Chapter 5.3, Problem 53PPSE

Solution

Find the length of the box.

The length of the box is $5\text{ m}$ .

Given:

The width of the box is $2$ meters less than the length.

The height of the box is $1$ meter less than the length.

The volume is $60{\text{ m}}^3$ .

Calculation:

Let the length of the box is $x$ .

Hence, width is $x-2$ and the height is $x-1$ .

Therefore, the volume of the box is given by

  $\begin{array}{c}v=lwh\\ =x\left(x-2\right)\left(x-1\right)\end{array}$

Now, the volume is given as $60{\text{ m}}^3$ . Thus, the equation is

  $\begin{array}{c}x\left(x-2\right)\left(x-1\right)=60\\ x^3-3x^2+2x-60=0\end{array}$

Solve this equation graphically. The x -intercept is the value for x.

  

Thus, $x=5$ .

Hence, the length of the box is $5\text{ m}$ .