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# Solution for Prove that a Conical Tent of Given Capacity Will Require the Least Amount of Canavas When the Height is √ 2 Times the Radius of the Base. - CBSE (Commerce) Class 12 - Mathematics

#### 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

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

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

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

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

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

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

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

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

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

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

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

$\Rightarrow 2 \pi^2 r^6 = 9 V^6$

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

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

$\Rightarrow 2 \pi^2 r^6 = \pi^2 r^4 h^2$

$\Rightarrow 2 r^2 = h^2$

$\therefore h = \sqrt{2} r$

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Solution Prove that a Conical Tent of Given Capacity Will Require the Least Amount of Canavas When the Height is √ 2 Times the Radius of the Base. Concept: Graph of Maxima and Minima.
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