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Show that the height of the cone of maximum volume

Question:

Show that the height of the cone of maximum volume that can be inscribed in a sphere of radius $12 \mathrm{~cm}$ is $16 \mathrm{~cm}$.

Solution:

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

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

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

Squaring both the sides, we get

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

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

Now,

$V=\frac{1}{3} \pi r^{2} h$

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

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

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

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

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

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

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

Now,

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

$\Rightarrow \frac{\pi}{3}(4 R-8 R)=0$

$\Rightarrow \frac{-4 \pi R}{3}<0$

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

$\Rightarrow h=\frac{4 \times 12}{3}=16 \mathrm{~cm}$