#### Question

Prove that a conical tent of given capacity will require the least amount of canavas when the height is \[\sqrt{2}\] times the radius of the base.

#### Solution

\[\text { Let the surface area of conical tent be S } = \pi r\sqrt{r^2 + h^2}\]

\[\text { Let the volume of the conical tent } V = \frac{1}{3} \pi r^2 h\]

\[ \Rightarrow h = \frac{3V}{\pi r^2}\]

\[ \therefore S = \pi r\sqrt{r^2 + \left( \frac{3V}{\pi r^2} \right)^2}\]

\[ \Rightarrow S = \frac{1}{r}\sqrt{\pi^2 r^6 + 9 V^6}\]

\[\text { Now differentiating with respect to r we get, }\]

\[\frac{dS}{dr} = \frac{d}{dr}\left[ \frac{1}{r}\sqrt{\pi^2 r^6 + 9 V^6} \right]\]

\[ = \frac{1}{r}\frac{6 \pi^2 r^5}{2\left( \sqrt{\pi^2 r^6 + 9 V^6} \right)} - \frac{\sqrt{\pi^2 r^6 + 9 V^6}}{r^2}\]

\[\text { For minima putting }\frac{dS}{dr} = 0 \text { we get, }\]

\[\frac{3 \pi^2 r^4}{\sqrt{\pi^2 r^6 + 9 V^6}} = \frac{\sqrt{\pi^2 r^6 + 9 V^6}}{r^2}\]

\[ \Rightarrow 3 \pi^2 r^6 = \pi^2 r^6 + 9 V^6 \]

\[ \Rightarrow 2 \pi^2 r^6 = 9 V^6 \]

\[\text { Substitutting the value of V we get }, \]

\[2 \pi^2 r^6 = 9 \left( \frac{1}{3} \pi r^2 h \right)^2 \]

\[ \Rightarrow 2 \pi^2 r^6 = \pi^2 r^4 h^2 \]

\[ \Rightarrow 2 r^2 = h^2 \]

\[ \therefore h = \sqrt{2} r\]