A certain company makes a laboratory microscopes. Setting up each production run

Question

A certain company makes a laboratory microscopes. Setting up each production run

costs $2,500. Insurance costs, based on the average number of microscopes in the warehouse,

mount to $20 per microscope per year. Storage costs, based on the maximum number of

microscopes in the warehouse, amount to $15 per microscope per year. If the company

expects to sell 1,600 microscopes at a fairly uniform rate throughout the year, determine the

number of production runs that will minimize the company's overall expenses.

Explanation / Answer

Suppose there's only one production run in a year (for 1600microscopes). Then the setup cost will be $2500*1. If sales areuniform during the year, then the average number of microscopes onhand during the year is 800, so the insurance cost will be $20*800.And the storage cost will be $15*1600 (we need a warehouse bigenough to hold the maximum number of microscopes we can have onhand at any time).

Suppose instead that there are two productions (of 800 microscopes)per year . Then the setup cost will be $2500*2. The average numberof microscopes on hand at any time will be 400, so the insurancecost will be $20*400. The max number of microscopes on hand is 800,so the storage cost will be $15*800.

Things get a bit more involved for 3 production runs. Proceeding asbefore, we get that the number of microscopes per run is 1600/3 =533.33. We can't make a fraction of a microscope. So for one of theproduction runs we will make 534 microscopes, and for the other tworuns we will make 533 microscopes. The average number ofmicroscopes on hand would be [(533/2) + (533/2) + (534/2)]/3 =800/3, so the insurance cost would be $20*800/3. The warehouse willneed to accommodate 534 microscopes, so the storage cost would be$15*534.

At this point I think the details just obscure the problem, solet's proceed as if these fraction distractions don't arise.

So let N be the number of production runs. Then the number ofmicroscopes produced per run is 1600/N. The setup cost for the Nruns is 2500*N, the insurance cost is 20*(1600/N)/2 = 16000/N, andthe storage cost is 15*(1600/N) = 24000/N, so the total cost is

C(N) = 2500*N + 40000/N

Your task is to find N that minimizes C. Happily, the answer youwill get sidesteps all of the "fraction of a scope" stuff. (It'salso fortunate that you will get an integer for N. Clearly, wecouldn't report "3.26 runs" as our answer.)

(And of course, in using a derivative to minimize C, we temporarilyignore the fact that N can only take on positive integer values;that is, that it can't vary continuously.)

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