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If a cylindrical vessel of given volume $V$ with no lid on the top is to be made from a sheet of metal, then the radius $(r)$ and height $(h)$ of the vessel so that the metal sheet used is minimum is

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$r = \sqrt[3]{\frac{V}{π}}, h = \sqrt[3]{\frac{V}{π}}$

$r = \sqrt{\frac{V}{π}}, h = \sqrt{\frac{V}{π}}$

$r = \sqrt{πV}, h = \sqrt{πV}$

$r = \sqrt[3]{\frac{π}{4}}, h = \sqrt[3]{\frac{π}{V}}$

answer is A.

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Detailed Solution

Volume V= ${πr}^{2} h$

$h = \frac{v}{{πr}^{2}}$

Surface area

$S = {πr}^{2} + 2 πrh = {πr}^{2} + 2 πr (\frac{V}{{πr}^{2}}) = {πr}^{2} + \frac{2 V}{r} S^{I} = 2 πr - \frac{2 V}{r^{2}}$

S is min or max

$\Rightarrow S^{I} = 0 2 πr = \frac{2 V}{r^{2}} r^{3} = (\frac{V}{π}) \Rightarrow r = {(\frac{V}{π})}^{1 / 3} h = \frac{V}{{πr}^{2}} = \frac{V}{π {(\frac{V}{π})}^{2 / 3}} = \frac{V^{1 / 3}}{π^{1 / 3}} = {(\frac{V}{π})}^{1 / 3}$

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