# Problem solving with linear functions.

1. Sep 10, 2011

### davie08

1. The problem statement, all variables and given/known data
The daily operation of a plant that produces LCD clocks entails a fixed cost of $150. Furthermore, the production of one clock requires$3.00 in raw materials,$0.25 in electricity and$1.50 in labor. If these are the only variable costs and the plant can produce up to 400 clocks per day write the daily production cost C as a function of output q, with the appropriate domain.

2. Relevant equations

3. The attempt at a solution
im not sure where to start id guess that you would add up the variables to give yourself the cost for a single clock, but im not sure what else i should do.

2. Sep 10, 2011

### LCKurtz

Please use proper English when posting to these forums. It is "I'm" not im and "I" not "i" etc.

1. How much would it cost if they make no clocks on a day?
2. Same question for 1 clock, 2 clocks, etc.

3. Sep 10, 2011

### davie08

Would it cost them $150 for 0 clocks, and$159.75 for one clock and 164.50 for 2 clocks. I'm still unsure of how I would write this out as a function. Sorry for the bad grammar and spelling.

4. Sep 10, 2011

OK for 0 clocks but how did you get the additional $9.75 for the first clock? Check your figures... 5. Sep 10, 2011 ### Ray Vickson No. It would cost them$0 for 0 clocks. The fixed cost is $150, while the variable cost is$(3.00+1.50+0.25) = $4.75 per clock. So, 1 clock costs 150 + 4.75 = 154.75. How much do you think it would cost to make 2 clocks? In this type of problem the cost function C(Q) (Q = quantity of production) has a discontinuity at Q = 0. RGV 6. Sep 10, 2011 ### davie08 okay would it be C=$4.75q + $150 , I'm getting pretty tired I've been working on math problems all day so please be patience with my ignorance lol. 7. Sep 10, 2011 ### LCKurtz I have never seen a problem of this type interpreted that way. Technically, of course, the function is defined only on the nonnegative integers. To me it seems clear that the question is asking for the total cost for producing 0 clocks, which includes the fixed cost. 8. Sep 10, 2011 ### LCKurtz Good job. Now go take a break. 9. Sep 10, 2011 ### davie08 I took a little break, the C=$4.75q + $150 wouldn't be the final answer would it, as it asks for the appropriate domain. 10. Sep 10, 2011 ### Ray Vickson It is standard in every treatment of fixed-charge problems in production/inventory management, etc. It is used all the time in Operations Research. The point is that in planning multiperiod production, you may not turn on the equipment at all in some periods, so those periods produce 0 items at zero cost, and it is important to have a way of distinguishing such periods from others, at least in cases where the fixed costs are large and significant. Of course, sometimes such problems require the introduction of binary variables in the optimization (produce = yes or no?) and that can make the solution of large-scale industrial-strength problems a real challenge. However, companies do it all the time. To be fair, though: it is not absolutely clear what the originator of the problem wants, so it might be best for the student to submit _two_ solutions, with an explanation that "if you mean A, then C(Q) = ... , otherwise, if you mean B then C(Q) = ... ". RGV Last edited: Sep 10, 2011 11. Sep 10, 2011 ### vela Staff Emeritus If this is an econ type of problem, "cost" refers to total economic cost, so even when Q=0, there'd still be an opportunity cost of letting the resources sit idle. Plus you have expenses, like rent for example, that you have to pay regardless of whether you produce any clocks or not. All that goes into the fixed cost. 12. Sep 10, 2011 ### Ray Vickson Those are not the type of fixed costs I was referring to. There sometimes are "lump-sum" costs that are variable in the sense that IF we produce in some period we pay a startup cost but if we don't produce we pay nothing. In addition, if we start the process there is a second variable cost that is a function of the amount produced. The startup costs have nothing at all to do with overhead, rent, or whatever. In some industries, starting a machine can cost$10,000 or more; or changing a paint mixing machine from one colour to another may entail loss of a whole day's output and the production of hundreds of gallons of useless, unsaleable paint, etc. When modelling such situations we, essentially, _neglect_ such fixed costs as overheads, rents, etc., that must be paid anyway as long as we stay in business; many, many examples can be given where incorporating such fixed overheads into the planning problem (for example, by increasing the unit production cost to account for overhead) can produce disastrously bad plans. Of course, if the costs are truly fixed, they are just a constant that is subtracted at the end, after all plans have been set and accounted for.

RGV

13. Sep 10, 2011

### davie08

the C=$4.75q +$150 wouldn't be the final answer would it, as it asks for the appropriate domain.

14. Sep 10, 2011

### LCKurtz

Right, so you just need to state the values of q that fit the requirements of the problem. These values of q are .....?

15. Sep 10, 2011

### LCKurtz

I don't know much, nothing really, about operations research. But I am pretty confident that the OP's question likely arises in something like a first year algebra class or maybe a "business" calculus class where the learning problem for the student is to figure out the appropriate linear function. The above discussion, while accurate, overthinks the problem in such a setting. As this thread indicates, figuring out the linear function was challenge enough.

16. Sep 10, 2011

### Ray Vickson

I agree that your guess as to what was wanted is likely correct. (However, if the question occurred in an Intro Operations Research class, that would probably not be the case.)

RGV

17. Sep 11, 2011

### davie08

would it be 0< q < 400 I'm really bad with this domain stuff, is this how it's supposed to look and can q be equal to 0.

18. Sep 11, 2011

### LCKurtz

Yes and yes with ≤ for both inequalities. And you might note that q takes on only integer values although that is sometimes overlooked.