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

### 2 Answers | Add Yours

Here we have to prove that the square has the greatest area for the same perimeter when a rectangle is considered.

So the perimeter P can be taken to be a constant P.

Use a variable x for the length. So the width is P/2 - x.

The area A = ( P/2 -x) *x.

Now to maximize the area, for a variable x, we have to find dA/dx.

dA/dx= P/2 - 2x.

Equate this to zero P/2 - 2x =0

=> x= P/4

So the length is P/4 and the width is P/2 - x = P/4.

**Therefore for a given perimeter of a rectangle the area is the largest if the shape is that of a square with each side equal to the perimeter divided by 4.**

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 .