Went over my head

It's easy to see how box A could be larger than box B. Just cut off a corner say 1/4 the length of the original square and use the same ratio to make box B. If the side of the original piece of tin is1, this will give us

$U = \frac{1}{16}, V =\frac{1}{1024}, U > V$

But the volume of box A can also be small. All we have to do is give it very small base by cutting out corners which are close to 1/2 (say 15/32), then use one of them with more suitable ratio, such as the one above, to make a box B.

$U = \frac{1}{16}\times\frac{1}{16}\times\frac{15}{32}=\frac{15}{8192}\approx0.0018$

$V=(\frac{15}{32})^3\times(\frac{1}{2})^2\times\frac{1}{4}\approx0.0064$

$U<V$

If either can be larger, and there is a continuous series of possibilities, there is bound to be one where U = V.

The fourth option is harder to understand. Value of U can go arbitrarily low as the area of the base goes down to zero. The maximum of V for a given size square is with the corners taking up 1/6 of the length of the side each. (Maximum of a function $f(x)=x\times(1-2x)^2$ where $x$ is the side of the corner and overall side of the square is 1.)