Your friend has a fabulous recipe for salsa, and he wants to start packing it up and selling it. He can rent the back room of a local restaurant any time he wants, complete with their equipment, for $100 per time. It costs him $2 a jar for the materials (ingredients for the salsa, jars, labels, cartons) and labor (you and a couple of friends of his) for each jar he makes. He can sell 12,000 jars of salsa each year (I told you it was a fabulous recipe!), with a constant demand (that is, it's not seasonal; it doesn't vary from week to week or month to month). It costs him $1 a year per jar to store the salsa in the warehouse he ships from. He wants to find the number of jars he should produce in each run in order to minimize his production and storage costs, assuming he'll produce 12,000 jars of salsa each year.

Question

oobleck · Answer

Maybe you can work out a continuous function that you can apply calculus methods to, but consider this:
the cost for ingredients and labor will not change, so you only have to consider the rent and storage costs.
Suppose he makes x batches, spaced evenly through the year, with the first batch ready to go on January 1, and assuming 1000 jars per month. The jars in storage will decline smoothly, but let's approximate with a step function -- the full 1000 jars is stored the whole month, and shipped at the end of the month, and the cost is pro-rated for fractions of a year.
So, if the 12000 jars are made in x batches, we have costs of
x=1: 100 + 1000 * 1 * (12/12 + 11/12 + 10/12 + ... + 1/12)
= 100 + 1000/12 ∑1..12 = 100 + 1000/12 * 78 = 6600

x=2: 200 + 1000/12 * 2∑1..6 = 200 + 1000/12 * 42 = 3700
x=3: 300 + 1000/12 * 3∑1..4 = 300 + 1000/12 * 30 = 2800
x=4: 400 + 1000/12 * 4∑1..3 = 400 + 1000/12 * 24 = 2400
x=6: 600 + 1000/12 * 6∑1..2 = 600 + 1000/12 * 18 = 2100
x=12: 1200 + 1000/12 * 12∑1..1 = 1200 + 1000/12 * 12 = 2200

Looks like his best bet is to make six batches per year. Why do I get the feeling that the optimal spacing is e months, instead of 2? Surely the continuous solution will involve 1/x in some way (storage time), thus a logarithmic solution.