Chapter 4.4, Problem 19E

a.

To determine

Write a formula $V(x)$ for the volume of the box.

Answer to Problem 19E

The volume of box $V(x)=2x\left(24-2x\right)\left(18-2x\right)$ cubic in.

Explanation of Solution

Given:

A 24 by 36 in. sheet cardboard from which congruent the squares of side length $x$ are cut from the corners of the folded rectangle sheet.

  

Calculation:

According to given figure,

The length of the box is $\left(24-2x\right)$ in.

The width of the box is $\left(18-2x\right)$ in.

The depth of the box is $2x$ in.

Volume $\left(V\right)$ of box = length $\times$ width $\times$ depth

  $\begin{array}{l}V(x)=\left(24-2x\right)\times \left(18-2x\right)\times 2x\\ V(x)=2x\left(24-2x\right)\left(18-2x\right)\end{array}$

Conclusion:

The volume of box $V(x)=2x\left(24-2x\right)\left(18-2x\right)$ cubic in.

b.

To determine

Find the domain of $V$ for the problem situation and graph $V$ over this domain.

Answer to Problem 19E

Domain for volume $V(x)$ is $\left(0,9\right)$ .

Explanation of Solution

Given:

From part (a):

Volume of box $V(x)=2x\left(24-2x\right)\left(18-2x\right)$ cubic in.

Calculation:

The volume of box cannot have negative values. So,

  $\begin{array}{c}V(x)>0\\ 2x\left(24-2x\right)\left(18-2x\right)>0\\ \text{If }2x>0\\ \text{then}x>0\\ \\ \text{If }\left(24-2x\right)>0\\ 24>2x\\ 2x<24\\ \text{then}x<12\\ \\ \text{If }\left(18-2x\right)>0\\ 18>2x\\ 2x<18\\ x<9.\end{array}$

Hence, Domain for volume $V(x)$ is $\left(0,9\right)$ .

Graph of $V(x)$:

  $Y$ -axis denotes $V(x)$ .

  $X$ -axis denotes variable $x$ .

  

Conclusion:

Domain for volume $V(x)$ is $\left(0,9\right)$ .

c.

To determine

Use analytic method to find the maximum volume and the value of $x$ .

Answer to Problem 19E

The maximum volume of box $V=1309.9547$ cubic in and $x=3.3944$ inch.

Explanation of Solution

Given:

From part (a):

  $V(x)=2x\left(24-2x\right)\left(18-2x\right)$

Calculation:

  $\begin{array}{c}V(x)=2x\left(24-2x\right)\left(18-2x\right)\\ V(x)=8\left(12x-x^2\right)\left(9-x\right)\\ \frac{d}{dx}V(x)=\frac{d}{dx}\left(8\left(12x-x^2\right)\left(9-x\right)\right)\\ \frac{dV}{dx}=8\frac{d}{dx}\left(\left(12x-x^2\right)\left(9-x\right)\right)\\ \frac{dV}{dx}=8\left(\left(12-2x\right)\left(9-x\right)+\left(12x-x^2\right)\left(-1\right)\right)\\ \frac{dV}{dx}=8\left(\left(12-2x\right)\left(9-x\right)-\left(12x-x^2\right)\right)\end{array}$

For extremum points put $\frac{dV}{dx}=0$ ,

  $\begin{array}{c}\frac{dV}{dx}=0\\ 8\left(\left(12-2x\right)\left(9-x\right)-\left(12x-x^2\right)\right)=0\\ x^2-14x+36=0\end{array}$

Apply quadratic formula,

  $\begin{array}{l}x=7\pm \sqrt{13}\\ x=3.3944,10.6056\\ x=3.3944\left(\because 10.6056\notin \left(0,9\right)\right)\end{array}$

The maximum occurs at $x=3.3944$ .

Volume:-

  $\begin{array}{l}V=2\left(3.3944\right)\left(24-2\left(3.3944\right)\right)\left(18-2\left(3.3944\right)\right)\\ V=1309.9547\end{array}$

Conclusion:

The maximum volume of box $V=1309.9547$ cubic in and $x=3.3944$ inch.

d.

To determine

Find the maximum volume graphically.

Answer to Problem 19E

The maximum volume of box $V=1309.9547$ cubic in and $x=3.3944$ inch.

Explanation of Solution

Given:

Volume of box

  $V(x)=2x\left(24-2x\right)\left(18-2x\right)$ .

Calculation:

Draw graph,

  

$Y$ -axis denotes $V(x)$ .

  $X$ -axis denotes variable $x$ .

Conclusion:

The maximum volume of box $V=1309.9547$ cubic in and $x=3.3944$ inch.

e.

To determine

Find the value of $x$ that yields a volume of $1120{\text{ in}}^3$ .

Answer to Problem 19E

The value of $x\text{ is }2\text{ inch and }5\text{ inch}.$

Explanation of Solution

Given:

Volume of box

  $V(x)=2x\left(24-2x\right)\left(18-2x\right)$ .

Calculation:

Given that $V=1120{\text{ in}}^3$ then

Put $V=1120{\text{ in}}^3$ in given equation,

  $\begin{array}{c}V(x)=2x\left(24-2x\right)\left(18-2x\right)\\ 1120=2x\left(24-2x\right)\left(18-2x\right)\\ 1120=8x\left(12-x\right)\left(9-x\right)\\ x\left(12-x\right)\left(9-x\right)=\frac{1120}{8}\\ x\left(12-x\right)\left(9-x\right)=140\end{array}$

On solving the equation,

  $x=2,5,14$ ,

But, $14\notin \left(0,9\right)$ . So, solution is $x=2,5\text{ inch}.$

Conclusion:

The value of $x\text{ is }2\text{ inch and }5\text{ inch}.$

f.

To determine

Write a paragraph describing the issues that arise in part (b).

Explanation of Solution

Given:

Volume of box

  $V(x)=2x\left(24-2x\right)\left(18-2x\right)$ .

Concept used:

Real world problem: In real world problem, the domain and range for function has positive values

Calculation:

The domain of $V(x)$ is all real numbers because $V(x)$ is defined at every real value. But, in real world problem volume cannot be negative and $V(x)$ denotes the volume of box. So, domain of $V(x)$ is $\left(0,9\right)$ .