Problem solving with linear functions.

In summary, the daily production cost C as a function of output q for a plant that produces LCD clocks is a function of the fixed cost and the variable cost. The fixed cost is $150, while the variable cost is $(3.00+1.50+0.25) = $4.75 per clock.
  • #1
davie08
115
0

Homework Statement


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.


Homework Equations





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 I am not sure what else i should do.
 
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  • #2
Please use proper English when posting to these forums. It is "I'm" not I am and "I" not "i" etc.

Start by answering these questions:
1. How much would it cost if they make no clocks on a day?
2. Same question for 1 clock, 2 clocks, etc.
 
  • #3
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
OK for 0 clocks but how did you get the additional $9.75 for the first clock? Check your figures...
 
  • #5
davie08 said:
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.

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
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
Ray Vickson said:
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

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
davie08 said:
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.

Good job. Now go take a break. :smile:
 
  • #9
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
LCKurtz said:
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.

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:
  • #11
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
vela said:
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.

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
the C=$4.75q + $150 wouldn't be the final answer would it, as it asks for the appropriate domain.
 
  • #14
davie08 said:
the C=$4.75q + $150 wouldn't be the final answer would it, as it asks for the appropriate domain.

Right, so you just need to state the values of q that fit the requirements of the problem. These values of q are ...?
 
  • #15
Ray Vickson said:
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

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
LCKurtz said:
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.

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

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
davie08 said:
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.

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

What is a linear function?

A linear function is a mathematical equation that can be represented by a straight line on a graph. It has the form y = mx + b, where m is the slope of the line and b is the y-intercept. Linear functions are commonly used to model relationships between two variables.

What are some real-life applications of linear functions?

Linear functions are used in many fields, including physics, economics, and engineering. Some examples of real-life applications of linear functions include calculating distance and time in physics, predicting sales and revenue in economics, and designing bridges and other structures in engineering.

What is problem solving with linear functions?

Problem solving with linear functions involves using mathematical equations to solve real-life problems. This typically involves identifying the variables involved, setting up a linear function to model the relationship between the variables, and using algebraic techniques to find the solution to the problem.

What are the steps to solving a problem with linear functions?

The steps to solving a problem with linear functions include: 1) identifying the variables involved, 2) writing a linear function to model the relationship between the variables, 3) plugging in known values and solving for the unknown variable, and 4) checking the solution to ensure it makes sense in the context of the problem.

How can I use linear functions to make predictions?

Linear functions can be used to make predictions by analyzing the relationship between two variables and using the equation to calculate the value of one variable based on the value of the other. This is useful in fields such as economics and finance, where linear functions can be used to make predictions about future trends and outcomes.

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