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Box Optimization problem

  1. Dec 8, 2008 #1
    1. The problem statement, all variables and given/known data

    Question:

    You are planning to make an open rectangular box from a 10 by 18 cm piece of cardboard by cutting congruent squares from the corners and folding up the sides.

    1) What are the dimensions of the box of largest volume you can make this way?
    2) What is its volume?

    v - volume
    l - length
    h - height
    w - width

    2. Relevant equations

    v = lwh

    3. The attempt at a solution

    Since v = lwh w = 10 - 2h and l = 18 - 2h.
    Therefore:

    v(h) = (18-2h)(10-2h)(h)
    v(h) = (4h^3) - (56h^2) + 4h^2

    Differentiation:

    dv/dh = (12h^2) - 112h + 180

    Finding critical numbers:

    (12h^2) - 112h + 180 = 0
    (12h^2) - 112h = -180
    h(12h - 112) = -180

    12h - 112 = -180
    12h = -68
    h = -68/12

    critical numbers: k{-180, -68/12}

    My question is, how can I have a negative height? Is this correct so far or have I made some error?
     
  2. jcsd
  3. Dec 8, 2008 #2
    for the purpose of this problem you cannot have negative heights, you made a simple error. You can fix it easily. the problem is your step near the end when you have
    (12h^2) - 112h + 180 = 0 dont subtract the 180. when you get to this part factor our a value of 4 you should get 4((3h^2)-28h+45) from here just use the quadratic formula to find your zero's. you don't need to factor out the 4 you can just go straight to the quadratic formula, but it makes it less tedious if you do. =) Study Hard.
     
  4. Dec 8, 2008 #3

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    ab= 180 does NOT mean "a= 180" or "b= 180" as you seem to think. If ab= 0, then either a= 0 or b= 0 but that is a specific property of 0 not of other numbers.
     
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