[Area and Volume (Extremum Problems).]

Consider all possible rectangular parallelepipeds whose bases are squares and each of whose lateral faces has a perimeter of 6. Find among them the parallelepiped with the greatest volume and calculate this volume.
Let's denote the side of the base of the rectangular parallelepiped by $x$. Then its lateral edge is $\frac{1}{2}(6-2 x) = 3 - x$. If $V(x)$ is the volume of the parallelepiped, then

$$
V(x) = x^2 (3 - x),
$$

so the problem reduces to finding the maximum value of the function $V(x) = x^2 (3 - x)$ on the interval $(0 ; 3)$.

![](https://cdn.mathpix.com/cropped/2024_05_06_2c7701213788eaf04894g-33.jpg?height=55&width=306&top_left_y=1133&top_left_x=10)

Let's find the critical points of the function $V(x) = x^2 (3 - x)$ on the interval $(0 ; 3)$. For this, we solve the equation

$$
V'(x) = (3x^2 - x^3)' = 6x - 3x^2 = 3x(2 - x) = 0
$$

The interval $(0 ; 3)$ contains the only root of this equation, $x = 2$. On this interval, $V'(x)$ is positive for $x < 2$ and negative for $x > 2$, so the function $V(x)$ is increasing on the interval $(0 ; 2)$ and decreasing on the interval $(2 ; 3)$. Therefore, $x = 2$ is the point of maximum of the function. Consequently, $V(2) = 4$ is the maximum volume of the parallelepiped.

## Second Method.

Applying the Cauchy inequality for three numbers, we get

$$
\begin{aligned}
V(x) & = x^2 (3 - x) = 4 \cdot \frac{1}{2} x \cdot \frac{1}{2} x \cdot (3 - x) \leq \\
& \leq 4 \cdot \left( \frac{\frac{1}{2} x + \frac{1}{2} x + (3 - x)}{3} \right)^3 = 4
\end{aligned}
$$

with equality if $\frac{1}{2} x = 3 - x$, i.e., when $x = 2$. Therefore, the maximum volume of the parallelepiped is 4.

## Answer

4.00