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Suppose all firms are monopolistically competitive with constant elasticity of substitution. Can they profit maximize at the quantity point $MC=AC$ (So $P \neq MC$ but $Q$ is set at where $MC=MR=AC$ is established)?

By constant elasticity of substitution, I mean that $\epsilon = -(\partial Q_j/\partial P_i)/(Q_j/P_i)$ is constant for $\forall i \neq j$ where $i,j$ refers to firms, $Q$ refers to quantity, and $P$ refers to price.

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What here has the "constant elasticity of substitution" property? Production? Demand? Supply? Revenues? Costs? Please clarify. –  Alecos Papadopoulos Dec 24 '14 at 19:57
Sorry for my omission. Now I clarified. –  Gantakata Dec 24 '14 at 20:01
Your title had a condition on MR, but the question does not. Can you clarify? –  Dimitriy V. Masterov Dec 25 '14 at 4:12
Clarified. It's just ordinary profit-maximization stuff - in almost all economic structure, firms set their quantity at $MC=MR$ point (it's just first-order condition) - though if firms are not perfect competition type, then $P \neq MC$ - I omitted in the main text because it was so obvious anyway. –  Gantakata Dec 25 '14 at 4:21

1 Answer 1

The long-run picture for a firm in monopolistically competitive industry should look like this:

enter image description here

The level of production is pinned down by where MR=MC. Price will be equal to AC at that point, so the firm has zero profits.

If the firm was producing at the point where MC=AC, it would have positive profits, which would encourage entry into the industry, shifting the demand in and reducing profits to zero.

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If by MC=AC, you are referring to the intersection of MC and AC, that would generate negative profits. AR<AC. –  Pburg Dec 26 '14 at 15:09
@Pburg I think profit maximization means that the MR curve would also have to pass trough the point where MC=AC, which means AR (or demand) must lie above it, so AR>AC at that point, indicating positive profits. –  Dimitriy V. Masterov Dec 26 '14 at 15:32
I see, my mistake. I thought you were referring to the MC=AC point on the same graph you have included. –  Pburg Dec 26 '14 at 19:13

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