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.

]]>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...]]>

Just for that you can check my work (my failure rate is about 1 in 10)

]]>good luck]]>

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!)

]]>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.

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