimum is absolute.

- Use symmetry to express the width as \(2x\). - The height is the function value \(f(x)\). - Differentiate the resulting rational area function. - Report the rectangle's side lengths, not just the value of \(x\).

1. Let the upper vertices be \((\pm x, f(x))\), where \(x>0\). The width is \(2x\), and the height is \(f(x)\). Thus \(A(x)=2x\left(\frac{4}{x^2+1}\right)=\frac{8x}{x^2+1}\). 2. Differentiate: \(A'(x)=\frac{8(1-x^2)}{(x^2+1)^2}\). The only positive critical number is \(x=1\). 3. The derivative is positive for \(0<x<1\) and negative for \(x>1\). Also, the area approaches \(0\) at both ends of the positive domain, so \(x=1\) gives the absolute maximum. 4. The width is \(2x=2\), and the height is \(f(1)=\frac{4}{2}=2\).

The maximum-area rectangle has side lengths \(2\) and \(2\).

- Find the secant slope from the two given points. - Subtract the function from the secant line to form the vertical-distance function. - Find the critical number of that difference. - Check the interval endpoints.

1. The endpoint values are \(f(1)=4\) and \(f(4)=1\). The secant slope is \(m=\frac{1-4}{4-1}=-1\). Thus \(s(x)=-x+5\). 2. The vertical distance is \(d(x)=s(x)-f(x)=-x+5-\frac{4}{x}\). 3. Differentiate: \(d'(x)=-1+\frac{4}{x^2}\). Setting the derivative equal to zero gives \(x=2\). 4. Since \(d''(x)=-\frac{8}{x^3}<0\) on the interval, the critical number gives a maximum. Also, \(d(1)=d(4)=0\), so it is the absolute maximum. 5. The maximum distance is \(d(2)=-2+5-2=1\).

a) \(s(x)=-x+5\) b) The maximum vertical distance occurs at \(x=2\), and the distance is \(1\).

- Find the line's slope from the two given points. - Subtract the line from the logarithmic function. - Differentiate using \(\frac{d}{dx}\ln(x+1)=\frac{1}{x+1}\). - Keep both the maximizing x-value and maximum difference exact.

1. The slope of the line through \((0, 0)\) and \((e-1, 1)\) is \(m=\frac{1}{e-1}\). Therefore, \(g(x)=\frac{x}{e-1}\). 2. The difference function is \(d(x)=\ln(x+1)-\frac{x}{e-1}\). 3. Differentiate: \(d'(x)=\frac{1}{x+1}-\frac{1}{e-1}\). Setting the derivative equal to zero gives \(x=e-2\). 4. Since \(d''(x)=-\frac{1}{(x+1)^2}<0\), this critical number gives the maximum. The endpoint values are both zero. 5. The exact maximum difference is \(d(e-2)=\ln(e-1)-\frac{e-2}{e-1}\).

a) \(g(x)=\frac{x}{e-1}\) b) The difference is greatest at \(x=e-2\). Its maximum value is \(\ln(e-1)-\frac{e-2}{e-1}\).

- Represent a general point on the graph as \((x, e^x)\). - Minimize the squared distance to avoid a square root. - Test a simple value such as \(x=0\) in the critical-number equation. - Use the second derivative to establish uniqueness.

1. A general point on the graph is \((x, e^x)\). Minimize the squared distance \(g(x)=(x-1)^2+e^{2x}\). 2. Differentiate: \(g'(x)=2(x-1)+2e^{2x}\). At \(x=0\), \(g'(0)=-2+2=0\). 3. The second derivative is \(g''(x)=2+4e^{2x}>0\) for every real \(x\). Thus \(g\) is strictly convex, so this critical number gives the unique absolute minimum. 4. Since \(f(0)=1\), the closest point is \((0, 1)\).

The closest point is \(Q=(0, 1)\).

- Find the second derivative of \(f\) first. - A vertical distance is the absolute difference of the two function values. - Check the sign of the difference on the stated domain before removing the absolute value. - Maximize the resulting distance function with a derivative test.

1. Differentiate twice: \(f'(x)=(-x^2+4x+1)e^{-x}\) and \(f''(x)=(x^2-6x+3)e^{-x}\). 2. The vertical distance is \(L(a)=|f(a)-f''(a)|\). The difference simplifies to \((4a-6)e^{-a}\). Since \(a>\frac{3}{2}\), this expression is positive, so \(L(a)=(4a-6)e^{-a}\). 3. Differentiate: \(L'(a)=(10-4a)e^{-a}\). The only critical number in the domain is \(a=\frac{5}{2}\). 4. The derivative is positive for \(\frac{3}{2}<a<\frac{5}{2}\) and negative for \(a>\frac{5}{2}\). Therefore, \(L\) reaches its absolute maximum at \(a=\frac{5}{2}\).

\(a=\frac{5}{2}\).

- Express the rectangle's area using the coordinates of its graph point. - Simplify the area function before differentiating. - Use the second derivative and end behavior to verify the absolute minimum. - Report both coordinates.

1. If the vertex is \((x, f(x))\), the area is \(A(x)=xf(x)=x^2+\frac{54}{x}\). 2. Differentiate: \(A'(x)=2x-\frac{54}{x^2}\). Setting the derivative equal to zero gives \(2x^3=54\), so \(x=3\). 3. Since \(A''(x)=2+\frac{108}{x^3}>0\) for \(x>0\), the area function is strictly convex. Also, \(A(x)\to\infty\) as \(x\to0^+\) or \(x\to\infty\), so the critical number gives the absolute minimum. 4. The y-coordinate is \(f(3)=\frac{27+54}{9}=9\).

The area is minimized when the vertex is \((3, 9)\).

- Write the line through \(P\) using its slope. - Express both intercepts in terms of the slope. - Use the condition that both intercepts are positive. - Minimize their sum and check the end behavior of the domain.

1. Let the line have slope \(m<0\). Its equation through \(P\) is \(y-1=m(x-4)\). The intercepts are \(x_0=4-\frac{1}{m}\) and \(y_0=1-4m\). 2. Their sum is \(S(m)=5-\frac{1}{m}-4m\), where \(m<0\). 3. Differentiate: \(S'(m)=\frac{1}{m^2}-4\). Setting the derivative equal to zero gives \(m=-\frac{1}{2}\). 4. Since \(S''(m)=-\frac{2}{m^3}>0\) at \(m=-\frac{1}{2}\), and \(S(m)\to\infty\) as \(m\to0^-\) or \(m\to-\infty\), this point gives the absolute minimum. 5. Therefore, \(x_0=4-\frac{1}{-1/2}=6\) and \(y_0=1-4\left(-\frac{1}{2}\right)=3\).

The minimizing intercepts are \(x_0=6\) and \(y_0=3\).

- Use the shown rectangle inside the isosceles triangle to identify a pair of similar triangles. - Relate the triangle base to its height. - Determine the domain for \(h\). - Minimize the resulting one-variable area function.

1. Let \(g\) be the triangle's base length. Similar triangles give \(\frac{h-4}{10}=\frac{h}{g}\). Thus \(g=\frac{10h}{h-4}\), where \(h>4\). 2. The triangle area is \(A(h)=\frac{1}{2}gh=\frac{5h^2}{h-4}\). 3. Differentiate: \(A'(h)=\frac{5h(h-8)}{(h-4)^2}\). The only critical number in the domain is \(h=8\,\text{cm}\). 4. The derivative changes from negative to positive at \(h=8\). Also, \(A(h)\to\infty\) as \(h\to4^+\) or \(h\to\infty\), so this is the absolute minimum.

The triangle's area is minimized when its height is \(h=8\,\text{cm}\).

- Use the fixed volume to eliminate the height from the cost. - Keep the radius restricted to positive values. - Use both the second derivative and the behavior at the ends of the positive domain to justify an absolute minimum.

1. The volume constraint \(\pi r^2h=27\pi\) gives \(h=\frac{27}{r^2}\), with \(r>0\). 2. The base and lid cost \(2\pi r^2\) cents, while the curved side costs \(4\pi rh\) cents. Thus \(C(r)=2\pi r^2+\frac{108\pi}{r}\). 3. Differentiate: \(C'(r)=4\pi r-\frac{108\pi}{r^2}\). The only positive critical number satisfies \(4\pi r^3=108\pi\), so \(r=3\,\text{in}\). 4. The second derivative is \(C''(r)=4\pi+\frac{216\pi}{r^3}>0\) for \(r>0\), so \(C\) is strictly convex. Also \(C(r)\to\infty\) as \(r\to0^+\) and as \(r\to\infty\). Therefore the critical number gives the absolute minimum. 5. The corresponding height is \(h=\frac{27}{3^2}=3\,\text{in}\).

The absolute minimum material cost occurs when \(r=3\,\text{in}\) and \(h=3\,\text{in}\).

- Use symmetry to find the triangle's base length. - Write the area as one-half times base times height. - Apply the product and chain rules. - Give both the maximizing x-value and the coordinates of \(P\).

1. The triangle has base \(2x\) and height \(f(x)\), so \(A(x)=\frac{1}{2}(2x)(6e^{-0.1x^2})=6xe^{-0.1x^2}\). 2. Differentiate using the product and chain rules: \(A'(x)=(6-1.2x^2)e^{-0.1x^2}\). Since the exponential factor is positive, the critical number satisfies \(6-1.2x^2=0\). Thus \(x=\sqrt{5}\approx2.24\). 3. The derivative changes from positive to negative at \(x=\sqrt{5}\). Also, the area approaches zero as \(x\to0^+\) or \(x\to\infty\), so this is the absolute maximum. 4. The y-coordinate is \(f(\sqrt{5})=6e^{-0.5}\approx3.64\).

The area is maximized at \(x=\sqrt{5}\approx2.24\). The point is \(P(\sqrt{5}, 6e^{-0.5})\approx(2.24, 3.64)\).

- Multiply \(x\) by \(f(x)\) to form the area function. - Determine the first-quadrant domain from the intercepts. - Use the quotient rule and solve the resulting quadratic equation. - Check the endpoints.

1. The area is \(A(x)=xf(x)=\frac{8x-2x^2}{x+2}\). The graph meets the x-axis at \(x=4\), so \(0\le x\le4\). 2. Differentiate: \(A'(x)=\frac{-2x^2-8x+16}{(x+2)^2}\). The numerator is zero when \(x^2+4x-8=0\). The only solution in the domain is \(x=2\sqrt{3}-2\approx1.46\). 3. The endpoint areas are zero, and the derivative changes from positive to negative at the interior critical number. Therefore, it gives the absolute maximum. 4. Substitution gives \(A_{\max}=16-8\sqrt{3}\approx2.14\).

The area is maximized at \(x=2\sqrt{3}-2\approx1.46\). The maximum area is \(16-8\sqrt{3}\approx2.14\) square units.

- Use the fixed volume to eliminate the cylindrical height. - Include the condition that the cylinder height must remain nonnegative. - After finding the critical radius, verify it on the entire feasible domain before simplifying the ratio.

1. The fixed volume is \(V=\pi r^2h+\frac23\pi r^3\), so \(h=\frac{V}{\pi r^2}-\frac23r\). 2. Feasibility requires \(r>0\) and \(h\ge0\), so \(0<r\le\sqrt[3]{\frac{3V}{2\pi}}\). 3. Let the unit cost for the cylindrical side and lid be \(k>0\). Ignoring the positive common factor \(k\), the weighted cost is \(C=2\pi rh+5\pi r^2\). 4. Substitute the height expression: \(C(r)=\frac{2V}{r}+\frac{11}{3}\pi r^2\). 5. Differentiate: \(C'(r)=-\frac{2V}{r^2}+\frac{22}{3}\pi r\). The critical-number equation gives \(V=\frac{11}{3}\pi r^3\). 6. Since \(C''(r)=\frac{4V}{r^3}+\frac{22}{3}\pi>0\), \(C\) is strictly convex, and \(C(r)\to\infty\) as \(r\to0^+\). Hence the feasible critical number gives the absolute minimum. 7. Substitute the critical-number relation into \(h=\frac{V}{\pi r^2}-\frac23r\) to get \(h=3r\).

The absolute minimum cost occurs when \(h:r=3:1\).

- Use the coordinates of the moving vertex as the rectangle's side lengths. - Factor the derivative before deciding which critical numbers lie in the feasible interval. - A sign change over the entire interval establishes the global maximum.

1. The rectangle has width \(u\) and height \(f(u)\), so \(A(u)=u f(u)=u^4-12u^2+16u\). 2. Differentiate: \(A'(u)=4u^3-24u+16=4(u-2)(u^2+2u-2)\). The only critical number in \((0,2)\) is \(u=\sqrt3-1\). 3. The derivative is positive for \(0<u<\sqrt3-1\) and negative for \(\sqrt3-1<u<2\), so this critical number gives the absolute maximum. Equivalently, \(A(u)\to0\) at both boundary values. 4. The y-coordinate is \(f(\sqrt3-1)=18-6\sqrt3\). Thus \(P=(\sqrt3-1,18-6\sqrt3)\).

a) \(A(u)=u^4-12u^2+16u\) b) \(P=(\sqrt3-1,18-6\sqrt3)\approx(0.732,7.608)\) c) The maximum is absolute because \(A'\) changes from positive to negative at the only interior critical number.

- Use \(2x\) for the base length. - The height is \(10-f(x)\). - Substitute \(z=x^2\) in the critical-number equation. - Evaluate the area at the maximizing value.

1. The horizontal base has length \(2x\), and the vertical height is \(10-f(x)\). Thus \(A(x)=\frac{1}{2}(2x)(10-f(x))=-x^5+6x^3+8x\). The condition \(f(x)\le10\) gives \(0<x\le\sqrt{3+\sqrt{17}}\). 2. Differentiate: \(A'(x)=-5x^4+18x^2+8\). Let \(z=x^2\). The critical-point equation becomes \(5z^2-18z-8=0\). Its positive solution is \(z=4\), so \(x=2\). 3. The second derivative satisfies \(A''(2)=-88<0\). The area approaches zero as \(x\to0^+\) and equals zero at \(x=\sqrt{3+\sqrt{17}}\), so \(x=2\) gives the absolute maximum. 4. The maximum area is \(A(2)=-32+48+16=32\).

The area is maximized when \(x=2\), and the maximum area is \(32\) square units.

- Separate the parameter-independent and parameter-dependent parts of the function. - In the supplied objective, determine which factors of the derivative can change sign. - Use both sides of the critical number and the positive parameter domain.

1. Rewrite \(f_k(x)=x+\frac{4x-x^2}{k}\). A point is independent of \(k\) when \(4x-x^2=0\), so \(x=0\) or \(x=4\). The common points are \((0,0)\) and \((4,4)\). 2. Differentiate the supplied objective: \(A'(k)=\frac{(k+4)^2(2k-4)}{6k^2}\). The only positive critical number is \(k=2\). 3. Since all factors except \(2k-4\) are positive, \(A'(k)<0\) for \(0<k<2\) and \(A'(k)>0\) for \(k>2\). Thus \(k=2\) gives the absolute minimum. 4. Also \(A(k)\to\infty\) as \(k\to0^+\) or \(k\to\infty\), confirming the global result. 5. \(A(2)=18\).

a) \((0,0)\) and \((4,4)\) b) The absolute minimum occurs at \(k=2\), with value \(18\) square units.

- Evaluate both graphs and their derivatives at the common point. - At the specified x-value, subtract the lower y-value from the upper y-value. - The domain of the parameter is positive, so only positive critical numbers are admissible. - Compare the critical number with the behavior as the parameter approaches both ends of its domain.

1. At \(x=1\), \(f_k(1)=g_k(1)=0\). Their slopes are \(f_k'(1)=k\) and \(g_k'(1)=-\frac1k\), whose product is \(-1\). Thus the tangent lines are perpendicular. 2. At \(x=e\), \(f_k(e)=k\) and \(g_k(e)=-\frac{e-1}{k}\). Therefore, \(D(k)=k+\frac{e-1}{k}\). 3. Differentiate: \(D'(k)=1-\frac{e-1}{k^2}\). The positive critical number is \(k=\sqrt{e-1}\). 4. Since \(D''(k)=\frac{2(e-1)}{k^3}>0\) for \(k>0\), and \(D(k)\to\infty\) as \(k\to0^+\) or \(k\to\infty\), this is the absolute minimum. 5. The minimum distance is \(D(\sqrt{e-1})=2\sqrt{e-1}\).

1) The tangent slopes are \(k\) and \(-\frac1k\), so the graphs meet orthogonally. 2) \(D(k)=k+\frac{e-1}{k}\) 3) \(k=\sqrt{e-1}\), with absolute minimum distance \(2\sqrt{e-1}\).

- When one object is above another, subtract the lower altitude from the higher altitude. - Simplify the difference before differentiating. - Find and classify the critical times. - Compare the values at the critical times with both endpoint values.

1. Subtract the altitude functions: \(d(t) = -t^3 + 9t^2 - 5t + 60 - (10t + 50) = -t^3 + 9t^2 - 15t + 10\). 2. Differentiate: \(d'(t) = -3t^2 + 18t - 15 = -3(t - 1)(t - 5)\). The critical times are \(t = 1\) and \(t = 5\). 3. Use \(d''(t) = -6t + 18\). Since \(d''(1) = 12 > 0\), \(t = 1\) is a local minimum. Since \(d''(5) = -12 < 0\), \(t = 5\) is a local maximum. 4. Evaluate the candidates: \(d(0) = 10\), \(d(1) = 3\), \(d(5) = 35\), and \(d(7) = 3\). 5. The greatest separation is \(35\,\text{m}\), reached after \(5\) minutes.

a) \(d(t) = -t^3 + 9t^2 - 15t + 10\) b) The maximum separation is \(35\,\text{m}\) at \(t = 5\,\text{min}\).

- Intersections satisfy \(f(x)=g(x)\). - Write the vertical distance as \(f(x)-g(x)\) and maximize it. - The derivative gives each tangent slope. - Relate each tangent slope to its direction angle. - Use the angle-between-lines formula for two known slopes.

1. Solving \(f(x)=g(x)\) gives \(\frac{7}{20}x^2-\frac{21}{10}x=0\), so \(x=0\) or \(x=6\). Both functions equal \(0\) there, giving tips \((0, 0)\) and \((6, 0)\). 2. The vertical width is \(h(x)=f(x)-g(x)=-\frac{7}{20}x^2+\frac{21}{10}x\). Since \(h'(x)=-\frac{7}{10}x+\frac{21}{10}\), its critical number is \(x=3\). The second derivative is negative, and \(h(3)=\frac{63}{20}=3.15\). The maximum width is \(3.15\,\text{cm}\) at \(x=3\,\text{cm}\). 3. At \(x=6\), the tangent slopes are \(f'(6)=-\frac{3}{2}\) and \(g'(6)=\frac{3}{5}\). Thus, \(\tan\theta=\left|\frac{-\frac{3}{2}-\frac{3}{5}}{1+(-\frac{3}{2})(\frac{3}{5})}\right|=21\). Therefore, \(\theta=\arctan(21)\approx87.3^\circ\).

a) \((0, 0)\) and \((6, 0)\) b) \(3.15\,\text{cm}\) at \(x=3\,\text{cm}\) c) \(87.3^\circ\)

- Express the triangle's base and height in terms of the same variable. - The exponential factor never changes sign. - Use the sign of the derivative on both sides of the critical number. - Compare the critical number with the limiting area near both ends of the domain.

1. The triangle has base \(u\) and height \(f(u)=6e^{-0.4u}\), so \(A(u)=\frac12\cdot u\cdot6e^{-0.4u}=3ue^{-0.4u}\). 2. Differentiate: \(A'(u)=(3-1.2u)e^{-0.4u}\). Since the exponential factor is positive, the only critical number is \(u=2.5\). 3. The derivative is positive for \(0<u<2.5\) and negative for \(u>2.5\), so the area increases and then decreases. The maximum area is \(A(2.5)=7.5e^{-1}=\frac{7.5}{e}\approx2.76\) square units. 4. Also, \(A(u)\to0\) as \(u\to0^+\) and as \(u\to\infty\). Therefore, the maximum at \(u=2.5\) is absolute.

a) \(A(u)=3ue^{-0.4u}\) b) \(u=2.5\); maximum area \(\frac{7.5}{e}\approx2.76\) square units c) It is absolute because \(A\) increases then decreases and approaches \(0\) at both ends of the domain.

- Smooth connections to horizontal platforms require matching heights and zero slopes. - Begin with a general cubic polynomial and its derivative. - Use the four endpoint conditions to determine the cubic. - The extremum of the slope occurs where the second derivative is \(0\). - Compare the maximum derivative value with \(0.6\).

1. Smooth horizontal connections give \(f(0)=0\), \(f'(0)=0\), \(f(4)=2\), and \(f'(4)=0\). Solving for a cubic gives \(f(x)=-\frac{1}{16}x^3+\frac{3}{8}x^2\). 2. The second derivative is \(f''(x)=-\frac{3}{8}x+\frac{3}{4}\), so the inflection point occurs at \(x=2\). Since \(f(2)=1\), the point is \(W=(2, 1)\). This is where the ramp is steepest. 3. The maximum slope is \(f'(2)=\frac{3}{4}=0.75\), or \(75\%\). Since \(75\%>60\%\), the design does not meet the requirement.

a) \(f(x)=-\frac{1}{16}x^3+\frac{3}{8}x^2\) b) \(W=(2, 1)\); the ramp is steepest there. c) No. The maximum grade is \(75\%\), which exceeds \(60\%\).

- Express the vertical difference between the curve and the line as a function. - Set the derivative of the harvest function equal to zero. - Convert from hundreds of deer to individual deer. - What does \(f(x) = x\) mean from one year to the next?

1. Write the harvest function: \(E(x) = -0.12x^2 + 1.44x\). 2. Differentiate: \(E'(x) = -0.24x + 1.44\). Set \(E'(x) = 0\): \(x = 6\). 3. Since \(E''(x) = -0.24 < 0\), this gives the absolute maximum. The optimal population is \(6\) hundred deer, or \(600\) deer. 4. Evaluate \(E(6) = 4.32\), which represents \(432\) deer per year. 5. The positive intersection satisfies \(f(x) = x\). Solving gives \(x = 12\), so a population of \(1200\) deer is an equilibrium: without harvesting or other changes, the model predicts the same population one year later.

a) \(600\) deer b) \(432\) deer per year c) The intersection at \(x = 12\) represents an equilibrium population of \(1200\) deer.

- The vertical intercept of a cost graph represents fixed cost. - Profit is the vertical distance between the revenue line and the cost curve. - Use the derivative of the profit function to find its maximum. - Convert hundreds of dollars to dollars in the final answer.

1. The fixed cost is \(C(0)=10\) hundreds of dollars, or \(\$1000\). 2. At \(x=8\), \(C(8)=0.2(8)^3-2(8)^2+8(8)+10=48.4\) hundreds of dollars, or \(\$4840\). 3. The revenue graph is the line \(R(x)=8x\), passing through \((0,0)\) and \((10,80)\). Its intersections with the cost graph occur at approximately \(x=2.6\) and \(x=9.4\), the break-even production levels. 4. Profit, in hundreds of dollars, is \(P(x)=R(x)-C(x)=-0.2x^3+2x^2-10\). 5. Differentiate: \(P'(x)=-0.6x^2+4x=-0.2x(3x-20)\). The relevant interior critical number is \(x=\frac{20}{3}\approx 6.67\). 6. The derivative changes from positive to negative there, so profit is maximized at \(x=\frac{20}{3}\). 7. The maximum profit is \(P\left(\frac{20}{3}\right)=\frac{530}{27}\approx 19.63\) hundreds of dollars, or approximately \(\$1962.96\).

a) The fixed cost is \(\$1000\), and the cost of producing \(8\) units is \(\$4840\). b) The break-even production levels are approximately \(x=2.6\) and \(x=9.4\). c) Profit is maximized at \(x_{\text{opt}}=\frac{20}{3}\approx 6.67\) units, with a maximum profit of approximately \(\$1962.96\).

- Use the sign of the first derivative over the complete displayed domain to locate the highest point. - For steepest descent, optimize the first derivative itself on the stated right-side interval. - Use the second derivative to determine where the slope changes from decreasing to increasing.

1. Differentiate: \(f'(x)=(-0.5x^2+2)e^{-0.5x}\) and \(f''(x)=(0.25x^2-x-1)e^{-0.5x}\). 2. Because the exponential factor is positive, \(f'(x)>0\) for \(-2<x<2\) and \(f'(x)<0\) for \(2<x\le8\). Thus \(f\) increases up to \(x=2\) and decreases afterward on the full displayed domain, so the absolute maximum occurs at \(x=2\). 3. The maximum height is \(f(2)=16e^{-1}\approx5.89\,\text{m}\). 4. On \(2<x\le8\), the steepest descent occurs where \(f'\) has its absolute minimum. Solve \(f''(x)=0\): \(x^2-4x-4=0\), giving \(x=2+2\sqrt2\approx4.83\). 5. The second derivative is negative on \((2,2+2\sqrt2)\) and positive on \((2+2\sqrt2,8]\), so \(f'\) decreases and then increases. Therefore this point gives the absolute minimum slope on the right side. 6. The slope is approximately \(-0.864\), corresponding to a downward grade of about \(86.4\%\).

a) The absolute maximum height is \(\frac{16}{e}\,\text{m}\approx5.89\,\text{m}\). b) The steepest descent occurs at \(x=2+2\sqrt2\approx4.83\,\text{m}\). The slope is approximately \(-0.864\), corresponding to a downward grade of about \(86.4\%\).

- Express the width and height of the rectangle in terms of \(u\). - Use area equals width times height. - Differentiate the area function using the product and chain rules. - Use the sign of the derivative and the end behavior to confirm an absolute maximum.

1. The rectangle has width \(2u\) and height \(g(u)=8e^{-u^2}\). Therefore, \(A(u)=2u\cdot8e^{-u^2}=16ue^{-u^2}\). 2. Using the product and chain rules, \(A^{\prime}(u)=16e^{-u^2}(1-2u^2)\). 3. Since \(e^{-u^2}>0\), \(A^{\prime}(u)=0\) for positive \(u\) only when \(u=\frac{1}{\sqrt{2}}\). 4. The derivative is positive for \(0<u<\frac{1}{\sqrt{2}}\) and negative for \(u>\frac{1}{\sqrt{2}}\). Also, \(A(u)\to0\) as \(u\to0^+\) and as \(u\to\infty\), so this critical point gives the unique absolute maximum. 5. The maximum area is \(A\left(\frac{1}{\sqrt{2}}\right)=8\sqrt{2}e^{-1/2}\approx6.86\) square units.

a) \(A(u)=16ue^{-u^2}\) b) \(u=\frac{1}{\sqrt{2}}\approx0.71\) c) \(A_{\max}=8\sqrt{2}e^{-1/2}\approx6.86\) square units

- Determine where the square root is defined and where the physical model begins. - Reflecting across the x-axis changes the sign of every output. - Use the derivative to find the tangent slope, then relate slope to angle. - First maximize the upper boundary; symmetry gives the full width.

1. a) The radicand must satisfy \(9-x\ge0\), so \(x\le9\). The modeled hull begins at the intercept \(x=-3\). Therefore, \(D_f=[-3,9]\). 2. b) Reflection across the x-axis changes each output to its opposite, so \(g(x)=-f(x)=-0.2(x+3)\sqrt{9-x}\). 3. c) By the product and chain rules, \(f'(x)=0.1\frac{15-3x}{\sqrt{9-x}}\). Thus, \(f'(-3)=\frac{2.4}{\sqrt{12}}=\frac{2\sqrt{3}}{5}\approx0.6928\). 4. The crossing angle is \(\alpha=\arctan\!\left(\frac{2\sqrt{3}}{5}\right)\approx34.7^\circ\). 5. d) The upper boundary is largest where \(f'(x)=0\), which gives \(15-3x=0\), so \(x=5\). The endpoint values are \(0\), while \(f(5)=0.2(8)(2)=3.2\,\text{cm}\). 6. By symmetry, the maximum width is \(2f(5)=6.4\,\text{cm}\).

a) \(D_f=[-3,9]\) b) \(g(x)=-0.2(x+3)\sqrt{9-x}\) c) \(\alpha\approx34.7^\circ\) d) \(6.4\,\text{cm}\)

- Minimize the squared distance instead of carrying the outer square root. - Completing the square can show whether the distance minimum is global. - Compare the slope from the fixed point to the closest point with the derivative at that point.

1. By the distance formula, \(d(x)=\sqrt{(x-3)^2+(2\sqrt{x})^2}=\sqrt{x^2-2x+9}\). 2. Since the square-root function is increasing, minimize \(h(x)=x^2-2x+9=(x-1)^2+8\) for \(x\ge0\). Its absolute minimum occurs at \(x=1\), so \(Q_E=(1,2)\). 3. The slope of \(\overline{PQ_E}\) is \(\frac{2-0}{1-3}=-1\). 4. Since \(f'(x)=\frac1{\sqrt{x}}\), the tangent slope at \(x=1\) is \(1\). The product of the two slopes is \(-1\), so the segment and tangent line are perpendicular.

a) \(d(x)=\sqrt{x^2-2x+9}\) b) \(Q_E=(1,2)\) c) The segment slope is \(-1\) and the tangent slope is \(1\), so the lines are perpendicular.

- Write the width and height of the rectangle in terms of \(u\). - Substitute the function value for the height. - Differentiate the area function using the product rule. - Use the sign of the derivative to confirm that the critical number gives a maximum.

1. The rectangle has width \(u\) and height \(f(u)\), so \(A(u)=u\cdot f(u)=12u^2e^{-0.5u}\). 2. Using the product and chain rules, \(A^{\prime}(u)=(24u-6u^2)e^{-0.5u}=6u(4-u)e^{-0.5u}\). 3. For \(u>0\), the only critical number is \(u=4\). The derivative is positive on \((0, 4)\) and negative on \((4, \infty)\). 4. Also, \(A(u)\to0\) as \(u\to0^+\) and as \(u\to\infty\), so \(u=4\) gives the unique absolute maximum. 5. The maximum area is \(A(4)=192e^{-2}\approx25.98\) square units.

The area is maximized when \(u=4\). The maximum area is \(192e^{-2}\approx25.98\) square units.

- What condition must the first derivative satisfy at an interior local extremum? - What does the sign of the first derivative tell you about total cost? - Profit equals revenue minus cost. - Keep the units in thousands of dollars consistent. - Use derivatives and endpoint comparisons to find the absolute maximum.

1. Differentiate the cost function: \(K'(x) = x^2 - 8x + 20\). 2. The discriminant of \(x^2 - 8x + 20 = 0\) is \((-8)^2 - 4(1)(20) = -16\), so \(K'(x)\) has no real zeros. Since \(K'(0) = 20 > 0\), the cost function is increasing everywhere and has no local extrema. This is reasonable because total cost generally rises as production increases. 3. Revenue is \(R(x) = 40x\), so \(P(x) = R(x) - K(x) = -\frac{1}{3}x^3 + 4x^2 + 20x - 50\). 4. Differentiate: \(P'(x) = -x^2 + 8x + 20\). Solving \(P'(x) = 0\) gives \(x = -2\) and \(x = 10\). Only \(x = 10\) is in the production interval. 5. Since \(P''(x) = -2x + 8\) and \(P''(10) = -12 < 0\), \(x = 10\) gives a local maximum. 6. Compare values: \(P(0) = -50\), \(P(10) = \frac{650}{3}\), and \(P(15) = 25\). Therefore, the maximum profit is \(\frac{650}{3}\) thousand dollars, or approximately \(\$216{,}666.67\), when \(1000\) units are produced.

a) \(K'(x) = x^2 - 8x + 20\) has no real zeros, so \(K\) has no local extrema and is increasing throughout the interval. b) \(P(x) = -\frac{1}{3}x^3 + 4x^2 + 20x - 50\). The maximum profit is approximately \(\$216{,}666.67\), achieved by producing \(1000\) units.

- Solve each quadratic equation before comparing the intercepts. - Evaluate both functions at \(x=0\) to obtain the vertical separation. - Use the parameter interval when choosing the critical number. - A derivative sign change can establish the absolute maximum on the whole interval.

1. Solving \(f_k(x)=0\) gives \(k(x^2-k^2)=0\). Since \(k>0\), the zeros are \(x=\pm k\). 2. Solving \(g_k(x)=0\) gives \(\frac1k(x^2-k^2)=0\), so its zeros are also \(x=\pm k\). Thus both graphs have x-intercepts \((-k,0)\) and \((k,0)\). 3. Both formulas contain only \(x^2\) and constants, so both functions are even and their graphs are symmetric about the y-axis. 4. On the y-axis, \(f_k(0)=-k^3\) and \(g_k(0)=-k\). Since \(0<k<1\), \(D(k)=(-k^3)-(-k)=k-k^3\). 5. Differentiate: \(D'(k)=1-3k^2\). The critical number in \((0,1)\) is \(k=\frac1{\sqrt3}\). 6. The derivative is positive before this value and negative after it, so the critical point gives the absolute maximum. The maximum distance is \(D\left(\frac1{\sqrt3}\right)=\frac{2}{3\sqrt3}=\frac{2\sqrt3}{9}\).

a) \(x=\pm k\) b) Both graphs have x-intercepts \((-k,0)\) and \((k,0)\). c) Both functions are even. d) The maximum y-axis separation occurs at \(k=\frac1{\sqrt3}\), and the maximum distance is \(\frac{2\sqrt3}{9}\).

- Treat the parameter as constant during the first optimization. - After locating the minimum in \(x\), use its y-coordinate as a new function of the parameter. - For each positive-domain minimum, compare the critical point with behavior near both ends of the domain.

1. Treat \(k>0\) as a constant and differentiate with respect to \(x\): \(f_k'(x)=k-\frac{1}{kx^2}\) and \(f_k''(x)=\frac{2}{kx^3}>0\). 2. The critical-point equation gives \(x=\frac1k\), since \(x>0\). Because \(f_k''>0\), this is the unique local minimum. Also \(f_k(x)\to\infty\) as \(x\to0^+\) or \(x\to\infty\), so it is the absolute minimum for each \(k\). 3. Its y-coordinate is \(H(k)=f_k\left(\frac1k\right)=k^2+\frac{110}{k}+2\), so the minimum point is \(\left(\frac1k,k^2+\frac{110}{k}+2\right)\). 4. Differentiate \(H\): \(H'(k)=2k-\frac{110}{k^2}\). The only positive critical value is \(k=\sqrt[3]{55}\approx3.80\). 5. Since \(H''(k)=2+\frac{220}{k^3}>0\) for \(k>0\), and \(H(k)\to\infty\) as \(k\to0^+\) or \(k\to\infty\), this parameter gives the absolute smallest possible minimum y-coordinate.

Minimum point for fixed \(k\): \(\left(\frac1k,k^2+\frac{110}{k}+2\right)\) Optimal parameter: \(k=\sqrt[3]{55}\approx3.80\)

- Complete the optimization in \(x\) before optimizing the resulting minimum height in \(a\). - Keep both variables restricted to the positive domain. - Use global behavior, not only a second-derivative value at the critical number.

1. Treat \(a>0\) as a constant and differentiate with respect to \(x\): \(g_a'(x)=a-\frac{a^3}{x^2}\) and \(g_a''(x)=\frac{2a^3}{x^3}>0\). 2. The critical-number equation gives \(x=a\). Since \(g_a''>0\), this is the unique local minimum. Also \(g_a(x)\to\infty\) as \(x\to0^+\) or \(x\to\infty\), so it is the absolute minimum for each \(a\). 3. The minimum y-coordinate is \(H(a)=g_a(a)=2a^2+\frac5a\), giving minimum point \(\left(a,2a^2+\frac5a\right)\). 4. Differentiate: \(H'(a)=4a-\frac5{a^2}\). The only positive critical number is \(a=\sqrt[3]{\frac54}\approx1.08\). 5. Since \(H''(a)=4+\frac{10}{a^3}>0\) for \(a>0\), and \(H(a)\to\infty\) as \(a\to0^+\) or \(a\to\infty\), this parameter gives the absolute smallest possible minimum y-coordinate.

Minimum point for fixed \(a\): \(\left(a,2a^2+\frac5a\right)\) Optimal parameter: \(a=\sqrt[3]{\frac54}\approx1.08\)

- Remember that the open tank has no top when writing its surface-area formula. - Use the fixed volume to express the depth in terms of the radius. - Find and verify the minimum of the resulting one-variable area function. - For the closed cylinder, include both circular ends and repeat the same substitution before comparing the two ratios.

1. The volume constraint is \(\pi r^2h=50\), so \(h=\frac{50}{\pi r^2}\), with \(r>0\). 2. Because the tank has no top, its interior area is \(A=\pi r^2+2\pi rh\). Substituting for \(h\) gives \(A(r)=\pi r^2+\frac{100}{r}\). 3. Differentiate: \(A'(r)=2\pi r-\frac{100}{r^2}\). Setting \(A'(r)=0\) gives \(2\pi r^3=100\), so \(r=\sqrt[3]{\frac{50}{\pi}}\approx2.515\,\text{m}\). 4. Since \(A''(r)=2\pi+\frac{200}{r^3}>0\) for \(r>0\), this critical number gives a local minimum. Also, \(A(r)\to\infty\) as \(r\to0^+\) or as \(r\to\infty\), so it is the absolute minimum. 5. From the critical-number equation, \(\pi r^3=50\). Therefore, \(h=\frac{50}{\pi r^2}=r\approx2.515\,\text{m}\). Thus the open tank has optimal ratio \(h:r=1:1\). 6. For a closed cylinder with the same volume, substitute \(h=\frac{50}{\pi r^2}\) into \(S=2\pi r^2+2\pi rh\) to get \(S(r)=2\pi r^2+\frac{100}{r}\). Then \(S'(r)=4\pi r-\frac{100}{r^2}=0\), so \(\pi r^3=25\). Hence \(h=\frac{50}{\pi r^2}=2r\), and the optimal closed-cylinder ratio is \(2:1\).

a) \(A(r)=\pi r^2+\frac{100}{r}\), for \(r>0\). b) \(r\approx2.515\,\text{m}\) and \(h\approx2.515\,\text{m}\). c) Open tank: \(h:r=1:1\). Closed cylinder: \(h:r=2:1\).

- Write formulas for the surface area and volume of a rectangular box with a square base. - Use the fixed surface area to express the height in terms of the base side length. - Create a one-variable volume function. - Find the largest value of that function. - Compare the maximum with \(80\,\text{cm}^3\) and use continuity to justify your conclusion.

1. Let \(x\) be the side length of the square base and \(h\) be the height. The surface-area constraint is \(2x^2+4xh=120\). 2. Solve for the height: \(h=\frac{120-2x^2}{4x}=\frac{30}{x}-\frac{x}{2}\). For positive dimensions, \(0<x<\sqrt{60}\). 3. The volume is \(V(x)=x^2h=x^2\left(\frac{30}{x}-\frac{x}{2}\right)=30x-\frac{1}{2}x^3\). 4. Differentiate: \(V'(x)=30-\frac{3}{2}x^2\). Setting \(V'(x)=0\) gives \(x^2=20\), so \(x=\sqrt{20}\). Since \(V''(x)=-3x<0\) for \(x>0\), this critical number is a local maximum. 5. The maximum volume is \(V(\sqrt{20})=20\sqrt{20}=40\sqrt{5}\approx 89.44\,\text{cm}^3\). Also, \(V(x)\to 0\) as \(x\to 0^+\) and as \(x\to\sqrt{60}^{\, -}\), so this is the absolute maximum. 6. The function \(V\) is continuous on \((0, \sqrt{60})\). Because its values rise from near \(0\) to about \(89.44\) and then return toward \(0\), the Intermediate Value Theorem guarantees at least one box with volume exactly \(80\,\text{cm}^3\). Therefore, the requested box is possible.

Yes. The greatest possible volume with the given surface area is \(40\sqrt{5}\approx 89.44\,\text{cm}^3\), which is greater than \(80\,\text{cm}^3\). Continuity then guarantees that a volume of exactly \(80\,\text{cm}^3\) occurs.

- Use the surface-area and volume formulas for a closed cylinder. - Express the height in terms of the radius using the fixed surface area. - Write the volume as a function of the radius alone. - Find the absolute maximum on the physically meaningful domain. - Compare that maximum with the target volume.

1. For a closed cylinder, \(2\pi r^2+2\pi rh=24\pi\). Dividing by \(2\pi\) gives \(r^2+rh=12\), so \(h=\frac{12-r^2}{r}\). Positive dimensions require \(0<r<\sqrt{12}\). 2. The volume as a function of \(r\) is \(V(r)=\pi r^2h=\pi r^2\left(\frac{12-r^2}{r}\right)=\pi(12r-r^3)\). 3. Differentiate: \(V'(r)=\pi(12-3r^2)\). Setting \(V'(r)=0\) gives \(r^2=4\), so \(r=2\,\text{m}\). Since \(V''(r)=-6\pi r<0\) for \(r>0\), this critical point is a local maximum. 4. Also, \(V(r)\to 0\) as \(r\to 0^+\) and as \(r\to\sqrt{12}^{\, -}\). Therefore, the critical point gives the absolute maximum: \(V(2)=\pi(12\cdot 2-2^3)=16\pi\,\text{m}^3\). 5. Because \(16\pi<20\pi\), the requested capacity cannot be reached.

No. The greatest possible volume is \(16\pi\,\text{m}^3\), which is less than \(20\pi\,\text{m}^3\).

- Use the fixed area to express the rectangle height in terms of the width. - Determine the feasible width range from the requirement that the rectangle height remain positive. - A derivative sign change across the only feasible critical number establishes the global minimum.

1. The outside perimeter is \(P=3b+2h\). The area constraint is \(bh+\frac{\sqrt3}{4}b^2=10\), so \(h=\frac{10}{b}-\frac{\sqrt3}{4}b\). 2. Physical dimensions require \(b>0\) and \(h>0\), hence \(0<b<\sqrt{\frac{40}{\sqrt3}}\). 3. Substitute: \(P(b)=\left(3-\frac{\sqrt3}{2}\right)b+\frac{20}{b}\). 4. Differentiate: \(P'(b)=3-\frac{\sqrt3}{2}-\frac{20}{b^2}\). The critical number is \(b=\sqrt{\frac{40}{6-\sqrt3}}\approx3.06\,\text{cm}\), which lies in the feasible interval. 5. Since \(P''(b)=\frac{40}{b^3}>0\), \(P'\) is strictly increasing and changes from negative to positive at this critical number. Thus it gives the absolute minimum perimeter on the feasible interval.

The optimal width is \(b=\sqrt{\frac{40}{6-\sqrt3}}\approx3.06\,\text{cm}\).

- Write formulas for the volume and for each part of the cylinder’s surface area. - Multiply each type of surface area by its corresponding price. - Use the fixed volume to eliminate the height. - Find and verify the minimum of the resulting one-variable cost function.

1. The volume constraint is \(\pi r^2h=2\), so \(h=\frac{2}{\pi r^2}\), with \(r>0\). 2. The top and bottom have total area \(2\pi r^2\), and the curved side has area \(2\pi rh\). Therefore, the cost is \(C(r, h)=10(2\pi r^2)+5(2\pi rh)=20\pi r^2+10\pi rh\). Substituting for \(h\) gives \(C(r)=20\pi r^2+\frac{20}{r}\). 3. Differentiate: \(C'(r)=40\pi r-\frac{20}{r^2}\). Set the derivative equal to zero: \(40\pi r^3=20\Rightarrow r^3=\frac{1}{2\pi}\). Thus \(r=\sqrt[3]{\frac{1}{2\pi}}\approx 0.542\,\text{m}\). 4. Since \(C''(r)=40\pi+\frac{40}{r^3}>0\) for \(r>0\), the critical number is a local minimum. Also, \(C(r)\to\infty\) as \(r\to 0^+\) or as \(r\to\infty\), so it is the absolute minimum. 5. The corresponding height is \(h=\frac{2}{\pi r^2}=2\sqrt[3]{\frac{4}{\pi}}\approx 2.168\,\text{m}\).

The minimum-cost dimensions are \(r=\sqrt[3]{\frac{1}{2\pi}}\approx 0.542\,\text{m}\) and \(h=2\sqrt[3]{\frac{4}{\pi}}\approx 2.168\,\text{m}\).

- Use the zeros and vertex of the parabola for the building dimensions. - Minimizing a nonnegative square root is equivalent to minimizing its radicand. - The interval is closed, so include its endpoints in the absolute-minimum comparison. - Substitute the minimizing x-values back into the wall equation to obtain complete points.

1. \(f(-8)=f(8)=0\), so the floor extends from \(x=-8\) to \(x=8\) and is \(16\,\text{m}\) wide. The parabola has vertex \((0,8)\), so the maximum height is \(8\,\text{m}\). 2. The distance formula gives \(d(x)=\sqrt{x^2+[f(x)]^2}=\sqrt{0.015625x^4-x^2+64}\). 3. Minimizing \(d\) is equivalent to minimizing its radicand \(q(x)=0.015625x^4-x^2+64\). Then \(q'(x)=0.0625x^3-2x=x(0.0625x^2-2)\), so the critical numbers are \(x=0\) and \(x=\pm4\sqrt2\). 4. Compare all candidates on \([-8,8]\): \(q(0)=64\), \(q(\pm4\sqrt2)=48\), and \(q(\pm8)=64\). Thus the absolute minimum distance is \(\sqrt{48}=4\sqrt3\approx6.93\,\text{m}\), attained at \(x=\pm4\sqrt2\). 5. Since \(f(\pm4\sqrt2)=4\), the closest points are \((-4\sqrt2,4)\) and \((4\sqrt2,4)\).

a) Floor width \(16\,\text{m}\); maximum height \(8\,\text{m}\) b) \(d(x)=\sqrt{0.015625x^4-x^2+64}\) c) Absolute minimum distance \(4\sqrt3\approx6.93\,\text{m}\), at \(x=\pm4\sqrt2\) d) \((-4\sqrt2,4)\) and \((4\sqrt2,4)\)

- Find each graph's local maximum in terms of \(k\), then eliminate the parameter. - For part b, hold \(x\) fixed and view the expression as a function of \(k\). - Find the greatest y-value the family can attain at that fixed x-coordinate.

1. Differentiate with respect to \(x\): \(g_k'(x)=e^{k-x}(1-x+k)\). The only critical number is \(x=k+1\). Since \(g_k''(k+1)=-e^{-1}<0\), the graph has a local maximum there. Its coordinates are \(\left(k+1, k+\frac1e\right)\). Since \(x=k+1\), we have \(k=x-1\), so the maxima lie on \(y=x-1+\frac1e\). 2. Fix \(x\) and vary \(k\). Let \(t=k-x\). Then \(g_k(x)=x+t-te^t\). As a function of \(t\), its derivative is \(1-(t+1)e^t\). This derivative is positive for \(t<0\), zero at \(t=0\), and negative for \(t>0\). Also, the expression approaches \(-\infty\) as \(t\to\pm\infty\). Therefore, its absolute maximum occurs at \(t=0\), or \(k=x\), and the maximum value is \(x\). Thus, for each fixed \(x\), the values attained by the family are exactly \((-\infty, x]\). A point lies on no graph exactly when \(y>x\).

a) \(y=x-1+\frac1e\) b) All points \((x, y)\) such that \(y>x\).

- Relate the width and height of the rectangular cross section to the diameter of the circular log. - Use the constraint to write the stiffness as a function of one variable. - Differentiate using the product and chain rules and compare the critical number with the endpoints. - For the general proof, keep the diameter as \(d\) rather than substituting \(40\).

1. A rectangle inscribed in a circle of diameter \(d\) has diagonal \(d\), so \(b^2+h^2=d^2\). Thus \(h^2=d^2-b^2\), and \(S(b)=kb(d^2-b^2)^{3/2}\), for \(0\le b\le d\). 2. Differentiate using the product and chain rules: \(S'(b)=k\sqrt{d^2-b^2}(d^2-4b^2)\). The interior critical number satisfies \(d^2-4b^2=0\Rightarrow b=\frac{d}{2}\). The stiffness is \(0\) at both endpoints of the domain and positive at this interior critical number, so it gives the absolute maximum. 3. For \(d=40\,\text{cm}\), \(b=20\,\text{cm}\) and \(h=\sqrt{40^2-20^2}=20\sqrt{3}\approx 34.64\,\text{cm}\). 4. In general, when \(b=\frac{d}{2}\), \(h^2=d^2-\frac{d^2}{4}=\frac{3d^2}{4}\), so \(h=\frac{\sqrt{3}}{2}d\). Therefore, \(\frac{h}{b}=\frac{\frac{\sqrt{3}}{2}d}{\frac{1}{2}d}=\sqrt{3}\). The claimed ratio is independent of \(d\).

a) \(b=20\,\text{cm}\) and \(h=20\sqrt{3}\approx 34.64\,\text{cm}\). b) For any diameter \(d\), the maximizing dimensions satisfy \(b=\frac{d}{2}\) and \(h=\frac{\sqrt{3}}{2}d\), so \(\frac{h}{b}=\sqrt{3}\).

- First use the log’s diameter to relate the width and height of the inscribed rectangle. - The area and load-capacity objectives lead to different one-variable functions. - Find the maximizing dimensions separately for each objective. - In the percent comparison, the proportionality constant \(k\) cancels.

1. The log has diameter \(60\,\text{cm}\). Because the rectangular cross section is inscribed in the circle, \(b^2+h^2=60^2=3600\). 2. For part a, the area is \(A(b)=b\sqrt{3600-b^2}\). Differentiating gives \(A'(b)=\frac{3600-2b^2}{\sqrt{3600-b^2}}\). The interior critical number satisfies \(b^2=1800\), so \(b=h=\sqrt{1800}=30\sqrt{2}\approx 42.43\,\text{cm}\). The area is \(0\) at the endpoints of the physical domain, so this critical number gives the absolute maximum. 3. For part b, \(T(b)=kb(3600-b^2)=k(3600b-b^3)\). Then \(T'(b)=k(3600-3b^2)\). The critical number satisfies \(b^2=1200\), so \(b=20\sqrt{3}\approx 34.64\,\text{cm}\) and \(h=\sqrt{3600-1200}=20\sqrt{6}\approx 48.99\,\text{cm}\). Again, the endpoint values are \(0\), so this is the absolute maximum. 4. For part c, the common factor \(k\) cancels. The two proportional load-capacity values are \(T_a=(30\sqrt{2})(1800)=54{,}000\sqrt{2}\) and \(T_b=(20\sqrt{3})(2400)=48{,}000\sqrt{3}\). Therefore, \(\frac{T_b-T_a}{T_a}\cdot 100\%=\left(\frac{8\sqrt{6}}{18}-1\right)100\%\approx 8.87\%\).

a) \(b=h=30\sqrt{2}\approx 42.43\,\text{cm}\). b) \(b=20\sqrt{3}\approx 34.64\,\text{cm}\) and \(h=20\sqrt{6}\approx 48.99\,\text{cm}\). c) The load capacity is about \(8.87\%\) greater.

- Use a cross section through the sphere’s center, the pyramid’s apex, and a diagonal of the square base. - Relate the base circumradius, the pyramid height, and the sphere radius with the Pythagorean theorem. - Express \(a^2\) in terms of \(h\) before substituting into the volume formula. - Compare the interior candidate with the endpoint cases.

1. The volume is \(V=\frac{1}{3}a^2h\). In a cross section through the pyramid axis and a diagonal of the base, the circumradius of the square base is \(r_b=\frac{a}{\sqrt{2}}\). The base plane is a distance \(|h-R|\) from the sphere’s center, so \(r_b^2+(h-R)^2=R^2\). 2. Therefore, \(\frac{a^2}{2}=R^2-(h-R)^2=2Rh-h^2\), and \(a^2=4Rh-2h^2\). 3. Substitute into the volume formula: \(V(h)=\frac{1}{3}(4Rh-2h^2)h=\frac{4}{3}Rh^2-\frac{2}{3}h^3\), for \(0\le h\le 2R\). 4. Differentiate: \(V'(h)=\frac{8}{3}Rh-2h^2=2h\left(\frac{4R}{3}-h\right)\). The nonzero critical number is \(h=\frac{4R}{3}\). The volume is \(0\) at \(h=0\) and \(h=2R\), so this interior critical number gives the absolute maximum. 5. Substituting the maximizing height into the constraint gives \(a^2=4R\left(\frac{4R}{3}\right)-2\left(\frac{4R}{3}\right)^2=\frac{16R^2}{9}\), so \(a=\frac{4R}{3}\).

The volume is maximized when \(h=\frac{4R}{3}\) and \(a=\frac{4R}{3}\).

- Relate the radius of the rotated solid to the altitude from the right angle to the hypotenuse. - Add the volumes of the two cones, noting that their heights sum to \(12\,\text{cm}\). - Use the Pythagorean theorem to eliminate one leg. - A substitution such as \(x=a^2\) simplifies the objective function.

1. The legs satisfy \(a^2+b^2=12^2=144\). The radius of the double cone is the altitude to the hypotenuse: \(r=\frac{ab}{12}\). 2. The two cone heights add to the hypotenuse length, so the total volume is \(V=\frac{1}{3}\pi r^2(12)=\frac{\pi}{36}a^2b^2\). Using \(b^2=144-a^2\), \(V(a)=\frac{\pi}{36}a^2(144-a^2)\). 3. Let \(x=a^2\), where \(0\le x\le 144\). Then \(V(x)=\frac{\pi}{36}(144x-x^2)\). Differentiate: \(V'(x)=\frac{\pi}{36}(144-2x)\). The critical number is \(x=72\). Since this quadratic is concave down, it gives the absolute maximum. 4. Therefore, \(a=b=\sqrt{72}=6\sqrt{2}\approx 8.49\,\text{cm}\). 5. The maximum volume is \(V=\frac{\pi}{36}(72)(72)=144\pi\approx 452.39\,\text{cm}^3\).

a) \(a=b=6\sqrt{2}\approx 8.49\,\text{cm}\). b) The maximum volume is \(144\pi\approx 452.39\,\text{cm}^3\).

- Use the two given relationships to express \(b\) and \(c\) in terms of \(a\). - For the area, maximizing its square avoids differentiating a square root. - When the triangle rotates about \(a\), identify which leg becomes the cone height and which becomes the radius. - Check that each critical number lies in the physical domain.

1. The constraints are \(a+c=12\) and \(a^2+b^2=c^2\). Thus \(c=12-a\), and \(b^2=(12-a)^2-a^2=144-24a\), so the physical domain is \(0<a<6\). 2. For part a, \(A(a)=\frac{1}{2}a\sqrt{144-24a}\). Because \(A\ge 0\), maximizing \(A\) is equivalent to maximizing \(A(a)^2=36a^2-6a^3\). Differentiate: \(\frac{d}{da}[A(a)^2]=72a-18a^2=18a(4-a)\). The interior critical number is \(a=4\). The squared area approaches \(0\) at both ends of the domain, so this point gives the absolute maximum. Then \(c=8\,\text{cm}\) and \(b=\sqrt{144-96}=4\sqrt{3}\approx 6.93\,\text{cm}\). 3. For part b, rotation about \(a\) makes \(a\) the cone height and \(b\) the radius. Therefore, \(V(a)=\frac{1}{3}\pi b^2a=\frac{1}{3}\pi(144-24a)a=48\pi a-8\pi a^2\). Differentiate: \(V'(a)=48\pi-16\pi a\). The critical number is \(a=3\). Since the quadratic volume function is concave down, this is the absolute maximum. Then \(c=9\,\text{cm}\) and \(b=\sqrt{144-72}=6\sqrt{2}\approx 8.49\,\text{cm}\).

a) \(a=4\,\text{cm}\), \(b=4\sqrt{3}\approx 6.93\,\text{cm}\), and \(c=8\,\text{cm}\). b) \(a=3\,\text{cm}\), \(b=6\sqrt{2}\approx 8.49\,\text{cm}\), and \(c=9\,\text{cm}\).

- Draw an axial cross section to relate the cone’s base radius, height, and the sphere radius. - Use the Pythagorean theorem to express the base radius in terms of \(h\) and \(R\). - Substitute into the cone-volume formula and optimize on \([0, 2R]\). - Compare the maximum cone volume with the sphere-volume formula.

1. The cone volume is \(V=\frac{1}{3}\pi r^2h\). In an axial cross section, the base radius, sphere radius, and cone height satisfy \(r^2+(h-R)^2=R^2\). Therefore, \(r^2=2Rh-h^2\). 2. Substitute into the volume formula: \(V(h)=\frac{\pi}{3}(2Rh^2-h^3)\), for \(0\le h\le 2R\). 3. Differentiate: \(V'(h)=\frac{\pi}{3}(4Rh-3h^2)=\frac{\pi h}{3}(4R-3h)\). The interior critical number is \(h=\frac{4R}{3}\). The volume is \(0\) at both endpoints of the domain, so this critical number gives the absolute maximum. 4. At this height, \(V_{\max}=\frac{32}{81}\pi R^3\). The sphere volume is \(V_{\text{sphere}}=\frac{4}{3}\pi R^3\). Thus \(\frac{V_{\max}}{V_{\text{sphere}}}=\frac{8}{27}\).

The maximizing height is \(h=\frac{4R}{3}\). The maximum cone volume is \(\frac{8}{27}\) of the sphere’s volume.

- Use a right triangle in an axial cross section to relate \(r\), \(h\), and \(R\). - Write the lateral surface area as circumference times height. - Squaring the nonnegative objective function can remove the square root before differentiation. - Substitute the optimal height back into the geometric constraint to find the ratio.

1. The lateral surface area is \(M=2\pi rh\). In an axial cross section, the cylinder’s diagonal is the sphere’s diameter, so \((2r)^2+h^2=(2R)^2\). Thus \(r=\sqrt{R^2-\frac{h^2}{4}}\). 2. Therefore, \(M(h)=2\pi h\sqrt{R^2-\frac{h^2}{4}}\), for \(0\le h\le 2R\). Since \(M\ge 0\), maximizing \(M\) is equivalent to maximizing \(M(h)^2=\pi^2(4R^2h^2-h^4)\). Ignoring the positive constant \(\pi^2\), let \(q(h)=4R^2h^2-h^4\). 3. Differentiate: \(q'(h)=8R^2h-4h^3=4h(2R^2-h^2)\). The interior critical number is \(h=\sqrt{2}R\). The lateral area is \(0\) at both endpoints, so this critical number gives the absolute maximum. 4. The corresponding radius is \(r=\sqrt{R^2-\frac{2R^2}{4}}=\frac{R}{\sqrt{2}}\). Therefore, \(\frac{h}{r}=\frac{\sqrt{2}R}{R/\sqrt{2}}=2\).

The lateral surface area is maximized when \(h=\sqrt{2}R\). At that point, \(r=\frac{R}{\sqrt{2}}\), so \(\frac{h}{r}=2\).

- Place the intact opposite corner at the origin and locate the endpoints of the break line. - Find the equation of the line through those endpoints. - Express the area of an axis-aligned rectangle whose upper-right corner lies on the break. - Compare the interior maximum with the cases at the ends of the break segment.

1. Place the origin at the intact corner opposite the break. Let the original sheet occupy \(0\le x\le 120\) and \(0\le y\le 80\). The break joins \((60, 80)\) and \((120, 40)\). 2. The break line has slope \(\frac{40-80}{120-60}=-\frac{2}{3}\), so its equation is \(y=-\frac{2}{3}x+120\), for \(60\le x\le 120\). 3. If the upper-right corner of the new rectangle lies on the break line, its area is \(A(x)=x\left(-\frac{2}{3}x+120\right)=-\frac{2}{3}x^2+120x\). Differentiate: \(A'(x)=-\frac{4}{3}x+120\). The critical number is \(x=90\,\text{cm}\). Since \(A''(x)=-\frac{4}{3}<0\), it is a maximum on the line segment. 4. The corresponding height is \(y=-\frac{2}{3}(90)+120=60\,\text{cm}\). For completeness, the largest rectangle whose right edge is before the break begins has area \(60\cdot 80=4800\,\text{cm}^2\), and the endpoint at \(x=120\) also gives \(4800\,\text{cm}^2\). The \(90\times 60\) rectangle has area \(5400\,\text{cm}^2\), so it is the absolute maximum.

The largest new glass rectangle measures \(90\,\text{cm}\times 60\,\text{cm}\).

- Use the volume and total surface-area formulas for a closed cylinder. - Treat the original surface area as the fixed constraint in part b. - Express the height in terms of the radius before maximizing volume. - Compare the two volumes using percent increase.

1. For part a, \(V_1=\pi(3.3)^2(11.5)\approx 393.44\,\text{cm}^3\). The total surface area is \(S=2\pi(3.3)^2+2\pi(3.3)(11.5)=97.68\pi\approx 306.87\,\text{cm}^2\). 2. For a can with this fixed surface area, \(S=2\pi r^2+2\pi rh\), so \(h=\frac{S}{2\pi r}-r\). The volume is therefore \(V(r)=\pi r^2h=\frac{S}{2}r-\pi r^3\). 3. Differentiate: \(V'(r)=\frac{S}{2}-3\pi r^2\). The positive critical point is \(r=\sqrt{\frac{S}{6\pi}}=\sqrt{16.28}\approx 4.035\,\text{cm}\). Since \(V''(r)=-6\pi r<0\) for \(r>0\), this point gives the maximum. The surface-area constraint gives \(h=2r\approx 8.070\,\text{cm}\). 4. The optimized volume is \(V_{\max}\approx 412.73\,\text{cm}^3\). The percent increase is \(\left(\frac{V_{\max}}{V_1}-1\right)100\%\approx 4.90\%\).

a) \(V\approx 393.44\,\text{cm}^3\) and \(S\approx 306.87\,\text{cm}^2\). b) \(r\approx 4.035\,\text{cm}\) and \(h\approx 8.070\,\text{cm}\). c) The optimized volume is about \(4.90\%\) greater.

- First solve the maximum-volume problem for a cylinder inside a general cone. - Use similar triangles to connect cylinder radius and height. - Check how the derivative behaves on the entire allowable height interval before reusing the optimal proportions for the smaller cone.

1. For a cylinder of height \(h\) inscribed with its base in a cone's base, similar triangles give \(r=R\left(1-\frac{h}{H}\right)\). Hence \(V(h)=\pi R^2h\left(1-\frac{h}{H}\right)^2\) for \(0\le h\le H\). 2. Differentiate and factor: \(V'(h)=\pi R^2\left(1-\frac{h}{H}\right)\left(1-\frac{3h}{H}\right)\). On \((0,H)\), the derivative is positive for \(0<h<\frac{H}{3}\) and negative for \(\frac{H}{3}<h<H\). Thus the absolute maximum occurs at \(h=\frac{H}{3}\), with \(r=\frac{2R}{3}\). 3. For the original cone, \(h_1=6\,\text{cm}\) and \(r_1=6\,\text{cm}\). The remaining cone has height \(12\,\text{cm}\) and radius \(6\,\text{cm}\), so its maximum-volume cylinder has \(h_2=4\,\text{cm}\) and \(r_2=4\,\text{cm}\). 4. The original cone volume is \(V_K=\frac13\pi(9)^2(18)=486\pi\,\text{cm}^3\). The cylinder volumes are \(V_1=216\pi\,\text{cm}^3\) and \(V_2=64\pi\,\text{cm}^3\). 5. Therefore \(\frac{V_1+V_2}{V_K}=\frac{280}{486}=\frac{140}{243}\approx0.576\).

1. First cylinder: \(r_1=6\,\text{cm}\), \(h_1=6\,\text{cm}\). Second cylinder: \(r_2=4\,\text{cm}\), \(h_2=4\,\text{cm}\). 2. The combined-volume ratio is \(\frac{140}{243}\approx0.576\).

- Use the volume condition to express the cylindrical height in terms of the radius. - Include only the lateral cylinder and conical roof in the fabric area. - Determine which positive radii keep the cylindrical height nonnegative, then verify the minimum on that feasible set.

1. The volume constraint is \(\pi r^2h+\frac13\pi r^3=600\), so \(h=\frac{600}{\pi r^2}-\frac{r}{3}\). Feasibility requires \(r>0\) and \(h\ge0\), so \(0<r\le\sqrt[3]{\frac{1800}{\pi}}\). 2. The roof slant height is \(s=r\sqrt2\). The fabric area is \(A=2\pi rh+\pi rs\), giving \(A(r)=\frac{1200}{r}+\pi r^2\left(\sqrt2-\frac23\right)\). 3. Differentiate: \(A'(r)=-\frac{1200}{r^2}+2\pi r\left(\sqrt2-\frac23\right)\). The critical point satisfies \(r^3=\frac{600}{\pi(\sqrt2-\frac23)}\), so \(r\approx6.345\,\text{m}\), which lies in the feasible interval. 4. Since \(A''(r)=\frac{2400}{r^3}+2\pi\left(\sqrt2-\frac23\right)>0\) for \(r>0\), \(A\) is strictly convex. Also \(A(r)\to\infty\) as \(r\to0^+\). Therefore the feasible critical point gives the absolute minimum. 5. The corresponding cylindrical height is \(h=\frac{600}{\pi(6.345331\ldots)^2}-\frac{6.345331\ldots}{3}\approx2.628\,\text{m}\).

The absolute minimum fabric use occurs at \(r\approx6.345\,\text{m}\) and \(h\approx2.628\,\text{m}\).

- For a fixed base size, use the largest permitted length. - Determine which length restriction is active for each range of \(a\). - Optimize the volume separately on the two ranges. - Compare the largest volumes from both cases.

1. The volume is \(V=a^2l\). For a fixed \(a\), the volume increases with \(l\), so choose \(l=\min(80, 160-4a)\). 2. When \(0<a\le 20\), the length cap is active and \(l=80\). Then \(V_1(a)=80a^2\), which is increasing. Its largest value on this interval occurs at \(a=20\) and is \(32{,}000\,\text{cm}^3\). 3. When \(20\le a<40\), the combined-size constraint is active and \(l=160-4a\). Then \(V_2(a)=a^2(160-4a)=160a^2-4a^3\). Differentiate: \(V_2'(a)=320a-12a^2=4a(80-3a)\). The interior critical number is \(a=\frac{80}{3}\,\text{cm}\). Since \(V_2''\left(\frac{80}{3}\right)=-320<0\), this critical number gives the maximum in the second interval. 4. The corresponding length is \(l=160-4\left(\frac{80}{3}\right)=\frac{160}{3}\,\text{cm}\). The volume is \(V=\left(\frac{80}{3}\right)^2\left(\frac{160}{3}\right)=\frac{1024000}{27}\approx 37{,}925.93\,\text{cm}^3\). This exceeds \(32{,}000\,\text{cm}^3\), so it is the global maximum.

The maximum occurs at \(a=\frac{80}{3}\approx 26.67\,\text{cm}\) and \(l=\frac{160}{3}\approx 53.33\,\text{cm}\). The maximum volume is \(\frac{1024000}{27}\approx 37{,}925.93\,\text{cm}^3\).

- Resolve each slanted side into horizontal and vertical components. - Use the feasible angle range before solving the critical-number equation. - A sign change of the derivative across the critical angle gives a global comparison on this interval.

1. The trapezoid height is \(10\sin\alpha\), and its upper base is \(10+20\cos\alpha\). Thus \(A(\alpha)=100\sin\alpha(1+\cos\alpha)\), for \(0<\alpha\le\frac{\pi}{2}\). 2. Differentiate: \(A'(\alpha)=100(\cos\alpha+\cos2\alpha)\). Using \(\cos2\alpha=2\cos^2\alpha-1\), the critical-number equation becomes \(2u^2+u-1=0\), where \(u=\cos\alpha\). 3. The feasible solution is \(u=\frac12\), so \(\alpha=\frac{\pi}{3}=60^\circ\). 4. Since \(A'(\alpha)>0\) for \(0<\alpha<\frac{\pi}{3}\) and \(A'(\alpha)<0\) for \(\frac{\pi}{3}<\alpha\le\frac{\pi}{2}\), this is the absolute maximum on the feasible interval. 5. \(A(60^\circ)=75\sqrt3\approx129.90\,\text{cm}^2\).

The optimal angle is \(\alpha=60^\circ\), and the maximum cross-sectional area is \(75\sqrt3\approx129.90\,\text{cm}^2\).

- Express the four triangular face areas using the slant height. - Relate slant height, vertical height, and half the base edge. - Maximizing the square of the nonnegative volume removes a square root. - Combine the critical-number equation with the fixed-area constraint.

1. The slant height of each triangular face is \(s=\sqrt{H^2+\frac{a^2}{4}}\). The lateral area is \(M=4\left(\frac{1}{2}as\right)=a\sqrt{4H^2+a^2}\). Squaring and solving for \(H^2\) gives \(H^2=\frac{M^2-a^4}{4a^2}\). 2. The volume is \(V=\frac{1}{3}a^2H\). Because \(V\ge 0\), maximize \(V^2=\frac{1}{36}(M^2a^2-a^6)\), for \(0<a\le\sqrt{M}\). 3. Ignoring the positive constant, differentiate \(f(a)=M^2a^2-a^6\): \(f'(a)=2M^2a-6a^5\). The positive critical number satisfies \(M^2=3a^4\). At this point, \(f''(a)=2M^2-30a^4=-24a^4<0\), so it is a local maximum. Also, \(V\to 0\) as \(a\to 0^+\), and \(V=0\) when \(a=\sqrt{M}\), so the critical number gives the absolute maximum. 4. Combining the critical-number equation with \(M^2=4a^2H^2+a^4\) gives \(3a^4=4a^2H^2+a^4\Rightarrow a^2=2H^2\). Thus \(\frac{H}{a}=\frac{1}{\sqrt{2}}\).

The volume is maximized when \(H=\frac{a}{\sqrt{2}}\), so the optimal ratio is \(H:a=1:\sqrt{2}\).

- First express the cone's slant height in terms of \(r\). - Write equations for the exterior surface area and total volume. - Use the fixed surface area to eliminate \(h\). - After finding the critical radius, check the whole feasible interval before claiming an absolute maximum.

1. The cone slant height is \(s=\sqrt{r^2+H^2}=\sqrt{r^2+\left(\frac{4}{3}r\right)^2}=\frac{5}{3}r\). 2. The exterior surface area is the circular base, the cylindrical side, and the cone's lateral surface: \(O=\pi r^2+2\pi rh+\pi rs=\frac{8}{3}\pi r^2+2\pi rh\). 3. The total volume is \(V=\pi r^2h+\frac{1}{3}\pi r^2H=\pi r^2h+\frac{4}{9}\pi r^3\). 4. Solve the surface-area equation for \(h\): \(h=\frac{O}{2\pi r}-\frac{4}{3}r\). 5. Substitute into the volume: \(V(r)=\frac{Or}{2}-\frac{8}{9}\pi r^3\). 6. Differentiate and set the derivative equal to zero: \(V'(r)=\frac{O}{2}-\frac{8}{3}\pi r^2=0\). Therefore, \(O=\frac{16}{3}\pi r^2\). 7. Substitute this relation into the expression for \(h\): \(h=\frac{\frac{16}{3}\pi r^2}{2\pi r}-\frac{4}{3}r=\frac{4}{3}r\). 8. The feasibility condition \(h\ge0\) gives \(0<r\le\sqrt{\frac{3O}{8\pi}}\). Also, \(V'(r)=\frac{O}{2}-\frac{8}{3}\pi r^2\) is positive before the critical radius and negative after it. Therefore the volume increases and then decreases across the entire feasible interval, so the critical number gives the absolute maximum.

The absolute maximum occurs when \(h=\frac{4}{3}r\). Thus the cylinder height equals the cone height, and the ratio is \(h:r=4:3\).

- Combine the cylinder and both cones in the volume and exterior-area models. - Eliminate the cylinder height while keeping its nonnegativity restriction. - Use the critical-number relation only after verifying that it gives the absolute minimum on the feasible domain.

1. The total volume is \(V=\pi r^2h+\frac23\pi r^3\). Each cone has slant height \(r\sqrt2\), so the exterior area is \(A=2\pi rh+2\sqrt2\pi r^2\). 2. Solve for \(h\): \(h=\frac{V}{\pi r^2}-\frac23r\). Feasibility requires \(0<r\le\sqrt[3]{\frac{3V}{2\pi}}\). 3. Substitute: \(A(r)=\frac{2V}{r}+\pi r^2\left(2\sqrt2-\frac43\right)\). 4. Differentiate: \(A'(r)=-\frac{2V}{r^2}+2\pi r\left(2\sqrt2-\frac43\right)\). The critical-number equation is \(V=\pi r^3\left(2\sqrt2-\frac43\right)\). 5. The second derivative \(A''(r)=\frac{4V}{r^3}+2\pi\left(2\sqrt2-\frac43\right)>0\), so \(A\) is strictly convex; also \(A(r)\to\infty\) as \(r\to0^+\). Thus the feasible critical number gives the absolute minimum. 6. Substitute the critical-number relation into the height formula: \(h=2(\sqrt2-1)r\).

The optimal ratio is \(h:r=2(\sqrt2-1):1\), so \(h=2(\sqrt2-1)r\approx0.828r\).

- Build the fixed-area constraint from the floor, four walls, and four roof faces. - Use the area constraint to determine the feasible range of base sizes and eliminate the prism height. - Verify the maximizing critical number by tracking the volume derivative across that full range.

1. The roof slant height is \(\frac{a}{\sqrt2}\), so the four triangular roof faces have total area \(\sqrt2a^2\). 2. The surface-area constraint is \(S=a^2(1+\sqrt2)+4ah\), and the volume is \(V=a^2h+\frac16a^3\). 3. Solve for \(h\): \(h=\frac{S}{4a}-\frac{1+\sqrt2}{4}a\). Feasibility gives \(0<a\le\sqrt{\frac{S}{1+\sqrt2}}\). 4. Substitute: \(V(a)=\frac S4a-\frac{1+3\sqrt2}{12}a^3\). 5. Differentiate: \(V'(a)=\frac S4-\frac{1+3\sqrt2}{4}a^2\). The critical number is \(a=\sqrt{\frac{S}{1+3\sqrt2}}\), which lies in the feasible interval. 6. The derivative is positive before this value and negative after it throughout the feasible interval, so this critical number gives the absolute maximum. 7. Substituting \(S=(1+3\sqrt2)a^2\) into the height expression gives \(h=\frac{\sqrt2}{2}a\).

The absolute maximum volume occurs for \(a=\sqrt{\frac{S}{1+3\sqrt2}}\) and \(h=\frac{\sqrt2}{2}a\).

- Use the Pythagorean theorem in an axial cross section. - Recall that a cylinder's lateral area does not include its bases. - Because the area is nonnegative, maximizing its square gives the same maximizing height. - Compare your maximum with the curved area of a hemisphere.

1. In an axial cross section, the cylinder radius \(r\), height \(h\), and hemisphere radius \(R\) form a right triangle: \(r^2+h^2=R^2\). Thus \(r=\sqrt{R^2-h^2}\), and the lateral area is \(M(h)=2\pi h\sqrt{R^2-h^2}\). 2. Because \(M(h)\ge 0\), maximize \(f(h)=\left(\frac{M}{2\pi}\right)^2=h^2(R^2-h^2)=R^2h^2-h^4\). Its derivative is \(f'(h)=2R^2h-4h^3=2h(R^2-2h^2)\). The interior critical number is \(h=\frac{R}{\sqrt{2}}\). Since \(f=0\) at \(h=0\) and \(h=R\), this critical number gives the maximum. 3. At this height, \(r=\sqrt{R^2-\frac{R^2}{2}}=\frac{R}{\sqrt{2}}\). Therefore, \(M_{\max}=2\pi\left(\frac{R}{\sqrt{2}}\right)\left(\frac{R}{\sqrt{2}}\right)=\pi R^2\). The hemisphere's curved surface area is \(2\pi R^2\), so \(\frac{M_{\max}}{2\pi R^2}=\frac{1}{2}\).

1. \(M(h)=2\pi h\sqrt{R^2-h^2}\) 2. The lateral area is maximized when \(h=\frac{R}{\sqrt{2}}\). 3. The maximum is \(M_{\max}=\pi R^2\), exactly one-half of the hemisphere's curved surface area \(2\pi R^2\).

- Use the coordinate placement and the broken edge in the figure to write the break line's equation. - Let the upper-left corner of the new rectangle move along that line. - Determine the allowed interval for that corner. - Check whether the critical number lies inside the interval; otherwise use the appropriate boundary.

1. Place the original slab in the coordinate plane with corners \((0, 0)\), \((b, 0)\), \((b, a)\), and \((0, a)\), and let the broken corner be \((0, a)\). The break runs from \((0, \frac{3a}{4})\) to \((\frac{a}{4}, a)\), so its equation is \(y=x+\frac{3a}{4}\), where \(0\le x\le\frac{a}{4}\). 2. If the new rectangle's upper-left corner is \((x, y)\) on the break, its width and height are \(w=b-x\) and \(h=y=x+\frac{3a}{4}\). Thus \(A(x)=(b-x)\left(x+\frac{3a}{4}\right)\). 3. Differentiate: \(A'(x)=-2x+b-\frac{3a}{4}\). The unconstrained critical number is \(x_0=\frac{b}{2}-\frac{3a}{8}\). Since \(b\ge a\), this value is positive. It lies in the allowed interval when \(\frac{b}{2}-\frac{3a}{8}\le\frac{a}{4}\), which is equivalent to \(b\le\frac{5a}{4}\). 4. If \(a\le b\le\frac{5a}{4}\), the concave quadratic is maximized at \(x_0\). Then \(w=h=\frac{b}{2}+\frac{3a}{8}\). 5. If \(b>\frac{5a}{4}\), the critical number lies beyond the allowed interval, and \(A\) is increasing throughout that interval. The maximum occurs at \(x=\frac{a}{4}\), giving \(w=b-\frac{a}{4}\) and \(h=a\).

If \(a\le b\le\frac{5a}{4}\), the maximum-area piece is a square with side length \(\frac{b}{2}+\frac{3a}{8}\). If \(b>\frac{5a}{4}\), the maximum-area rectangle has dimensions \(b-\frac{a}{4}\) by \(a\).

- Write the prism volume using its square base edge and height. - Use similar triangles to find the cone's radius at the height of the prism's upper face. - Relate that radius to half the square's diagonal. - Check the critical numbers and endpoints.

1. The prism volume is \(V=a^2h\). At height \(h\), the cone's cross-sectional radius is related by similar triangles. The distance from the center of the square to a vertex is \(\frac{a\sqrt{2}}{2}\), so \(\frac{a\sqrt{2}/2}{12}=\frac{18-h}{18}\). Therefore, \(a=\frac{2\sqrt{2}}{3}(18-h)\). 2. Substitute into the volume: \(V(h)=\frac{8}{9}(18-h)^2h\), where \(0\le h\le18\). 3. Differentiate: \(V'(h)=\frac{8}{3}(h^2-24h+108)\). The critical numbers are \(h=6\) and \(h=18\). The endpoint volumes are \(V(0)=V(18)=0\), while the interior critical number \(h=6\,\text{cm}\) gives positive volume. Therefore \(h=6\,\text{cm}\) gives the absolute maximum. 4. The base edge length is \(a=\frac{2\sqrt{2}}{3}(18-6)=8\sqrt{2}\,\text{cm}\approx 11.31\,\text{cm}\). 5. The maximum volume is \(V=(8\sqrt{2})^2(6)=768\,\text{cm}^3\).

The maximum-volume prism has base edge length \(a=8\sqrt{2}\,\text{cm}\approx 11.31\,\text{cm}\) and height \(h=6\,\text{cm}\). Its maximum volume is \(768\,\text{cm}^3\).

- Use a right triangle whose hypotenuse is the distance \(s\). - Use the cylinder volume to eliminate one variable. - Minimize \(s^2\) instead of \(s\). - Rewrite the critical-number equation using \(V=\pi r^2h\).

1. The distance satisfies \(s^2=r^2+h^2\). Because \(s>0\), minimizing \(s\) is equivalent to minimizing \(s^2\). 2. The fixed-volume condition is \(V=\pi r^2h\), so \(h=\frac{V}{\pi r^2}\). 3. Substitute into the squared distance: \(f(r)=r^2+\frac{V^2}{\pi^2r^4}\). 4. Differentiate and set the derivative equal to zero: \(f'(r)=2r-\frac{4V^2}{\pi^2r^5}=0\). Thus \(r^6=\frac{2V^2}{\pi^2}\). 5. Since \(V=\pi r^2h\), substitute \(V^2=\pi^2r^4h^2\): \(r^6=2r^4h^2\). Therefore, \(r^2=2h^2\), so \(\frac{r}{h}=\sqrt{2}\). The objective approaches infinity at both ends of its positive domain, so the critical number gives the absolute minimum.

The optimal ratio is \(r:h=\sqrt{2}:1\), or \(r=\sqrt{2}h\).

- Write the pyramid volume and total surface-area formulas. - Express each triangular face's slant height using \(a\) and \(h\). - Squaring the nonnegative volume can simplify the optimization. - Determine the physically possible interval for the base edge.

1. The volume is \(V=\frac{1}{3}a^2h\). The total surface area is \(a^2+a\sqrt{4h^2+a^2}=432\). 2. Isolate the radical and square: \((432-a^2)^2=a^2(4h^2+a^2)\). This gives \(h^2=\frac{432^2}{4a^2}-216\). 3. Because \(V\ge0\), maximize \(V^2\). Let \(u=a^2\). Then \(V^2=\frac{1}{9}a^4h^2=5184u-24u^2\). 4. Differentiate: \(\frac{d}{du}(V^2)=5184-48u\). The critical point is \(u=108\). The physical domain is \(0<u\le216\), corresponding to nonnegative height. Since \(V^2=5184u-24u^2\) is a concave-down quadratic and its vertex \(u=108\) lies inside this domain, it gives the absolute maximum. 5. Therefore, \(a=\sqrt{108}=6\sqrt{3}\,\text{cm}\approx 10.39\,\text{cm}\). Also, \(h^2=\frac{432^2}{4(108)}-216=216\), so \(h=6\sqrt{6}\,\text{cm}\approx 14.70\,\text{cm}\).

The volume is maximized when \(a=6\sqrt{3}\,\text{cm}\approx 10.39\,\text{cm}\) and \(h=6\sqrt{6}\,\text{cm}\approx 14.70\,\text{cm}\).

- Use similar triangles to express each inner solid's height in terms of its horizontal dimension. - For the square prism, relate the square's side length to the radius of its circumscribed circle. - Differentiate each one-variable volume function. - In the final ratio, \(R^2H\) cancels.

1. For a cylinder of radius \(r\) and height \(h\), similar triangles give \(\frac{r}{R}=\frac{H-h}{H}\), so \(h=H\left(1-\frac{r}{R}\right)\). The volume is \(V(r)=\pi Hr^2\left(1-\frac{r}{R}\right)\). Differentiate: \(V'(r)=\pi H\left(2r-\frac{3r^2}{R}\right)\). The interior critical number is \(r=\frac{2R}{3}\), which gives \(h=H\left(1-\frac{2}{3}\right)=\frac{H}{3}\). The endpoint volumes are zero, so this critical number gives the maximum. 2. Let \(s\) be the square base edge. The distance from the square's center to a vertex is \(\frac{s\sqrt{2}}{2}\). Similar triangles give \(h=H\left(1-\frac{s\sqrt{2}}{2R}\right)\). Thus \(V(s)=Hs^2\left(1-\frac{s\sqrt{2}}{2R}\right)\). Differentiating gives the interior critical point \(s=\frac{2\sqrt{2}}{3}R\). At this value, \(h=\frac{H}{3}\), so \(V_{\text{prism, max}}=\left(\frac{2\sqrt{2}}{3}R\right)^2\left(\frac{H}{3}\right)=\frac{8}{27}R^2H\). 3. The maximum cylinder volume is \(V_{\text{cyl, max}}=\pi\left(\frac{2R}{3}\right)^2\left(\frac{H}{3}\right)=\frac{4\pi}{27}R^2H\). Therefore, \(\frac{V_{\text{cyl, max}}}{V_{\text{prism, max}}}=\frac{\pi}{2}\approx1.57\). The cylinder has the larger maximum volume.

1. The maximum-volume cylinder has height \(h=\frac{H}{3}\). 2. The maximum square-prism volume is \(V_{\text{prism, max}}=\frac{8}{27}R^2H\). 3. The ratio is \(\frac{V_{\text{cyl, max}}}{V_{\text{prism, max}}}=\frac{\pi}{2}\approx1.57\), so the cylinder uses the available space more efficiently.

- Write the cone volume and lateral-area formulas. - Relate \(r\), \(h\), and the slant height \(s\). - Use the fixed lateral area to eliminate \(s\). - Maximize the expression under the square root.

1. The volume is \(V=\frac{1}{3}\pi r^2h\). The fixed lateral area is \(M=\pi rs\), where \(s=\sqrt{r^2+h^2}\). 2. From the area constraint, \(s=\frac{M}{\pi r}\). Therefore, \(h=\sqrt{\frac{M^2}{\pi^2r^2}-r^2}\). 3. Substitute into the volume: \(V(r)=\frac{1}{3}\sqrt{M^2r^2-\pi^2r^6}\). Because the square root is increasing, maximize \(f(r)=M^2r^2-\pi^2r^6\). 4. Differentiate: \(f'(r)=2M^2r-6\pi^2r^5\). At the positive critical number, \(M^2=3\pi^2r^4\), so \(M=\pi r^2\sqrt{3}\). 5. Since \(M=\pi rs\), it follows that \(s=r\sqrt{3}\). Thus \(h=\sqrt{s^2-r^2}=r\sqrt{2}\). The objective is zero at the endpoints of its physical domain, so this critical number gives the maximum.

The volume is maximized when \(\frac{h}{r}=\sqrt{2}\), or \(h:r=\sqrt{2}:1\).

- Rewrite \(\frac{k}{x}\) using a negative exponent before differentiating. - At a critical point, use the derivative equation to replace \(k\). - Use the distance formula and square both sides. - A substitution for \(x^2\) turns the resulting equation into a quadratic.

1. Differentiate: \(f_k'(x)=2x-\frac{k}{x^2}\) and \(f_k''(x)=2+\frac{2k}{x^3}>0\). The critical-number equation gives \(2x^3=k\), so \(x=\sqrt[3]{\frac{k}{2}}\). Thus the critical point is a local minimum. 2. At this point, \(k=2x^3\), so \(y=x^2+\frac{2x^3}{x}=3x^2\). Therefore, the local minimum is \(\left(\sqrt[3]{\frac{k}{2}}, 3\left(\frac{k}{2}\right)^{2/3}\right)\), and the locus is \(y=3x^2\) for \(x>0\). 3. On the locus, the squared distance from the origin is \(x^2+y^2=x^2+9x^4\). Setting this equal to \(10\) gives \(9x^4+x^2-10=0\). 4. Let \(u=x^2\). Then \(9u^2+u-10=(u-1)(9u+10)=0\). Since \(x>0\), \(u=1\) and \(x=1\). From \(k=2x^3\), \(k=2\).

a) \(\left(\sqrt[3]{\frac{k}{2}}, 3\left(\frac{k}{2}\right)^{2/3}\right)\) b) \(y=3x^2\) for \(x>0\) c) \(k=2\)

- Differentiate the family with respect to \(x\) first. - Express the two critical numbers in terms of the parameter. - The distance between the two roots can be simplified before optimizing it. - A positive second derivative can establish the required minimum.

1. Expand and differentiate: \(f_a(x)=x^3-(a+1)x^2+ax\), so \(f_a'(x)=3x^2-2(a+1)x+a\). 2. The critical numbers are \(x=\frac{a+1-\sqrt{a^2-a+1}}{3}\) and \(x=\frac{a+1+\sqrt{a^2-a+1}}{3}\). 3. Their distance is \(d(a)=\frac{2}{3}\sqrt{a^2-a+1}\). Minimizing \(d\) is equivalent to minimizing \(q(a)=a^2-a+1\). 4. Since \(q'(a)=2a-1\), the only critical number is \(a=\frac12\). Also \(q''(a)=2>0\), so \(q\), and therefore \(d\), has an absolute minimum there on \(0<a<1\). 5. At \(a=\frac12\), \(d=\frac{2}{3}\sqrt{\frac34}=\frac{\sqrt3}{3}\).

The critical numbers are closest when \(a=\frac12\). The minimum distance is \(\frac{\sqrt3}{3}\).

- Solve the zero equation before considering parameter ranges. - Analyze each piece of the supplied objective on its own domain. - Compare the best value from each parameter range to obtain the global minimum.

1. The zero satisfies \(e^x=t\), so \(x=\ln t\). It lies in \([0,1]\) exactly when \(1\le t\le e\). 2. On \(0<t\le1\), \(A(t)=e-1-t\) is decreasing, so its smallest value there is \(A(1)=e-2\). 3. On \(1\le t\le e\), \(A'(t)=2\ln t-1\). The critical number is \(t=\sqrt e\), and \(A''(t)=\frac2t>0\), so it is the minimum on this middle interval. 4. On \(t\ge e\), \(A(t)=t-e+1\) is increasing, so its smallest value there is \(A(e)=1\). 5. \(A(\sqrt e)=e+1-2\sqrt e\approx0.421\), which is less than \(e-2\) and \(1\). Therefore, this is the absolute minimum for \(t>0\).

a) The zero is \(x=\ln t\), and it lies in \([0,1]\) for \(1\le t\le e\). b) The absolute minimum occurs at \(t=\sqrt e\), with value \(e+1-2\sqrt e\approx0.421\).

- Write the squared distance from \((x, f(x))\) to the origin. - Differentiate using the chain rule and the derivative of \(\ln x\). - Study the monotonicity of the expression inside the factored derivative. - Check the behavior as \(x\to0^+\) and \(x\to\infty\).

1. A point on the graph is \((x, 1-\ln x)\). Minimize the squared distance \(g(x)=x^2+(1-\ln x)^2\). 2. Differentiate: \(g'(x)=2x-\frac{2}{x}+\frac{2\ln x}{x}=\frac{2}{x}(x^2-1+\ln x)\). 3. Let \(q(x)=x^2-1+\ln x\). Then \(q'(x)=2x+\frac{1}{x}>0\) for \(x>0\), so \(q\) is strictly increasing. Since \(q(1)=0\), the only critical number is \(x=1\). 4. Also, \(g(x)\to\infty\) as \(x\to0^+\) or \(x\to\infty\). Therefore, \(x=1\) gives the absolute minimum. 5. The corresponding y-coordinate is \(f(1)=1\).

The closest point is \(P=(1, 1)\).

- Write the tangent line at \(x=a\). - Find each axis intercept by setting one coordinate equal to \(0\). - Use an absolute value when writing the rectangle's area. - For a maximum, find critical points and check the behavior at the ends of the interval.

1. Since \(g'(x)=\frac{2}{x}\), the tangent line at \(x=a\) is \(y=\frac{2}{a}(x-a)+2\ln(a)=\frac{2}{a}x-2+2\ln(a)\). 2. Set \(x=0\) to find the y-intercept: \(R=(0, 2\ln(a)-2)\). 3. Set \(y=0\) to find the x-intercept: \(0=\frac{2}{a}x-2+2\ln(a)\), so \(S=(a(1-\ln(a)), 0)\). 4. For \(0<a<e\), \(x_S>0\) and \(y_R<0\). Therefore, \(A(a)=|x_Sy_R|=2a(1-\ln(a))^2\). 5. Differentiate: \(A'(a)=2(1-\ln(a))^2-4(1-\ln(a))=2(1-\ln(a))(-1-\ln(a))\). 6. The critical numbers satisfy \(1-\ln(a)=0\) or \(-1-\ln(a)=0\). The first gives \(a=e\), which is not in the interval. The interior critical number is \(a=\frac{1}{e}\). 7. Also, \(A(a)\to0\) as \(a\to0^+\) and as \(a\to e^-\). Thus the interior critical number gives the global maximum. Its value is \(A\left(\frac{1}{e}\right)=\frac{8}{e}\approx2.943\) square units.

a) \(S=(a(1-\ln(a)), 0)\); \(R=(0, 2\ln(a)-2)\) b) \(A(a)=2a(1-\ln(a))^2\) c) The maximum area is \(\frac{8}{e}\approx2.943\) square units at \(a=\frac{1}{e}\).

- Convert the glide time to seconds before computing altitude loss. - Remove only positive constant factors when simplifying the optimization. - Check the behavior of the simplified time function near both ends of its positive domain. - For the tangent condition, compare its slope and intercept with the point specified in the problem.

1. The glide time is \(\frac{3600s}{c}\) seconds. The altitude lost is \(g(c)\frac{3600s}{c}\), so the climbing time is \(\frac{g(c)}{v_{\mathrm{climb}}}\frac{3600s}{c}\). Adding gives \(T(c)=\frac{3600s}{c}\left(1+\frac{g(c)}{v_{\mathrm{climb}}}\right)\). 2. Positive constant factors do not affect the minimizing speed, so minimize \(F(c)=\frac{v_{\mathrm{climb}}+g(c)}{c}=0.0005c-0.08+\frac{7.2}{c}\). 3. Differentiate: \(F'(c)=0.0005-\frac{7.2}{c^2}\). The only positive critical point satisfies \(c^2=14{,}400\), so \(c=120\,\text{km/h}\). 4. Since \(F''(c)=\frac{14.4}{c^3}>0\) for \(c>0\), \(F\) is strictly convex. Also \(F(c)\to\infty\) as \(c\to0^+\) and as \(c\to\infty\). Therefore \(c=120\) gives the absolute minimum. 5. At \(c=120\), \(g(120)=1.6\) and \(g'(120)=0.04\). The tangent line is \(y=0.04(c-120)+1.6=0.04c-3.2\), so it passes through \((0,-3.2)\).

a) \(T(c)=\frac{3600s}{c}\left(1+\frac{g(c)}{v_{\mathrm{climb}}}\right)\) b) \(c=120\,\text{km/h}\) c) The tangent line at \((120,1.6)\) is \(y=0.04c-3.2\), so it passes through \((0,-3.2)\).

- Use the two-point slope formula to relate the intercepts. - Apply the Pythagorean theorem to the intercept triangle. - Simplify the derivative before solving the critical-number equation. - Use the monotonicity of \(z^2\) for \(z>0\).

1. The line through \((x_0, 0)\) and \((1, 8)\) has slope \(m=\frac{8}{1-x_0}\). Its y-intercept is \(y_0=\frac{8x_0}{x_0-1}\). Therefore, by the Pythagorean theorem, \(L^2=x_0^2+y_0^2=x_0^2+\frac{64x_0^2}{(x_0-1)^2}\). 2. Let \(f(x_0)=L^2(x_0)\). Its derivative simplifies to \(f'(x_0)=2x_0\left(1-\frac{64}{(x_0-1)^3}\right)\). Since \(x_0>1\), the critical number satisfies \((x_0-1)^3=64\), so \(x_0=5\). 3. Also, \(f(x_0)\to\infty\) as \(x_0\to1^+\) or \(x_0\to\infty\), and this is the only critical number. Thus it gives the absolute minimum. 4. Because \(L(x_0)>0\) and the squaring function is strictly increasing for positive inputs, \(L\) and \(L^2\) attain their minima at the same value of \(x_0\).

a) \(L^2(x_0)=x_0^2+\frac{64x_0^2}{(x_0-1)^2}\) b) The hypotenuse is minimized when \(x_0=5\). c) Since \(L>0\) and squaring is strictly increasing on positive numbers, minimizing \(L^2\) also minimizes \(L\).

- Use the fixed total area to eliminate the rectangular height. - Determine which radii keep that height nonnegative. - Apply the different frame costs to straight and curved lengths, then verify the minimum across the feasible domain.

1. Let \(r=\frac b2\). The area constraint is \(2rh+\frac12\pi r^2=27\), so \(h=\frac{13.5}{r}-\frac{\pi r}{4}\). 2. Feasibility requires \(r>0\) and \(h\ge0\), so \(0<r\le\sqrt{\frac{54}{\pi}}\). 3. The straight frame length is \(2h+2r\), and the curved length is \(\pi r\). Thus \(C=12(2h+2r)+18\pi r\). 4. Substitution gives \(C(r)=\frac{324}{r}+24r+12\pi r\). 5. Differentiate: \(C'(r)=-\frac{324}{r^2}+24+12\pi\). The critical number satisfies \(r^2=\frac{27}{2+\pi}\), so \(r\approx2.29\,\text{ft}\). 6. Since \(C''(r)=\frac{648}{r^3}>0\), \(C\) is strictly convex, and \(C(r)\to\infty\) as \(r\to0^+\). Thus the feasible critical number gives the absolute minimum. 7. Therefore \(b=2r\approx4.58\,\text{ft}\) and \(h\approx4.09\,\text{ft}\).

The absolute minimum frame cost occurs when the rectangular part is approximately \(4.58\,\text{ft}\) wide and \(4.09\,\text{ft}\) high.

- Expand the equation of \(C_t\) and subtract the two circle equations. - Write the resulting line in standard form. - Use the point-to-line distance formula for the origin. - Differentiate the distance function and apply the second derivative test.

1. Write the circle equations as \(x^2+y^2=32\) and \((x-t)^2+(y-t)^2=16\). Subtracting the equations eliminates the quadratic terms and gives \(2tx+2ty-2t^2=16\). Since \(t>0\), the radical axis is \(x+y=t+\frac8t\). 2. In standard form, the line is \(x+y-t-\frac8t=0\). Its distance from the origin is \(d(t)=\frac{t+\frac8t}{\sqrt2}\). 3. Differentiate: \(d'(t)=\frac1{\sqrt2}\left(1-\frac8{t^2}\right)\). Setting the derivative equal to zero gives \(t=2\sqrt2\). 4. The second derivative is \(d''(t)=\frac{16}{\sqrt2\, t^3}\). Therefore, \(d''(2\sqrt2)=\frac12>0\), so the critical number gives a local minimum. Since \(d(t)\to\infty\) as \(t\to0^+\) or \(t\to\infty\), this is the absolute minimum. 5. The minimum distance is \(d(2\sqrt2)=4\) units.

a) \(x+y=t+\frac8t\) b) The distance is minimized at \(t=2\sqrt2\). c) The distance is \(4\) units. Since \(d''(2\sqrt2)=\frac12>0\), the minimum is absolute.

- Find the cone's slant height from its radius and height. - Use the fixed volume to eliminate \(h\). - Include the bottom, cylinder side, and cone side in the area. - Rewrite the critical-number equation as a ratio.

1. The total volume is \(V=\pi r^2h+\frac{1}{3}\pi r^3\). Thus \(h=\frac{V}{\pi r^2}-\frac{r}{3}\). Because \(h\ge0\), the feasible radius satisfies \(0<r\le\sqrt[3]{\frac{3V}{\pi}}\). 2. The exterior surface consists of the bottom, cylindrical side, and conical lateral surface. The cone slant height is \(r\sqrt{2}\), so \(A=(1+\sqrt{2})\pi r^2+2\pi rh\). 3. Substitute for \(h\): \(A(r)=\left(\frac{1}{3}+\sqrt{2}\right)\pi r^2+\frac{2V}{r}\). 4. Differentiate and set the derivative equal to zero: \(A'(r)=2\left(\frac{1}{3}+\sqrt{2}\right)\pi r-\frac{2V}{r^2}=0\). Therefore, \(V=\left(\frac{1}{3}+\sqrt{2}\right)\pi r^3\). This critical radius is feasible because \(\frac{V}{\pi(\frac{1}{3}+\sqrt{2})}<\frac{3V}{\pi}\). 5. The second derivative is \(A''(r)=2\left(\frac{1}{3}+\sqrt{2}\right)\pi+\frac{4V}{r^3}>0\). Thus \(A\) is strictly convex on the feasible domain, so its feasible critical number gives the absolute minimum. 6. Substitute the critical-number relation into the height formula: \(h=\left(\frac{1}{3}+\sqrt{2}\right)r-\frac{r}{3}=\sqrt{2}r\).

The optimal ratio is \(\frac{h}{r}=\sqrt{2}\), or \(h:r=\sqrt{2}:1\).

- Minimize the square of the distance instead of the distance itself. - Check what happens to that squared distance near both ends of the positive domain. - After finding the closest point, compare the slope from the origin with the tangent slope there.

1. Minimize the squared distance \(D^2(x)=x^2+\left(\frac9x\right)^2=x^2+81x^{-2}\) for \(x>0\). 2. Differentiate: \((D^2)'(x)=2x-162x^{-3}\). The only positive critical number satisfies \(x^4=81\), so \(x=3\). 3. Since \((D^2)''(x)=2+486x^{-4}>0\), the squared-distance function is strictly convex. Also \(D^2(x)\to\infty\) as \(x\to0^+\) and as \(x\to\infty\). Thus \(x=3\) gives the absolute minimum distance. Since \(f(3)=3\), \(P=(3,3)\). 4. The slope of \(\overline{OP}\) is \(1\). Since \(f'(x)=-\frac9{x^2}\), the tangent slope at \(x=3\) is \(-1\). 5. Their slopes multiply to \(-1\), so the segment and tangent line are perpendicular.

1. \(P=(3,3)\) 2. The slope of \(\overline{OP}\) is \(1\), while the tangent slope is \(-1\); therefore the segment and tangent line are perpendicular.

- Minimize the squared distance from a general graph point to \(P\). - Differentiate and clear the denominators. - Test small positive integer roots of the resulting polynomial. - Use convexity and end behavior to verify the global minimum.

1. A general graph point is \((x, 4/x)\). Minimize the squared distance \(D(x)=(x+1)^2+\left(\frac{4}{x}+1\right)^2\). 2. Differentiate: \(D'(x)=2x+2-\frac{8}{x^2}-\frac{32}{x^3}\). Setting the derivative equal to zero and multiplying by \(x^3\) gives \(x^4+x^3-4x-16=0\). The positive solution \(x=2\) satisfies this equation. 3. The second derivative is \(D''(x)=2+\frac{16}{x^3}+\frac{96}{x^4}>0\) for \(x>0\). Thus \(D\) is strictly convex, so the critical number is unique. Since \(D(x)\to\infty\) as \(x\to0^+\) or \(x\to\infty\), it is the absolute minimum. 4. The y-coordinate is \(f(2)=2\).

The closest point is \(Q=(2, 2)\).

- Relate the triangle height, base, and inradius using a right-triangle cross section. - The height must be greater than twice the radius. - Minimize the square of the positive area to remove a square root. - Check the behavior at both ends of the domain.

1. Let \(b\) be the base length. The inradius geometry gives \(b^2=\frac{4r^2h}{h-2r}\), where \(h>2r\). With \(r=3\), \(b^2=\frac{36h}{h-6}\). 2. The triangle area is \(A=\frac{1}{2}bh\). Since \(A>0\), minimize its square: \(A^2=\frac{1}{4}b^2h^2=\frac{9h^3}{h-6}\). 3. Differentiate: \(\frac{d}{dh}(A^2)=\frac{18h^2(h-9)}{(h-6)^2}\). The only critical number in the domain is \(h=9\,\text{cm}\). 4. The derivative changes from negative to positive at \(h=9\). Also, \(A^2\to\infty\) as \(h\to6^+\) or \(h\to\infty\), so this is the absolute minimum.

The triangle area is minimized when \(h=9\,\text{cm}\).

- Find where the parabola meets the positive x-axis. - Express the rectangle’s width and height in terms of \(u\) and \(a\). - Treat \(a\) as a constant while optimizing with respect to \(u\). - Set the resulting maximum-area formula equal to the given value.

1. The positive x-intercept satisfies \(a-x^2=0\), so \(x=\sqrt a\). Thus the rectangle parameter satisfies \(0<u<\sqrt a\). 2. The rectangle has width \(u\) and height \(a-u^2\), so \(A(u)=au-u^3\). Differentiating gives \(A'(u)=a-3u^2\). The positive critical number is \(u=\sqrt{a/3}\). 3. The derivative changes from positive to negative there, and the area approaches \(0\) at both ends of the feasible interval. Hence the absolute maximum is \(A_{\max}(a)=\frac{2a\sqrt a}{3\sqrt3}\). 4. Set \(\frac{2a\sqrt a}{3\sqrt3}=\frac{16}{3\sqrt3}\). Then \(a^{3/2}=8\), so \(a=4\).

a) \(x=\sqrt a\) b) \(A_{\max}(a)=\frac{2a\sqrt a}{3\sqrt3}\) c) \(a=4\)

- Evaluate both functions and their derivatives at the intersection input. - Use the slope relationship for perpendicular lines to eliminate one parameter. - The absolute values simplify because the signs of both parameters are given. - Check the behavior of the score near both ends of its positive domain.

1. Since \(f_a(0)=g_b(0)=1\), both graphs pass through \(P=(0,1)\). Their tangent slopes there are \(a\) and \(b\). Perpendicular tangent lines require \(ab=-1\). 2. Because \(a>0\), \(b<0\), and \(b=-\frac1a\), the score is \(S(a)=a+\frac1a\), for \(a>0\). 3. Differentiate: \(S'(a)=1-\frac1{a^2}\). The only positive critical number is \(a=1\). Also \(S''(a)=\frac2{a^3}>0\) for \(a>0\), and \(S(a)\to\infty\) as \(a\to0^+\) or \(a\to\infty\). Thus the absolute minimum occurs at \(a=1\). 4. The orthogonality condition gives \(b=-1\), and \(S_{\min}=1+1=2\).

1) \(ab=-1\) 2) \(S(a)=a+\frac1a\) 3) \(a=1\), \(b=-1\), and the absolute minimum score is \(2\).

- Write the tangent line using the point and the derivative at the same input. - Isolate the constant term of the line as a function of \(a\). - Find all critical numbers of that intercept function. - Compare those values with the function's end behavior.

1. Differentiate: \(f'(x)=(x^2+4x+2)e^x\). 2. The tangent line at \(x=a\) is \(y=f'(a)(x-a)+f(a)\). Simplifying gives \(t_a:y=(a^2+4a+2)e^a x+(-a^3-3a^2)e^a\). 3. Thus the y-intercept is \(n(a)=(-a^3-3a^2)e^a\). 4. Differentiate: \(n'(a)=-a(a^2+6a+6)e^a\). The critical numbers are \(a=0\) and \(a=-3\pm\sqrt3\). 5. The value at \(a=-3-\sqrt3\) is positive, \(n(0)=0\), and \(n(-3+\sqrt3)<0\). Also \(n(a)\to0\) as \(a\to-\infty\) and \(n(a)\to-\infty\) as \(a\to\infty\). Therefore, the absolute maximum occurs at \(a=-3-\sqrt3\).

a) \(t_a:y=(a^2+4a+2)e^a x+(-a^3-3a^2)e^a\) b) The y-intercept is greatest at \(a=-3-\sqrt3\).

- Use the derivative to write the tangent line at a general input. - Treat the y-intercept as its own function of the parameter. - Solve for every critical value of that function. - End behavior is needed to justify an absolute maximum on an unbounded domain.

1. Differentiate: \(f'(x)=(2x-x^2)e^{-x}\). 2. The tangent line at \(x=a\) is \(y=f'(a)(x-a)+f(a)\). Simplifying gives \(t_a:y=(2a-a^2)e^{-a}x+(a^3-a^2)e^{-a}\). 3. Thus the y-intercept is \(n(a)=(a^3-a^2)e^{-a}\). 4. Differentiate: \(n'(a)=-a(a^2-4a+2)e^{-a}\). The critical values are \(a=0\) and \(a=2\pm\sqrt2\). 5. The value at \(a=2+\sqrt2\) is positive, \(n(0)=0\), and \(n(2-\sqrt2)<0\). Also \(n(a)\to-\infty\) as \(a\to-\infty\) and \(n(a)\to0\) as \(a\to\infty\). Therefore, the absolute maximum occurs at \(a=2+\sqrt2\).

a) \(t_a:y=(2a-a^2)e^{-a}x+(a^3-a^2)e^{-a}\) b) The y-intercept is greatest at \(a=2+\sqrt2\).

- Substitute the given time and amount into the model to solve for \(k\). - In part b), look for a factor that cancels from both sides. - In the schematic for part c), identify a horizontal base and a vertical height for triangle \(ABC\). - Use the product rule on the resulting area function. - Because \(a>5\), the horizontal base length is positive.

1. Use \(g_k(10)=500\): \(k(1-e^{-2})=500\). Thus \(k=\frac{500}{1-e^{-2}}\approx 578.26\). 2. To reach \(75\%\) of the saturation value, set \(g_k(t)=0.75k\): \(k(1-e^{-0.2t})=0.75k\). Since \(k>0\), divide by \(k\) to obtain \(e^{-0.2t}=0.25\). The equation no longer contains \(k\), so the time is independent of \(k\). 3. Solve for time: \(-0.2t=\ln(0.25)\), so \(t=5\ln(4)\approx 6.93\). 4. Since \(g_k'(0)=0.2k\), the tangent line at the origin is \(y=0.2kt\). It meets \(y=k\) when \(t=5\), so \(A=(5,k)\). Also, \(B=(a,k)\) and \(C=(a,k(1-e^{-0.2a}))\). 5. For \(a>5\), the base length is \(a-5\), and the vertical height is \(ke^{-0.2a}\). Therefore, \(F(a)=\frac{k}{2}(a-5)e^{-0.2a}\). 6. Differentiate: \(F'(a)=\frac{k}{2}e^{-0.2a}(2-0.2a)\). 7. Setting \(F'(a)=0\) gives \(a=10\). The derivative changes from positive to negative there, so the triangle has maximum area when \(a=10\).

a) \(k=\frac{500}{1-e^{-2}}\approx 578.26\) b) \(t=5\ln(4)\approx 6.93\) c) \(a=10\)

- Differentiate the product carefully before simplifying the difference. - A signed difference and an actual distance are different objective functions. - The smallest possible value of an absolute value is worth checking before differentiating it.

1. Differentiating twice gives \(f''(x)=(x^2+4x-1)e^x\). Hence \(d(x)=(4x+2)e^x\). 2. Differentiate: \(d'(x)=(4x+6)e^x\). The only critical number is \(x=-\frac32\). Since \(d''\left(-\frac32\right)=4e^{-3/2}>0\), \(d\) has a local minimum there, with value \(-4e^{-3/2}\). 3. The distance \(D(x)=|d(x)|\) is always nonnegative. It equals \(0\) exactly when \(d(x)=0\). Since \(e^x>0\), \((4x+2)e^x=0\) at \(x=-\frac12\). Thus the absolute minimum distance is \(0\) at \(x=-\frac12\). 4. Minimizing the signed difference \(d\) makes it as negative as possible locally; minimizing the vertical distance means minimizing \(|d|\), whose smallest possible value is \(0\).

a) \(f''(x)=(x^2+4x-1)e^x\), so \(d(x)=(4x+2)e^x\). b) \(d\) has a local minimum at \(\left(-\frac32,-4e^{-3/2}\right)\). c) The absolute minimum vertical distance is \(0\), attained at \(x=-\frac12\).

- Use the distance formula for two points in the coordinate plane. - Minimizing a positive distance is equivalent to minimizing its square. - Differentiate the squared-distance function and find its critical numbers. - Use the derivative of \(f\) to find the tangent slope. - Two nonvertical lines are perpendicular when the product of their slopes is \(-1\).

1. The distance from \(Q=(x, 0.5x^2)\) to \(P(6, 0)\) is \(d(x)=\sqrt{(x-6)^2+(0.5x^2)^2}\). Squaring and expanding gives \(d^2(x)=(x-6)^2+0.25x^4=0.25x^4+x^2-12x+36\). 2. Minimize \(g(x)=0.25x^4+x^2-12x+36\) on \(x\ge0\). Its derivative is \(g'(x)=x^3+2x-12=(x-2)(x^2+2x+6)\). The quadratic factor has no real zeros, so the only critical number is \(x=2\). Because \(g''(x)=3x^2+2>0\) for every \(x\), \(g\) is strictly convex and this critical number gives the absolute minimum. Since \(f(2)=2\), the connection point is \(Q_E=(2, 2)\). 3. The slope of the tangent to the bypass at \(Q_E\) is \(f'(2)=2\). The slope of the access road is \(m=\frac{2-0}{2-6}=-\frac12\). Their product is \(2\cdot\left(-\frac12\right)=-1\), so the two lines are perpendicular.

a) \(d(x)=\sqrt{(x-6)^2+(0.5x^2)^2}\), and \(d^2(x)=0.25x^4+x^2-12x+36\). b) \(Q_E=(2, 2)\) c) The tangent slope is \(2\), and the access-road slope is \(-\frac12\). Their product is \(-1\), so the roads are perpendicular at \(Q_E\).

- Translate each endpoint and horizontal-tangent statement into an equation. - A horizontal tangent means the first derivative is \(0\). - Use the second derivative to find the inflection point. - Convert slope magnitude to percent and use inverse tangent for the angle.

1. The equation \(f(10)=0\) means the slide reaches ground level after \(10\,\text{m}\) horizontally. The equation \(f'(0)=0\) means the slide begins horizontally. 2. Let \(f(x)=ax^3+bx^2+cx+d\). The conditions \(f(0)=8\), \(f'(0)=0\), \(f(10)=0\), and \(f'(10)=0\) give \(d=8\), \(c=0\), \(a=\frac{2}{125}\), and \(b=-\frac{6}{25}\). Thus, \(f(x)=\frac{2}{125}x^3-\frac{6}{25}x^2+8\). 3. Since \(f''(x)=\frac{12}{125}x-\frac{12}{25}\), the inflection point occurs at \(x=5\), where \(f(5)=4\). Thus, \(W=(5, 4)\). The slide is steepest there and changes concavity. 4. The slope at \(x=5\) is \(f'(5)=-\frac{6}{5}\). Therefore, the maximum downward grade is \(120\%\), and the angle of descent is \(\arctan(\frac{6}{5})\approx50.2^\circ\).

a) The slide reaches the ground at \(x=10\) and begins horizontally at \(x=0\). b) \(f(x)=\frac{2}{125}x^3-\frac{6}{25}x^2+8\) c) \(W=(5, 4)\); this is the steepest point and where concavity changes. d) \(120\%\); \(50.2^\circ\)

- Write the area of each trapezoid before combining them. - Substitute the fixed endpoint heights before differentiating. - Use the sign of the second derivative on the whole feasible interval to justify the type of extremum.

1. The two trapezoids have areas \(A_1=\frac{g(0)+g(u)}{2}u\) and \(A_2=\frac{g(u)+g(4)}{2}(4-u)\). Since \(g(0)=2\) and \(g(4)=6\), their sum simplifies to \(A(u)=4\sqrt{u}-2u+16\). 2. Differentiate: \(A'(u)=\frac{2}{\sqrt{u}}-2\). Setting \(A'(u)=0\) gives \(u=1\). 3. Since \(A''(u)=-u^{-3/2}<0\) for \(0<u<4\), \(A\) is strictly concave down throughout its domain, so the critical point gives the absolute maximum. 4. Since \(g(1)=4\), \(R=(1,4)\), and \(A(1)=18\) square units.

a) \(A(u)=4\sqrt{u}-2u+16\) b) \(u=1\), \(R=(1,4)\), and the maximum pentagon area is \(18\) square units. c) The maximum is absolute because \(A''(u)<0\) throughout \((0,4)\).

- First determine whether the line ever intersects the parabola. - Express the point-to-line distance as a function of the parabola's x-coordinate. - Connect the minimizing derivative condition to equality of tangent slopes.

1. The parabola lies entirely above the line because \(f(x)-g(x)=0.25x^2-0.5x+3>0\) for all \(x\). 2. The perpendicular distance from \((x,f(x))\) to the line \(0.5x-y-1=0\) is \(d(x)=\frac{f(x)-0.5x+1}{\sqrt{1.25}}\). The numerator is the strictly convex quadratic \(0.25x^2-0.5x+3\), so its unique critical point gives the absolute minimum distance. 3. Differentiating the numerator gives \(0.5x-0.5=0\), so \(x=1\). Equivalently, \(f'(1)=0.5\), which equals the slope of \(g\); hence the tangent to the parabola is parallel to \(g\). 4. Since \(f(1)=2.25\), the closest point is \(P=(1,2.25)\). 5. The minimum distance is \(d=\frac{|0.5(1)-2.25-1|}{\sqrt{1.25}}=\frac{11}{2\sqrt5}\approx2.46\) units.

a) At the closest point, the normal to the parabola is perpendicular to line \(g\), so the tangent is parallel to \(g\). b) \(P=(1,2.25)\), and the absolute minimum distance is \(\frac{11}{2\sqrt5}\approx2.46\) units.