We have given,
$V=$ volume of cylindrical vessel
$r=$ radius and
$h=$ height


As we know for cylindrical vessel
$$
v=\pi r^2 h
$$

Let $S$ be the area of metal sheet used to form a cylindrical vessel. Then,
$$
\begin{aligned}
& S=2 \pi r h+\pi r^2=2 \pi r \frac{V}{\pi r^2}+\pi r^2 \quad \text { [using Eq. }(i) \text { ] } \\
& \Rightarrow \quad S(r)=\frac{2 V}{r}+\pi r^2
\end{aligned}
$$
For minimum value of $S(r), S^{\prime}(r)=0$
$$
\begin{aligned}
& \Rightarrow \quad \frac{-2 V}{r}+2 \pi r=0 \Rightarrow \frac{2 V}{r^2}=2 \pi r \\
& \Rightarrow r^3=\frac{V}{\pi} \Rightarrow r=\sqrt[3]{\frac{V}{\pi}}
\end{aligned}
$$
Now, by using the value of $r$ in Eq. ( $i)$
$$
\begin{aligned}
V & =\pi\left(\frac{V}{\pi}\right)^{\frac{2}{3}} \cdot h \\
\Rightarrow \quad & h=\frac{V}{\pi} \times \frac{\pi^{2 / 3}}{v^{2 / 3}} \Rightarrow h=\frac{V^{1 / 3}}{\pi^{1 / 3}}=\sqrt[3]{\frac{V}{\pi}}
\end{aligned}
$$
Hence, the required radius is $r=\sqrt[3]{\frac{V}{\pi}}$ and height is
$$
h=\sqrt[3]{\frac{V}{\pi}}
$$