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Show that semi-verticle angle of a cone of maximum volume and given slant height is $\cos^{-1}\left ( \frac{1}{\sqrt{3}} \right )$ .

Answers (1)

Given slant height AB $l=\sqrt{r^2+h^2}$

$\Rightarrow l^2-h^2=r^2\; \; -(i)$

Volume of the cone, $V=\frac{1}{3}\pi r^2h$

$\Rightarrow V=\frac{1}{3}\pi \left [ l^2-h^2 \right ]h$ [by using (i)]

$\Rightarrow V=\frac{1}{3}\pi \left [ l^2h-h^3 \right ]$

On diff. wrt h both sides, $\frac{\mathrm{d} V}{\mathrm{d} h}=\frac{1}{3}\pi \left [ l^2-3h^2 \right ]$

Again diff wrt h both sides, $\frac{\mathrm{d}^2 V}{\mathrm{d} h^2} =-2\pi h$

For local points of maxima and minima

$\frac{\mathrm{d} V}{\mathrm{d} h}=0$

$\Rightarrow \frac{1}{3}\pi \left [ l^2 -3h^2\right ] =0\Rightarrow h=\frac{l}{\sqrt{3}}\Rightarrow \frac{d^2V}{dh^2}=-2\pi \left ( \frac{l}{\sqrt{3}} \right )<0$

$\therefore \left [ \frac{\mathrm{d} V}{\mathrm{d} h^2} \right ]_{at \; h=\frac{l}{3}}=-2\pi \left [ \frac{l}{\sqrt{3}} \right ]<0$

So V is maximum at $h=\frac{l}{\sqrt{3}}$

Now in $\triangle AOB, \cos \theta =\frac{h}{l}=\frac{1}{\sqrt{3}},$

where $\theta$ is the semi verticle angle of cone.

$\therefore \theta =\cos^{-1}\left ( \frac{1}{\sqrt{3}} \right )$

Posted by

Ravindra Pindel

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