Question

the owner of an office building is expanding the length and width of a parking lot by the same amount. the lot currently measures 120 ft by 80 ft, and the expansion will increase its area by 4,400 ft². by how many feet should the length of the parking lot be increased?
a = lw
○ 1.2 ft
○ 20 ft
○ 66.3 ft
○ 220 ft

Explanation:

Step1: Define the variable

Let \( x \) be the number of feet by which both the length and the width are increased.

Step2: Find the original area

The original length \( l = 120 \) ft and original width \( w = 80 \) ft. Using the formula \( A = lw \), the original area \( A_{original}=120\times80 = 9600\) \( \text{ft}^2 \).

Step3: Find the new area

The new length is \( (120 + x) \) ft and the new width is \( (80 + x) \) ft. So the new area \( A_{new}=(120 + x)(80 + x) \).
We know that the expansion increases the area by \( 4400 \) \( \text{ft}^2 \), so \( A_{new}=A_{original}+ 4400=9600 + 4400=14000\) \( \text{ft}^2 \).

Step4: Set up the equation and solve

We have the equation \( (120 + x)(80 + x)=14000 \).
Expand the left - hand side:

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Divide the entire equation by \( 20 \) to simplify: \( \frac{x^{2}}{20}+\frac{200x}{20}-\frac{4400}{20}=0\), which gives \( \frac{x^{2}}{20}+10x - 220 = 0\), or multiply through by \( 20 \) to get \( x^{2}+200x - 4400 = 0\). We can also use the quadratic formula \( x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a} \) for the quadratic equation \( ax^{2}+bx + c = 0 \). Here, \( a = 1 \), \( b = 200 \), \( c=- 4400 \).

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We have two solutions for \( x \):
\( x=\frac{-200 + 240}{2}=\frac{40}{2}=20 \) and \( x=\frac{-200 - 240}{2}=\frac{-440}{2}=-220 \).
Since the length of increase cannot be negative, we reject \( x=-220 \).

Answer:

20 ft