To determine by how many feet the length of the parking lot should be increased, let's proceed step by step:
1. Define Variables:
- Let [tex]\( x \)[/tex] be the amount by which both the length and the width of the parking lot are to be increased, in feet.
2. Initial Dimensions:
- The current length of the parking lot is [tex]\( 120 \, \text{ft} \)[/tex].
- The current width of the parking lot is [tex]\( 80 \, \text{ft} \)[/tex].
3. Existing Area Calculation:
- The current area of the parking lot is:
[tex]\[
\text{Current Area} = \text{length} \times \text{width} = 120 \times 80 = 9600 \, \text{ft}^2
\][/tex]
4. New Area after Expansion:
- The new area of the parking lot after both the length and the width are increased by [tex]\( x \)[/tex] feet is:
[tex]\[
\text{New Area} = (120 + x) \times (80 + x)
\][/tex]
5. Equation Setup:
- The increase in area is given as [tex]\( 4400 \, \text{ft}^2 \)[/tex].
- Therefore, the equation representing the increase in area is:
[tex]\[
(\text{New Area}) = (\text{Current Area}) + 4400
\][/tex]
- Substituting the values:
[tex]\[
(120 + x) \times (80 + x) = 9600 + 4400
\][/tex]
[tex]\[
(120 + x) \times (80 + x) = 14000
\][/tex]
6. Expand and Simplify:
- Expanding the left side of the equation:
[tex]\[
(120 + x)(80 + x) = 120 \times 80 + 120x + 80x + x^2
\][/tex]
[tex]\[
9600 + 200x + x^2 = 14000
\][/tex]
7. Form a Quadratic Equation:
- Rearrange to form a quadratic equation:
[tex]\[
x^2 + 200x + 9600 = 14000
\][/tex]
[tex]\[
x^2 + 200x - 4400 = 0
\][/tex]
8. Solve the Quadratic Equation:
- Using the quadratic formula [tex]\( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)[/tex], where [tex]\( a = 1 \)[/tex], [tex]\( b = 200 \)[/tex], and [tex]\( c = -4400 \)[/tex]:
[tex]\[
x = \frac{-200 \pm \sqrt{200^2 - 4 \cdot 1 \cdot (-4400)}}{2 \cdot 1}
\][/tex]
[tex]\[
x = \frac{-200 \pm \sqrt{40000 + 17600}}{2}
\][/tex]
[tex]\[
x = \frac{-200 \pm \sqrt{57600}}{2}
\][/tex]
[tex]\[
x = \frac{-200 \pm 240}{2}
\][/tex]
- This gives us two potential solutions:
[tex]\[
x_1 = \frac{-200 + 240}{2} = 20
\][/tex]
[tex]\[
x_2 = \frac{-200 - 240}{2} = -220
\][/tex]
Since a negative length does not make sense in this context, we discard [tex]\( x = -220 \)[/tex].
9. Select the Valid Solution:
- The valid solution is [tex]\( x = 20 \)[/tex].
Hence, the length of the parking lot should be increased by [tex]\( 20 \, \text{ft} \)[/tex].
Therefore, the correct answer is [tex]\( 20 \, \text{ft} \)[/tex].
