Arectangle has a perimeter of 80cm. if its width is x, express its length and its area in the terms of x. what is the maximum area of the rectangle?

Perimeter of a rectangle=2(length)+2(width)
length=y
width=x
Then:
80=2x+2y
40=x+y
y=40-x

Then:
width=x
length=(40-x)

Area of a rectangle=length x width.
A(x)=x(40-x)
A(x)=40x-x²

We have to find the maximums or minimums of this function
1) we calculate the first derivative:
A´(x)=40-2x

2) we have to find the values of "x" when A´(x)=0
40-2x=0
-2x=-40
x=-40/-2
x=20

3) We have to calculate the second derivative.
A´´(x)=-2
Because A´´(x)<0; then we have a maximum at x=20

Therefore:
width:  x=20
length:  40-x=40-20=20

This rectangle with maximum area is a square:
Area=(20 cm)(20 cm)=400 cm²

The lenght expressed in the terms of x would be: 40-x; and the maximun area would be 400 cm².