If the second question is to consider the additional volume of the box due to the thickness of the wall, then here is the solution.
Let a, b, c be the height, width and length of the box, measured from inside. Let d be the thickness of the box wall. The internal volume of the box (or box capacity) is abc, and the volume of the box wall is
2(ab+bc+ca). We wish to minimize this latter quantity. Use the inequality "arithmetic average greater than geometric average", we have
(ab+bc+ca)/3>=f(ab*bc*ca*), where function f denotes taking the 3rd root. (开三次方)。
The equality in the above inequality holds when ab=bc=ca. That corresponds to a=b=c. That is, wall volume is minimized when the box is a cube, in which case, the wall volume is f(V)*f(V)*6. Here V denotes the internal volume of the box.