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

(N/A) Let $OC = r$ be the radius of the cone and $OA = h$ be its height. Let a cylinder with radius $OE = x$ be inscribed in the given cone. The height $QE$ of the cylinder is given by:
$\frac{QE}{OA} = \frac{EC}{OC}$ (since $\Delta QEC \sim \Delta AOC$)
$\frac{QE}{h} = \frac{r - x}{r}$
$QE = \frac{h(r - x)}{r}$
Let $S$ be the curved surface area of the cylinder. Then:
$S(x) = 2 \pi x \cdot QE = 2 \pi x \cdot \frac{h(r - x)}{r} = \frac{2 \pi h}{r}(rx - x^2)$
To find the maximum surface area,we find the derivative $S'(x)$:
$S'(x) = \frac{2 \pi h}{r}(r - 2x)$
Setting $S'(x) = 0$ gives $r - 2x = 0$,so $x = \frac{r}{2}$.
Now,find the second derivative $S''(x)$:
$S''(x) = \frac{2 \pi h}{r}(-2) = -\frac{4 \pi h}{r}$
Since $S''(x) < 0$ for all $x$,the function $S(x)$ has a maximum at $x = \frac{r}{2}$.
Thus,the radius of the cylinder of greatest curved surface area is half of the radius of the cone.