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Math Help - Cost optimization problem?

  1. #1
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    Cost optimization problem?

    A rectangular storage container with an open top is to have a volume of 30 cubic meters. The length of its base is twice the width. Material for the base costs 13 dollars per square meter. Material for the sides costs 9 dollars per square meter. Find the cost of materials for the cheapest such container.
    Total cost = ________________dollars

    i got about $234.

    i am pretty sure I did all the steps right.

    First I got the equation which is C=13Lw+6LW+6WH

    got it to one variable

    26w^2 + 18w(30/2w^2)

    derived and set equal to zero

    Found W

    and then plugged it in.

    Did I do something wrong?
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  2. #2
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    Did it on the fly, but I get \displaystyle C = 26w^2+\frac{90}{w}
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  3. #3
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    Quote Originally Posted by softballchick View Post
    A rectangular storage container with an open top is to have a volume of 30 cubic meters. The length of its base is twice the width. Material for the base costs 13 dollars per square meter. Material for the sides costs 9 dollars per square meter. Find the cost of materials for the cheapest such container.
    Total cost = ________________dollars

    i got about $234.

    i am pretty sure I did all the steps right.

    First I got the equation which is C=13Lw+6LW+6WH

    got it to one variable

    26w^2 + 18w(30/2w^2)

    derived and set equal to zero

    Found W

    and then plugged it in.

    Did I do something wrong?
    Let's see...

    \displaystyle V = 30 = lwh and \displaystyle l = 2w, so

    \displaystyle 30 = 2w^2 h

    \displaystyle 15 = w^2h

    \displaystyle h= \frac{15}{w^2}.


    The area of the base is \displaystyle lw = 2w^2 and the area of the sides are \displaystyle 2lh + 2wh = \frac{60}{w} + \frac{30}{w} = 90w^{-1}.

    Since the cost is \displaystyle \$ 13 per square metre for the base and \displaystyle \$ 9 per square metre for the sides,

    \displaystyle C = 13\cdot 2w^2 + 9\cdot 90w^{-1} = 26w^2 + 810w^{-1}.


    Now minimise the cost function.
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  4. #4
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    I minimized it and got w=(810/52)^(1/3)

    Now what? where do I plug this into?
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  5. #5
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    Now you plug this into your original equation for the cost, that is:

    \displaystyle C = 26w^2 + \dfrac{810}{w}
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