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algebra

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I have a piece of cardboard that is twice as long as it is wide .I f I cut a 1-inch by 1-inch square from each corner and fold up the resulting flaps ,I get a box with a volume of 40 cubic inches.what are the dimensions of the cardboard?

  • algebra -

    width = w
    length = 2w

    (w-2)(2w-2)(1) = 40
    (w-1)(w-2) = 20

    20 = 4x5, so w = 6

    The box is 6 x 12

    check: cut off the 1" corners and you get a box 4x10

  • algebra -

    width --- x
    length ---- 2x

    after cut-out
    width = x-2
    length = 2x-2
    height = x

    volume = x(x-2)(2x-2) = 40
    2x^3 - 6x^2 + 4x = 40
    x^3 - 3x^2 + 2x - 20 = 0
    tried ±1, ±2, ±4, ±5 and could not find a "nice" solution, so I ran it through
    Wolfram ,,,,,,
    http://www.wolframalpha.com/input/?i=x%5E3+-+3x%5E2+%2B+2x+-+20+%3D+0
    and got
    x = appr. 3.837

    so the cardboard was 3.837 by 7.674

    check:
    width of box = 3.837-2 = 1.837
    length of box =7.674-2 = 5.674
    height = 3.837
    vol = 1.837x5.674x3.837 =39.99 , not bad

  • go with Steve - algebra -

    forget my solution,
    I was thinking of an entirely different question.

  • algebra -

    thank you

  • algebra - to Reiny -

    oddly enough, I did exactly what you did! Then when I saw the messy answer, I reread the problem.

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