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Optimization of box, varied material cost.

  1. Jul 4, 2011 #1
    1. The problem statement, all variables and given/known data
    Stewart Calculus 6E: 4.7 #14
    A rectangular storage container with an open top is to have a volume of 10m³. The length of it's base is twice the width. Material for the base costs $10 per square meter. Material for the sides cost $6 per square meter. Find the cost of materials for the cheapest such container.

    2. Relevant equations

    3. The attempt at a solution
    I let the width of the box be W, the length be 2W, and the height be H.

    Since:
    [tex]2W^{2}H=10[/tex]
    I let [itex]H=\frac{10}{2W^{2}}[/itex]

    Then I claim that the cost of the base is given by:
    [tex](2W^{2})10 = 20W^{2}[/tex]

    The cost of the sides are given by:
    [tex](2WH + 4WH)6[/tex]

    So the total cost for the box could be written as:
    [tex]20W^{2} + (2WH + 4WH)6[/tex]
    Substituting in [itex]H=\frac{10}{2W^{2}}[/itex], I get cost as a function of width?
    [tex]C(W) =(2W(\frac{10}{2W^{2}})+4W(\frac{10}{2W^{2}}))6 + 20W^{2}[/tex]
    [tex]C(W) = \frac{180+20W^{3}}{W}[/tex]

    So I can minimize that function, and I find that the minimum is when the width is 1.65, and the cost of the box is about $163.50. So $163.50 is the solution? Does anyone see a problem with this?
     
  2. jcsd
  3. Jul 4, 2011 #2

    ideasrule

    User Avatar
    Homework Helper

    Nope, that's right. :)
     
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