Question

Show that among rectangles of given area, the square has least perimeter.

Answer 1

Let x be the length and y be the breadth of the rectangle whose area is A sq units (which is given as constant).

Then xy = A

∴ y = `"A"/x`                               ...(1)
Let P be the perimeter of the retangle.

Then P = 2(x + y)

= `2(x + "A"/x)`                   ...[By(1)]

∴ `"dP"/dx = 2.d/dx(x + "A"/x)`

= 2[1 + A(– 1)x^–2]

= `2(1 - "A"/x^2)`
and
`(d^2P)/(dx^2) = 2d/dx(1 - "A"/x^2)`

= 2[0 – A(– 1)x^–3]

= `(4"A")/x^3`

Now, `"dp"/dx  0, "gives" 2(1 - "A"/x^2)` = 0

∴ x² – a = 0
∴ x² = A
∴ x = `sqrt("A")`                    ...[∵ x > 0]
and
`((d^2P)/(dx^2))_("at"  x = dsqrt("A")`

= `(4"A")/(sqrt("A"))^3 > 0`

∴ P is minimum when x = `sqrt("A")`

If x = `sqrt("A"), "then"  y = "A"/x = "A"/sqrt("A") = sqrt("A")`

∴ x = y
∴  rectangle is a square.
Hence, among rectangles of given area, the square has least perimeter.