 Mathematics
12 December, 21:52

The Friendly Sausage Factory (FSF) can produce hot dogs at a rate of 5,500 per day. FSF supplies hot dogs to local restaurants at a steady rate of 260 per day. The cost to prepare the equipment for producing hot dogs is \$66. Annual holding costs are 46 cents per hot dog. The factory operates 296 days a year. a. Find the optimal run size. (Do not round intermediate calculations. Round your answer to the nearest whole number.) Optimal run size b. Find the number of runs per year. (Round your answer to the nearest whole number.) Number of runs c. Find the length (in days) of a run. (Round your answer to the nearest whole number.) Run length (in days)

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1. 12 December, 23:33
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a. 5218 hot dogs

b. 18 runs per year

c. 1 day per run

Step-by-step explanation:

(a) Let n represent the number of hotdogs in an optimal run. The number of runs needed in a year is ...

r = (260 hotdogs/day) · (365 days/year) / (n hotdogs/run) = 94900/n runs/year

The number of hotdogs in storage decreases from n at the end of a run to 0 just before a run. The decrease is linear, so the average number in storage is n/2.

We can go to the trouble to write the equation for total cost, then differentiate it to find the value of n at the minimum, or we can just jump to the solution. That solution is ...

holding cost = setup cost

(n/2) ·0.46 = (94900/n) ·66

n^2 = 94900·66·2/0.46 = 27,232,173.9

n ≈ 5218.446

The optimal number of hotdogs in a run is 5218.

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(b) The number of runs needed per year is (from part (a)) ...

r = 94900/5218 ≈ 18.187

The number of runs per year is 18.

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(c) The run lenght is 5218 hotdogs, and the run produces them at the rate of 5500 hotdogs/day, so the run length in days is ...

(5218 hotdogs) / (5500 hotdogs/day) = 0.9487 days

The run length is 1 day.