```Question 548213
The original square of cardboard, with the cutting and folding lines would look like this:
{{{drawing(300,300,-1,10,-1,10,
rectangle(0,0,9,9), blue(rectangle(2,2,7,7)),
red(line(0,2,2,2)),red(line(0,7,2,7)),
red(line(2,0,2,2)),red(line(7,0,7,2)),
red(line(9,2,7,2)),red(line(9,7,7,7)),
red(line(2,9,2,7)),red(line(7,9,7,7))
)}}}Cutting lines are red, folding lines are blue.
You see 4 congruent squares cut out of the corners. If the length of the sides of those squares is 8 inches, then the height of the box will be 8 inches.
Let x be the length (in inches) of the side of the blue square (the bottom of the box).
The surface area of the bottom (in square inches) is {{{x^2}}},
and the volume of the box, calculated as area of the bottom times height, (in cubic inches) is
{{{8x^2}}}
So {{{8x^2=64}}} ---> {{{x^2=64/8=8}}} ---> {{{x=sqrt(8)=2sqrt(2)}}} or about 2.83 inches.
(That is going to be a tall box with a very narrow base, too easily tipped over.
I would have cut squares with side length 4 inches from the corner, and that would have given me a nice cube.)
So if the length of the side of the base is {{{2sqrt(2)}}}, with 8 inches of length added to each side, the length of the side of the original cardboard square must have been {{{8+8+2sqrt(2)=16+2sqrt(2)}}}, or about 18.83 inches.
I still think we should have made a cubic box with 4 inch sides by cutting squares with 4 inch sides from a 12 inch by 12 inch square of cardboard. It would have used less material, wasted less material in the cutouts, for a box with the same volume, less surface, and much better stability.```