Question

Show that the altitude of the right circular cone of maximum volume that can be inscribed in a sphere of radius $r$ is $\frac{4r}{3}$. Also find maximum volume in terms of volume of the sphere
OR
Find the intervals in which $f\left(x\right)=\sin 3x-\cos 3x,0<x<\pi$, is strictly increasing or strictly decreasing

Answer 1

We imagine a cone such that its boundaries touch the sphere.

Let $\theta$ be the angle made by the slant height of the cone with the height of the cone.

Thus, the central angle becomes $2\theta$ by the property in a circle.

We therefore get the height of the cone to be $R+Rcos2\theta$

Similarly, the radius of the cone becomes $R\sin 2\theta$

The volume of the cone is given by $\pi (Radius{)}^2\times Height$

$\therefore V=\pi (R\sin 2\theta {)}^2\times (R+R\cos 2\theta )$

$=\pi R^3\times {\sin }^{2}2\theta (1+\cos 2\theta )$

Differentiating, we get $4\sin 2\theta \cos 2\theta (1+\cos 2\theta )-2{\sin }^{3}2\theta =0$

$\Rightarrow sin2\theta \times [2\cos 2\theta +2{\cos }^{2}2\theta -1+{\cos }^{2}2\theta ]=0$

$\Rightarrow \sin 2\theta \times [2\cos 2\theta +3{\cos }^{2}2\theta -1]=0$

$\Rightarrow \sin 2\theta \times (\cos 2\theta +1)\times (3\cos 2\theta -1)=0$

$\Rightarrow \sin 2\theta =0$ or $\cos 2\theta =-1$ or $\cos 2\theta =\frac{1}{3}$

For which $V=0$ or $V=0$ or $V=\frac{32}{27}\pi R^3$

Thus, we get the maximum volume when altitude is $\frac{4R}{3}$

Volume of sphere is $\frac{4\pi }{3}{R}^3$, which implies the volume of cone is $\frac{8}{9}$ times that of sphere.