The height of the cone of maximum volume inscribed in a sphere of radius $R$ is

Question

Aicharya · Accepted Answer

Let the height of the cone $=h$ and the radius of the cone $=r$ Given, radius of the sphere $=R$ Now, In $\triangle O P B$

$\Rightarrow \quad R^2=r^2+(h-R)^2$ $\Rightarrow \quad r^2=R^2-(h-R)^2$ $=(R+h-R)(R-h+R)$ $\Rightarrow \quad r^2=h(2 R-h)$ The volume of the cone is $V=\frac{1}{3} \pi r^2 h$ $\Rightarrow \quad V=\frac{1}{3} \pi h(2 R-h) h$ $\Rightarrow \quad V=\frac{\pi}{3}\left(2 R h^2-h^3\right)$ 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 $\frac{d V}{d h}=0$ $\Rightarrow \quad \frac{\pi}{3}\left(4 R h-3 h^2\right)=0$ $\Rightarrow \quad h(4 R-3 h)=0$ $\Rightarrow \quad h=0, h=\frac{4 R}{3} \quad$ (Not possible) Now, $\quad \frac{d^2 V}{d h^2}=\frac{\pi}{3}(4 R-6 h)$ $\left(\frac{d^2 V}{d h^2}\right)_{\left(\text {at } h=\frac{4 R}{3}\right)}=\frac{\pi}{3}\left(4 R-6 \cdot \frac{4 R}{3}\right)$ $=\frac{\pi}{3}(4 R-8 R)=-\frac{4 \pi}{3} R \Rightarrow$ Negative ie., Maximum Hence, the height of the cone of maximum volume is $\left(\frac{4 R}{3}\right)$.

The height of the cone of maximum volume inscribed in a sphere of radius $R$ is

Solution