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Math Help - Box Volume/Calculus problem

  1. #1
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    Box Volume/Calculus problem

    Trying to find the maximum volume achievable for a container to be made from a steel sheet measuring 3.4m X 1.9m.

    This is my workings so far,where am i going wrong?

    v = l x w x h

    V(h) (304 - 2h)(1.9 - 2h) X h
    (6.46 - 13.6h + 4h^2) x h
    4h^3 - 13.6h^2 + 6.46h
    v(h) 4h^3 - 13.6h^2 + 6.46h
    v(h) 12h^2 - 27.2h + 6.46
    0 = 12h^2 - 27.2h + 6.46 divide both sides by 12
    0 = 2.266 + 0.538 = 2.804 I Think I am supposed to factor around here!
    h = 2.804
    V = 4h^3 -27.2 + 6.46 = 1390.2

    Any advice appreciated

    Thanks.....
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  2. #2
    MHF Contributor
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    Quote Originally Posted by manich44 View Post
    Trying to find the maximum volume for a container to be made from a steel sheet measuring 3.4m X 1.9m.

    This is my workings so far,where am i going wrong?

    v = l x w x h

    V(h) (304 - 2h)(1.9 - 2h) X h
    (6.46 - 13.6h + 4h^2) x h
    4h^3 - 13.6h^2 + 6.46h
    v(h) 4h^3 - 13.6h^2 + 6.46h
    v(h) 12h^2 - 27.2h + 6.46
    0 = 12h^2 - 27.2h + 6.46 divide both sides by 12
    0 = 2.266 + 0.538 = 2.804 I Think I am supposed to factor around here!
    h = 2.804
    V = 4h^3 -27.2 + 6.46 = 1390.2

    Any advice appreciated

    Thanks.....
    Your work is quite "messy".
    Try to be clearer: keep your teacher in good mood

    Multiply dimensions by 10: 34 by 19; easier to handle.

    V = h(19 - 2h)(34 - 2h)
    V = 4h^3 - 106h^2 + 646h ; divide by 2:
    V = 2h^3 - 53h^2 + 323h
    Still with me? OK:
    6h^2 - 106h + 323 = 0
    Solve using quadratic; you'll get h = ~13.75 or ~3.92

    13.75 impossible, so h = 3.92; or .392 due to original multiplication by 10

    So h = .392, w = 1.9 - 2(.392), l = 3.4 - 2(.392)

    Finish it...
    Last edited by Wilmer; June 1st 2010 at 07:47 AM. Reason: none
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