## Exercises3.3Optimization

Optimization problems in calculus are always asking to maximize or minimize some quantity. Typical phrases that indicate an optimization problem include: "Find the largest ..." or "Find the minimum ..."

To solve an optimization problem you need to do the following:

1. Find a function of one variable that models the situation described in the given problem.

2. Use the Calculus tools to find the critical numbers and determine whether they correspond to a local maximum or minimum.

Solve the following optimization problems:

###### 1.

Find the absolute maximum and minimum values of $f(x)=3x^2-9x$ on the interval $[-1,2]\text{.}$

The maximum value is $f(-1)=12$ and the minimum value is $\ds f\left( \frac{3}{2}\right) =-\frac{27}{4}\text{.}$
Solution

Note that the function $f$ is continuous on the closed interval $[-1,2]\text{.}$ By the Intermediate Value Theorem the function $f$ attains its maximum and minimum values on $[-1,2]\text{.}$ To find those global extrema we evaluate and compare the values of $f$ at the endpoints and critical numbers that belong to $(-1,2)\text{.}$ From $f'(x)=6x-9=3(2x-3)$ we conclude that the critical number is $\ds x=\frac{3}{2}\text{.}$ From $f(-1)=12\text{,}$ $f(2)=-6\text{,}$ and $\ds f\left( \frac{3}{2}\right) =-\frac{27}{4}$ we conclude that the maximum value is $f(-1)=12$ and the minimum value is $\ds f\left( \frac{3}{2}\right) =-\frac{27}{4}\text{.}$

###### 2.

Find the absolute maximum and minimum values of $f(x)=x^3-12x-5$ on the interval $[-4,6]\text{.}$ Clearly explain your reasoning.

The global minimum value is $f(-4)=f(2)=-21\text{,}$ and the global maximum value is $f(6)=139\text{.}$ Note that $f(2)=-21$ is also a local minimum and that $f(-2)$ is a local maximum. (Reminder: By our definition, for $x=c$ to be a local extremum of a function $f$ it is necessary that $c$ is an interior point of the domain of $f\text{.}$ This means that there is an open interval $I$ contained in the domain of $f$ such that $c\in I\text{.}$)

###### 3.

If $a$ and $b$ are positive numbers, find the maximum value of $\ds f(x)=x^a(1-x)^b\text{.}$

The maximum value of $f$ is $\ds f\left( \frac{a}{a+b}\right) =\left( \frac{a}{a+b}\right)^a\left( \frac{b}{a+b}\right)^b\text{.}$
Solution

From $f^\prime(x)=ax^{a-1}(1-x)^b-bx^a(1-x)^{b-1}=x^{a-1}(1-x)^{b-1}(a-(a+b)x)$ and the fact that $a$ and $b$ are positive conclude that $\ds x=\frac{a}{a+b}\in (0,1)$ is a critical number of the function $f\text{.}$ Since $f(0)=f(1)=0$ and $f(x)>0$ for all $x\in (0,1)$ it follows that the maximum value of $f$ is $\ds f\left( \frac{a}{a+b}\right) =\left( \frac{a}{a+b}\right)^a\left( \frac{b}{a+b}\right)^b\text{.}$

###### 4.

Find all critical points of the function $f(x)=|3x-5|$ on the interval $[-3,2]\text{.}$ Also find all maxima and minima of this function on $[-3,2]\text{,}$ both local and global.

The global and local minimum is $\ds f\left( \frac{5}{3}\right) =0$ and that the global maximum is $f(-3)=14\text{.}$
Solution

From $\ds f(x)=\left\{ \begin{array}{rr} 3x-5\amp \mbox{if } x\geq \frac{5}{3}\\ -3x+5\amp \mbox{if } x\lt \frac{5}{3} \end{array} \right.$ we conclude that $\ds f'(x)=\left\{ \begin{array}{rr} 3\amp \mbox{if } x> \frac{5}{3}\\ -3\amp \mbox{if } x\lt \frac{5}{3} \end{array} \right.\text{.}$ Thus, for $\ds x\not= \frac{5}{3}\text{,}$ $f'(x)\not= 0$ and the derivative of $f$ is not defined at $\ds x= \frac{5}{3}\text{.}$ We conclude that the only critical number of the function $f$ on the interval $[-3,2]$ is $\ds x= \frac{5}{3}\text{.}$ Clearly, $\ds f\left( \frac{5}{3}\right) =0\text{.}$ From $f(-3)=14$ and $f(1)=2$ it follows that the global and local minimum is $\ds f\left( \frac{5}{3}\right) =0$ and that the global maximum is $f(-3)=14\text{.}$

###### 5.

The sum of two positive numbers is 12. What is the smallest possible value of the sum of their squares? Show your reasoning.

$f(6)=72\text{.}$
Solution

The question is to find the minimum value of the function $f(x)=x^2+(12-x)^2\text{,}$ $x\in (0,12)\text{.}$ From $f'(x)=4(x-6)$ it follows that $x=6$ is the only critical number. From $f"(6)=4>0\text{,}$ by the second derivative test, it follows that $f(6)=72$ is the minimum value of the function $f\text{.}$

###### 6.

If $a$ and $b$ are positive numbers, find the $x$ coordinate which gives the absolute maximum value of $f(x)=x^a(1-x)^b$ on the interval $[0,1]\text{.}$

$\ds x=\frac{a}{a+b}\text{.}$
Solution

Note that $f(0)=f(1)=0$ and that $f(x)>0$ for $x\in (0,1)\text{.}$ Thus by the Intermediate Value Theorem there is $c\in (0,1)$ such that $f(c)$ is the maximum value of $f\text{.}$ Since $f$ is differentiable on $(0,1)\text{,}$ $c$ must be a critical point. Note that $f'(x)=x^{a-1}(1-x)^{b-1}(a-(a+b)x)\text{.}$ Since $a$ and $b$ are both positive we have that $\ds x=\frac{a}{a+b}\in (0,1)\text{.}$ Thus $\ds x=\frac{a}{a+b}$ is the only critical point of the function $f$ in the interval $(0,1)$ and $\ds f\left( \frac{a}{a+b}\right) =\frac{a^ab^b}{(a+b)^{a+b}}$ is the maximum value.

###### 7.

Find the point on the curve $x+y^2=0$ that is closest to the point $(0,-3)\text{.}$

$(-1,-1)\text{.}$
Solution

The distance between a point $(x,y)$ on the curve and the point $(0,-3)$ is $\ds d=\sqrt{(x-0)^2+(y-(-3))^2}=\sqrt{y^4+(y+3)^2}\text{.}$ The question is to minimize the function $f(y)= y^4+(y+3)^2\text{,}$ $y\in \mathbb{R}\text{.}$ From $f'(y)=2(2y^3+y+3)=2(y+1)(2y^2-2y+3)$we conclude that $y=-1$ is the only critical number of the function $f\text{.}$ From $f"(-1)=10>0\text{,}$ by the second derivative test we conclude that $f(-1)=5$ is the (local and global) minimum value of $f\text{.}$ Thus the closest point is $(-1,-1)\text{.}$

###### 8.

A straight piece of wire 40 cm long is cut into two pieces. One piece is bent into a circle and the other is bent into a square. How should wire be cut so that the total area of both the circle and square is minimized?

The two pieces should be of the length $\ds \frac{40\pi }{\pi +4}$ and $\ds \frac{160 }{\pi +4}\text{.}$
Solution

Let $x$ be the radius of the circle. The question is to minimize the function $\ds f(x)=\pi x^2+\left( \frac{40-2\pi x}{4}\right) ^2\text{,}$ $\ds x\in \left( 0,\frac{20}{\pi }\right)\text{.}$ (We are given that there are TWO pieces.) The only critical number of the function $f$ is $\ds x=\frac{20}{\pi +4}\text{.}$ To minimize the total area the two pieces should be of the length $\ds \frac{40\pi }{\pi +4}$ and $\ds \frac{160 }{\pi +4}\text{.}$

###### 9.

A straight piece of wire 28 cm long is cut into two pieces. One piece is bent into a square (i.e. dimensions $x$ times $x\text{.}$) The other piece is bent into a rectangle with aspect ratio three (i.e. dimensions $y$ times $3y\text{.}$) What are the dimensions, in centimetres, of the square and the rectangle such that the sum of their areas is minimized?

Hint
The question is to minimize the function $\ds f(x)=x^2+\frac{3(7-x)^2}{4}\text{,}$ $x\in (0,7)\text{.}$ (We are given that there are TWO pieces.)
$\ds x=3$ and $y=2\text{.}$
###### 10.

With a straight piece of wire 4m long, you are to create an equilateral triangle and a square, or either one only. Suppose a piece of wire of length $x$ metres is bent into a triangle and the remainder is bent into a square. Find the value of $x$ which maximizes the total area of both the triangle and the square.

The maximum total area is obtained when only the square is constructed.
Solution

The question is to maximize the function $\ds f(x)=\frac{x^2\sqrt{3}}{36}+\frac{(4-x)^2}{16}\text{,}$ $x\in [0,4]\text{.}$ Note that $f"(x)>0$ for $x\in (0,4)$ and conclude that the maximum value must occur at $x=0$ and/or $x=4\text{.}$ Since $f(4)\lt f(0)\text{,}$ the maximum total area is obtained when only the square is constructed.

###### 11.

Show that a perimeter of $2\sqrt{5}R$ is the largest possible perimeter of a rectangle inscribed in a semicircle of radius $R\text{,}$ with one side of the rectangle lying along the diameter of the semicircle. Hint

Maximize the function $P(x)=2x+\sqrt{R^2-x^2}\text{,}$ where $2x$ represents the width of a rectangle inscribed in the semicircle.

Solution
Let $x$ and $y$ be as follows: Then the perimeter is given by the function $P(x,y) = 4x + 2y \text{,}$ for $0 \leq x \leq R$ and $0 \leq y \leq R\text{.}$ From the constraint, we see that $R^2 = x^2 + y^2 \text{,}$ and so we can write $y = \sqrt{R^2 - x^2} \text{.}$ Hence, we get
\begin{equation*} P(x) = 4x + 2\sqrt{R^2-x^2}, \ \ 0 \leq x \leq R. \end{equation*}
We find one interior critical point at $x= \frac{2R}{\sqrt{5}}.$ Since
\begin{equation*} P(0) = 2R, \ \ P(R) = 4R, \ \ P\left(\frac{2R}{\sqrt{5}}\right) = \frac{10R}{\sqrt{5}} = 2\sqrt{5}R. \end{equation*}
Therefore, we see that the largest possible perimeter is $2\sqrt{5}R\text{,}$ as desired.
###### 12.

A rectangle with sides parallel to the coordinate axes is to be inscribed in the region enclosed by the graphs of $y=x^2$ and $y=4$ so that its perimeter has maximum length.

1. Sketch the region under consideration.

2. Supposing that the $x$-coordinate of the bottom right vertex of the rectangle is $a\text{,}$ find a formula which expresses $P\text{,}$ the length of the perimeter, in terms of $a\text{.}$

3. Find the value of $a$ which gives the maximum value of $P\text{,}$ and explain why you know that this value of $a$ gives a maximum.

4. What is the maximum value of $P\text{,}$ the length of the perimeter of the rectangle?

1. 2. $P=4a+2(4-a^2)=2(4+2a-a^2)\text{,}$ $a\in (0,2)\text{.}$

3. $\ds P(1)\text{.}$

4. $\ds P(1)=10\text{.}$

Solution
1. 2. We let $(a, a^2)$ be bottom right vertex of the rectangle for $a\in (0,2)\text{.}$ Then the remaining three vertices are given by $(-a, a^2)\text{,}$ $(a,4)$ and $(-a,4) \text{.}$ Therefore, the perimeter of the rectangle is given by

\begin{equation*} P(a) = 2(2a) + 2(4-a^2) = 2(4+2a-a^2), \end{equation*}

for $a\in (0,2)\text{.}$

3. From $\ds \frac{dP}{da}=4(1-a)$ it follows that $\ds a=1$ is the only critical number. The fact that $\ds P^{\prime\prime}(a) =-4\lt 0$ for all $a\in (0,2)$ implies, by the second derivative test, that $\ds P(1)$ is the maximum value.

4. We have that $\ds P(1)=2(4+2-1)=10\text{.}$

###### 13.

Find the dimensions of the rectangle of largest area that has its base on the $x$-axis and its other two vertices above the $x$-axis and lying on the parabola $y=12-x^2\text{.}$

The length of the rectangle with the largest area is 4 and its height is 8.
Solution

Let $(x,0)$ be the bottom right vertex of the rectangle. The question is to maximize $f(x)=2x(12-x^2)\text{,}$ $x\in (0,2\sqrt{3})\text{.}$ The only critical number is $x=2\text{.}$ The length of the rectangle with the largest area is 4 and its height is 8.

###### 14.

A farmer has 400 feet of fencing with which to build a rectangular pen. He will use part of an existing straight wall 100 feet long as part of one side of the perimeter of the pen. What is the maximum area that can be enclosed?

$f(150)=15000$ ft$^2\text{.}$
Solution

Let $x$ be the length of one side of the fence that is perpendicular to the wall. Note that the length of the side of the fence that is parallel to the wall equals $400-2x$ and that this number cannot be larger than 100.The question is to maximize the function $f(x)=x(400-2x)\text{,}$ $x\in [150,400)$ . The only solution of the equation $f'(x)=4(100-x)=0$ is $x=100$ but this value is not in the domain of the function $f\text{.}$ Clearly $f'(x)\lt 0$ for $x\in [150,400)$ which implies that $f$ is decreasing on its domain. Therefore the maximum area that can be enclosed is $f(150)=15000$ ft$^2\text{.}$

###### 15.

A $10\sqrt{2}$ ft wall stands 5 ft from a building. Find the length $L$ of the shortest ladder, supported by the wall, that reaches from the ground to the building. Hint
To minimize $L^2=(x+5)^2+y^2\text{,}$ use the fact that $\ds \frac{x}{10\sqrt{2}}=\frac{x+5}{y}$ and the first derivative.
$L=15\sqrt{3}\text{.}$
###### 16.

An attacking player (Gretzky) is skating with the puck along the boards as shown. As Gretzky proceeds, the apparent angle $\alpha$ between the opponent's goal posts first increases, then decreases. 1. Using dimensions given in the Figure above, find an expression for $\alpha$ in terms of the distance $x$ from Gretzky to the goal line.

2. Assume that Gretzky's chance of scoring is greatest when $\alpha$ is maximum (this may be the case if the opposing team has “pulled” their goalie). At which distance $x$ from the goal line should Gretzky shoot the puck? It is clear that $\alpha$ is very small when $x=35$ and $x=0\text{,}$ so there is no need to check the endpoints of the domain $[0,35]\text{.}$

1. $\ds \alpha =\tan^{-1}\left(\frac{7}{x}\right)- \tan^{-1}\left(\frac{5}{x}\right)\text{.}$

2. $x=\sqrt{35}\text{.}$

###### 17.

In an elliptical sport field we want to design a rectangular soccer field with the maximum possible area. The sport field is given by the graph of $\ds \frac{x^2}{a^2}+\frac{y^2}{b^2}=1\text{.}$ Find the length $2x$ and width $2y$ of the pitch (in terms of $a$ and $b$) that maximize the area of the pitch. [Hint: express the area of the pitch as a function of $x$ only.]

To maximize the area of the soccer field its length should be $a\sqrt{2}$ and its width should be $b\sqrt{2}\text{.}$
Solution

Let $\ds (x,y)=\left( x,\frac{b}{a}\sqrt{a^2-x^2}\right)$ be the upper right vertex of the rectangle. The question is to maximize the function $\ds f(x)=\frac{4b}{a}x\sqrt{a^2-x^2}\text{,}$ $x\in (0,a)\text{.}$ From $\ds f'(x)=\frac{4b}{a}\frac{a^2-2x^2}{\sqrt{a^2-x^2}}$ we conclude that the only critical number is $\ds x=\frac{a}{\sqrt{2}}\text{.}$ By the first derivative test, there is a local maximum at this critical number. Since $\ds \lim _{x\to 0^+}f(x)= \lim _{x\to a^-}f(x)=0\text{,}$ it follows that $\ds f\left( \frac{a}{\sqrt{2}}\right) =2ab$ is the maximum value of the function $f\text{.}$ Thus to maximize the area of the soccer field its length should be $a\sqrt{2}$ and its width should be $b\sqrt{2}\text{.}$

###### 18.

The top and bottom margins of a poster are each 6 cm, and the side margins are each 4 cm. If the area of the printed material on the poster (that is, the area between the margins) is fixed at 384 cm$^2\text{,}$ find the dimensions of the poster with the smallest total area.

The dimensions of the poster with the smallest area are $x=24$ cm and $y=36$ cm.
Solution

Let $a$ be the length of the printed material on the poster. Then the width of this area equals $\ds b=\frac{384}{a}\text{.}$ It follows that the length of the poster is $x=a+8$ and the width of the poster is $\ds y=b+12=\frac{384}{a}+12\text{.}$ The question is to minimize the function $\ds f(a)=xy=(a+8)\left( \frac{384}{a}+12\right) =12\left( 40+a+\frac{256}{a}\right)\text{.}$ It follows that the function has a local minimum at $\ds a=16\text{.}$ The dimensions of the poster with the smallest area are $x=24$ cm and $y=36$ cm.

###### 19.

A poster is to have an area of 180 in$^2$ with 1-inch margins at the bottom and the sides and a 2 inch margin at the top. What dimensions will give the largest printed area? Hint
Maximize the function $\ds A(x)=(x-2)\left( \frac{180}{x}-3\right)\text{,}$ where $x$ represents the length of the poster.
$2\sqrt{30}\cdot 3\sqrt{30}\text{.}$
###### 20.

Each rectangular page of a book must contain 30 cm$^2$ of printed text, and each page must have 2 cm margins at top and bottom, and a 1 cm margin at each side. What is the minimum possible area of such a page?

$(\sqrt{15}+2)\times (2\sqrt{15}+4)\text{.}$
###### 21.

Maya is 2 km offshore in a boat and wishes to reach a coastal village which is 6 km down a straight shoreline from the point on the shore nearest to the boat. She can row at 2 km/hr and run at 5 km/hr. Where should she land her boat to reach the village in the least amount of time?

Maya should land her boat $\ds \frac{4}{3}$ km from the point initially nearest to the boat.
Solution

Let $P$ be the point on the shore where Maya lands her boat and let $x$ be the distance from $P$ to the point on the shore that is closest to her initial position. Thus to reach the village she needs to row the distance $z=\sqrt{4+x^2}$ and run the distance $y=6-x\text{.}$ Time needed to row the distance $z$ is given by $\ds T_1=\frac{z}{2}$ and time she needs to run is $\ds T_2=\frac{y}{5}\text{.}$ Therefore the question is to minimize the function $\ds T=T(x)=T_1+T_2=\frac{\sqrt{4+x^2}}{2}+\frac{6-x}{5}\text{,}$ $x\in [0,6]\text{.}$ From $\ds f'(x)=\frac{x}{2\sqrt{4+x^2}}-\frac{1}{5}$ it follows that the only critical number is $\ds x=\frac{4}{3}\text{.}$ From $\ds T(0)=\frac{11}{5}=2.2\text{,}$ $T(6)=\sqrt{10}\text{,}$ and $\ds T\left( \frac{4}{3}\right) \approx 2.135183758$ it follows that the minimum value is $\ds T\left( \frac{4}{3}\right)\text{.}$ Maya should land her boat $\ds \frac{4}{3}$ km from the point initially nearest to the boat.

###### 22.

A hiker is in the woods at a distance of 3 km from a straight road. She would like to reach a supply store which is located 10 km down the road from the nearest point on the road to her. Assuming that she hikes through the woods at a rate of 2 km/hr and walks along the road at a rate of 4 km/h, what point on the road should she hike through the woods to so as to minimize her travel time? Hint
Minimize the function $T(x)=\frac{1}{2}\cdot \sqrt{9+(10-x)^2}+\frac{x}{4}\text{,}$ where $x$ is the distance between a point on the road and the store.
$x=10-\sqrt{3}\text{.}$
###### 23.

A rectangular box has a square base with edge length $x$ of at least 1 unit. The total surface area of its six sides is 150 square units.

1. Express the volume $V$ of this box as a function of $x\text{.}$

2. Find the domain of $V(x)\text{.}$

3. Find the dimensions of the box in part (a) with the greatest possible volume. What is this greatest possible volume?

1. Express the volume $V$ of this box as a function of $x\text{.}$

2. $[1, 5\sqrt{3})\text{.}$

3. $V(5)=125$ cube units.

Solution
1. Let $y$ be the height of the box. Then the surface area is given by $S=2x^2+4xy\text{.}$ From $S=150$ it follows that $\ds y=\frac{1}{2}\left( \frac{75}{x}-x\right)\text{.}$ Therefore the volume of the box is given by $\ds V=V(x)=\frac{x}{2}\left( 75-x^2\right)\text{.}$

2. From the fact that $\ds y=\frac{1}{2}\left( \frac{75}{x}-x\right) >0$ it follows that the domain of the function $V=V(x)$ is the interval $[1, 5\sqrt{3})\text{.}$

3. Note that $\ds \frac{dV}{dx}=\frac{3}{2}(25-x^2)$ and that $\ds \frac{d^2V}{dx^2}=-3x\lt 0$ for all $x\in (1, 5\sqrt{3})\text{.}$ Thus the maximum value is $V(5)=125$ cube units.

###### 24.

An open-top box is to have a square base and a volume of 10 m$^3\text{.}$ The cost per square metre of material is $\ 5$ for the bottom and $\ 2$ for the four sides. Let $x$ and $y$ be lengths of the box's width and height respectively. Let $C$ be the total cost of material required to make the box.

1. Express $C$ as a function of $x$ and find its domain.

2. Find the dimensions of the box so that the cost of materials is minimized. What is this minimum cost?

1. Note that $y=\frac{10}{x^2}\text{.}$ The cost function is given by $\ds C(x)=5x^2+2\cdot 4\cdot x\cdot \frac{10}{x^2}=5x^2+\frac{80}{x}\text{,}$ $x>0\text{.}$

2. $\ds 2\times 2\times \frac{5}{2}\text{.}$ The minimum cost is $C(2)=\ 60\text{.}$

###### 25.

An open-top box is to have a square base and a volume of 13500 cm$^3\text{.}$ Find the dimensions of the box that minimize the amount of material used.

$x=30\text{.}$
Solution

Let $x$ be the length and the width of the box. Then its height is given by $\ds y=\frac{13500}{x^2}\text{.}$ It follows that the surface area is $\ds S=x^2+\frac{54000}{x}$ cm$^2\text{,}$ $x>0\text{.}$ The question is to minimize $S\text{.}$ From $\ds \frac{dS}{dx}=2x-\frac{54000}{x^2}$ and $\ds \frac{d^2S}{dx^2}=2+\frac{3\cdot 54000}{x^3}>0$ for all $x>0$ it follows that the function $S$ has a local and global minimum at $x=30\text{.}$

###### 26.

A water trough is to be made from a long strip of tin 6 ft wide by bending up at the angle $\theta$ a 2 ft strip at each side. What angle $\theta$ would maximize the cross sectional area, and thus the volume, of the trough? $\ds \theta =\frac{\pi }{3}\text{.}$
Solution

We need to maximize the area of the trapezoid with parallel sides of lengths $a=2$ and $c=2+2\cdot 2\cos \theta=2+4\cos\theta$ and the height $h=2\sin \theta\text{.}$ Thus we maximize the function $\ds A=A(\theta )=\frac{2+(2+4\cos \theta )}{2}\cdot 2\sin \theta=4(\sin \theta +\sin \theta \cos\theta)\text{,}$ $\theta \in (0,\pi )\text{.}$ From $\ds \frac{dA}{d\theta }=4(\cos \theta +\cos ^2\theta-\sin ^2\theta)=4(2\cos ^2\theta +\cos \theta -1)=4(2\cos \theta -1)(\cos \theta +1)$ we obtain the critical number $\ds \theta =\frac{\pi }{3}\text{.}$ The First Derivative Test confirms that $\ds \theta =\frac{\pi }{3}$ maximizes the cross sectional area of the trough.

###### 27.

Find the dimensions of the right circular cylinder with greatest volume that can be inscribed in a right circular cone of radius 8 cm and height 12 cm. $\ds r=\frac{16}{3}$ cm, $h=4$ cm.
###### 28.

Find the dimension of the right circular cylinder of maximum volume that can be inscribed in a right circular cone of radius $R$ and height $H\text{.}$

The dimensions are $\ds r=\frac{2R}{3}$ and $\ds h=\frac{H}{3}\text{.}$
Solution

Let $r$ be the radius of the base of a cylinder inscribed in the cone and let $h$ be its height. From $\ds \frac{H}{R}=\frac{h}{R-r}\text{,}$ we conclude that $\ds h=\frac{H(R-r)}{R}\text{.}$ Thus the volume of the cylinder is $\ds V=V(r)=\frac{\pi H}{R}r^2(R-r)\text{,}$ $r\in (0,R)\text{.}$ From $\ds \frac{dV}{dr}=\frac{\pi H}{R}r(2R-3r)$ and $\ds \frac{d^2V}{dr^2}=\frac{2\pi H}{R}(R-3r)$ it follows that the maximum value of the volume of the cylinder is $\ds V\left( \frac{2R}{3}\right)=\frac{4\pi HR^2}{27}\text{.}$ The dimensions are $\ds r=\frac{2R}{3}$ and $\ds h=\frac{H}{3}\text{.}$

###### 29.

A hollow plastic cylinder with a circular base and open top is to be made and 10 m$^2$ of plastic is available. Find the dimensions of the cylinder that give the maximum volume and find the value of the maximum volume.

$\ds r=\sqrt{\frac{10}{3\pi }}$ m, $\ds h=\left( \frac{5}{\pi}\sqrt{\frac{3\pi}{10 }}-\frac{1}{2}\sqrt{\frac{10}{3\pi }}\right)$ m, $\ds V=\frac{10}{3}\sqrt{\frac{10}{3\pi }}$ m$^3\text{.}$

###### 30.

An open-topped cylindrical pot is to have volume 250 cm$^3\text{.}$The material for the bottom of the pot costs 4 cents per cm$^2\text{;}$ that for its curved side costs 2 cents per cm$^2\text{.}$ What dimensions will minimize the total cost of this pot?

$\ds r=\frac{5}{\sqrt{\pi }}\text{.}$
Solution

Let $r$ be the radius of the base of the pot. Then the height of the pot is $\ds h=\frac{250}{\pi r^2}\text{.}$ The cost function is $\ds C(r)=4\pi r^2+\frac{1000}{r}\text{,}$ $r>0\text{.}$ The cost function has its minimum at $\ds r=\frac{5}{\sqrt{\pi }}\text{.}$

###### 31.

A cylindrical can without a top is made to contain 1,000 cm$^2$ of liquid. Find the dimensions that will minimize the cost of the material to make the can.

$\ds R=\frac{10}{\sqrt{\pi}}$ cm, $\ds h=\frac{10}{\sqrt{\pi}}$ cm.

###### 32.

Cylindrical soup cans are to be manufactured to contain a given volume $V\text{.}$ No waste is involved in cutting the material for the vertical side of each can, but each top and bottom which are circles of radius $r\text{,}$ are cut from a square that measures $2r$ units on each side. Thus the material used to manufacture each soup can has an area of $A=2\pi rh+8r^2$ square units.

1. How much material is wasted in making each soup can?

2. Find the ratio of the height to diameter for the most economical can (i.e. requiring the least amount of material for manufacture.)

3. Use either the first or second derivative test to verify that you have minimized the amount of material used for making each can.

1. The amount of material wasted is $2(4-\pi )r^2\text{.}$

2. $\ds \frac{h}{r}=\frac{4}{\pi }\text{.}$

3. Second Derivative Test.

Solution
1. The surface area of the can is $S=2\pi rh+2\pi r^2\text{.}$ The amount of material wasted is $A-S=2(4-\pi )r^2\text{.}$

2. From $V=\pi r^2h$ it follows that the amount of material needed to make a can of the given volume $V$ is $\ds A=A(r)=\frac{2V}{r}+8r^2\text{.}$ This function has its minimum at $\ds r=\frac{\sqrt{V}}{2}\text{.}$ The ratio of the height to diameter for the most economical can is $\ds \frac{h}{r}=\frac{4}{\pi }\text{.}$

3. $\ds A"(r)=\frac{4V}{r^3}+8>0$ for $r>0\text{.}$ Hence, by the Second Derivative Test, the amount of material is minimized.

###### 33.

A storage container is to be made in the form of a right circular cylinder and have a volume of $28\pi$ m$^3\text{.}$ Material for the top of the container costs $\ 5$ per square metre and material for the side and base costs $\ 2$ per square metre. What dimensions will minimize the total cost of the container?

$\ds r=\sqrt{20}\text{,}$ $\ds h=\frac{14}{\sqrt{50}}\text{.}$ Minimize the cost function $\ds C=C(r)=7r^2\pi +\frac{280\pi }{r}\text{.}$
###### 34.

Show that the volume of the largest cone that can be inscribed inside a sphere of radius $R$ is $\ds \frac{32\pi R^3}{81}\text{.}$

Solution

From the Figure below, we conclude that $\ds r^2=R^2-(h-R)^2=h(4R-h)\text{.}$ Then the volume of the cone as a function of $h$ is given by $\ds V=\frac{\pi }{3}h^2(4R-h)\text{.}$ Maximize. ###### 35.

The sound level measured in watts per square metre, varies in direct proportion to the power of the source and inversely as the square of the distance from the source, so that is given by $\ds y=kPx^{-2}\text{,}$ where $y$ is the sound level, $P$ is the source power, $x$ is the distance form the source, and $k$ is a positive constant. Two beach parties, 100 metres apart, are playing loud music on their portable stereos. The second party's stereo has 64 times as much power as the first. The music approximates the white noise, so the power from the two sources arriving at a point between them adds, without any concern about whether the sources are in or out of phase. To what point on the line segment between the two parties should I go, if I wish to enjoy as much quiet as possible? Demonstrate that you have found an absolute minimum, not just a relative minimum.

Solution

Let $P$ be the source power of the first party's stereo and let $x$ be the distance between the person and the first party. Since the power of the second party's stereo is $64P\text{,}$ the sound level is $L(x)=kPx^{-2}+64kP(100-x)^{-2}\text{,}$ $x\in (0,100)\text{.}$ From $ds \frac{dL}{dx}=2kP\left( \frac{64}{(100-x)^3}-\frac{1}{x^3}\right)$ it follows that $x=20$ is the only critical number for the function $L\text{.}$ Since for $x\in (0,100)$

\begin{equation*} L'(x)>0\Leftrightarrow \frac{64}{(100-x)^3}-\frac{1}{x^3}>0\Leftrightarrow 64x^3>(100-x)^3 \Leftrightarrow 4x>100-x \Leftrightarrow x>20 \end{equation*}

the function $L$ is strictly increasing on the interval $(20,100)$ and strictly decreasing on the interval $(0,20)\text{.}$ Therefore, $L(20)$ is the absolute minimum.