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
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.