# Math Help - Optimaization

1. ## Optimaization

Each machine at a certain factory can produce 50 units per hour. The setup cost is $80 per machine, and the operating cost is$5 per hour. How many machines should be used to produce 8,000 units at the least possible cost? (answer should be a whole Number)

2. Originally Posted by lisa1984wilson
Each machine at a certain factory can produce 50 units per hour. The setup cost is $80 per machine, and the operating cost is$5 per hour. How many machines should be used to produce 8,000 units at the least possible cost? (answer should be a whole Number)
Suppose $n>0$ machines are used. $n$ machines will produce $N$ units in $N/(50n)$ hours. Cost to produce $N$ units is:

$C(n,N)=80n + \frac{N}{50n}\times5n=80n+\frac{N}{10}$

This is minimised when $1$ machine is used for the entire batch, since for any positive $N$ this is an increasing function of $n$.

CB

3. set up cost per hr=operating cost per hr=$5, when machine use 16 hrs 16 hr can produce 16x50=800 units 8000 unts/800 units per machine= 10 machines Is this correct? 4. I notice you posted your second post 10 hours after Captain Black posted his response. Is there any reason why you simply ignored his response? Captain Black said that you should use only one machine. If you use just the one machine to make all you have to pay one$80 set up charge. It will take that single machine 800/50= 16 hours, at $5 per hour to make 800 items for a total cost of 80+ 16(5)= 80+ 80=$160.

If you use your 16 machines, each machine will have to make only 5 items requiring only 1 hour: 16 set up costs plus 16 machines running for one hour, 16(80)+ 16(5)= $1360! No, it doesn't look like your solution is best. Can you explain why you want the "setup cost per hour" (for one machine) to equal the "cost of production per hour"? 5. I calculated this after I posted my first question and I did not see his post till this afternoon because my baby was sick last night...sorry about that, I just wanted to see if I was close! Again I was not ignoring his post I wrote what he has down and am trying to solve the problem that way! 6. Guys I think this question may have been misunderstood. I think CaptainBlack and HallsofIvy have taken the common sense approach to this question and taken the operating cost of$5 per hour to mean per machine however you'll note there is no per machine in that part of the question whilst there is in the other parts.

You see, I think this question is poorly written but I believe that as it is currently interpreted it's not much of a question in terms of optimisation. However I think the intended meaning of the question is that the operating cost is the overall operating cost and so then there is a trade off as having too few machines will result in too long a time to produce the units thus driving up the cost.

In this case then (using CB's notation)

$C(N,n) = 80 n + \frac{N}{50 n} 5 = 80 n + \frac{N}{10 n}$

so then the derivative needs to be calculated and set to zero to find the maximum i.e. (with $N=8000$)

$\frac{dC}{dn} = 80 \left(1 -\frac{10}{n^2} \right) = 0$ when $n^2 = 10$ .

By computing the second derivative we find this corresponds to a minimum and then trying the nearest integers $n=3$ and $n=4$ we find $n=3$ has the lowest cost. So then the answer is then 3 machines.

Lisa, was this part of a problem sheet related to derivatives or is what I've just done alot more complicated than what you've studied?

Guys, if I'm misinterpreting and overcomplicating this then I apologise but just thought I'd share the thought with you just in case as the question does seem too easy otherwise.

7. Yes this part of derivatives and also additional applied optimization...I got this wrong on my exam and I have been trying to see how it was done..I'm sorry if I confused anyone!

8. Originally Posted by the_doc
Guys I think this question may have been misunderstood.

I think CaptainBlack and HallsofIvy have taken the common sense approach to this question and taken the operating cost of $5 per hour to mean per machine however you'll note there is no per machine in that part of the question whilst there is in the other parts. That the operating cost is$5 per hour per machine is a legitimate interpretation of:

The setup cost is $80 per machine, and the operating cost is$5 per hour.
though it is possible that the per machine pertains only to the set up cost, but it makes little sense that it costs the same to run 100 machines for 100 hours as it does to run 1 machine for 100 hours (and the previous interpretation is the way costs usually accumulate, energy use, operator time and most maintence charges accumulate per machine per hour of operation).

Now I think that the running costs is per machine per hour is the intended meaning, but then we have a pointless question. Which is why I think this question had no response for 11 hours. It is also why I replied, to provoke some form of clarification.

CB

9. I don't think I made myself very clear in my last post.

I wasn't disagreeing with your interpretation. I agree it is a valid interpretation.

I was simply stating that there is another possible interpretation (which I agree goes against common sense - that more machines should cost more to run per hour) when you take into account the context of the question.

As you stated it does seem to be a bit of a pointless question but then maybe it's just posted in the wrong forum section.

10. Originally Posted by lisa1984wilson
Yes this part of derivatives and also additional applied optimization...I got this wrong on my exam and I have been trying to see how it was done..I'm sorry if I confused anyone!
I didn't see this when I wrote my last post .

So perhaps the alternative interpretation I suggested is the way to go given the context (despite being nonsensical in terms of real world thinking)?