The Friendly Sausage Factory (FSF) can produce hot dogs at a rate of 5,750 per day. FSF supplies hot dogs to local restaurants at a steady rate of 270 per day. The cost to prepare the equipment for producing hot dogs is $67. Annual holding costs are 47 cents per hot dog. The factory operates 297 days a year. Find A) The optimal run size.
B) The number of runs per year.
C) The length (in days) of a run.

A) 22812 hotdogs per run

B) 75 runs/yr

C) 4 days in a run

Step-by-step explanation:

We are given;

Production rate;p = 5750 per day

Steady Usage rate;u = 270 per day

Setup cost of hotdog;S = $67

Annual carrying cost (H) = 47 cents = $0.47 per hot dog

No. of Production days; d = 297 days

A) Let's first find the annual demand given by the formula;

Annual demand;D = pd

D = 5750 × 297

D = 1707750 hot dogs/yr

Now, formula for optimal run size is given by;

Q_o = √[(2DS/H) × (p/(p - u))]

Plugging in the relevant values gives;

Q_o = √[(2 × 1707750 × 67/0.47) × (5750/(5650 - 270))]

Q_o =√520375454.7971209

Q_o = 22812 hotdogs per run

B) formula for Number of runs per year is given as;

No. of Runs = D/Q₀

Thus;

no. of runs = 1707750/22812

no. of runs ≈ 75 runs/yr

C) Length of a run is given by the formula;

Length = Q₀/p

Length = 22812/5750

Length ≈ 4 days in a run

75 tickets have been sold so far.

step-by-step explanation:

in this question it is clearly mentioned that 45 children's tickets have been sold and this number is 60% of the total tickets sold.

now let's assume the total number of tickets sold be x.

then 60% of x = 45

⇒

⇒ x(.6) = 45

by dividing with .6 on both the sides of the equation

⇒ x = 75

therefore 75 tickets have been sold so far.