Explore BrainMass

Mathematics - Calculus - Maximizing Area

Not what you're looking for? Search our solutions OR ask your own Custom question.

This content was COPIED from BrainMass.com - View the original, and get the already-completed solution here!

A developer wants to enclose a rectangular grassy lot that borders a city street for parking. If the developer has 640 feet of fencing and does not fence the side along the street, what is the largest area that can be enclosed?

https://brainmass.com/math/basic-calculus/maximizing-area-fence-221913

SOLUTION This solution is FREE courtesy of BrainMass!

Let the length of the lot be x and width be y. Then total length of fence required =
x + y + y = x + 2y
Therefore, x + 2y = 640  x = 640 - 2y  (1). Also, differentiating wrt x, we get,
1 = -2.(dy/dx)  dy/dx = -1/2  (2)
Area of the enclosure A = xy = (640 - 2y)y = 640y - 2y^2 [From (1)]
Differentiating wrt x, we get dA/dx = 640.(dy/dx) - 4y.(dy/dx) = 0
 640(-1/2) - 4y(-1/2) = 0  y = 160. So, x = 640 - 2(160) = 320.

The desired dimensions are length = 320 ft and width = 160 ft.
Maximum area that can be enclosed = 320 x 160 = 51200 sq.ft.

This content was COPIED from BrainMass.com - View the original, and get the already-completed solution here!