A high school has a rectangular parking lot that measures 600 ft. long by 400 ft. wide. The School Board wants to double the area of the lot by increasing both its length and width by the same amount, x ft.
The Board also wants to build a fence around the new lot. The cost to
expand the lot is estimated to be $2 per square foot of new space.
The cost to fence the lot is
estimated to be $30 per foot of fencing. Costs include materials and labor.
Write an equation that you can use to find x. Solve the equation for x.
Let A be the area of the original parking lot.
Area of original parking lot = length x width
A = 600 ft x 400 ft
A = 240,000 sq ft
Let the increase in length and width be x ft.
New length = 600 ft + x ft
New width = 400 ft + x ft
New area = (600 + x) ft x (400 + x) ft
New area = 240,000 sq ft + 600x + 400x + x^2
New area = 240,000 sq ft + 1000x + x^2
The Board wants to double the area, so:
2A = 240,000 sq ft
480,000 sq ft = 240,000 sq ft + 1000x + x^2
0 = 240,000 sq ft + 1000x + x^2 - 480,000 sq ft
0 = 1000x + x^2 - 240,000 sq ft
Solving for x:
x^2 + 1000x - 240,000 = 0
Using the quadratic formula: x = (-b ± √(b^2 - 4ac)) / 2a
Where a = 1, b = 1000, and c = -240,000
x = (-1000 ± √(1000^2 - 4(1)(-240,000))) / 2(1)
x = (-1000 ± √(1,000,000 + 960,000)) / 2
x = (-1000 ± √1,960,000) / 2
x = (-1000 ± 1400) / 2
x = (-1000 + 1400) / 2 or x = (-1000 - 1400) / 2
x = 400 / 2 or x = -2400 / 2
x = 200 or x = -1200
Since x represents an increase in length and width, the negative value is not applicable in this context. Therefore, x = 200 ft.
So, the Board should increase both the length and width of the parking lot by 200 ft in order to double its area.