show all rectangles of given perimeter square has largest area

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.