How many DVDs should the company make to minimize total storage and setup cost?

Question

Sandy and Sam's multimedia company anticipates that there will be a demand for 20,000 copies of a certain DVD during the next year. It will cost the company $0.50 to store a DVD for one year. Each time it must make additional DVD's, it costs $200 to set up the production equipment. How many DVDs should the company make during each production run to minimize its total storage and setup cost? Assume that each production starts once the previous storage runs out and clearly justify your answer is finding the minimum cost by showing the appropriate calculus work.

Answer 1

The thing that makes this problem click is realizing there are two costs pulling in opposite directions. If you make a huge batch you barely ever pay the $200 setup, but you are sitting on a warehouse full of DVDs all year. If you make tiny batches you store almost nothing, but you are paying $200 over and over. The minimum sits somewhere in the middle, and calculus is what finds it.

So, let x be the number of DVDs made in each production run.

For the setup side, you need 20,000 copies over the year and each run makes x of them, so you do 20,000/x runs, and each one costs $200:

setup cost = 200(20,000/x) = 4,000,000/x

For the storage side, this is the part most people get wrong, so it is worth going slowly. Each run starts with x DVDs in storage and they sell off steadily until you are at 0, and then the next run begins. Since the level drops in a straight line from x down to 0, the AVERAGE number in storage is x/2, not x. At $0.50 each per year that gives:

storage cost = 0.50(x/2) = 0.25x

Adding them gives the total cost function:

C(x) = 4,000,000/x + 0.25x

Now we can do the calculus. Rewrite the first term as 4,000,000x⁻¹ so the power rule applies cleanly:

C′(x) = −4,000,000/x² + 0.25

Set that equal to zero to find the critical number:

0.25 = 4,000,000/x², so x² = 16,000,000 and x = 4,000

We throw out x = −4,000 since you cannot produce a negative number of DVDs.

The problem asks you to justify that this really is a minimum, so use the second derivative:

C″(x) = 8,000,000/x³

For any positive x this is positive, so the curve is concave up everywhere on the domain and the critical number at x = 4,000 is an absolute minimum, not just a local one. That is a stronger statement than the first derivative test would give you here, which is why the second derivative is the cleaner tool for this problem.

So, the company should make 4,000 DVDs per production run, which works out to 20,000/4,000 = 5 runs during the year.

One last thing worth noticing, because it is a great way to check yourself on any problem shaped like this one. At x = 4,000 the setup cost is 4,000,000/4,000 = $1,000 and the storage cost is 0.25(4,000) = $1,000, for a total of $2,000. The two costs come out exactly equal at the minimum. That is not a coincidence, it happens every time you minimize something of the form a/x + bx, and if you ever solve one of these and the two pieces do not match, you know to go back and hunt for an arithmetic slip.