# Why min AC = min SRAC at the minima of AC curve?

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?

• – tdm
May 8 at 9:27
• NO @tdm ! It's completely different. May 8 at 10:30
• @tdm I know that they are tangnet, but they are tangent at their minimas at the minima of LRAC May 8 at 10:31
• At the other points, say $q_0$, the $SMC(k_0)$ is below the $SAC(k_0)$. This means the $SAC(k_0)$ is decreasing at that point. So the tangent point can not be at the minimum.
– tdm
May 8 at 17:36

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 $$q$$.

• 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. May 8 at 15:47
• 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 $\min_{k_0} SAC(k_0,q)$. May 8 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.

See:

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