Minimizing Cost

• Dec 2nd 2009, 05:27 PM
seuzy13
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...
• Dec 3rd 2009, 08:25 AM
nehme007
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 $x = -\sqrt{2}$.