Using calculus prove that the square has the greatest area for the same perimeter when a rectangle is considered.

Let p be the perimeter , x be the length of the
rectangle.

Then the width of the rectangle is
(P-2x)/2.

Therefore the area of the rectangle with the
perimeter p and the side x and (p-2x)/2 is given by:

A(x) =
x*(p-2x)/2 = px/2 - x^2.

So by calculus, area A(x) is a
function of perimeter p and A(x) is maximum, for x = c for which A'(c) = 0 and
A"(c) < 0.

A'(x) = {px/2 - x^2}' =  (px/2)' - (x^2)'
.

So A'(x) = p/2 -2x. Equating to zero, we get p/2 - 2x =
0. Or

2x= p/2

x
= p/4.

Again find A''(p/2)
:

A'(x) =
p/2-2x.

Differentiate:

A"(x) =
0-2 , which is negative for all x and so for x = p/4
also.

So A"(p/2)  is negative for x= p/2 for which f'(p/2)
= 0.

So, for a given perimeter p, x = p/4 gives the maximum
area.

So , if p is the perimeter, the maximum area formed
among the set of rectangles is the square with sides of length p/4
.