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#1 2005-12-01 18:36:38
- Saitenji
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Optimization Problems (Calculus AB)
I was absent for several days on the account of me being sick...And so, I have no other resource but the MathIsFun forums in order to receive help for my homework during this late night... (I can only hope someone is online and able to help right at this moment...)
I would kindly appreciate anyone who is willing to assist me by informing me on how to work out this problem...My mind has pulled a complete blank...I've tried things that we've went over in the past, but to no avail...It seems that this is new material...material of which I have missed due to my absences.
In any case...here is the problem:
A company estimates that the cost (in dollars) of producing x units of a certain product is given by C= 800 + 0.04x + 0.0002x². Find the production level that minimizes the average cost per unit.
Again...I have very little clue on how to do this, so any form of help would be very much appreciated.
Thanks in advance.
Last edited by Saitenji (2005-12-01 18:37:17)
#2 2005-12-01 19:29:39
- MathsIsFun
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Re: Optimization Problems (Calculus AB)
Hi Saitenji,
You don't need a reason to ask for help.
Cost = 800 + 0.04x + 0.0002x²
Cost/Unit = 800/x + 0.04 + 0.0002x
To find a minimum/maximum we usually differentiate and find where the rate of change is zero (I hope you are OK with a little calculus)
derivative of 800/x + 0.04 + 0.0002x = 800 (-1/x²) + 0 + 0.0002
And you want that to be zero, so: -800/x² + 0.0002 = 0
Rearranging: 800/x² = 0.0002
Rearranging: x = √(800/0.0002) = 2000
(This is where the slope of the line is zero, it could be a minimum or maximum, I am guessing it is the minumum!)
"The physicists defer only to mathematicians, and the mathematicians defer only to God ..." - Leon M. Lederman
#3 2005-12-01 19:32:22
- darthradius
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Re: Optimization Problems (Calculus AB)
I can't remember...but isn't it usually the place where the derivative is equal to zero...those critical points are either maximums or minimums, and so I suppose that would be a place to start.....??
good luck
The greatest challenge to any thinker is stating the problem in a way that will allow a solution.
-Bertrand Russell
#4 2005-12-01 19:35:08
- MathsIsFun
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Re: Optimization Problems (Calculus AB)
LOL, you must have written that while I was finishing off my reply!
Just for that you can check my work 
(my failure rate is about 1 in 10)
"The physicists defer only to mathematicians, and the mathematicians defer only to God ..." - Leon M. Lederman
#5 2005-12-01 20:15:18
- darthradius
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Re: Optimization Problems (Calculus AB)
It looks right to me!
And, it is definitely a minimum...the value of the derivative function around x=2000 confirms it...plus the second derivative [ 1600/(x^3) ]at x=2000 is positive...
The greatest challenge to any thinker is stating the problem in a way that will allow a solution.
-Bertrand Russell
#6 2005-12-02 08:41:20
- irspow
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Re: Optimization Problems (Calculus AB)
Technically speaking, setting the first derivative equal to zero only finds local extrema. To be sure that the minimum or maximum is truly the global extrema one needs to find the limit of the function at its end points.
If you do this in this case it indeed proves that the local minimum found is global also. There is no need to test in the negative direction because negative units do not exist, but....
lim 800/x + .04 + .0002x = + ∞
x⇒+∞
Many people forget to do this, but many functions continue to change after a local extrema and unless the value changes direction again the first derivative will not point it out.
Last edited by irspow (2005-12-02 12:26:01)