1021. Among all rectangular parallelepipeds having a given volume $\mathrm{V}$, find the one with the smallest total surface area.
Solution. Let $x, y, z$ be the dimensions of the given parallelepiped. Then its volume is $\mathrm{V}=x y z$ (the constraint equation). The total surface area of the parallelepiped is

$$
S=2 x y+2(x+y) z
$$

From the constraint equation, we find $z=\frac{V}{x y}$ and substitute this value into the expression for the function $S$ to be investigated for an extremum:

$$
S=2 x y+2(x+y) \frac{v}{x y}=2\left(x y+\frac{V}{x}+\frac{v}{y}\right)
$$

We find the partial derivatives

$$
\frac{\partial S}{\partial x}=2\left(y-\frac{V}{x^{2}}\right), \frac{\partial S}{\partial y}=2\left(x-\frac{V}{y^{2}}\right)
$$

and form the system of equations

$$
\left\{\begin{array} { l } 
{ y - \frac { V } { x ^ { 2 } } = 0 } \\
{ x - \frac { V } { y ^ { 2 } } = 0 }
\end{array} \quad \text { or } \left\{\begin{array}{l}
x^{2} y-V=0 \\
x y^{2}-V=0
\end{array}\right.\right.
$$

from which $x=y=\sqrt[3]{V}$. Substituting the values of $x$ and $y$ into the constraint equation, we get $z \sqrt[3]{V^{2}}-V$, from which $z=\sqrt[3]{V}$.

Therefore, the sought parallelepiped is a cube.

Investigate the following functions for extrema: