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Of all the closed right circular cylindricalcans ofvolume 128 Ď cm, find the dimensionsofthe can which has minimum surface area |
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Answer» Given pi r^2h = 128 piSo h = 128 / r^2Total surface area S = 2 pi r^2+ 2 pi r hPlug hSo we get S = 2 pi r^2+ 2 pi * 128 / rDifferentiating w r t r we get dS/dr = 4 pi r - 256 pi / r^2As S is to be minimum then dS/dr has to be 0Plugging 0 we have r^3= 64So r = 16 cm |
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