A right circular cone, having maximum volume is inscribed in a sphere of radius $3 \mathrm{cm},$ then the curved surface area $($in ${\mathrm{cm}}^2)$ of this cone is

$$\begin{gathered} 9={\left(\mathrm{h}-3\right)}^2+{\mathrm{r}}^2 \\ \Rightarrow {\mathrm{r}}^2=6\mathrm{h}-{\mathrm{h}}^2 \end{gathered}$$

Volume $$\begin{gathered} \mathrm{V}=1/3{\mathrm{\pi r}}^2\mathrm{h} \\ =1/3\mathrm{\pi }\left(6\mathrm{h}-{\mathrm{h}}^2\right)\mathrm{h} \\ =\mathrm{\pi }/3\left(6{\mathrm{h}}^2-{\mathrm{h}}^3\right) \\ {\mathrm{V}}^{\mathrm{I}}=\mathrm{\pi }/3\left(12\mathrm{h}-3{\mathrm{h}}^2\right) \end{gathered}$$

V is Min or max

$$\begin{gathered} \Rightarrow {\mathrm{V}}^{\mathrm{I}}=0 \\ 12\mathrm{h}=3{\mathrm{h}}^2\Rightarrow \mathrm{h}=4 \\ {\mathrm{r}}^2=6\mathrm{h}-{\mathrm{h}}^2=24-16=8 \end{gathered}$$

Surface area

$$\begin{gathered} \mathrm{S}=\mathrm{\pi rl} \\ =\mathrm{\pi r}\sqrt{{\mathrm{r}}^2+{\mathrm{h}}^2} \\ =\mathrm{\pi }\left(2\sqrt{2}\right)\sqrt{8+16} \\ =\mathrm{\pi }2\sqrt{2} 2\sqrt{6} \\ =8\mathrm{\pi }\sqrt{3} \end{gathered}$$