Ok. I have a solution. My solution finds the largest box (maximum value) with a known surface area x. the problem is, im pretty sure there isnt a way to cut the 6 pieces needed for the box out of a square. I didnt really understand the conditions for making the square- can you cut it to small pieces and then put some of them together? do you even need to be able to cut the pieces needed for the box, or can you assume you can shape the surface area into any shape you want?
So, my solution assumes we can shape the square (our surface area) into any shape we want. suppose the area is 'x'.
so, our box has 12 edges- lets call those a, b and c (4 that are 'a' centimeters/meters/whatever long, 4 that are 'b'..... and so one). that means the volume is a*b*c. the surface area is 2(ab+bc+ca) and we know there is a constant x that is also that surface area (2(ab+bc+ca)=x)
2(ab+bc+ca)=x | divide by 2
ab+bc+ca=x/2=ab+c(a+b) | subtract ab
c(a+b)=x/2-ab | divide by (a+b)
c=(x/2-ab)/(a+b)
so, we can express the length 'c' using the 2 other lengthes and x, that is a constant.
so, What we need to do now, is do some calculus. unfortunately, it involves a function with 2 variables- 'a' and 'b'.
our function is: a*b*(x/2-ab)/(a+b) and we have to find its maximum value (x is a constant, remember. only a and b are the variables). Because im lazy, i wont find its derivative, but the function reaches its maximum point when a=b=(x/6)1/2. that means c is also (x/6)1/2. and that means that our box is actually a cube.
This is not the full answer to your question, but i believe after knowing this you can answer questions 'a' and 'b'
