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An industry has a total cost function : TC=$4Q^2+100Q+100$ . Where $Q$ is the quantity produced. They are asking me to find the long run equilibrium price. How do I find it? What I've found is that i calculate the sratc(short run average total cost) and then solve for Q (while equating the derivative of sratc to 0), then find Price after substituting Q. Is this the way to go?

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  • $\begingroup$ Depends on the model. In a model with no entry costs and perfect competition, total costs equal price long term. $\endgroup$ – Don Thousand May 8 at 1:00
  • $\begingroup$ yes its perfectly competitive and no entry cost. So P=TC? $\endgroup$ – The Poor Jew May 8 at 1:07
  • $\begingroup$ @ThePoorJew I had uploaded a picture based on a wrong function. The current picture is the right one. $\endgroup$ – callculus May 8 at 14:42
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Hint: In a competitive market the firms produce on the long run at a level where the average total cost function has its minimum. Thus you have to find the minimum of

$$\frac{TC(Q)}{Q}=4Q+100+\frac{100}{Q}$$

The minimum can be found by setting the derivative equal to $0$. In this case it is good to remember that $Q$ is defined for non-negative values only. The picture below shows the course of the average cost function.

enter image description here

The price of the product will be equal the solution.

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    $\begingroup$ Okay so this means I was right at the beginning? Thanks for the answer $\endgroup$ – The Poor Jew May 8 at 16:32
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    $\begingroup$ @ThePoorJew Maybe. The term "short run" has confused me. But you know by yourself better if you were right at the beginning. You´re welcome. $\endgroup$ – callculus May 8 at 16:38
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In a perfectly competitive market, the long-run equilibrium price is the price at which the firm earns zero economic profit, which will happen at $MC=ATC$. Hence: $$MC=TC'=(4Q^2+100Q+100)'=8Q+100\\ ATC=\frac{TC}{Q}=4Q+100+\frac{100}{Q}\\ 8Q+100=4Q+100+\frac{100}{Q} \Rightarrow 4Q^2=100 \Rightarrow Q=5\\ P=MR=MC(5)=8\cdot 5+100=140.$$ Verify: $$TR=140Q; TC=4Q^2+100Q+100\\ TR(5)-TC(5)=140\cdot 5-(4\cdot 5^2+100\cdot 5+100)=0.$$

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  • $\begingroup$ yes, i got the same answers $\endgroup$ – The Poor Jew May 8 at 21:58

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