Chapter 3, Problem 22RE

a)

To determine

To express: the volume $V$ of the shelter as a function of the depth $x$ .

Answer to Problem 22RE

  $V=600x-x^3$

Explanation of Solution

Given information:

A small shelter for delicate plants is to be constructed of thin plastic material. It will have square ends and a rectangular top and back, with an open bottom and front, as shown in the figure. The total area of the four plastic sides is to be $1200{\text{in}}^{\text{2}}$ .

Calculation:

Consider the diagram:

  

In this diagram, there are four plastic sides. Two sides are rectangular and two are square.

Area of the four plastic sides is $1200{\text{in}}^{\text{2}}$ .

That is,

  $x^2+xy+x^2+xy=1200$

Here $x$ is the width and $y$ is the length of the rectangles.

  $\begin{array}{c}2x^2+2xy=1200\\ x^2+xy=600\end{array}$

To express this equation in terms of $y$ ;

  $\begin{array}{c}{x}^2+xy-x^2=600-x^2\\ xy=600-x^2\\ \frac{xy}{x}=\frac{600-x^2}{x}\\ y=\frac{600-x^2}{x}\end{array}$

To express the volume $V$ of the shelter, take the cuboid volume.

  $V=l\cdot w\cdot h$

Here, $w=x=h$ and $l=y$ ;

  $\begin{array}{c}V=x^{2}y\\ V=x^2\left(\frac{600-x^2}{x}\right)\\ =x\left(600-x^2\right)\\ V=600x-x^3\end{array}$

b)

To determine

To draw: a graph of $V$ .

Explanation of Solution

Given information:

A small shelter for delicate plants is to be constructed of thin plastic material. It will have square ends and a rectangular top and back, with an open bottom and front, as shown in the figure. The total area of the four plastic sides is to be $1200{\text{in}}^{\text{2}}$ .

Calculation:

Consider the diagram:

  

To draw a graph of a function $V=600x-x^3$ ;

  

c)

To determine

To find: the dimensions will maximize the volume of the shelter.

Answer to Problem 22RE

The dimensions of the shelter is $x=14.1421\text{in}$ and $y=28.2843\text{in}$ .

Explanation of Solution

Given information:

A small shelter for delicate plants is to be constructed of thin plastic material. It will have square ends and a rectangular top and back, with an open bottom and front, as shown in the figure. The total area of the four plastic sides is to be $1200{\text{in}}^{\text{2}}$ .

Calculation:

Consider the diagram:

  

Consider the function $V=600x-x^3$ ;

Taking the derivative with respect to $x$ ;

  $\frac{dV}{dx}=600-3x^2$

To maximize the volume of the shelter, equate the derivative zero.

  $\begin{array}{c}600-3x^2=0\\ 3x^2=600\\ x^2=200\\ x=10\sqrt{2}\\ x\approx 14.1421\end{array}$

To find the value of $y$ , use the formula $y=\frac{600-x^2}{x}$ ;

  $\begin{array}{c}y=\frac{600-200}{14.1421}\\ =28.2843\end{array}$

Therefore, the dimensions of the shelter is $x=14.1421\text{in}$ and $y=28.2843\text{in}$ .