+233 What if you were to cut a square piece of paper into 4 smaller congruent squares and from each construct a box with the maximum volume? How would the combined volume of these 4 rectangular prisms compare with the rectangular prism with the maximum volume constructed from the original square?
a) form and record your hypothesis before investigating the problem.
b) Now investigate the problem using whatever approach you believe would be appropriate. How will you modify your hypothesis in light of the results of your investigation?
+9479 This is the most efficient way that I know of to create a box from a square. First cut along the pink lines. Use two of the cut off squares as bases. Then fold the paper along all the horizontal lines. And then glue/tape the bases on.
Now it can be made into a prism with length, width, and height as shown.
length = s/4
width = s/4
height = 3s/4
volume of original prism = lwh
volume of original prism = (s/4)(s/4)(3s/4)
volume of original prism = 3s3/64
If we divide the original square into 4 squares, the side length of each square will halve.
volume of littler prism = 3(s/2)3 / 64 = 3s3/512
There are 4 littler prisms, so total volume of all 4 littler prisms = 4(3s3/512) = 3s3/128
volume of original prism : volume of all littler prisms = \(\frac{3s^3}{64}:\frac{3s^3}{128}=1:\frac12\)
So..the volume of all 4 littler prisms together is half the volume of the 1 big prism 
*edited because my first answer was messed up*
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'
+9479 Hmmmmm..I think that yours is a much better approach than mine..
Mine does assume that you can cut a piece and glue it on. Why not cut off a bunch of pieces and use them all to make a box..instead of throwing paper away. And on mine, for the four littler prisms, you could use some of the old paper that you cut off from other prisms to make the bases.
