# Mathematics - Calculus - Maximizing Area

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?

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## SOLUTION This solution is **FREE** courtesy of BrainMass!

Please see the attachment.

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 ïƒ x = 640 - 2y ïƒ (1). Also, differentiating wrt x, we get,

1 = -2.(dy/dx) ïƒ dy/dx = -1/2 ïƒ (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

ïƒ 640(-1/2) - 4y(-1/2) = 0 ïƒ 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.

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