Perimeter of rectangle and area of square

To verify that amongst all the rectangles of the same perimeter, the square has the maximum area

Let the length and breadth of rectangle be x and y.

The perimeter of the rectangle P = 48 cm.

2 (x + y) = 48

or

x + y=24 or y=24-x

Let A (x) be the area of rectangle, then

A (x) = xy

= x (24 - x)

= 24x - x2

A′ (x) = 24 – 2x

A′ (x) = ⇒ 24 – 2x = 0 ⇒ x = 12

A′′ (x) = – 2

A′′ (12) = – 2, which is negative

Therefore, area is maximum when x = 12

y = x = 24 – 12 = 12

So, x = y = 12

Hence, amongst all rectangles, the square has the maximum area.