Chapter 1.3, Problem 72PS

Solution

To find: the find the distance x by which the parking lot must be expanded so that it doubles in the area.

  $x=60\text{ft}$ .

Given information:

The grocery store wants to double the area of its parking lot by expanding the old parking lot as shown:

  

Formula Used:

  $\text{Area of rectangle}=\text{Length}\times \text{Width}$

Property Used:

Zero Product property: If product of two expressions is zero, then either one or both the expressions equal zero.

Calculation:

From the given picture, it is clear that the grocery is rectangular-shaped. And its length of is 300 ft, while its width is 165 ft.

So,

  $\begin{array}{c}\text{Area of the store}=\text{Length×Width}\\ =300\times 165\\ =49500{\text{ft}}^2\end{array}$

The area of the grocery store and the old parking lot combined is:

  $\begin{array}{c}\text{Area of the store and old lot}=\text{Length×Width}\\ =\left(300+75\right)\times \left(165+75\right)\\ =375\times 240\\ =90000{\text{ft}}^2\end{array}$

So, the area of the old parking lot is:

  $\begin{array}{c}\text{Area of old parking lot = Area of the store and old lot}-\text{Area of store}\\ =90000{\text{ft}}^2-49500{\text{ft}}^2\\ =40500{\text{ft}}^2\end{array}$

The grocery store wants to double the area of the parking lot. So, the area of the new parking lot is:

  $\begin{array}{c}\text{New parking lot Area}=2\times 40500\\ =81000{\text{ft}}^2\end{array}$

Thus, the area of the grocery store combined with the area of the new parking lot is:

  $\begin{array}{c}\text{New Area = Area of grocery store + New parking lot Area}=2\times 40500\\ =49500+81000\\ =130,500{\text{ft}}^2\end{array}$

Now, by looking the given diagram, it can be seen that the after expanding by x , the new area that is the combined area of the grocery store and the new parking lot is:

So, the new area is:

  $\begin{array}{c}\text{New Area}=\text{New Length × New Width}\\ =\left(300+75+x\right)\times \left(165+75+x\right)\\ =\left(375+x\right)\left(240+x\right)\end{array}$

But, as obtained above, the area of the grocery store combined with the area of the new parking lot is $130,500{\text{ft}}^2$ . So,

  $\left(x+375\right)\left(x+240\right)=130,500$

Multiply using FOIL,

  $x^2+240x+375x+90000=130500$

Combine the like terms,

  $x^2+615x+90000=130500$

Subtract 130500 from both sides,

  $x^2+615x-40500=0$

Factor the above equation,

  $\begin{array}{c}{x}^2+675x-60x-40500=0\\ x\left(x+675\right)-60\left(x+675\right)=0\\ \left(x+675\right)\left(x-60\right)=0\end{array}$

Then, by zero product property,

  $x+675=0$ or $x-60=0$

Solve for x ,

  $x=-675$ or $x=60$

Since the grocery store wants to double the area of the parking lot, so the value of x can’t be negative. So, the only solution to the given problem is:

  $x=60\text{ft}$ .

Thus, the length and the width must be increased by 60 feet.