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could you help me to solve the task, please? The task is following. Total costs of the сompetitive firm are $TC = Q^3 +6$ (Q in thousands \$). If the production is stopped then monthly costs are 4000\$. At what prices the company that maximising profit should shut off the production?

I know that if $P < minAVC$, then the company should stop producing. So $VC = Q^3$ then $AVC = \frac{Q^3}{Q} = Q^2$ and $minAVC = 0$ So I get $P<0$ that sounds stupid. And I don't even know how to use that 4000\$. The answer should be $P<3$ Could you help me to find out how to solve this?
Thank you in advance.

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  • $\begingroup$ Realistically, you can get VC=0 only if Q=0.But if Q=0, then you do NOT have AVC, because you would divide by zero. $\endgroup$
    – user161005
    Jan 8 '18 at 13:27
  • $\begingroup$ "(I made a mistake, there was 'AC=Q^3', but still, not ATC)" AC is just shorthand for ATC, as far as I know. $\endgroup$
    – user161005
    Jan 8 '18 at 13:30
  • $\begingroup$ oh, didn't know that. it was just a typo. But still, any thoughts what should I do to get a correct answer? :D $\endgroup$
    – elfinorr
    Jan 8 '18 at 13:37
  • $\begingroup$ One more thing, I think it's strange that according to function of TC the fixed costs are $6000 (when Q=0), but then you say that fixed costs will be 4000 dollars if the firm stops. $\endgroup$
    – user161005
    Jan 8 '18 at 13:53
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I think I understand the problem now, your AVC(Q) is wrong. Your fixed costs are 4000, because it's what the firm will pay after it stops production. VC=TC-4=Q^3+2. Thus AVC=Q^2+2/Q. Now you should use your calculus in order to find where AVC has local positive minimum (because it doesn't have global minimum due to its hyperbolic nature).

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  • $\begingroup$ ... and that is exactly at p=3 Thank you very much, you are a genious! ^^ $\endgroup$
    – elfinorr
    Jan 8 '18 at 15:27
  • $\begingroup$ @elfinorr Thanks, but are you sure that you calculated everything correctly? Shutdown price is when P=min AVC, and my calculations make me believe, that Q=3 is NOT its local minimum $\endgroup$
    – user161005
    Jan 9 '18 at 8:45
  • $\begingroup$ But Q = 1 is, and AVC(1) = 3. It seems to be okay. $\endgroup$
    – elfinorr
    Jan 9 '18 at 16:56
  • $\begingroup$ Ooops, sorry, my mistake. Everything is OK. $\endgroup$
    – user161005
    Jan 9 '18 at 19:36

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