Glenview Community College wants to build a rectangular parking lot on land bordered on one side by a highway. It has 380 feet of fencing to use on the other three sides. What should be the dimensions of the lot if the enclosed area is to be a maximum?
190 x 95 feet gives you the most area.
It gives you a 18,050 ft^2 area.
191 x 94.5 gives you 18,049.5 ft^2
189 x 95.5 gives you 18,049.5 ft^2
You can use calculus to get the result. The lot dimensions are L x 190 - L/2. L is the side parallel to the highway.
Area A = L*(190 - L/2) = 190L - L^2/2
dA/dL = 0 = 190 -L for a maximum
L = 190
To find the dimensions of the rectangular parking lot that will maximize the enclosed area, we can use the concept of optimization. The enclosed area can be represented as the product of the length and width of the rectangular lot.
Let's assume that the length of the parking lot is x feet. Since the lot is bordered on one side by a highway, the width of the parking lot will be y feet.
We know that the length and the two widths (top and bottom) combined will require 380 feet of fencing. This gives us the equation:
2x + y = 380
To find the dimensions that maximize the enclosed area, we need to express the area in terms of a single variable, either x or y. Since the equation above already expresses y in terms of x, we can solve for y:
y = 380 - 2x
Now, we can represent the area A as:
A = x * y
Substituting the value for y we just found:
A = x * (380 - 2x)
A = 380x - 2x^2
The goal is to maximize the enclosed area A, so we need to find the maximum point of the equation. We can do this by finding the derivative of A with respect to x, setting it equal to zero, and solving for x.
dA/dx = 380 - 4x
Setting dA/dx = 0:
380 - 4x = 0
4x = 380
x = 380/4
x = 95
Now that we have the value of x, we can substitute it back into the equation for y to find its value:
y = 380 - 2x
y = 380 - 2 * 95
y = 380 - 190
y = 190
Therefore, the dimensions of the parking lot that maximize the enclosed area are a length of 95 feet and a width of 190 feet.