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Show that of all therectangles of given area, the square has the smallest perimeter. |
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Answer» Let A be the fixed area. Consider a rectangle with sides x and y, and area A Let P be the perimeter of the this rectangle Then, `A = xy rArr y = (A)/(x)`..(i) And, `P = 2x + 2y = 2x + (2A)/(x)`[using (i)] Now, `P = (2x + (2A)/(x)) rArr (dP)/(dx) = (2 - (2A)/(x^(2))) and (d^(2)P)/(dx^(2)) = (4A)/(x^(3))` Now, `(dP)/(dx) = 0 rArr 2 - (2A)/(x^(2)) = 0 rArr x = sqrtA` and, `[(d^(2)P)/(dx^(2))]_(x = sqrtA) = (4A)/(A^(3//2)) = (4)/(sqrtA) gt 0` So, `x = sqrtA` is a point of minimum. Moreover, `x = sqrtA rArr y = (A)/(x) = (A)/(sqrtA) = sqrtA = x` So, when the perimeter is smallest, the rectangle is a square. |
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