At the lowest point of long run average cost curve AC, the SAC is also at its minimum and that is not the case with any other SAC curve bounded by the envelope. Why is that?

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Short-run implies that some decision variable cannot be free set, it is fixed for a time (in the short-run). In the long-run all variables may be freely set. Let us denote the fixed variable by $x$. This can take different values, so there is a family of short run cost functions is $SC(x,q)$.

Given a value $q$ let us denote the cost minimizing value of $x$ by $x^*$, i.e. $$ x^* = \arg\min_x SC(x,q). $$ Cost minimization implies that in the long-run this will be the chosen value of $x$, thus $$ C(q) = SC(x^*,q) = \min_x SC(x,q). $$ Dividing by $q$ we get $$ AC(q) = SAC(x^*,q) = \min_x SAC(x,q). $$ The above holds for any $q$.

  • $\begingroup$ Thanks for the answer, but your final conclusion doesn't agree with other parts of the graph right? At other places, the SAC's minima is not tangent to the AC. $\endgroup$ May 8 '21 at 15:47
  • $\begingroup$ I am not sure what you mean, as I don't think I wrote anything about this, but note that the graph displays minima with respect to $q$, not $x$. (Your graph denotes variable $x$ by $k$). That is the lowest point of curve $SAC(k_0)$ on the graph has height $\min_q SAC(k_0,q)$ and not $\min_{k_0} SAC(k_0,q)$. $\endgroup$
    – Giskard
    May 8 '21 at 16:36

@Giskard and @tdm have given you perfectly valid responses. I am not convinced that your question is perfectly clear.

Are you asking why only at $q_1$ (in your graph) an SRAC and AC are tangent at the min of an SRAC? If so, that's a result of the underlying assumptions about the production functions that lead to your graph.

In short, you have economies of scale (i.e., AC has a negative slope), a minimum efficient scale at $q_1$ (i.e., AC's slope is 0 at a single point), and diseconomies of scale (i.e., AC has a positive slope). SRACs have slope 0 at their minimum, while AC has slope 0 only at $q_1$ (at its minimum efficient scale). Hence, only at $q_1$ can the min of AC be tangent with the min of an SRAC.

Now had you made different assumptions about the production function the situation would be different. AC is the lower envelope of your SRACs. So, for example, if your minimum efficient scale was a range (instead of a point), or if your production function had constant returns to scale, then you would have infinite number of tangent points at min SRAC and min AC.


  1. https://www.economics.utoronto.ca/osborne/2x3/tutorial/C2LSFRM.HTM

  2. Varian (1992): Microeconomic Analysis p.p.71-72.

  3. https://uh.edu/~ghong/fina3334/chap_07.doc

Edit: fixed username


This result comes from the "zero-profit" condition used to define long run equilibrium in this model.

Recall the fact that in the "firm" part of the model, the marginal cost curve intersects with the average total cost curve at the minimum of the average total cost curve. Since marginal cost is the individual firm's supply curve, it follows that when price equals the minimum of total cost, the firm produces the quantity at the crossing point and has zero profit. A higher price will give the firm positive profit since P > ATC, so Revenues > Costs. A lower price will give the firm negative profit since P < ATC, so Revenues < Costs.

Long run equilibrium is defined by the number of firms being at the level that gives each firm zero profit. Thus the long run supply curve must be at the zero-profit price level, i.e. the minimum of average total cost.

  • 1
    $\begingroup$ (-1) This is not true. The result holds even without assuming prices or perfect competition. $\endgroup$
    – Giskard
    May 12 '21 at 7:52
  • $\begingroup$ Oh! That was not clear from the poster's original question, and the post was tagged "competitive equilibrium" so I interpreted it in that context. Could you point to the part of the question that tells you that this information? $\endgroup$
    – Jonathan
    May 12 '21 at 10:26
  • $\begingroup$ "Could you point to the part of the question that tells you that this information?" What do you mean? $\endgroup$
    – Giskard
    May 12 '21 at 10:27
  • $\begingroup$ You are right about competitive equilibrium being an irrelevant tag, I removed it. $\endgroup$
    – Giskard
    May 12 '21 at 10:28
  • 1
    $\begingroup$ I am not going to argue with you about this, you corrected me about the setting and so I accepted that you are correct and changed my position in response. Your answer above uses cost minimization, and I agree with it. Sorry if that was unclear. $\endgroup$
    – Jonathan
    May 12 '21 at 10:45

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