Question

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

Answer 1

Let $\theta$ be the semi vertical angle, $l$ be the given slant height, then radius of base $=l\sin \theta$ and height$=l\cos \theta$

$V=\frac{1}{3}\pi \left(l\sin \theta {)}^2\right(l\cos \theta ),(V=\frac{1}{3}\pi r^{2}h)$

$=\frac{1}{3}\pi l^3{\sin }^2\theta \cos \theta$

$\frac{dV}{d\theta }=\frac{\pi }{3}{l}^3[2\sin \theta {\cos }^2\theta +{\sin }^2\theta (-\sin \theta \left)\right]$

$=\frac{\pi }{3}{l}^3\sin \theta [-(1-{\cos }^2\theta )+2{\cos }^2\theta ]$

$=\frac{\pi }{3}{l}^3\sin \theta (3{\cos }^2\theta -1)$

For maximum volume, differentiating volume w.r.t. $\theta$,
$\frac{dV}{d\theta }=0$

$\Rightarrow 3{\cos }^2\theta -1=0$
$\Rightarrow \cos \theta =\frac{1}{\sqrt{3}}$
$\Rightarrow \theta ={\cos }^{-1}\left(\frac{1}{\sqrt{3}}\right)$

$\frac{d^{2}V}{d{\theta }^2}=\frac{\pi }{3}{l}^3\left[\cos \theta \right(3{\cos }^2\theta -1)+\sin \theta (6\cos \theta \left)\right(-\sin \theta \left)\right]$

$=\frac{\pi }{3}{l}^3[3{\cos }^3\theta -\cos \theta -6\cos \theta (1-{\cos }^2\theta \left)\right]$

$=\frac{\pi }{3}{l}^3(9{\cos }^3\theta -7\cos \theta )$

$=\frac{\pi }{3}{l}^3\left(\cos \theta \right)(9{\cos }^2\theta -7)$

When $\cos \theta =\frac{1}{\sqrt{3}}$, then
$\frac{d^{2}V}{d{\theta }^2}=\frac{\pi }{3}{l}^3\left(\frac{1}{\sqrt{3}}\right)\times (3-7)<0$

$\therefore$ When $\theta ={\cos }^{-1}\left(\frac{1}{\sqrt{3}}\right)$,given volume is maximum
$\therefore V$ is absolutely maximum for $\theta ={\cos }^{-1}\left(\frac{1}{\sqrt{3}}\right)$.