Volume of an Open-Top Box with 3-Sided Squares

AI Thread Summary
An open-top box can be created by cutting equal squares from the corners of a rectangular piece of cardboard, leading to a volume equation based on the dimensions of the cardboard. The discussion reveals that the same volume is achieved when squares of side one or two are cut out, prompting analysis of the volume when squares of side three are removed. Two equations emerge from the volume relationships: 0 = xy + 6x + 6y + 28 and 1 = 2xy + 4x + 4y + 8. The algebraic manipulation of these equations suggests a complex relationship between the dimensions x and y. Further exploration is needed to determine the specific volume when squares of side three are cut out.
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Homework Statement



An open-top box can be formed from a rectangular piece of cardboard by cutting equal squares from the four corners and then folding up the four sections that stick out. For a particular-sized piece of cardboard, the same volume results whether squares of side one or squares of side two have been cut out. What is the resulting volume if squares of side three are cut out?
 
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More information is needed. Is there any relation between the sides of the rectangle? Any perimeters? any areas? Just the equality of two particular starter volumes gives this:

V = (x-2)(y-2) = 2(x-4)(y-4)
and then
steps of algebra,
0 = xy + 6x + 6y + 28, apparently not factorable, not useful alone;

Or since the second volume expression is factorable,
(x-2)(y-2) = 2(x+2)(x-2)(y+2)(y-2)
which yields
1 = 2(x+2)(y+2)

Interesting. That gives two different possible useable equations in x and y:
0=xy+6x+6y+28
AND
1=2xy+4x+4y+8
 
With continued steps, I obtain
16x + 16y + 63 =0
 
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