Practice Problems

1. (a) You have 200 feet of fencing to enclose a rectangular vegetable garden. What should the dimensions of your garden be in order to enclose the largest area?
(b) Show that if you have $P$ feet of fencing available, the garden of greatest area is a square.
(c) What are the dimensions of the largest rectangular garden you can enclose with $\mathrm{P}$ feet of fencing if one edge of the garden borders a straight river and does not need to be fenced?
(d) Just thinking - calculus will not help with this one: What do you think is the shape of the largest garden which can be enclosed with $P$ feet of fencing if we do not require the garden to be rectangular? What do you think is the shape of the largest garden which can be enclosed with $P$ feet of fencing if one edge of the garden borders a river and does not need to be fenced?

3. You have $120$ feet of fencing to construct a pen with $4$ equal sized stalls.
(a) If the pen is rectangular and shaped like the Fig. 8, what are the dimensions of the pen of largest area and what is that area?

Fig. 8

(b) The square pen in Fig. 9 uses $120$ feet of fencing and encloses a larger area ($400$ square feet) than the best design in part (a). Design a pen which uses only $120$ feet of fencing and has $4$ equal sized stalls but which encloses even more than $400$ square feet. (Suggestion: don't use rectangles and squares.)

Fig. 9

5. You have a $10$ inch by $10$ inch piece of cardboard which you plan to cut and fold as shown in Fig. 11 to form a box with a top. Find the dimensions of the box which has the largest volume.

Fig. 11

7. (a) Determine the dimensions of the least expensive cylindrical can which will hold $100$ cubic inches if the materials cost $2¢, 5¢$ and $3¢$ respectively for the top, bottom and sides.
(b) How do the dimensions of the least expensive can change if the bottom material costs more than $5¢$ per square inch?

9. You are a lifeguard standing at the edge of the water when you notice a swimmer in trouble (Fig. 13). Assuming you can run about $8$ meters per second and swim about $2 \mathrm{~m} / \mathrm{s}$, how far along the shore should you run before diving into the water in order to reach the swimmer as quickly as possible?

Fig. 13

11. You have been asked to determine where a water works should be built along a river between Chesterville and Denton (see Fig. 15 ) to minimize the total cost of the pipe to the towns.
(a) Assume that the same size (and cost) pipe is used to each town. (This part can be done quickly without using calculus.)
(b) Assume that the pipe to Chesterville costs $\$ 3000$ per mile and to Denton it costs $\$ 7000$ per mile.

Fig. 15

13. U.S. postal regulations state that the sum of the length and girth (distance around) of a package must be no more than 108 inches. (Fig. 17)
(a) Find the dimensions of the acceptable box with a square end which has the largest volume.
(b) Find the dimensions of the acceptable box which has the largest volume if its end is a rectangle twice as long as it is wide.
(c) Find the dimensions of the acceptable box with a circular end which has the largest volume.

Fig. 17

15. Two sides of a triangle are $7$ and $10$ inches respectively. What is the length of the third side so the area of the triangle will be greatest? (This problem can be done without using calculus. How? If you do use calculus, consider the angle $\theta$ between the two sides.)

17. Find the dimensions of the rectangle with the largest area if the base must be on the $\mathrm{x}$-axis and its other two corners are on the graph of
(a) $\mathrm{y}=16-\mathrm{x}^{2}$ on $[-4,4]$
(b) $x^{2}+y^{2}=1$ on $[-1,1]$
(c) $|x|+|y|=1$ on $[-1,1]$
(d) $\mathrm{y}=\cos (\mathrm{x})$ on $[-\pi / 2, \pi / 2]$

19. You have a long piece of $12$ inch wide metal which you are going to fold along the center line to form a V-shaped gutter (Fig. 20). What angle $\theta$ will give the gutter which holds the most water (the largest cross-sectional area)?

Fig. 20

21. You have a $6$ inch diameter circle of paper which you want to form into a drinking cup by removing a pie-shaped wedge and forming the remaining paper into a cone (Fig. 22). Find the height and top radius of the cone so the volume of the cup is as large as possible.

Fig. 22

23. You own a small airplane which holds a maximum of $20$ passengers. It costs you $\$ 100$ per flight from St. Thomas to St. Croix for gas and wages plus an additional $\$ 6$ per passenger for the extra gas required by the extra weight. The charge per passenger is $\$ 30$ each if $10$ people charter your plane ( $10$ is the minimum number you will fly), and this charge is reduced by $\$ 1$ per passenger for each passenger over 10 who goes (that is, if $11$ go they each pay $\$ 29$, if $12$ go they each pay $\$ 28$, etc.). What number of passengers on a flight will maximize your profits?

25. Profit is revenue minus expenses. Assume that revenue and expenses are differentiable functions and show that when profit is maximized, then marginal revenue $(\mathrm{d} \mathrm{R} / \mathrm{d} \mathrm{x})$ equals marginal expense $(\mathrm{dE} / \mathrm{dx})$.

27. After the table was wiped and the potato chips dried off, the question remained: "Just how far could a can of cola be tipped before it fell over?"

(i) For a full can or an empty can the answer was easy: the center of gravity $(cg)$ of the can is at the middle of the can, half as high as the height of the can, and we can tilt the can until the $\mathrm{cg}$ is directly above the bottom rim. (Fig. 25a) Find $\theta$.

(ii) For a partly filled can more thinking was needed. Some ideas you will see in chapter 5 let us calculate that the $\mathrm{cg}$ of a can containing $\mathrm{x}$ $\mathrm{cm}$ of cola is $\mathrm{C}(\mathrm{x})=\frac{360+9.6 \mathrm{x}^{2}}{60+19.2 \mathrm{x}} \mathrm{cm}$ above the bottom of the can. Find the height $\mathrm{x}$ of cola in the can which will make the $\mathrm{cg}$ as low as possible.

(iii) Assuming that the cola is frozen solid (so the top of the cola stays parallel to the bottom of the can), how far can we tilt a can containing $\mathrm{x}$ $\mathrm{cm}$ of cola. (Fig. 25b)

(iv) If the can contained $\mathbf{x}$ $\mathrm{cm}$ of liquid cola, could we tilt it more or less far than the frozen cola before it would fall over?

Fig. 25

29. (a) Find the dimensions of the rectangle with the greatest area that can be built so the base of the rectangle is on the $x$-axis between $0$ and $1(0 \leq x \leq 1)$ and one corner of the rectangle is on the curve $\mathrm{y}=\mathrm{x}^{2}$ (Fig. 27a). What is the area of this rectangle?
(b) Generalize the problem in part (a) for the parabola $\mathrm{y}=\mathrm{Cx}^{2}$ with $\mathrm{C} > 0$ and $0 \leq \mathrm{x} \leq 1$ (Fig. 27b).
(c) Generalize for the parabola $\mathrm{y}=\mathrm{Cx}^{2}$ with $\mathrm{C} > 0$ and $0 \leq \mathrm{x} \leq \mathrm{B}$ (Fig. 27c).

Fig. 27

31. (a) The base of a right triangle is $50$ and the height is $20$ (Fig. 28a). Find the dimensions and area of the rectangle with the greatest area that can be enclosed in the triangle if the base of the rectangle must lie on the base of the triangle.
(b) The base of a right triangle is $\mathrm{B}$ and the height is $\mathrm{H}$ (Fig. 28b). Find the dimensions and area of the rectangle with the greatest area that can be enclosed in the triangle if the base of the rectangle must lie on the base of the triangle.
(c) State your general conclusion from part (b) in words.

Fig. 28

33. Determine the dimensions of the least expensive cylindrical can which will hold $\mathrm{V}$ cubic inches if the top material costs $\$A$ per square inch, the bottom material costs $\$ \mathrm{~B} / \mathrm{in}^{2}$, and the side material costs $\mathrm{\$ C} / \mathrm{in}^{2}$.

Source: Dale Hoffman, https://s3.amazonaws.com/saylordotorg-resources/wwwresources/site/wp-content/uploads/2012/12/MA005-4.5-Applied-Maximum-and-Minimum.pdf
This work is licensed under a Creative Commons Attribution 3.0 License.

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 $F$.