10. As shown in the figure, equal squares are cut from the four corners of a square sheet of iron with a side length of $60 \mathrm{~cm}$, and then its edges are folded along the dotted lines to make an open-top box with a square base. What should the side length of the box's base be to maximize the volume of the box? What is the maximum volume?
10. Let the length of the box bottom be $x$, then the height of the box $h=\frac{60-x}{2}$. The volume of the box $V(x)=x^{2} h=\frac{60 x^{2}-x^{3}}{2}(0<x<60)$, $V^{\prime}(x)=60 x-\frac{3}{2} x^{2}$. Let $V^{\prime}(x)=60 x-\frac{3}{2} x^{2}=0$, solving gives $x=0$ (discard), $x=40$, and we find $V(40)=$ $16000\left(\mathrm{~cm}^{3}\right)$. From the problem, we know that when $x$ is too small (close to 0) or too large (close to 60), the volume of the box is very small, therefore, $16000\left(\mathrm{~cm}^{3}\right)$ is the maximum volume.