JEE Main · 2019 · 08 Apr 2019 · Shift 2 · Application of Derivatives

Inscribe the cylinder in the sphere of radius $3$ and let the half-angle at the centre be $\theta$, so the radius is $r=3\sin \theta$ and the height is $h=2\cdot 3\cos \theta =6\cos \theta$. The volume is $V=\pi r^{2}h=54\pi {\sin }^2\theta \cos \theta$. Setting $\frac{dV}{d\theta }=0$ gives $2\sin \theta {\cos }^2\theta -{\sin }^3\theta =0$, so $2{\cos }^2\theta ={\sin }^2\theta$; with ${\cos }^2\theta +{\sin }^2\theta =1$ this gives ${\sin }^2\theta =\frac{2}{3}$ and ${\cos }^2\theta =\frac{1}{3}$. The height of the largest cylinder is $h=6\cos \theta =\frac{6}{\sqrt{3}}=2\sqrt{3}$.