PUC Karnataka Science Class 12Department of Pre-University Education, Karnataka
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A given quantity of metal is to be cast into a half cylinder with a rectangular base and semicircular ends. cylinder to the diameter of its semi-circular ends is π : ( π + 2 ) . - PUC Karnataka Science Class 12 - Mathematics

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Question

A given quantity of metal is to be cast into a half cylinder with a rectangular base and semicircular ends. Show that in order that the total surface area may be minimum the ratio of the length of the cylinder to the diameter of its semi-circular ends is \[\pi : (\pi + 2)\] .

Solution

\[\text { Volume },V = \frac{1}{2}\pi l \left( \frac{D}{2} \right)^2 \]

\[ \Rightarrow V = \frac{\pi D^2 l}{8}\]

\[ \Rightarrow l = \frac{8V}{\pi D^2} ..............\left( 1 \right)\]

\[\text { Total surface area } = \frac{\pi D^2}{4} + lD + \frac{\pi Dl}{2}\]

\[ \Rightarrow S = \frac{\pi D^2}{4} + \frac{8V}{\pi D} + \frac{8V}{2D} .............\left[ \text { From eq .} \left( 1 \right) \right]\]

\[ \Rightarrow \frac{dS}{dD} = \frac{\pi D}{2} - \frac{8V}{\pi D^2} - \frac{8V}{2 D^2}\]

\[\text { For maximum or minimum values of S, we must have }\]

\[\frac{dS}{dD} = 0\]

\[ \Rightarrow \frac{\pi D}{2} - \frac{8V}{\pi D^2} - \frac{8V}{2 D^2} = 0\]

\[ \Rightarrow \frac{\pi D}{2} = \frac{8V}{D^2}\left( \frac{1}{\pi} + \frac{1}{2} \right)\]

\[ \Rightarrow D^3 = \frac{16V}{\pi}\left( \frac{1}{\pi} + \frac{1}{2} \right)\]

\[\text { Now, }\]

\[\frac{d^2 S}{d D^2} = \frac{\pi}{2} + \frac{16V}{D^3}\left( \frac{1}{\pi} + \frac{1}{2} \right)\]

\[ \Rightarrow \frac{d^2 S}{d D^2} = \frac{\pi}{2} + \pi > 0\]

\[l = \frac{8V}{\pi D^2}\]

\[ \Rightarrow l = \frac{8}{\pi D^2}\left[ \frac{\pi D^3}{16}\left[ \frac{2\pi}{\pi + 2} \right] \right]\]

\[ \Rightarrow l = D\left( \frac{\pi}{\pi + 2} \right)\]

\[ \Rightarrow \frac{l}{D} = \frac{\pi}{\pi + 2}\]

\[\text { Hence proved}.\]

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Solution A given quantity of metal is to be cast into a half cylinder with a rectangular base and semicircular ends. cylinder to the diameter of its semi-circular ends is π : ( π + 2 ) . Concept: Graph of Maxima and Minima.
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