Chapter 4.3, Problem 44E

(a)

To determine

To Find: The polynomial function to model the volume of the box.

Answer to Problem 44E

The volume of the box is $V\left(x\right)=2x^3-38x^2+180x$ .

Explanation of Solution

Given:

The cardboard sheet is of 18 inch by 20 inch.

The given diagram is shown in Figure 1

  

Figure 1

Calculation:

Consider the box is shown in Figure 2

  

Figure 2

The length of the box is $l=\frac{20}{2}-x$ that is $10-x$ .

The width of the box is $b=18-2x$ .

The height of the box is $h=x\text{in}$ .

The volume of the box is $V=lbh$ .

Consider the volume of the box is obtained as,

  $\begin{array}{c}V\left(x\right)=\left(10-x\right)\left(18-2x\right)x\\ =\left(180-20x-18x+2x^2\right)\\ =2x^3-38x^2+180x\end{array}$

(b)

To determine

To Find: Thegraph of the function.

Answer to Problem 44E

The graph of the function is shown in Figure 1

Explanation of Solution

Consider the function is $V\left(x\right)=2x^3-38x^2+180x$ .

The graph of the function is shown in Figure 1

  

Figure 1

(c)

To determine

To Find: The equation to model the situation when the company wants the box to have the volume of 224 cubic inches.

Answer to Problem 44E

The required equation is $224=2x^3-38x^2+180x$ .

Explanation of Solution

Consider the volume is,

  $V\left(x\right)=2x^3-38x^2+180x$

Then,

  $224=2x^3-38x^2+180x$

(d)

To determine

To Find: The value of $x$ .

Answer to Problem 44E

The value of the positive integer is $2$ .

Explanation of Solution

Consider the given equation is,

  $224=2x^3-38x^2+180x$

Then,

  $\begin{array}{l}2x^3-38x^2+180x-224=0\\ x^3-19x^2+90x-112=0\end{array}$

The graph of the equation is shown in Figure 4

  

Figure 4

From the above graph the value of the positive integer is $2$ .