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an open box contains 80cm^3 and is made from a square piece of tinplate with 3cm squares cut from each of its 4 corners. Find the dimensions of the original piece of tinplate.

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2 answers

  1. I see the equation
    3(x-6)(x-6) = 80
    3(x^2 - 12x + 36) = 80
    3x^2 - 36x + 108 - 80 = 0
    3x^2 - 36x + 28 = 0

    Use the quadratice formula to find x

    I got x=11.164

    check 11.164-6 = 5.164

    so the box is 5.164x5.164x3
    which gives 80

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  2. Let the side of the square plate be x cm.
    The height of the box is 3 cm.
    The length & width are both (x-6) cm.
    volume = length*width*height
    =3*(x-6)*(x-6)
    =80
    x can then be obtained by solving
    3(x-6)(x-6)=80

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