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Cube volume

  1. Oct 6, 2007 #1
    1. The problem statement, all variables and given/known data

    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?
     
  2. jcsd
  3. Oct 6, 2007 #2

    symbolipoint

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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, apparantly 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
     
  4. Oct 6, 2007 #3

    symbolipoint

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    With continued steps, I obtain
    16x + 16y + 63 =0
     
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