Mathematics

This is an optimization problem where you need to express volume as a function of one variable, then find its maximum. When you cut squares of side $x$ from each corner and fold up, the base of the box becomes $(30-2x)$ by $(30-2x)$ and the height is $x$—so volume is $V(x) = x(30-2x)^2$. The key insight is that $x$ is constrained (it can't be too large or the base disappears), and the maximum occurs where $\frac{dV}{dx} = 0$, not at the boundary.

This is an optimization problem where you need to express volume as a function of one variable, then find its maximum. When you cut squares of side $x$ from each corner and fold up, the base of the box becomes $(30-2x)$ by $(30-2x)$ and the height is $x$—so volume is $V(x) = x(30-2x)^2$. The key insight is that $x$ is constrained (it can't be too large or the base disappears), and the maximum occurs where $\frac{dV}{dx} = 0$, not at the boundary.