7. There is a rectangular iron sheet of size $80 \times 50$. Now, a square of the same size is to be cut from each corner, and then it is to be made into an open box. What should be the side length of the square to be cut so that the volume of this open box is maximized?
7. Let the side length of the cut-out square be $x$, then the volume of the made box is $V=x(80-2 x)(50-2 x)$. Introducing parameters $\alpha, \beta>0$, we have
$$
\begin{aligned}
V & =\frac{1}{\alpha \beta}(\alpha x)[\beta(50-2 x)](80-2 x) \\
& \leqslant \frac{1}{\alpha \beta}\left[\frac{(50 \beta+80)+(\alpha-2 \beta-2) x}{3}\right]^{3} .
\end{aligned}
$$

Let $\alpha-2 \beta-2=0$, and $\alpha x=50 \beta-2 \beta x=80-2 x$, then $\alpha=6, \beta=2, x=10$. Hence $V$ has a maximum value of 18000, at which the side length of the cut-out square should be 10.