Question
The height of a closed cylinder of given volume and the minimum surface area is :
Answer: Option A
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V = `pi r^2 h and S = 2 pi r h + 2 pi r^2`
`rArr S = 2 pi r ( h + r), where h =( V)/( pi r^2)`
`rArr S = 2 pi r ((V)/(pi r^2) + r) = (2V)/(r) + 2pi r^2`
`rArr (ds)/(dr) = (-2V)/(r^2) + 4 pi r and (d^2S)/(dr^2) = ((4v)/(r^3) + 4pi)` > 0
`:.` S is minimum when `(ds)/(dr) = 0`
`hArr (-2v)/(r^2) + 4 pi r = 0`
`hArr V = 2 pi r^3` `hArr pi r^2 h = 2 pi r^3`
`hArr h = 2r`.
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