$\left[\begin{array}{l}{[\text { Problems on maximum and minimum (miscellaneous). }} \\ {[\quad \underline{\text { Rectangular parallelepipeds }}]}\end{array}\right]$

Consider all possible rectangular parallelepipeds, each with a volume of 4, and whose bases are squares. Find among them the parallelepiped with the smallest perimeter of the lateral face and calculate this perimeter.
Let $x$ be the side of the base of a rectangular parallelepiped. Then its lateral edge is $\frac{4}{x^{2}}$. If $P(x)$ is the perimeter of the lateral face of the parallelepiped, then

$$
P(x)=\frac{B}{x^{2}}+2 x
$$

Thus, the problem reduces to finding the minimum value of the function $P(x)=\frac{8}{x^{2}}+2 x$ on the ray $(0 ;+\infty)$.

## First Method.

We find the critical points of the function $P(x)=\frac{8}{x^{2}}+2 x$ on the ray $(0 ;+\infty)$.

$$
P^{\prime}(x)=-\frac{16}{x^{3}}+2=\frac{2\left(-x^{3}\right)}{x^{3}}=0
$$

The ray $(0 ;+\infty)$ contains the unique root of this equation $x=2$. On this ray, when $x>2$, the derivative is positive, so the function $P(x)$ decreases on the interval $(0 ; 2)$ and increases on the interval $(2; +\infty)$. Therefore, $x=2$ is a point of minimum of the function. Consequently, $P(2)=6$ is the minimum value of the perimeter of the lateral face of the parallelepiped.

## Second Method.

Applying the Cauchy inequality to three numbers, we get that

$$
P(x)=\frac{8}{x^{2}}+2 x=\frac{8}{x^{2}}+x+x \geqslant 3 \sqrt[3]{\frac{8}{x^{2}} \cdot x \cdot x}=6
$$

with equality if $\frac{8}{x^{2}}=x$, i.e., when $x=2$. Therefore, the minimum value of the perimeter of the lateral face of the parallelepiped is 6.

## Answer

The parallelepiped, whose base side is 2, has a lateral edge of 1; the sought perimeter is 6.