Calculus Maximization problem Please help!?

Question

Calculus Maximization problem Please help!?
a) A rectangular parking is to be fenced on three sides leaving the fourth side open to the street. If there is 800 meters of fencing available, determine the dimensions that would produce the maximum area.
How would you create the derivatives.

Answer 1

Hello,

Let w = the width
Let l = the length

so p = 2l + 2w

now p = 800 so 800 = 2l + 2w solve for l giving us l = (800-2w)/2 or l = 400 - w now the area is a = l*w so substituting for l we have a = (400 - w) * w giving us
a = 400w - w^2 Now take the derivative and we have
da/dw = 400 -2w since we want a max set it equal to 0

0 = 400 -2w so w = 200 meters. Then l = 200 meters. This is a square.

Hop This Helps!!

Answer 2

f(x) = xy, the fenced area
2x+y = 800
f(x) = x(800-2x)
f'(x) = 0 => x = 200
the maximum area. = f(200) = 200(800-400) = 80,000 m^2