Production Inc. has two types of machines to produce beer funnels. The old machi

Question

Production Inc. has two types of machines to produce beer funnels. The old machine costs S500/ day to operate and produces 700 funnels/day. The new machines (NM) cost y to operate and produce 1,000 funnels/day. They have to operate 10 OM because they on them that cannot be broken. They must operate at least a total of 25 machines. $7 have a lease e to keep operating costs/day to a minimum but they face several and constraints. The must produce 40,000 funnels a day to meet dem Hint: Watch signs of constraints carefully, 1. How many OM and NM machines should they operate? What does th 2. Suppose they obtain a concession on the lease of the O Ms so that they now c ost $450/day to operate. Also we now need 50,000 funnels/day How much does this change the optimum solution and total cost? The OMs are breaking down and you can't get parts. As a result, only 45 are available. Assume that Part 2 conditions are the starting point for the question. What happens to your costs and why does it happen? Suppose you can now get the NMs for $600. Assume that Part 3 conditions are the starting point for the question. What happens to your costs and number of machines used and why does this happen? 4.

Explanation / Answer

1. The primery targets are operating cost per day to be minimum and 40,000 funnels to be produced

Given-

1 (old machine) costs- $500/ day , capacity -700funnels/day

1 (new machine) costs- $700/day ,capacity- 1000funnels /day

Let's take x (OM) and y (NM) operates

Funnel equation-

700x+1000y=40000

7x+10y=400

Cost equation -

Z= 500x+700y=100(5x+7y)

And x>=10

x+y>=25

If we put x= 10 then y= 33 (to satisfy minimum funnel production requirement)

And thus the operating cost would be $ 100(5*10+7*33)= $28100 ,also x+y=43>=25.

2. Equations are now-

7x+10y= 500

Z=100(4.5x+7y)

x+y>=25

x=10,y=43 total operation cost involved is z=$100(45+301)

Z=$34600

3. If OMs are breaking down and 45 are remaining but we got the answers from x=10 as the constraint for minimum operations cost and required funnel production hence forth answer should remain same as we have to take ground conditions in question 2 .

4. New equation are

7x+10y=500

Z=100(4.5x+6y)

Putting x=10 and y=43 , the cost of production=$100(45+258)

Z=$30300

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