Chapter 3.1, Problem 75E

To determine

To calculate:

The expanded distance $x$ for the expansion of the parking

Answer to Problem 75E

The expanded distance $x$ for the expansion of the parking is $60ft$

Explanation of Solution

Given information:

  

Calculation:

The area of the school is calculated as:

  $\begin{array}{l}{A}_s=165\left(300\right)\\ A_s=49500f{t}^2\end{array}$

The area of the parking lot with the school is calculated as:

  $\begin{array}{l}{A}_{p+s}=\left(165+75\right)\left(300+75\right)\\ A_{p+s}=90000f{t}^2\end{array}$

This means the area of just the parking lot is:

  $\begin{array}{l}{A}_p=90000-49500\\ A_p=40500f{t}^2\end{array}$

The area of the new expanded lot will be doubled; this means it is $2\left(40500\right)=81000f{t}^2$ adding back the area of the school, this means the total lot is $81000+49500=130500f{t}^2$ .

The width of the rectangle is $375+x$ and length $240+x$ which gives the area as:

  $130500=\left(375+x\right)\left(240+x\right)$

Multiplying the binomials gives that:

  $\begin{array}{l}130500=90000+375+240x+x^2\\ 130500=90000+615x+x^2\\ x^2+615x-40500=0\end{array}$

Use the quadratic formula to get values of $x$ :

  $\begin{array}{l}x=\frac{-615\pm \sqrt{{615}^2-4\left(1\right)\left(-40500\right)}}{2\left(1\right)}\\ x=\frac{-615\pm 735}{2}\\ x=\frac{-615+735}{2}orx=\frac{-615-735}{2}\\ x=60orx=-675\end{array}$

One cannot use the negative number so, $x=60ft$ is correct. Therefore, the expanded distance $x$ for the expansion of the parking is $60ft$ .