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