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 April 26th, 2011, 02:33 PM #1 Newbie   Joined: Apr 2011 Posts: 5 Thanks: 0 Average cost differentiation Hi i need some help to derive the AC (average cost) function and find the output which minimises the AC I've been given TC= 24 +0.3q² TR= 297 - 13q²
 April 26th, 2011, 06:41 PM #2 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 521 Math Focus: Calculus/ODEs Re: Average cost differentiation The average cost AC is found by dividing the total cost TC by the output quantity q: $AC=\frac{TC}{q}=\frac{24+0.3q^2}{q}=24q^{-1}+0.3q$ Now, to find the output that minimizes AC, we must differentiate with respect to q, and equate to 0: $\frac{d}{dq}AC=-24q^{-2}+0.3=\frac{3$$q^2-80$$}{10q^2}=0$ Since this is equal to zero when the numerator is equal to zero, we have: $q^2-80=0$ Taking the positive root: $q=\sqrt{80}=4\sqrt{5}$ To ensure that this is a minimum, we note 8 < q < 9 and: $AC'( < 0" /> and $AC'(9) > 0$ Now, if q is in units, then we should round to q = 9. This would have been a nicer problem if we had been given TC = 24.3 + 0.3q². Something else we may use is "When average cost is neither rising nor falling (at a minimum or maximum), marginal cost equals average cost." To see why this works, consider: $AC'=0$ $TC'=AC=\frac{TC}{q}$ $AC'=\frac{q\cdot TC#39;-TC}{q^2}=\frac{q\cdot AC-q\cdot AC}{q^2}=0$ Now, marginal cost MC is the derivative of TC with respect to q: $MC=\frac{d}{dq}$$24+0.3q^2$$=0.6q$ $AC=24q^{-1}+0.3q$ Equating the two, we have: $0.6q=24q^{-1}+0.3q$ $0.3q=24q^{-1}$ $0.3q^2=24$ $q^2=80\:\therefore\:q=4\sqrt{5}$ as we found before.
 April 26th, 2011, 10:41 PM #3 Newbie   Joined: Apr 2011 Posts: 5 Thanks: 0 Re: Average cost differentiation Thanks that really helped alot
 April 26th, 2011, 10:48 PM #4 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 521 Math Focus: Calculus/ODEs Re: Average cost differentiation You're welcome and welcome to the forum! Does this question have more to it? I'm wondering where the total revenue function comes into play.
 April 26th, 2011, 11:56 PM #5 Newbie   Joined: Apr 2011 Posts: 5 Thanks: 0 Re: Average cost differentiation oh yeah the TR is then used to; set up the firms profit function and then find the output that will maximize the profit.
 April 27th, 2011, 12:00 AM #6 Newbie   Joined: Apr 2011 Posts: 5 Thanks: 0 Re: Average cost differentiation but as far as i could see i didnt need AC to work out a profit function as ?= TR-TC my maximizing output was 11.165 or 11 rounded i wonder if i was right
 April 27th, 2011, 12:15 AM #7 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 521 Math Focus: Calculus/ODEs Re: Average cost differentiation I've never seen ? used to represent profit, but any character can be used! $\pi=TR-TC=$$297-13q^2$$-$$24+0.3q^2$$=273-13.3q^2$ Obviously, the profit function has its maximum at q = 0...are you sure the TR function is correct?
 April 27th, 2011, 12:29 AM #8 Newbie   Joined: Apr 2011 Posts: 5 Thanks: 0 Re: Average cost differentiation Its what i got on my sheet... but the TR and TCs numbers are based on my student ID number so everyones numbers are meant to be different...
 April 27th, 2011, 12:37 AM #9 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 521 Math Focus: Calculus/ODEs Re: Average cost differentiation How did you get 11.165?

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# differentiate average total cost

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