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Q.
The height of the cone of maximum volume inscribed in a sphere of radius $R$ is
EAMCETEAMCET 2010
Solution:
Let the height of the cone $=h$
and the radius of the cone $=r$
Given, radius of the sphere $=R$
Now, In $\Delta O P B$
$\Rightarrow R^{2}=r^{2}+(h-R)^{2}$
$\Rightarrow r^{2}=R^{2}-(h-R)^{2}$
$=(R +h -R)(R- h+ R)$
$\Rightarrow r^{2}=h(2 R-h)$
The volume of the cone is
$V=\frac{1}{3} \pi r^{2} h$
$\Rightarrow V=\frac{1}{3} \pi h(2 R-h) h$
$\Rightarrow V=\frac{\pi}{3}\left(2 R h^{2}-h^{3}\right)$
Differentiating with $r$ to $h$
$\frac{d V}{d h}=\frac{\pi}{3}\left(4 R h-3 h^{2}\right)$
For maximum or minimum value of volume
$\frac{d V}{d h}=0$
$\Rightarrow \frac{\pi}{3}\left(4 R h-3 h^{2}\right)=0$
$\Rightarrow h(4 R-3 h)=0$
$\Rightarrow h=0, h=\frac{4 R}{3}$ (Not possible)
Now, $\frac{d^{2} V}{d h^{2}}=\frac{\pi}{3}(4 R-6 h)$
$\left(\frac{d^{2} V}{d h^{2}}\right)_{\left(\text {at } h=\frac{4 R}{3}\right)}= \frac{\pi}{3}\left(4 R-6 \cdot \frac{4 R}{3}\right)$
$=\frac{\pi}{3}(4 R-8 R)=-\frac{4 \pi}{3} R \Rightarrow $ Negative
ie., Maximum Hence, the height of the cone of maximum volume is
$\left(\frac{4 R}{3}\right)$.
