Houston Community College is planning to construct a rectangular parking lot on land
bordered on one side by a highway. The plan is to use 640 ft of fencing to fence off the
other three sides. What should the dimensions of the lot be if the enclosed area is to be a
maximum?
The side parallel to the highway be x ft
let each of the other two sides by y
2x + y = 640
y = 640 - 2x
area = xy = x(640-2x)
= -2x^2 + 640x
I assume you know how to find the vertex of this downwards parabola, and that vertex is
(160 , 51200)
if x = 160
y = 640 - 2(160) = 320
the lot should be 320 ft by 160 ft
To find the dimensions of the rectangular parking lot that will maximize the enclosed area, we can use calculus.
Let's assume the length of the rectangular lot is L and the width is W. We know that the lot is bordered on one side by a highway, so there is no need to fence that side. The remaining three sides will require 640 feet of fencing.
We can label the two sides that are perpendicular to the highway as L (length) and the other side as 2W (width) since there will be two equal sides of width.
The perimeter of the lot is given by:
Perimeter = 2L + 2W = 640
We can rewrite this equation to solve for L in terms of W:
L = 320 - W
Now, we need to find the area of the rectangular lot. The area can be calculated by multiplying the length and width:
Area = L * W = (320 - W) * W
To find the maximum area, we need to find the critical points of the area function. We can do this by taking the derivative of the area function with respect to W and setting it equal to zero:
d/dW [Area] = (320 - W) - 2W = 320 - 3W
Now, set 320 - 3W = 0 and solve for W:
320 - 3W = 0
3W = 320
W = 320/3
W = 106.67 ft (rounded to 2 decimal places)
Now that we have the value of W, we can substitute it back into our equation for L:
L = 320 - W
L = 320 - 106.67
L = 213.33 ft (rounded to 2 decimal places)
Therefore, the dimensions of the lot that will maximize the enclosed area are approximately 213.33 ft by 106.67 ft.