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Prove that the Semi-vertical Angle of the Right Circular Cone of Given Volume and Least Curved Surface is Cot − 1 ( √ 2 ) . - CBSE (Science) Class 12 - Mathematics

Question

Prove that the semi-vertical angle of the right circular cone of given volume and least curved surface is $\cot^{- 1} \left( \sqrt{2} \right)$ .

Solution

Let:
Radius of the base = r,
Height = h,
Slant height = l,
Volume = V,
Curved surface area = C

$\text { As, Volume }, V = \frac{1}{3}\pi r^2 h$

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

$\text { Also, the slant height, l } = \sqrt{h^2 + r^2}$

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

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

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

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

$\text { Now, }$

$\text { CSA, C } = \pi rl$

$\Rightarrow C\left( r \right) = \pi r\frac{\sqrt{9 V^2 + \pi^2 r^6}}{\pi r^2}$

$\Rightarrow C\left( r \right) = \frac{\sqrt{9 V^2 + \pi^2 r^6}}{r}$

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

$= \frac{\left[ \frac{3 \pi^2 r^6 - \left( 9 V^2 + \pi^2 r^6 \right)}{\sqrt{9 V^2 + \pi^2 r^6}} \right]}{r^2}$

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

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

$\text { For maxima or minima, C }'\left( r \right) = 0$

$\Rightarrow \frac{2 \pi^2 r^6 - 9 V^2}{r^2 \sqrt{9 V^2 + \pi^2 r^6}} = 0$

$\Rightarrow 2 \pi^2 r^6 - 9 V^2 = 0$

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

$\Rightarrow V^2 = \frac{2 \pi^2 r^6}{9}$

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

$\Rightarrow V = \frac{\pi r^3 \sqrt{2}}{3} or \ r = \left( \frac{3V}{\pi\sqrt{2}} \right)^\frac{1}{3}$

$\text { So,} h = \frac{3}{\pi r^2} \times \frac{\pi r^3 \sqrt{2}}{3}$

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

$\Rightarrow \frac{h}{r} = \sqrt{2}$

$\Rightarrow \cot\theta = \sqrt{2}$

$\therefore \theta = \cot^{- 1} \left( \sqrt{2} \right)$

$\text { Also },$

$\text { Since, for } r < \left( \frac{3V}{\pi\sqrt{2}} \right)^\frac{1}{3} , C'\left( r \right) < 0 \text { and for } r > \left( \frac{3V}{\pi\sqrt{2}} \right)^\frac{1}{3} , C'\left( r \right) > 0$

$\text { So, the curved surface for r }= \left( \frac{3V}{\pi\sqrt{2}} \right)^\frac{1}{3} or V = \frac{\pi r^3 \sqrt{2}}{3}\text { is the least } .$

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Solution Prove that the Semi-vertical Angle of the Right Circular Cone of Given Volume and Least Curved Surface is Cot − 1 ( √ 2 ) . Concept: Graph of Maxima and Minima.
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