Paul has 400 yards of fencing to enclose a rectangular area. Find the dimensions of the rectangle that maximize the enclosed area. What is the maximum area?

A rectangle that maximizes the enclosed area has a length of _ nothing yards and a width of _ nothing yards.

The maximum area is _ nothing square yards.

P=2 (x+y) perimeter equals two time the sum of x and y dimensions ...

We are told that P=400 so

2 (x+y) = 400

x+y=200

x=200-y ...

Now area is:

A=xy, using x found above we have:

A = (200-y) y

A=200y-y^2

The rate of change of the area is dA/dy

dA/dy=200-2y

The maximum area occurs when the rate of change is zero or dA/dy=0 so

200-2y=0

2y=200

y=x=100yd

So the maximum area enclosed is when y=x=100 which is a perfect square (as is always the case for a four sided enclosure with a given amount of material)

A=100^2=10000 yd^2