Chapter 6.5, Problem 15CYU

To determine

To find: the width of the boardwalk.

Answer to Problem 15CYU

  $x=-40,5$ .

Explanation of Solution

Given:

The width of the boardwalk around a rectangular pond is $xft$

The dimension of pond is $30ft\text{by}40ft$ .

Area of the pond is $2000f{t}^2$ .

Calculation:

According to the question,

The given equation is $\left(2x+30\right)\left(2x+40\right)=2000$ .

In order to find the width of the boardwalk, factor the equation. Therefore,

  $\begin{array}{l}\left(2x+30\right)\left(2x+40\right)=2000\left\{\text{given equation}\right\}\\ \left\{\left(x+a\right)\left(x+b\right)=x^2+\left(a+b\right)x+ab\right\}\\ 4x^2+\left(40+30\right)2x+1200=2000\left\{\text{simplify}\right\}\\ 4x^2+140x-800=0\end{array}$

  $\begin{array}{l}{x}^2+35x-200=0\left\{\text{divide both sides by 4}\right\}\\ x^2-5x+40x-200=0\left\{\text{simplified}\right\}\\ x\left(x-5\right)+40\left(x-5\right)=0\left\{\text{use distributive law}\right\}\\ \left(x-5\right)\left(x+40\right)=0\left\{\text{factor the given equation}\right\}\\ \left(x-5\right)=0;\left(x+40\right)=0\left\{\text{use zero product property law}\right\}\end{array}$

When $\left(x+40\right)=0$ Then, $x=-40$

When, $\left(x-5\right)=0$ Then, $x=5$

Hence, the the width of the boardwalk is $x=-40,5$ .