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A closed cylinder hasvolume 2156cm3. What will be the radius of its base so that itstotal surface area is minimum? |
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Answer» Let r be the radius and h the height of the cylinder. Then, `pi r^(2) h = 2156 rArr h = ((2156)/(pir^(2)))`. Let S be its total surface area. Then, `S = (2pi r^(2) + 2 pi r h) = (2pi r^(2) + 2pi r xx (2156)/(pi r^(2))) = (2pi r^(2) + (4312)/(r))` `:. (dS)/(dr) = (4pi r - (4312)/(r^(2))) and (d^(2)S)/(dr^(2)) = (4pi + (8624)/(r^(3)))` Now, `(dS)/(dr) = 0 rArr (4pi r - (4312)/(r^(2))) = 0` `rArr 4 pi r^(3) = 4312 rArr r = ((4312)/(4pi))^(1//3) = ((4312 xx 7)/(4 xx 22))^(1//3) = 7` And, `[(d^(2)S)/(dr^(2))]_(r =7) = (4pi + (8624)/(343)) gt 0 " " :. S` is maximum when r = 7 |
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