Volume of a container with unknown side and known surface area We have to make a closed container.  All side surfaces (walls, ceiling and bottom) must be rectangles and stand perpendicular to each other. One of the sides should be $3 \,m$ long.  The surface area of ​​the container shall be $32 \,m^2$. Let one of the other two sides be $x$ meters long. Show that the volume is given by the formula: $$V = \frac{48x-9x^2}{x + 3}$$

Unless we let $x=y$ the formula does not have enough information to identify volume.  If, however, we let $x=y$, then we can show that the formula works, with the restriction that $x=-8$ or $x=2$.
$$
2(3\cdot x)+2(3\cdot y) + 2(x\cdot y) = 32 \\
2 x y + 6 x + 6 y = 32\\
x = \dfrac{16 - 3 y}{y + 3},
y=x\implies x,y\in\big\{-8,2\big\}\\
$$
\begin{align*}
x=y\in\big\{-8,2\big\}\implies V=3\cdot (-8)^2=192\\
V = 3\cdot (2)^2=12
\end{align*}
$$
V = \frac{48x-9x^2}{x + 3}, x=-8 \implies V=192\\
V = \frac{48x-9x^2}{x + 3}, x=2 \implies V=12
$$
