3. (10 points) From a sheet of metal with an area of $2 a^{2}$, it is required to manufacture a closed box in the shape of a parallelepiped with the maximum volume $V_{m}=\max V$.
1) What are the dimensions of the edges of such a box? The metal can be bent along straight lines and cut, but different parts of the cut sheet cannot be joined (glued or welded).
2) The same question about $\operatorname{maxV}$ and the lengths of the edges for an open box (without a lid) from a rectangular sheet of metal with dimensions $b$ by $c$. Find $\operatorname{maxV}$ given the perimeter of this sheet.
Solution.

Let $x, y, z$ be the edges of the desired rectangular parallelepiped, then $V = xyz$, and the surface area of the parallelepiped is $S = 2(xy + xz + yz) = 2a^2$. We need to find $\max V = \max (xyz)$ under the condition $xy + xz + yz = a^2 = \text{const}$. However, $\max V^2 = \max (xyz)^2 = \max [(xy)(xz)(yz)]$ under the condition that $xy + xz + yz = a^2$. Then we have the system $\{\max [(xy)(xz)(yz)] = ?$

$\left\{xy + xz + yz = a^2\right.$

By the theorem of means (arithmetic and geometric) for three numbers, $\max V^2 = \max (xyz)^2 = \max [(xy)(xz)(yz)]$ is achieved under the condition $xy = xz = yz = \frac{a^2}{3}$, $x = y = z = \sqrt{\frac{a^2}{3}} = \frac{a \sqrt{3}}{3}$.

However, such a cube can only be folded if the original sheet of metal has the shape of a cube net in the form of a "cross" $4x3$ or the letter T.

If $h = x$ is the height of the open box, then the area of the base (see figure) is calculated by the formula $S = (b - 2x)(c - 2x) = 4x^2 - 2(b + c)x + bc$. Therefore, its volume $V = S \cdot h = x[4x^2 - 2(b + c)x + bc]$, and the necessary condition for $\max V$ is obtained from the derivative $V' = \frac{dV}{dx} = 12x^2 - 4(b + c)x + bc = 0$.

The critical points are obtained from the equation (1):

$x_{1,2} = \frac{b + c \pm \sqrt{b^2 + c^2 - bc}}{6}$ and, justifying the choice of the sign,

at $h = \frac{b + c - \sqrt{b^2 + c^2 - bc}}{6}$, $\max V$ of the open box will be achieved.

The lengths of the edges are $h, b - 2h, c - 2h$.

If the perimeter of the original sheet of metal $2(b + c) = \text{const}$, then

$h = \frac{b + c - \sqrt{b^2 + c^2 - bc}}{6} \leq \frac{b + c - \sqrt{bc}}{6} = \frac{b}{6}$.

The last transition is valid by the theorem of means:

![](https://cdn.mathpix.com/cropped/2024_06_04_c790142104ba3f7af959g-34.jpg?height=622&width=345&top_left_y=240&top_left_x=1552)

$\frac{b^2 + c^2}{2} \geq \sqrt{bc} \Leftrightarrow b^2 + c^2 \geq 2bc$, and the equality sign holds when $b^2 = c^2 \Leftrightarrow b = c$.

Therefore, we obtain the lengths of the edges of the box:

$\left\{\begin{array}{c}h = \frac{b}{6} \\ b - 2h = \frac{2b}{3} \\ c - 2h = \frac{2b}{3}\end{array} \Leftrightarrow V_{\text{max}} = \frac{2b^3}{27}\right.$

Answer: 1) $\frac{a \sqrt{3}}{3}$. 2) $h = \frac{b + c - \sqrt{b^2 + c^2 - bc}}{6}$; $b - 2h$; $c - 2h$; $V_{\text{max}} = \frac{2b^3}{27}$.