943. A cylindrical tank must hold $V \Omega$ of water. What should its dimensions be so that the surface area (without the lid) is the smallest?
Solution.

$$
V=\pi x^{2} y \text { (equation of connection), }
$$

where $x$ is the radius of the base of the cylinder, and $y$ is its height (Fig. 72). The surface area

$$
S=\pi x^{2}+2 \pi x y
$$

(function to be minimized). From equation (a), we determine $y=\frac{V}{\pi x^{2}}$. Substituting the value of $y$ into formula (b), we get:

$$
\begin{gathered}
S(x)=\pi x^{2}+\frac{2 V}{x} ; S^{\prime}(x)=2 \pi x-\frac{2 V}{x^{2}}=0 ; x=\sqrt[3]{\frac{V}{\pi}} \\
S^{\prime \prime}(x)=2 \pi+\frac{4 V}{x^{3}}, \quad S^{\prime \prime}\left(\sqrt[3]{\frac{V}{\pi}}\right)>0
\end{gathered}
$$

i.e., at the point $x=\sqrt[3]{\frac{V}{\pi}}$, the function $S(x)$ has a minimum.

Thus, the radius of the cylinder $x=\sqrt[3]{\frac{V}{\pi}}$, the height $y=\frac{V}{\pi x^{2}}=$ $=\frac{V}{\pi \sqrt[3]{\left(\frac{V}{\pi}\right)^{2}}}=\sqrt[3]{\frac{V}{\pi}}$, therefore, $x=y=\sqrt[3]{\frac{V}{\pi}}$.