AP Calculus AB Optimization Worksheet Solutions (PDF)

AP Calculus AB math worksheet solutions: Optimization – Maths PDF

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This optimization practice solutions for AP Calculus AB covers optimization (Common Core AP CALC AB 5.10, AP CALC AB 5.11), with a free PDF to print.

This answer key gives a full worked solution for every problem in the optimization worksheet. Compare each step, not only the final answer, because AP graders award credit for a correct setup and clear justification. Final answers are in bold. Redo any problem you missed before moving on to the test.

AP Calculus AB · Common Core State Standards: AP CALC AB 5.10, AP CALC AB 5.11

Optimization practice solutions: exercises 1 to 10

Exercise 1 – Maximum product

Let the numbers be $x$ and $20-x$. Then $P(x)=x(20-x)=20x-x^{2}$ on $0 < x < 20$.

$P^{\prime}=20-2x=0$ gives $x=10$. $P^{\prime\prime}=-2 < 0$, so it is a maximum.

Final answer: $10\text{ and }10,\ \text{product }100$.

Exercise 2 – Minimum sum of squares

$S(x)=x^{2}+(10-x)^{2}=2x^{2}-20x+100$.

$S^{\prime}=4x-20=0$ gives $x=5$. $S^{\prime\prime}=4 > 0$: minimum.

Final answer: $5\text{ and }5,\ \text{sum of squares }50$.

Exercise 3 – Maximum area, fixed perimeter

Let the sides be $x$ and $100-x$. $A=x(100-x)$.

$A^{\prime}=100-2x=0$ gives $x=50$, a $50\times 50$ square. $A^{\prime\prime}=-2 < 0$.

Final answer: $50\text{ ft by }50\text{ ft, area }2500\text{ ft}^{2}$.

Area of a rectangle with perimeter 200 ft.

Exercise 4 – Fence against a wall

$2x+y=120$ gives $y=120-2x$. $A=x(120-2x)$ on $0 < x < 60$.

$A^{\prime}=120-4x=0$ gives $x=30$, $y=60$. $A^{\prime\prime}=-4 < 0$.

Final answer: $30\text{ ft by }60\text{ ft, area }1800\text{ ft}^{2}$.

Exercise 5 – Minimum fence, fixed area

$xy=800$, so $y=\frac{800}{x}$ and $P(x)=2x+\frac{1600}{x}$ for $x > 0$.

$P^{\prime}=2-\frac{1600}{x^{2}}=0$ gives $x^{2}=800$, so $x=20\sqrt{2}\approx 28.28$ and $y=20\sqrt{2}$.

$P^{\prime\prime}=\frac{3200}{x^{3}} > 0$: minimum.

Final answer: $\text{a square of side }20\sqrt{2}\approx 28.28\text{ ft; fence }80\sqrt{2}\approx 113.1\text{ ft}$.

Exercise 6 – Open box from a square sheet

The base is $12-2x$ by $12-2x$ and the height is $x$: $V=x(12-2x)^{2}=4x^{3}-48x^{2}+144x$, $0 < x < 6$.

$V^{\prime}=12x^{2}-96x+144=12(x-2)(x-6)$. Only $x=2$ is in the domain.

$V^{\prime\prime}(2)=24(2)-96=-48 < 0$: maximum. $V(2)=2\cdot 8^{2}=128$.

Final answer: $x=2\text{ in, maximum volume }128\text{ in}^{3}$.

Volume of the open box in cubic inches.

Exercise 7 – Open box from a rectangle

$V=x(10-2x)(16-2x)=4x^{3}-52x^{2}+160x$ for $0 < x < 5$.

$V^{\prime}=12x^{2}-104x+160=4(3x^{2}-26x+40)=4(3x-20)(x-2)$. In the domain: $x=2$ (since $\frac{20}{3} > 5$).

$V(2)=2\cdot 6\cdot 12=144$.

Final answer: $x=2\text{ in, volume }144\text{ in}^{3}$.

Exercise 8 – Can with least material

$V=\pi r^{2}h=500$ gives $h=\frac{500}{\pi r^{2}}$. $S=2\pi r^{2}+2\pi rh=2\pi r^{2}+\frac{1000}{r}$.

$S^{\prime}=4\pi r-\frac{1000}{r^{2}}=0$ gives $r^{3}=\frac{250}{\pi}$, so $r\approx 4.3$ cm and $h=2r\approx 8.6$ cm.

$S^{\prime\prime}=4\pi+\frac{2000}{r^{3}} > 0$: minimum.

Final answer: $r\approx 4.3\text{ cm},\ h\approx 8.6\text{ cm}$.

Surface area of the can in square centimeters.

Exercise 9 – Closest point on a parabola

Minimize the square of the distance: $D^{2}=x^{2}+(x^{2}-1)^{2}=x^{4}-x^{2}+1$.

$4x^{3}-2x=2x(2x^{2}-1)=0$: $x=0$ or $x=\pm\frac{1}{\sqrt{2}}$. $D^{2}(0)=1$ and $D^{2}\left(\pm\frac{1}{\sqrt{2}}\right)=\frac{3}{4}$.

Final answer: $\left(\pm\frac{1}{\sqrt{2}},\frac{1}{2}\right),\ \text{distance }\frac{\sqrt{3}}{2}$.

Exercise 10 – Closest point on a line

$D^{2}=x^{2}+(2x+1)^{2}=5x^{2}+4x+1$. $(D^{2})^{\prime}=10x+4=0$ gives $x=-\frac{2}{5}$.

$y=2\left(-\frac{2}{5}\right)+1=\frac{1}{5}$. $D^{2}=\frac{1}{5}$.

Final answer: $\left(-\frac{2}{5},\frac{1}{5}\right),\ \text{distance }\frac{1}{\sqrt{5}}$.

Exercises 11 and up

Exercise 11 – Price for maximum revenue

$R(p)=p(500-10p)=500p-10p^{2}$. $R^{\prime}=500-20p=0$ gives $p=25$. $R^{\prime\prime}=-20 < 0$.

$R(25)=25(250)=6250$.

Final answer: $p=\$25,\ \text{revenue }\$6250\text{ per week}$.

Exercise 12 – Maximizing profit

$P=xp-C=x(20-0.01x)-(50+4x)=-0.01x^{2}+16x-50$.

$P^{\prime}=-0.02x+16=0$ gives $x=800$. $P^{\prime\prime}=-0.02 < 0$.

$P(800)=-6400+12800-50=6350$.

Final answer: $800\text{ pies, profit }\$6350$.

Exercise 13 – Rectangle in a semicircle

With corners at $(\pm x,\sqrt{25-x^{2}})$, $A=2x\sqrt{25-x^{2}}$ on $0 < x < 5$.

It is easier to maximize $A^{2}=4x^{2}(25-x^{2})=100x^{2}-4x^{4}$: $200x-16x^{3}=0$ gives $x^{2}=\frac{25}{2}$, $x=\frac{5}{\sqrt{2}}$.

$A=2\cdot\frac{5}{\sqrt{2}}\cdot\frac{5}{\sqrt{2}}=25$.

Final answer: $25\text{ square units}$.

Exercise 14 – Rectangle under a parabola

$A=2x(12-x^{2})=24x-2x^{3}$ for $0 < x < \sqrt{12}$.

$A^{\prime}=24-6x^{2}=0$ gives $x=2$. $A^{\prime\prime}=-12x < 0$.

$A(2)=2\cdot 2\cdot 8=32$.

Final answer: $32\text{ square units}$.

Exercise 15 – Fencing with two prices

Let the \$5 sides have length $x$ and the \$10 sides length $y=\frac{800}{x}$. Cost: $C=2(5x)+2(10y)=10x+\frac{16000}{x}$.

$C^{\prime}=10-\frac{16000}{x^{2}}=0$ gives $x=40$, $y=20$. $C^{\prime\prime} > 0$: minimum.

$C(40)=400+400=800$.

Final answer: $40\text{ ft by }20\text{ ft, cost }\$800$.

Exercise 16 – Poster margins

Let the printed width be $x$, so its height is $\frac{200}{x}$. The paper is $(x+4)\left(\frac{200}{x}+8\right)=232+8x+\frac{800}{x}$.

$8-\frac{800}{x^{2}}=0$ gives $x=10$. Printed area: $10\times 20$. Paper: $14\times 28=392$.

Final answer: $14\text{ cm by }28\text{ cm, area }392\text{ cm}^{2}$.

Exercise 17 – Open-top box

$x^{2}h=1000$ gives $h=\frac{1000}{x^{2}}$. $S=x^{2}+4xh=x^{2}+\frac{4000}{x}$.

$S^{\prime}=2x-\frac{4000}{x^{2}}=0$ gives $x^{3}=2000$, $x\approx 12.6$ in and $h=\frac{1000}{x^{2}}\approx 6.3$ in.

$S^{\prime\prime}=2+\frac{8000}{x^{3}} > 0$: minimum.

Final answer: $x\approx 12.6\text{ in},\ h\approx 6.3\text{ in}$.

Exercise 18 – Closest point to (4, 0)

$D^{2}=(x-4)^{2}+x$. $(D^{2})^{\prime}=2(x-4)+1=0$ gives $x=\frac{7}{2}$.

$D^{2}=\frac{1}{4}+\frac{7}{2}=\frac{15}{4}$, so $D=\frac{\sqrt{15}}{2}\approx 1.94$.

Final answer: $\left(\frac{7}{2},\sqrt{\frac{7}{2}}\right)$.

Exercise 19 – Two adjacent pens

The fence has three sides of length $x$ (two outside and one middle) and two sides of length $y$: $3x+2y=1000$, so $y=500-\frac{3x}{2}$.

$A=xy=500x-\frac{3}{2}x^{2}$, $A^{\prime}=500-3x=0$ gives $x=\frac{500}{3}\approx 166.7$ and $y=250$.

$A=\frac{500}{3}\cdot 250\approx 41{,}667$ square feet.

Final answer: $x=\frac{500}{3}\text{ ft},\ y=250\text{ ft},\ A\approx 41{,}667\text{ ft}^{2}$.

Exercise 20 – Ticket price

With $x$ dollars of increase: $R(x)=(12+x)(200-10x)=2400+80x-10x^{2}$.

$R^{\prime}=80-20x=0$ gives $x=4$. $R^{\prime\prime}=-20 < 0$. Price $12+4=16$; $R(4)=16\cdot 160=2560$.

Final answer: $\$16\text{ per ticket, revenue }\$2560$.

Key facts behind this optimization practice solutions

Key facts

Name the variables, write an objective function, and use a constraint to reduce it to one variable.
Find critical numbers by solving f′ = 0 on the allowed domain.
Justify the extreme with a sign change of f′, the sign of f″, or by comparing endpoint values.
Check the domain and give the final answer in context with units.

How to use this optimization practice solutions

Cover each solution, solve the exercise again, then uncover it and compare every line. If your answer differs, find the first step that changed and redo the exercise from there.

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