Let x and y be the side lengths of the rectangele. If the perimeter is 80, then 2x+2y=80

x+y=40

y=40-x

The are of the rectangle is A=x*y=x*(40-x)=40x-x²

Since it is a maximization problem the first derivative of the area function (A) must be equal to zero and the second derivative must be negative.

A''(x)=-2<0 (Second derivative rule satisfied)

A'(x)=40-2x=0, 40=2x, 20=x, y=40-x=20 So the area A=20*20=400