Chapter 4, Problem 71EP

$\text{(a)}$

To determine

To find: The value of $a$ and $b$ that will make the cylinder of maximum volume.

Answer to Problem 71EP

The answer: $V=\frac{500}{\pi }$

Explanation of Solution

Given information:

Two portions of length $a$ and four parts of length $b$ are taken from a piece of wire that is $60$ inches long. In order to create a frame for a model of a right circular cylinder, each of the two pieces of length $a$ is bent into the shape of a circle. The circles are then connected by the four sections of length $b$ , as illustrated in the accompanying image.

Calculation:

  $a$ is the circumference of the circle So,

  $\begin{array}{c}a=2\pi r\\ \to r=\frac{a}{2\pi }\end{array}$

The height of the cylinder,

  $\text{h=b}$

The volume in terms of $a$ and $b$ :

  $\begin{array}{c}V=\pi r^{2}h\\ =\pi {\left(\frac{a}{2\pi }\right)}^{2}b\\ =\frac{a^2}{4\pi }b\end{array}$

The constraint:

  $\begin{array}{c}2a+4b=60\\ \to a+2b=30\\ \to b=15-\frac{a}{2}\end{array}$

The volume in terms of one variable:

Let, express the volume to use $a$ : $V=\frac{a^2}{4\pi }b$

Then,

  $\begin{array}{c}V=\frac{a^2}{4\pi }\left(15-\frac{a}{2}\right)\\ \frac{dV}{da}=\frac{a}{2\pi }\left(15-\frac{a}{2}\right)-\frac{1}{2}\frac{a^2}{4\pi }\\ =\left(\frac{30a-a^2}{4\pi }\right)-\frac{a^2}{8\pi }\\ =\left(\frac{60a-3a^2}{8\pi }\right)\\ =\frac{3a(20-a)}{8\pi }\\ \frac{dV}{da}=0\\ \to a=0\\ \text{or }a=20\end{array}$

Hence,

  $\frac{dV}{da}>0$ on $(0,20)$

And,

  $\frac{dV}{da}<0$ on $(20,30)$

If

  $a=0$

Or

  $a=30$

then the volume will be zero

  $\text{V=0}$

So, the maximum occurs when

  $\text{a=20}$

For,

  $\text{a=20}$ ,

The volume will be maximum and equal to:

  $\begin{array}{c}V=\frac{{20}^2}{4\pi }\left(15-\frac{20}{2}\right)\\ =\frac{500}{\pi }\end{array}$

Therefore, the required value of $a$ and $b$ that will make the cylinder of maximum volume is:

  $V=\frac{500}{\pi }$

$\text{(b)}$

To determine

To explain: The answer to justify in part $\text{(a)}$ .

Answer to Problem 71EP

  $a=20$ is a critical number for $V(a)$ , now show that $V^{\prime \prime }(20)<0$ .

So, the volume $V$ must be maximum at $a=20$ .

Explanation of Solution

Given information:

Two portions of length $a$ and four parts of length $b$ are taken from a piece of wire that is $60$ inches long. In order to create a frame for a model of a right circular cylinder, each of the two pieces of length $a$ is bent into the shape of a circle. The circles are then connected by the four sections of length $b$ , as illustrated in the accompanying image.

Explain:

From part $\text{(a)}$ :

  $\begin{array}{c}V=\frac{a^2}{4\pi }\left(15-\frac{a}{2}\right)\\ =\frac{15}{4\pi }{a}^2-\frac{1}{8\pi }{a}^3\end{array}$

Also found that,

  $\begin{array}{c}a=20\text{ inches}\\ b=5\end{array}$

To use the second derivative test,

  $\begin{array}{c}V(a)=\frac{15}{4\pi }{a}^2-\frac{1}{8\pi }{a}^3\\ \to V^{\prime }(a)=\frac{15}{2\pi }a-\frac{3}{8\pi }{a}^2\\ \to V^{\prime \prime }(a)=\frac{15}{2\pi }-\frac{3}{4\pi }a\end{array}$

Now put

  $a=20$ into $V^{\prime \prime }(a)$ ,

So,

  $\begin{array}{c}{V}^{\prime \prime }(20)=\frac{15}{2\pi }-\frac{3}{4\pi }(20)\\ =\frac{15}{2\pi }-\frac{15}{\pi }\\ =\frac{15}{2\pi }-\frac{30}{2\pi }\\ =-\frac{15}{2\pi }<0\end{array}$

Therefore, $a=20$ is a critical number for $V(a)$ , now show that $V^{\prime \prime }(20)<0$ .

So, the volume $V$ must be maximum at $a=20$ .