If the curved surface of right circular cylinder inscribed in a sphere of radius ‘$r$’ is maximum, then the height of the cylinder is $\sqrt{2}r$.

Let $h, R$ be height, radius of cylinder

Let ‘$r$’ be radius of sphere

$\mathrm{△}OAB$ is right angled at ‘$A$’

$r^2={\left(\frac{h}{2}\right)}^2+R^2\Rightarrow R^2=r^2-\frac{h^2}{4}\to \left(1\right)$

C.S.A of cylinder $\left(S\right)=2\pi Rh$

S.O.B.S, $S^2=4{\pi }^2{R}^2{h}^2=4{\pi }^2\left[r^2-\frac{h^2}{4}\right]h^2$

$=4{\pi }^2{r}^2{h}^2-{\pi }^2{h}^4$

Let $f\left(h\right)=4{\pi }^2{r}^2{h}^2-{\pi }^2{h}^4$

Diff w.r.to ‘$h$’

$f^{\mathrm{\prime }}\left(h\right)=4{\pi }^2{r}^2\left(2h\right)-{\pi }^2\left(4h^3\right)=8{\pi }^2{r}^{2}h-4{\pi }^2{h}^3$

Again diff w.r.to ‘$h$’

$f^{\prime \prime }\left(h\right)=8{\pi }^2{r}^2-12{\pi }^2{h}^2$

Consider $f^{\mathrm{\prime }}\left(h\right)=0\text{; }$

$8{\pi }^2{r}^{2}h-4{\pi }^2{h}^3=0;4{\pi }^{2}h\left(2r^2-h^2\right)=0$

$4{\pi }^{2}h=0$          $2r^2-h^2=0$

$h=0$                    $h^2=2r^2\Rightarrow h=\sqrt{2}r$

When $h=\sqrt{2}r$     $f^{\prime \prime }\left(h\right)=8\pi r^2-12{\pi }^2\left(2r^2\right)$

$f^{11}\left(h\right)=8\pi r^2-24{\pi }^2{r}^2=-16{\pi }^2{r}^2<0$

maximum

$\therefore$ it has maximum volume at $h=\sqrt{2}r$