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Math Help - Minimizing Cost

  1. #1
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    Minimizing Cost

    The cost of inventory depends on the ordering and storage costs according to the inventory model C = (\frac{Q}{x})s + (\frac{x}{2})r . Determine the order size that will minimize the cost, assuming that sales occur at a constant rate, Q is the number of units sold per year, r is the cost of storing one unit for 1 year, s is the cost of placing an order, and x is the number of units per order.

    I differentiated and got:
    x = \sqrt{\frac{2Qs}{r}}
    Not sure if this is right, and not sure how to verify it. I am always supposed to explain why there is a min at x = whatever, and I can't tell if the derivative is positive or negative because of all the variables...
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  2. #2
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    So you should get  C' = -Q s x^{-2} + r/2 for the first derivative and  C'' = 2 Q s x^{-3} for the second derivative. Setting the first derivative equal to zero and solving for x yields
     x = \pm \sqrt{\frac{2Qs}{r}} .
    We only care about the positive solution since we can't order a negative number of units. The positive solution corresponds to a local minimum, since when x is positive, the second derivative is just the product of a bunch of positive quantities (Q and s can't be negative given what they represent, and  x^{-3} has the same sign as x) and is therefore positive. The negative solution doesn't make sense, but even if it did, it corresponds to a local maximum since the second derivative is negative if x is negative.

    Quite often you can write the derivatives as the product of a bunch of quantities. If this is the case, determining the sign of the derivative is simply a matter of determining how many of those quantities are negative.

    If you wanted to check all this, I'd make up values for Q, s, and r and plot the cost function. If, for example, you let Q = s = r = 1, you'd expect to see a local min at  x = \sqrt{2} and a local max at  x = -\sqrt{2} .
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