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Math Help - this is a problem dealing with inventory and orders, Please help!!

  1. #1
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    this is a problem dealing with inventory and orders, Please help!!

    This is the problem im stuck on, any help is appreciated!

    A wine warehouse expects to sell 30,000 bottles of wine in a year. Each bottle
    costs $9, and there is a fixed charge of $200 per order. If it costs $3 to store a
    bottle for a year, how many bottles should be ordered at a time and how many
    orders should the warehouse place in a year to minimize inventory costs.
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  2. #2
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    please read http://www.mathhelpforum.com/math-he...y-problem.html

    and your course notes, and then post as much of the answer as you can do yourself
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  3. #3
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    I found something similar to what your post was about in my book and course notes, these are the answers I came up with, please let me know if they are correct:

    C=200y + 1.5x

    xy= 30,000, thus y=30000/x

    So, C(x)= 200(30000/x)+1.5x

    Then, C(x)= 6,000,000/x + 1.5x

    C '(x)= -6,000,000/x^2 + 1.5

    -6,000,000/x^2 + 1.5 = 0

    x^2= 6,000,000/1.5

    Divide, then square root to cancel out
    x^2= 4,000,000, the square root of 4,000,000= 2,000
    Thus, x=2,000, so y= 30,000/2,000 = 15

    The final answer:
    The company will minimize its inventory cost by ordering 2,000 bottles, 15 times during the year at a time.
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  4. #4
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    its the right answer, some minor comments in red

    you have not defined any of your variables so i have to guess what you mean...

    i assume:
    C = total annual cost
    y = number of orders per year
    x = number of bottles per order


    In that case:

    C=200y+ 1.5x
    You have not included the purchase price of the bottles in your cost function. This is actually ok, because the annual cost is a constant (30000*9) that is not affected by your choice of x and y. Since it is a constant, it will disappear when you differenciate anyway. Different professors will have their own preferences about whether this fixed cost should be included or not


    you have to do quite a bit of simplification to get the cost of storage, 1.5x
    • if there are y orders per year, then the time between orders is 1/y.
    • So the average time each bottle spends in storage is 0.5(1/y) = 1/2y
    • So the cost of storage between orders is x * 3 * (1/2y) = 3x/2y
    • Multiply by the number of orders: y * (3x/2y) = 3x/2 = 1.5x




    xy = 30000, so y=30000/x

    C(x) = 200(30000/x) + 1.5x = 6000000x^{-1} + 1.5x
    C'(x) = -6000000x^{-2} + 1.5

    Solve C'(x) =0
    6000000 = 1.5x^2
    = +/-2000
    Discard negative result, so x=2000

    y=30000/2000=15
    Last edited by SpringFan25; June 25th 2010 at 01:11 PM.
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