Chapter 4.4, Problem 31E

a.

To determine

To find: To find the volume of the cylinder.

Answer to Problem 31E

The volume of the cylinder is $\frac{1}{4\pi }(x^2)(18-x)$ .

Explanation of Solution

Given:

The Perimeter of the rectangle sheet is $36cm$ .

Calculation:

Let the dimensions of the sheet is $x$ and $y$ . Then,

  $\begin{array}{c}2(x+y)=36\\ x+y=18\\ y=18-x\end{array}$

Since the sheet is turned as cylinder let $r$ is the radius of the base.

  $\begin{array}{c}2\pi r=x\\ r=\frac{x}{2\pi }\\ h=y=18-x\end{array}$

Let the volume of the cylinder is V

  $\begin{array}{l}V=(\pi )(r^2)(h)\\ V=(\pi )({\frac{x}{2\pi }}^2)(18-x)\\ V=\frac{1}{4\pi }(x^2)(18-x)\end{array}$

b.

To determine

To find: To find the dimensions of the maximum volume.

Answer to Problem 31E

The dimensions of the maximum volume are $x=12\text{ cm}$ and $y=6\text{ cm}$ .

Explanation of Solution

Given:

The volume $V=(\pi )(x^2)(18-x)$.

Calculation:

In order to maximize the volume both parts require same parameters.

Let,

  $\begin{array}{l}f(x)=(x^2)(18-x)\\ f(x)=18x^2-x^3\end{array}$

Let any function , $f(x)$ the condition of minimum and maximum is obtained by finding

  $f\text{'}(x)=0$ , critical points and end points.

  $\begin{array}{l}f(x)=18x^2-x^3\\ f\text{'}(x)=36x-3x^2\end{array}$

Critical points occur at,

  $\begin{array}{c}f\text{'}(x)=0,0<x<18\\ 36x-3x^2=0\\ x=\frac{36}{3}\\ x=12\end{array}$

We get that critical point occurs at $x=12$ .

  $f\text{'}(x)>0$ for $0<r<12$ and $f\text{'}(x)<0$ for $12<x<18$ this corresponds to maximum volume.

For

  $\begin{array}{l}x=12\\ y=18-12\\ y=6\end{array}$

Thus we get, for maximum volume, the dimensions are $x=12\text{ cm}$ and $y=6\text{ cm}$ .