Practice Problems

1. (a) $2 \mathrm{x}+2 \mathrm{y}=200$ so $\mathrm{y}=100-\mathrm{x}$. Maximize $\mathrm{A}=\mathrm{x} \cdot \mathrm{y}=\mathrm{x} \cdot (100-\mathrm{x})=100 \mathrm{x}-\mathrm{x}^{2}$. $\mathrm{A}^{\prime}=100-2 \mathrm{x}$ and $\mathrm{A}^{\prime}=0$ when $\mathrm{x}=50(\mathrm{y}=100-\mathrm{x}=50)$. $\mathrm{A}^{\prime \prime}=-2 < 0$ so $\mathrm{x}=50$ yields the maximum enclosed area. When $\mathrm{x}=50$ $\mathrm{A}=50(100-50)=2500$ square feet.
(b) $2 \mathrm{x}+2 \mathrm{y}=\mathrm{P}$ so $\mathrm{y}=\mathrm{P} / 2-\mathrm{x}$. Maximize $\mathrm{A}=\mathrm{x} \cdot \mathrm{y}=\mathrm{x} \cdot(\mathrm{P} / 2-\mathrm{x})=(\mathrm{P} / 2) \mathrm{x}-\mathrm{x}^{2}$ $A^{\prime}=P / 2-2 x$ and $A^{\prime}=0$ when $x=P / 4$ (then $y=P / 2-x=P / 4$ ). $A^{\prime \prime}=-2 < 0$ so $\mathrm{x}=\mathrm{P} / 4$ yields the maximum enclosed area.
This garden is a $\mathrm{P} / 4$ by $\mathrm{P} / 4$ square.
(c) $2 \mathrm{x}+\mathrm{y}=\mathrm{P}$ so $\mathrm{y}=\mathrm{P}-2 \mathrm{x}$. Maximize $\mathrm{A}=\mathrm{xy}=\mathrm{x}(\mathrm{P}-2 \mathrm{x})=\mathrm{Px}-2 \mathrm{x}^{2}$ $A^{\prime}=P-4 x$ and $A^{\prime}=0$ when $x=P / 4$ (then $y=P-2 x=P / 2$)
(d) A circle. A semicircle.

Fig. 3.5P1

3. (a) $120=2 x+5 y$ so $y=24-\frac{2}{5} x$. Maximize $A=x y=x\left(24-\frac{2}{5} x\right)=24 x-\frac{2}{5} x^{2}$. $A^{\prime}=24-\frac{4}{5} x$ and $A^{\prime}=0$ when $x=30$ (then $y=12$). $A^{\prime \prime}=-4 / 5 < 0$ so $\mathrm{x}=30$ yields the maximum enclosed area. Area is $(30 \mathrm{ft})(12 \mathrm{ft})=360$ square feet.
(b) A circular pen divided into $4$ equal stalls by two diameters shown in diagram (a) does a better job than a square with $400$ square feet. If the radius is $\mathrm{r}$, then $4 \mathrm{r}+2 \pi \mathrm{r}=120$ so $\mathrm{r}=120 /(4+2 \pi) \approx 11.67$.
The resulting enclosed area is $\mathrm{A}=\pi \mathrm{r}^{2} \approx \pi(11.67)^{2} \approx 427.8 \mathrm{sq}. \mathrm{ft}$.
The pen shown in diagram (b) does even better. If each semicircle has radius $\mathrm{r}$, then the figure uses $4 \sqrt{2} \mathrm{r}+4 \pi \mathrm{r}=120$ feet of fence so $\mathrm{r}=120 /(4 \sqrt{2}+4 \pi) \approx 6.585$. The resulting enclosed area is $\mathrm{A}=(\text { square })+(\text { four semicircles })=(2 \mathrm{r})^{2}+4\left(\frac{1}{2} \pi \mathrm{r}^{2}\right) \approx 445.90 \text { sq. } \mathrm{ft}$

5. $2 \mathrm{x}+2 \mathrm{y}=10$ so $\mathrm{y}=5-\mathrm{x}$
Maximize $\mathrm{V}=\mathrm{xy}(10-2 \mathrm{x})=\mathrm{x}(5-\mathrm{x})(10-2 \mathrm{x})=50 \mathrm{x}-20 \mathrm{x}^{2}+2 \mathrm{x}^{3}$
$\mathrm{V}^{\prime}=50-40 \mathrm{x}+6 \mathrm{x}^{2}=2(3 \mathrm{x}-5)(\mathrm{x}-5)$ and $\mathrm{V}^{\prime}=0$ when $\mathrm{x}=5$ and $\mathrm{x}=5 / 3$. When $\mathrm{x}=5$, then $\mathrm{V}=0$, clearly not a maximum, so $x=5 / 3$. The dimensions of the box with the largest volume are $5 / 3,10 / 3$, and $20 / 3$.

7. (a) $\mathrm{V}=\pi \mathrm{r}^{2} \mathrm{~h}=100$ so $\mathrm{h}=\frac{100}{\pi \mathrm{r}}^{2}$.

$\begin{aligned}\text { Minimize } C &=2(\text { top area })+5(\text { bottom area })+3(\text { side area }) \\&=2\left(\pi r^{2}\right)+5\left(\pi r^{2}\right)+3(2 \pi r h)=7 \pi r^{2}+6 \pi r\left(\frac{100}{\pi r^{2}}\right)=7 \pi r^{2}+\frac{600}{r}\end{aligned}$

$C^{\prime}=14 \pi r-\frac{600}{r^{2}}$ and $C^{\prime}=0$ when $r=\sqrt[3]{600 /(14 \pi)} \approx 2.39$ (then $\mathrm{h}=\frac{100}{\pi r^{2}} \approx 5.57$)
(b) Let $\mathrm{k}=$ top $+$ bottom rate $=2¢ +$ the bottom rate $> 2¢ + 5¢ =7¢$. Minimize $\mathrm{C}=\mathrm{k} \pi \mathrm{r}^{2}+\frac{600}{\mathrm{r}}$. $\mathrm{C}^{\prime}=2 \mathrm{k} \pi \mathrm{r}-\frac{600}{\mathrm{r}^{2}}$ and $\mathrm{C}^{\prime}=0$ when $\mathrm{r}=\sqrt[3]{600 /(2 \mathrm{k} \pi)}$. If $\mathrm{k}=8$, then $\mathrm{r} \approx 2.29$
If $\mathrm{k}=9$, then $\mathrm{r} \approx 2.20$. If $\mathrm{k}=10$, then $\mathrm{r} \approx 2.12$. As the cost of the bottom material increases, the radius of the least expensive cylindrical can decreases: the least expensive can becomes narrower and taller

9. Time $=$ distance/rate. Run distance $= x$($0 \leq x \leq 60$ Why?) so run time $=x / 8$.
Swim distance $=\sqrt{40^{2}+(60-x)^{2}}$ so swim time $=\frac{1}{2} \sqrt{40^{2}+(60-x)^{2}}$ and the total time is

$\begin{aligned}&T=\frac{x}{8}+\frac{1}{2} \sqrt{40^{2}+(60-x)^{2}} \\&T^{\prime}=\frac{1}{8}+\frac{1}{2} \frac{1}{2}\left(40^{2}+(60-x)^{2}\right)^{-1 / 2} \cdot 2 \cdot(60-x) \cdot(-1)=\frac{1}{8}-\frac{60-x}{2 \sqrt{40^{2}+(60-x)^{2}}}\end{aligned}$

$\mathrm{T}^{\prime}=0$ when $\mathrm{x}=60 \pm \frac{40}{\sqrt{15}}$. The value $\mathrm{x}=60+\frac{40}{\sqrt{15}} > 60$ so the least total time occurs when $\mathrm{x}=60-\frac{40}{\sqrt{15}} \approx 49.7$ meters. In this situation, the lifeguard should run about $5 / 6$ of the way along the beach before going into the water.

11. (a) Consider a similar problem with a new town $D^*$ located at the "mirror image" of $\mathrm{D}$ across the river (Fig. 3.5P11a). If the water works is built at any location $\mathrm{W}$ along the river, then the distances are the same from $\mathrm{W}$ to $\mathrm{D}$ and to $\mathrm{D}^{*}$: dist $(\mathrm{W}, \mathrm{D})=\operatorname{dist}\left(\mathrm{W}, \mathrm{D}^{*}\right)$.
Then $\operatorname{dist}(\mathrm{C}, \mathrm{W})+\operatorname{dist}(\mathrm{W}, \mathrm{D})=\operatorname{dist}(\mathrm{C}, \mathrm{W})+\operatorname{dist}\left(\mathrm{W}, \mathrm{D}^{*}\right)$. The shortest distance from $\mathrm{C}$ to $\mathrm{D}^{*}$ is a straight line (Fig. 3.5P11b), and this straight line gives similar triangles with equal side ratios: $\frac{x}{3}=\frac{10-x}{5}$ so $\mathrm{x}=15 / 4=3.75$ miles. A consequence of this "mirror image" view of the problem is that "at the best location $\mathrm{W}$ the angle of incidence $\alpha$ equals the angle of reflection $\beta^{\prime  \prime}$
(b) Minimize $\mathrm{C}=3000 \operatorname{dist}(\mathrm{C}, \mathrm{W})+7000 \operatorname{dist}(\mathrm{W}, \mathrm{D})=3000 \sqrt{\mathrm{x}^{2}+9}+7000 \sqrt{(10-\mathrm{x})^{2}+25}$. $\mathrm{C}^{\prime}=\frac{3000 \mathrm{x}}{\sqrt{\mathrm{x}^{2}+9}}+\frac{-7000(10-\mathrm{x})}{\sqrt{(10-\mathrm{x})^{2}+25}}$ so $\mathrm{C}^{\prime}=0$ when $\frac{3 \mathrm{x}}{\sqrt{\mathrm{x}^{2}+9}}=\frac{7(10-\mathrm{x})}{\sqrt{(10-\mathrm{x})^{2}+25}}$ and $\mathrm{x} \approx 7.82$ miles.
As it becomes relatively more expensive to build the pipe from a point $\mathrm{W}$ on the river to $\mathrm{D}$, the cheapest route tends to shorten the distance from $\mathrm{W}$ to $\mathrm{D}$.

Fig. 3.5P11

13. (a) Let $\mathrm{x}$ be the length of one edge of the square end. Then $\mathrm{V}=\mathrm{x}^{2}(108-4 \mathrm{x})=108 \mathrm{x}^{2}-4 \mathrm{x}^{3}$ $\mathrm{V}^{\prime}=216 \mathrm{x}-12 \mathrm{x}^{2}=6 \mathrm{x}(18-\mathrm{x})$ so $\mathrm{V}^{\prime}=0$ when $\mathrm{x}=0$ or $\mathrm{x}=18$. The dimensions of the greatest volume acceptable box with a square end are $18$ by $18$ by $36$ inches: $V=11,664\ \mathrm{in}^{3}$,
(b) Let $x$ be the length of the shorter edge of the end. Then $V=2 x^{2}(108-6 x)=216 x^{2}-12 x^{3}$ $\mathrm{V}^{\prime}=432 \mathrm{x}-36 \mathrm{x}^{2}=36 \mathrm{x}(12-\mathrm{x})$ so $\mathrm{V}^{\prime}=0$ when $\mathrm{x}=0$ or $\mathrm{x}=12$. The dimensions of the largest box acceptable box with this shape are $12$ by $24$ by $36$ inches: $V=10,368\ \mathrm{in}^{3}$.
(c) Let $x$ be the radius of the circular end. Then $V=\pi x^{2}(108-2 \pi x)=108 \pi x^{2}-2 \pi^{2} x^{3}$. $\mathrm{V}^{\prime}=216 \pi \mathrm{x}-6 \pi^{2} \mathrm{x}^{2}=6 \pi \mathrm{x}(36-\pi \mathrm{x})$ so $\mathrm{V}^{\prime}=0$ when $\mathrm{x}=0$ or $\mathrm{x}=36 / \pi \approx 11.46$ inches. The dimensions of the largest box acceptable box with a circular end are a radius of $36 / \pi \approx 11.46$ and a length of $36$ inches: $\mathrm{V} \approx 14,851\ \mathrm{in}^{3}$.

15. Without calculus: The area of the triangle is $\frac{1}{2}$ (base)(height) $=\frac{1}{2}(7)$ (height) and the height is maximum when the angle between the sides is a right angle.
Using calculus: Let $\theta$ be the angle between the sides. Then the area of the triangle is
$A=\frac{1}{2}$ (base) (height) $=\frac{1}{2}(7)$ (height) $=\frac{1}{2}(7)(10 \sin \theta)=35 \sin \theta$. $A^{\prime}=35 \cos \theta$ so $A^{\prime}=0$
when $\theta=\pi / 2$, and the triangle is a right triangle with sides $7$ and $10$.
Using either approach, the maximum area of the triangle is $\frac{1}{2}(7)(10)=35$ square inches, and the other side is the hypotenuse with length $\sqrt{7^{2}+10^{2}}=\sqrt{149} \approx 12.2$ inches.

17. (a) $\mathrm{A}=2 \mathrm{x}\left(16-\mathrm{x}^{2}\right)=32 \mathrm{x}-2 \mathrm{x}^{3}$. Then $\mathrm{A}^{\prime}=32-6 \mathrm{x}^{2}$ so $\mathrm{A}^{\prime}=0$ when $\mathrm{x}=\sqrt{32 / 6} \approx 2.31$. The dimensions are $2 \sqrt{32 / 6} \approx 4.62$ and $16-(\sqrt{32 / 6})^{2}=64 / 6 \approx 10.67$.
(b) $\mathrm{A}=2 \mathrm{x}\left(\sqrt{1-\mathrm{x}^{2}}\right)$. Then $\mathrm{A}^{\prime}=2\left(\sqrt{1-\mathrm{x}^{2}}-\frac{\mathrm{x}^{2}}{\sqrt{1-\mathrm{x}^{2}}}\right)$ so $\mathrm{A}^{\prime}=0$ when $\mathrm{x}=1 / \sqrt{2} \approx 0.707$. The dimensions are $2(1 / \sqrt{2}) \approx 1.414$ and $1 / \sqrt{2} \approx 0.707$.
(c) The graph of $|\mathrm{x}|+\mid \mathrm{yl}=1$ is a "diamond" (a square) with corners at $(1,0),(0,1),(-1,0)$ and $(0,-1)$. For $0 \leq x \leq 1, A=2 x \cdot 2(1-x)$ so $A=4 x-4 x^{2}$. Then $A^{\prime}=4-8 x$ and $A^{\prime}=0$ when $\mathrm{x}=1 / 2$. $\mathrm{A}^{\prime \prime}=-8$ so we have a local $\max$. The dimensions are $2(1 / 2)=1$ and $2(1-1 / 2)=1$.
(d) $\mathrm{A}=2 \mathrm{x} \cos (\mathrm{x})(0 \leq \mathrm{x} \leq \pi / 2)$. Then $\mathrm{A}^{\prime}=2 \cos (\mathrm{x})-2 \mathrm{x} \cdot \sin (\mathrm{x})$ so $\mathrm{A}^{\prime}=0$ when $\mathrm{x} \approx 0.86$. The dimensions are $2(0.86)=1.72$ and $\cos (0.86) \approx 0.65$.

19. $A=6 \cdot\ \sin (\theta / 2) \cdot 6 \cdot\ \cos (\theta / 2)=36 \cdot \frac{1}{2}\ \sin (\theta)=18\ \sin (\theta)$ and this is a maximum when $\theta=\pi / 2$. Then the maximum area is $\mathrm{A}=18\ \sin (\pi / 2)=18$ square inches. (This problem is similar to problem 15.)

21. $\mathrm{V}=\frac{1}{3} \pi \mathrm{r}^{2} \mathrm{~h}$ and $\mathrm{h}=\sqrt{9-\mathrm{r}^{2}}$ so $\mathrm{V}=\frac{1}{3} \pi \mathrm{r}^{2} \sqrt{9-\mathrm{r}^{2}}=\frac{\pi}{3} \sqrt{9 \mathrm{r}^{4}-\mathrm{r}^{6}}$. Then $\mathrm{V}^{\prime}=\frac{\pi}{6} \frac{36 r^{3}-6 r^{5}}{\sqrt{9 r^{4}-r^{6}}}$, and $\mathrm{V}^{\prime}=0$ when $36 \mathrm{r}^{3}=6 \mathrm{r}^{5}$ so $\mathrm{r}=\sqrt{6} \approx 2.45$ inches and $\mathrm{h}=\sqrt{9-\mathrm{r}^{2}}=\sqrt{3} \approx 1.73$ inches.

23. Let $\mathrm{n} \geq 10$ be the number of passengers. The income is $\mathrm{I}=\mathrm{n}(30-(\mathrm{n}-10))=40 \mathrm{n}-\mathrm{n}^{2}$. The cost is $C=100+6 n$ so the profit is $P=$ Income $-$ Cost $=\left(40 n-n^{2}\right)-(100+6 n)=34 n-n^{2}-100$.
$P^{\prime}=34-2 \mathrm{n}$ and $P^{\prime}=0$ when $\mathrm{n}=17.17$ passengers on the flight maximize your profit. (This is an example of treating a naturally discrete variable, the number of passengers, as a continuous variable.)

25. Apply the result of problem 24 with $R=f$ and $E=g$.

27. (i) Let $\mathrm{D}=$ diameter of the base of the can, and let $\mathrm{H}=$ the height of the can.
Then $\theta=\arctan \left(\frac{\text { radius of can }}{\text { height of } \mathrm{cg}}\right)=\arctan \left(\frac{\mathrm{D} / 2}{\mathrm{H} / 2}\right)$.
For this can, $\mathrm{D}=5 \mathrm{~cm}$ and $\mathrm{H}=12 \mathrm{~cm}$ (sorry this should be in the statement of the problem) so $\theta=\arctan (2.5 / 6)=\arctan (0.42) \approx 0.395$ which is about $22.6^{\circ}$. The can can be tilted about $22.6^{\circ}$ before it falls over.
(ii) $C(x)=\frac{360+9.6 x^{2}}{60+19.2 x}$ so $C^{\prime}(x)=\frac{(60+19.2 x)(19.2 x)-\left(360+9.6 x^{2}\right)(19.2)}{(0+19.2 x)^{2}} \cdot C^{\prime}(x)=0$ when $(19.2)\left(9.6 \mathrm{x}^{2}+60 \mathrm{x}-360\right)=0$ so $\mathrm{x}=3.75$: the height of the cola is $\mathrm{h}=3.75 \mathrm{~cm}$.
(iii) $\mathrm{C}(3.75)=3.75$ (The center of gravity is exactly at the top edge of the cola. It turns out that when the $\mathrm{cg}$ of a can and liquid system is as low as possible then the cg is at the top edge of the liquid.) Then $\theta=\arctan$ (radius/(height of $\mathrm{cg}$)$=\arctan (2.5 / 3.75)=\arctan (0.667) \approx 0.588$ which is about $33.7^{\circ}$. In this situation, the can can be tilted about $33.7^{\circ}$ before it falls over.
(iv) Less.

29. (a) A (base)(height) = $(1-x)\left(x^{2}\right)=x^{2}-x^{3}$ for $0 \leq x \leq 1$. $A^{\prime}=2 x-3 x^{2}=0$ if $x=2 / 3$.
(Clearly the endpoints $\mathrm{x}=0$ and $\mathrm{x}=1$ will not give the largest area.) Then $A=\left(\frac{1}{3}\right)\left(\frac{2}{3}\right)^{2}=\frac{4}{27}$.
(b) A (base)(height) = $(1-\mathrm{x})\left(\mathrm{Cx}^{2}\right)=\mathrm{Cx}^{2}-\mathrm{Cx}^{3}$ for $0 \leq \mathrm{x} \leq 1$
$\mathrm{A}^{\prime}=2 \mathrm{Cx}-3 \mathrm{Cx}^{2}=0$ if $\mathrm{x}=2 / 3$. Then $\mathrm{A}=\left(\frac{1}{3}\right)(\mathrm{C})\left(\frac{2}{3}\right)^{2}=\frac{4 \mathrm{C}}{27}$.
(c) A=(base)(height) = $(\mathrm{B}-\mathrm{x})\left(\mathrm{Cx}^{2}\right)=\mathrm{BCx}^{2}-\mathrm{Cx}^{3}$ for $0 \leq \mathrm{x} \leq 1$.
$\mathrm{A}^{\prime}=2 \mathrm{BCx}-3 \mathrm{Cx}^{2}=0$ if $\mathrm{x}=\frac{2}{3} \mathrm{~B}$. Then $\mathrm{A}=\left(\frac{\mathrm{B}}{3}\right)(\mathrm{C})\left(\frac{2 \mathrm{~B}}{3}\right)^{2}=\frac{4}{27} \mathrm{~B}^{3} \mathrm{C}$.

31. (a) $\mathrm{y}=20-\frac{20}{50} \mathrm{x} \cdot \mathrm{A}$ = (base)height) = $\mathrm{xy}=\mathrm{x}\left(20-\frac{2}{5} \mathrm{x}\right)=20 \mathrm{x}-\frac{2}{5} \mathrm{x}^{2}$. $\mathrm{A}^{\prime}=20-\frac{4}{5} \mathrm{x}=0$ when $\mathrm{x}=25$. Then $\mathrm{y}=10$ and Area = $250$.
(b) $\mathrm{y}=\mathrm{H}-\frac{\mathrm{H}}{\mathrm{B}} \mathrm{x}$. $\mathrm{A}$ = (base)height) = $\mathrm{x}\left(\mathrm{H}-\frac{\mathrm{H}}{\mathrm{B}} \mathrm{x}\right)=\mathrm{Hx}-\frac{\mathrm{H}}{\mathrm{B}} \mathrm{x}^{2}$. $A^{\prime}=H-\frac{2 H}{B} x=0$ when $x=\frac{B}{2}$. Then $y=\frac{H}{2}$ and Area = $\frac{B H}{4}$.

33. $F$ = $\operatorname{cost}$ = (top cost) + (bottom) + (sides) = $\left(\pi r^{2}\right) A+\left(\pi r^{2}\right) B+(2 \pi r h) C$ But we know $\mathrm{V}=\pi \mathrm{r}^{2} \mathrm{~h}$ so $\mathrm{h}=\frac{\mathrm{V}}{\pi \mathrm{r}^{2}}$ so $\mathrm{F}=\pi \mathrm{r}^{2}(\mathrm{~A}+\mathrm{B})+\frac{2 \mathrm{CV}}{\mathrm{r}}$.
Then $F^{\prime}=2 \pi r(A+B)-\frac{2 C V}{r^{2}}=0$ when $r=\sqrt[3]{\frac{C V}{\pi(A+B)}}$. Now you can find $\mathrm{h}$ and $F$.