Answer to Question #168946 in Calculus for Vishal

Question

Prove that of all rectangles with given perimeter, the square has the greatest area.

Answer 1

Let P be the fixed perimeter of a rectangle with length x and height y

Hence, P = 2x + 2y and Area A = $xy$

Now, P = 2x -2y

y = $\dfrac{P}{2} - x$

Putting value of y in area A

We get,

A = $x(\dfrac{P}{2} - x) = (\dfrac{P}{2})x - x^2$

Let $A = f(x)$

f(x) = $(\dfrac{P}{2})x - x^2$

Differentiating with respect to 'x'

$f'(x) = \dfrac{P}{2} - 2x$

$f'(x) = 0$ when $x = \dfrac{P}{4}$

Hence, $x = \dfrac{P}{4}$ gives a local maximum value of the area.

Hence, the square of length $\dfrac{P}{4}$ has the greatest area.