A merchant wants to create a rectangular display area in front of her store. The area is to be enclose by fencing in three sides that cost $12 per foot and fencing on one side costs $25 per foot.

Question

She has $2000 to spend for the construction of the fence. What dimensions should she use for the fence that would maximize the area enclosed?

Answer 1

Let x and y be the dimensions of the rectangle.

Area, A = xy

Cost, C = 12(2x) + 12y + 25y = 24x + 37y = 2000

y = (2000-24x)/37

A(x) = x(2000-24x)/37 = (2000x - 24x²)/37

To maximize, take the derivative of A(x), then set that to zero and solve for x.

A'(x) = 2000/37 - 24x/37

2000/37 - 48x/37 = 0

48x = 2000

x = 41.67 feet

y = (2000-1000)/37 = 27.03 feet