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. Find the dimensions of the rectangle of perimeter 36 cm which will sweep out a volume as large as possible, when revolved about one of its sides. Also find the maximum volume.
Dear studentLet’s consider x and y to be the length and breadth of given rectangle ABCD.According to the question, the rectangle will be resolved about side AD which making a cylinder with radius x and height y.So, the volume of the cylinder V =π r^2 h =π x^2 y.... (1)Now, perimeter of rectangle P = 2(x + y)36 = 2(x + y)18 = x + yy = 18–x..... (ii)Putting the value of y in the equation (i), V =π x^2 (18–x)=π(18x^2–x^3)Differentiating both sides w.r.t. xdV/dx =π(36x–3x^2)................ (iii)For local maxima and local minima dV/dx = π(36x–3x^2) = 036x–3x^2=03x(12 - x) = 0x≠0 and 12–x = 0⇒x = 12From equation (ii), we havey = 18–12 = 6Differentiating equation (iii) w.r.t. x, we getd^2V/dx^2= π(36–6x)At x = 12,d^2 V/dx^2=π(36–6 x 12) =π(36–72) = -36π< 0 maximaNow, volume of the cylinder so formed =π x^2 y=πx (12)^2x (6)=π(144)^2x 6= 864πcm^3Therefore, the required dimension are 12 cm and 6 cm and the maximum volume is 864πcmThanks
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