Prove that the radius right circular cylinder of greatest curved surface area which can be inscribed in a given cone is half of that of the cone.

Le $h, r$ be height, radius of cone, let ‘$x$’ be radius of cylinder. Let ‘$u$’ height of cylinder

$\mathrm{△}AOC\approx \mathrm{△}QEC$

$\frac{QE}{OA}=\frac{EC}{OC}\Rightarrow \frac{u}{h}=\frac{r-x}{r}\Rightarrow u=\frac{h(r-x)}{r}$

C.S.A of cylinder $S=2\pi rh=2\pi xu$

$=2\pi x\frac{h(r-x)}{r}=\frac{2\pi h}{r}\left[rx-x^2\right]$

Let $f\left(x\right)=\frac{2\pi h}{r}\left(rx-x^2\right);f^{\mathrm{\prime }}\left(x\right)=\frac{2\pi h}{r}(r-2x)$

$\Rightarrow f^{\prime \prime }\left(x\right)=\frac{2\pi h}{r}(-2)=-\frac{4\pi h}{r}<0(max)$

Clearly it has greatest C.S.A

Consider $f^{\mathrm{\prime }}\left(x\right)=0;\frac{2\pi h(r-2x)}{r}=0$

$\Rightarrow r-2x=0\Rightarrow x=\frac{r}{2}$

$\therefore$radius of cylinder of greatest C.S.A which can be inscribed in a given cone is $r/2$