Let's find out which of the square columns has the largest volume, given that the sum of the edges is given.
I. Solution: More generally, we will show that among the cuboid-shaped bodies, the cube has the largest volume when the sum of the edges meeting at a vertex is given.

Let $a, b, c$ denote the edges meeting at a vertex. According to the condition, their sum is given. The volume of the cuboid is $a b c$. Our statement follows from the fact that the arithmetic mean of three positive numbers (or any number of numbers) is greater than or equal to the geometric mean, with equality only if the three numbers are equal. In our case, the arithmetic mean is $\frac{a+b+c}{3}=s$, which is constant by the condition. $\sqrt[3]{a b c} \leq s$, or $a b c \leq s^{3}$, according to the statement, so the maximum volume is $s^{3}$, and this is only achieved by the cuboid when $a=b=c=s$, i.e., the cube is the sought cuboid.

The inequality used can be proven as follows: if not all three numbers are equal, then there must be one smaller than $s$ and one larger than $s$. Let $a=s-\alpha<0$, $b=s+\beta>0$, and $c=s+\alpha-\beta>0$. Then

\[
\begin{gathered}
a b c=(s-\alpha)(s+\beta)(s+\alpha-\beta)=[s(s-\alpha+\beta)-\alpha \beta](s+\alpha-\beta)< \\
<s(s-\alpha+\beta)(s+\alpha-\beta)=s^{3}-(\alpha-\beta)^{2} s<s^{3}
\end{gathered}
\]

This proves our statement.

II. Solution: Let the edges of the square base of the column be denoted by $x$, and its height by $y$. Then its volume is $K=x^{2} y, y=a-2 x$, where $a$ (the sum of the edges meeting at a vertex) is constant by the condition.

\[
\frac{d K}{d x}=2 x y+x^{2} y^{\prime}=2 x(y-x)
\]

since $y^{\prime}=-2$. Since $x=0$ is not feasible, only $y=x=\frac{a}{3}$, so the cube can be the solution to the problem, and the function indeed has a maximum here.