Show that the cone of the greatest volume


Show that the cone of the greatest volume which can be inscribed in a given spher has an altitude equal to $2 / 3$ of the diameter of the sphere.


Let $h, r$ and $R$ be the height, radius of base of the cone and radius of the sphere, respectively. Then,



$\Rightarrow h^{2}+R^{2}-2 h r=R^{2}-r^{2}$

$\Rightarrow r^{2}=2 h R-h^{2}$         ....(1)

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

$\Rightarrow V=\frac{1}{3} \pi h\left(2 h R-h^{2}\right)$         [From eq. (1)]

$\Rightarrow V=\frac{1}{3} \pi\left(2 h^{2} R-h^{3}\right)$

$\Rightarrow \frac{d V}{d h}=\frac{\pi}{3}\left(4 h R-3 h^{2}\right)$

For maximum or minimum values of $V$, we must have

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

$\Rightarrow \frac{\pi}{3}\left(4 h R-3 h^{2}\right)=0$

$\Rightarrow 4 h R=3 h^{2}$

$\Rightarrow h=\frac{4 R}{3}$

Substituting the value of $y$ in eq. (1), we get


$\Rightarrow x^{2}=4\left(r^{2}-\frac{r^{2}}{2}\right)$

$\Rightarrow x^{2}=4\left(\frac{r^{2}}{2}\right)$

$\Rightarrow x^{2}=2 r^{2}$

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


$\frac{d^{2} V}{d h^{2}}=\frac{\pi}{3}(4 R-6 h)$

$\Rightarrow \frac{d^{2} V}{d h^{2}}=\frac{\pi}{3}\left(4 R-6 \times \frac{4 R}{3}\right)$

$\Rightarrow \frac{d^{2} V}{d h^{2}}=\frac{-4 \pi R}{3}<0$

So, the volume is maximum when $h=\frac{4 R}{3} .$

$\Rightarrow h=\frac{2}{3}$ (Diameter of sphere)

Hence proved.

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