Name: Local Maxima and Minima of a Function
Uploaded: 2019-07-19
Duration: 14 min 17 s
Description: Do you know what is minima or maxima of a function? This video also explains how to find local maxima and minima by differentiation in a detailed manner.

Question

Prove that the semi-vertical angle of the right circular cone of given volume and least curved surface is

\cot^{- 1} (\sqrt{2})

.

Accepted Answer

Let:

Radius of the base $= r$ ,

Height $= h$ ,

Slant height $= l$ ,

Volume $= V$ ,

Curved surface area $= C$

As, Volume, $V = \frac{1}{3} {πr}^{2} h$

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

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

$= \sqrt{{(\frac{3 V}{{πr}^{2}})}^{2} + r^{2}}$

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

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

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

Now,

CSA, $C = πrl$

$\Rightarrow C (r) = πr \frac{\sqrt{9 V^{2} + π^{2} r^{6}}}{{πr}^{2}}$

$\Rightarrow C (r) = \frac{\sqrt{9 V^{2} + π^{2} τ^{5}}}{r}$

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

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

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

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

For maxima or minima, $C^{'} (r) = 0$

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

$\Rightarrow 2 π^{2} r^{6} - 9 V^{2} = 0$

$\Rightarrow 2 π^{2} r^{6} = 9 V^{2}$

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

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

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

So, $h = \frac{3}{{πr}^{2}} \times \frac{{πr}^{3} \sqrt{2}}{3}$

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

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

$\Rightarrow cotθ = \sqrt{2}$

$∴ θ = \cot^{- 1} (\sqrt{2})$

Also,

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

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

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

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Prove that the semi-vertical angle of the right circular cone of given volume and least curved surface is cot-12 .

Important Questions on Maxima and Minima

Learn and Prepare for any exam you want !

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