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

For a given curved surface of a right circular cone when the volume is maximum, prove that the semi vertical angle is

\sin^{- 1} \frac{1}{\sqrt{3}}

.

Accepted Answer

Let the radius of the cone be $r$ , height be $h$ and slant height be $l$ , then

$C =$ Curved surface Area $= π r l$

$\Rightarrow l = \frac{C}{π r} \Rightarrow l^{2} = \frac{C^{2}}{π^{2} r^{2}} . . . . (i)$

And, volume of cone is

$V = \frac{1}{3} π r^{2} h$

$\Rightarrow V = \frac{1}{3} π r^{2} \sqrt{l^{2} - r^{2}}$

$[∵ r^{2} + h^{2} = l^{2}]$

$\Rightarrow V = \frac{1}{3} π r^{2} \sqrt{\frac{C^{2}}{π^{2} r^{2}} - r^{2}}$

$\Rightarrow V = \frac{1}{3} π r^{2} \sqrt{\frac{C^{2} - π^{2} r^{4}}{π^{2} r^{2}}}$

$\Rightarrow V = \frac{1}{3} r \sqrt{C^{2} - π^{2} r^{4}}$

Differentiating both sides w.r.t. $r$ , we get

$\frac{d V}{d r} = \frac{1}{3} [1 \cdot \sqrt{C^{2} - π^{2} r^{4}} + r (\frac{- 4 π^{2} r^{3}}{2 \sqrt{C^{2} - π^{2} r^{4}}})]$

$\Rightarrow \frac{d V}{d r} = \frac{1}{3} [\sqrt{C^{2} - π^{2} r^{4}} - \frac{2 π^{2} r^{4}}{\sqrt{C^{2} - π^{2} r^{4}}}]$

$\Rightarrow \frac{d V}{d r} = \frac{1}{3} [\frac{C^{2} - π^{2} r^{4} - 2 π^{2} r^{4}}{\sqrt{C^{2} - π^{2} r^{4}}}]$

$\Rightarrow \frac{d V}{d r} = \frac{1}{3} [\frac{C^{2} - 3 π^{2} r^{4}}{\sqrt{C^{2} - π^{2} r^{4}}}]$

For maximum volume,

$\frac{d V}{d r} = 0$

$\Rightarrow \frac{1}{3} [\frac{C^{2} - 3 π^{2} r^{4}}{\sqrt{C^{2} - π^{2} r^{4}}}] = 0$

$\Rightarrow C^{2} - 3 π^{2} r^{4} = 0$

$\Rightarrow r^{4} = \frac{C^{2}}{3 π^{2}}$

$\Rightarrow r^{2} = \frac{C}{\sqrt{3} π}$

By $(i)$ , $l^{2} = \frac{C^{2}}{π^{2} r^{2}} \Rightarrow l^{2} = \frac{\sqrt{3} C}{π}$

And,

$\frac{d^{2} V}{d r^{2}} = \frac{1}{3} (\frac{- 12 π^{2} r^{3} \sqrt{C^{2} - π^{2} r^{4}} - (C^{2} - 3 π^{2} r^{4}) (\frac{- 4 π^{2} r^{3}}{2 \sqrt{C^{2} - π^{2} r^{4}}})}{C^{2} - π^{2} r^{4}})$

$\Rightarrow \frac{d^{2} V}{d r^{2}} = \frac{1}{3} (\frac{- 12 π^{2} r^{3} (C^{2} - π^{2} r^{4}) + 2 π^{2} r^{3} (C^{2} - 3 π^{2} r^{4})}{{(C^{2} - π^{2} r^{4})}^{\frac{3}{2}}})$

$\Rightarrow \frac{d^{2} V}{d r^{2}} = \frac{1}{3} (\frac{- 12 π^{2} r^{3} C^{2} + 12 π^{4} r^{7} + 2 π^{2} r^{3} C^{2} - 6 π^{4} r^{7}}{{(C^{2} - π^{2} r^{4})}^{\frac{3}{2}}})$

$\Rightarrow \frac{d^{2} V}{d r^{2}} = \frac{1}{3} (\frac{- 10 π^{2} r^{3} C^{2} + 6 π^{4} r^{7}}{{(C^{2} - π^{2} r^{4})}^{\frac{3}{2}}})$

$\Rightarrow \frac{d^{2} V}{d r^{2}} = \frac{1}{3} π^{2} r^{3} (\frac{- 10 C^{2} + 6 π^{2} r^{4}}{{(C^{2} - π^{2} r^{4})}^{\frac{3}{2}}})$

$\Rightarrow {(\frac{d^{2} V}{d r^{2}})}_{r^{2} = \frac{C}{π \sqrt{3}}} = \frac{1}{3} π^{2} {(\frac{C}{\sqrt{3} π})}^{\frac{3}{2}} (\frac{- 10 C^{2} + \frac{6 C^{2}}{3}}{{(C^{2} - \frac{C^{2}}{3})}^{\frac{3}{2}}})$

$\Rightarrow {(\frac{d^{2} V}{d r^{2}})}_{r^{2} = \frac{C}{π \sqrt{3}}} = \frac{1}{3} π^{2} {(\frac{C}{\sqrt{3} π})}^{\frac{3}{2}} (\frac{- 24 C^{2}}{3 {(\frac{2 C^{2}}{3})}^{\frac{3}{2}}}) < 0$

Hence, volume is maximum.

Now,

$\sin^{2} θ = \frac{r^{2}}{l^{2}} = \frac{C}{\sqrt{3} π} \times \frac{π}{C \sqrt{3}}$

$\Rightarrow \sin^{2} θ = \frac{1}{3}$

$\Rightarrow θ = \sin^{- 1} (\frac{1}{\sqrt{3}})$

For a given curved surface of a right circular cone when the volume is maximum, prove that the semi vertical angle is $\sin^{- 1} \frac{1}{\sqrt{3}}$ .

Important Questions on Application of Derivatives

Learn and Prepare for any exam you want !

For a given curved surface of a right circular cone when the volume is maximum, prove that the semi vertical angle is sin-113.

Important Questions on Application of Derivatives

Learn and Prepare for any exam you want !

For a given curved surface of a right circular cone when the volume is maximum, prove that the semi vertical angle is $\sin^{- 1} \frac{1}{\sqrt{3}}$ .