Let $\mathrm{r}$ and $\mathrm{h}$ be the radius and height of the cone.
Volume $\mathrm{V}=\frac{1}{3} \pi \mathrm{r}^2 \mathrm{~h}=\frac{\pi \mathrm{k}}{3} \Rightarrow \mathrm{h}=\frac{\mathrm{k}}{\mathrm{r}^2} \quad \ldots(i)$
Surface $\mathrm{S}=\pi \mathrm{r} \ell=\pi \mathrm{r} \sqrt{\mathrm{h}^2+\mathrm{r}^2}$


Put $\mathrm{h}=\frac{\mathrm{k}}{\mathrm{r}^2}$
$\begin{aligned}
&\therefore \mathrm{S}=\frac{\pi \sqrt{\mathrm{k}^2+\mathrm{r}^6}}{\mathrm{r}} \\
&\Rightarrow \frac{\mathrm{dS}}{\mathrm{dr}}=\frac{2 \mathrm{r}^6-\mathrm{k}^2}{\mathrm{r}^2 \sqrt{\mathrm{r}^6+\mathrm{k}^2}}
\end{aligned}$
for minimum curved surface area
$\begin{aligned}
&\Rightarrow \frac{\mathrm{dS}}{\mathrm{dr}}=0 \Rightarrow \mathrm{k}^2=2 \mathrm{r}^6 \Rightarrow \mathrm{r}^3=\frac{\mathrm{k}}{\sqrt{2}} \quad \ldots (ii)\\
&\frac{\mathrm{d}^2 \mathrm{~S}}{\mathrm{dr}^2}=\frac{\mathrm{r}^2 \sqrt{\mathrm{r}^6+\mathrm{k}^2}\left(12 \mathrm{r}^5\right)-\left(2 \mathrm{r}^6-\mathrm{k}^2\right)\left[\frac{\mathrm{r}^2 \times 5 \mathrm{r}^5}{2 \sqrt{\mathrm{r}^6+\mathrm{k}^2}}+2 \mathrm{r} \sqrt{\mathrm{r}^6+\mathrm{k}^2}\right]}{\mathrm{r}^4\left(\mathrm{r}^6+\mathrm{k}^2\right)}
\end{aligned}$
At $\mathrm{r}^3=\frac{\mathrm{k}}{\sqrt{2}}, \frac{\mathrm{d}^2 \mathrm{~S}}{\mathrm{dr}^2}>0$
So it attains minimum curved surface area.