Question

Prove that among all the rectangles of the given perimeter, the square has the maximum area

Answer 1

Let x, y be the length and breadth of a rectangle and given perimeter is P  ....(say)

ie. 2(x + y) = P

y = `"P"/2 - x`

Area of a rectangle ‘A’ = xy

A = `x("P"/2 -  x) = "P"/2 x - x^2`

`"dA"/("d"x) = "P"/2 - 2x`

For maximum or minimum,

`"dA"/("d"x)` = 0

⇒ `"P"/2 - 2x` = 0

x = `"P"/4`

Now, `("d"^2"A")/("d"x^2)` = – 2

At x = `"P"/4, ("d"^2"A")/("d"x^2) < 0`

∴ Area of the rectangle is maximum when x = `"P"/4`

Now, y = `"P"/2 - x = "P"/2 - "P"/4 = "P"/4`

∴ Length of a rectangle = `"P"/4`

Breadth of a rectangle =  `"P"/4`

Since Length = Breadth, the rectangle is a square.

Hence Proved.