Chapter 7.4, Problem 21Q

To determine

The width of the bunker at ground level.

Explanation of Solution

Given information:

The front of the storage bunker can be modeled by the expression,

  $y=-\frac{5}{216}(x-72)(x+72)$ .

Calculations:

Here the front side of the bunker is modeled by the expression,

  $y=-\frac{5}{216}(x-72)(x+72)$

Now to get the width of the bunker along the base, we have to take, $y=0$

Since X-axis shows the base and Y axis shows the height of the bunker

Thus,

  $\begin{array}{l}0=-\frac{5}{216}(x-72)(x+72)\\ or,-\frac{5}{216}(x-72)(x+72)=0\end{array}$ .

So solving the expression we get,

  $\begin{array}{l}-\frac{5}{216}(x-72)(x+72)=0\\ \Rightarrow (x-72)(x+72)=0\\ \Rightarrow x^2-{72}^2=0\\ \Rightarrow x^2={72}^2\\ \Rightarrow x=\sqrt{{72}^2}\\ \Rightarrow x=\pm 72\end{array}$

Thus the width of the base is $72+72=144inch$ .