Chapter 4.1, Problem 69PPSE

Solution

To state: The quadratic function that can be used to find the maximum area of tennis court also find the maximum area, and the lengths of the sides of the resulting fence if an athletic club has 225 feet of fencing to enclose a tennis court.

The quadratic function that represents the area of all rectangles is $A=\frac{225}{2}w-w^2$ , the maximum area is 3164.06 square feet and the lengths are $\frac{225}{4}$ each.

Given information:

An athletic club has 225 feet of fencing to enclose a tennis court.

Explanation:

The shape of a tennis court is rectangular.

Then the perimeter of the rectangle is: 225ft

Let’s assume l as the length of the tennis court and w as the width.

Then the perimeter of the rectangular tennis court will be:

  $\begin{array}{c}2l+2w=225\\ l+w=\frac{225}{2}\\ l=\frac{225}{2}-w\end{array}$

Now the area of the rectangular tennis court is:

  $\begin{array}{l}A=lw\\ A=\left(\frac{225}{2}-w\right)w\\ A=\frac{225}{2}w-w^2\end{array}$

This is an equation of a parabola and it has its maximum at vertex:

  $\begin{array}{l}x=\frac{-b}{2a}\\ x=\frac{-\frac{225}{2}}{-2}\\ x=\frac{225}{4}\end{array}$

Therefore, one side of the tennis court is $w=\frac{225}{4}$ .

Then the length is:

  $\begin{array}{l}l=\frac{225}{2}-\frac{225}{4}\\ l=\frac{225}{4}\end{array}$

Now the maximum area will be:

  $\begin{array}{l}A=lw\\ A=\frac{225}{4}\cdot \frac{225}{4}\\ A=3164.06{\text{ft}}^2\end{array}$

Therefore, the maximum area is 3164.06 square feet.