Practice | Hashtable

1. The height of a closed cylinder of given volume and the minimum surface area is :

A. Equal to its diameter

B. Half of its diameter

C. Double of its diameter

D. None of these

Solution:

\begin{aligned} V = π r^{2} h and \\ S = 2 π r h + 2 π r^{2} \\ = 2 π r (h + r) \\ Where, h = \frac{V}{π r^{2}} \\ \Rightarrow S = 2 π r (\frac{V}{π r^{2}} + r) \\ \Rightarrow S = \frac{2 V}{r} + 2 π r^{2} \\ \Rightarrow \frac{d S}{d r} = \frac{- 2 V}{r^{2}} + 4 π r and \\ \frac{d^{2} S}{d r^{2}} = (\frac{4 V}{r^{3}} + 4 π) > 0 \end{aligned}

∴ S is minimum when :

\begin{aligned} \frac{d S}{d r} = 0 \\ \Rightarrow \frac{- 2 V}{r^{2}} + 4 π r = 0 \\ \Rightarrow V = 2 π r^{3} \\ \Rightarrow π r^{2} h = 2 π r^{3} \\ \Rightarrow h = 2 r \end{aligned}

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