Question

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

Answer 1

Let:

Radius of the base = r,

Height = h,

Slant height = l,

Volume = V,

Curved surface area = C

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

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

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

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

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

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

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

\[\text { Now, }\]

\[\text { CSA, C } = \pi rl\]

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

\[ \Rightarrow \frac{h}{r} = \sqrt{2}\]

\[ \Rightarrow \cot\theta = \sqrt{2}\]

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

\[\text { Also }, \]

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

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

Answer 2

Volume V = `1/3πr^2h`   ...(1)

Also l² = h² + r²   ...(2)

∵ Volume V is given, so V is constant.

Curved area of cone

c = πrl = `πrsqrt(h^2 + r^2)`   ...[By (2)]

c² = π²r²(h² + r²)

c² = `π^2r^2 {(9V^2)/(π^2r^2) + r^2}`   ...`{By (1), h = (3V)/(πr^2)}`

c² = `(9V^2)/r^2 + π^2r^4`

Differentiate w.r. to ‘r’

`d/(dr) (c^2) = (-18V^2)/r^3 + 4π^2r^3`   ...(3)

and `(d^2(c^2))/(dr^2) = (54V^2)/r^2 + 12π^2r^2`  ...(4)

For least curved area `(d(c^2))/(dr)` = 0

`\implies (-18V^2)/r^3 + 4π^2r^3` = 0

`\implies` 9V² = 2π²r⁶

`\implies` `9 xx 1/9 π^2r^4h^2` = 2π²r⁶ 

`\implies` h² = 2r²

h = `sqrt(2)r`

Also by (4) `(d^2(c^2))/(dr^2) > 0`

∴ c² is minimum

`\implies` c is also minimum.

Hence c is least when h = `sqrt(2)r`

`\implies h/r = sqrt(2)`

`\implies` cot α = `sqrt(2)`

α = `cot^-1 (sqrt(2))`