Mathematically show that MC cross AC can only occur on upward sloping part of MC curve

Question as in title, does anyone know how to show it? with mean value theorem?

I can prove that AC is minimised at MC=AC but I'm not sure how to show MC can only cross AC at upward-sloping point of MC

My first part simply takes AC = c(q)/q and AC' eventually after simplifying leads to c'(q) = c(q)/q [given AC=0 at min point] so thus c'(q) = MC and therefore at the min point AC = MC

I am thinking how to show using mean value theorem that AC & MC can only cross at the upward sloping MC curve (where MC' >0); I know graphically how and why but not how to start off to show? • When you proved AC is minimised at MC=AC, did you look at the second order conditions? If yes, can you please edit your work into your question? Aug 30 '21 at 14:31
• You need to make some assumptions on your cost function for this to be true Aug 30 '21 at 15:45
• @MichaelGreinecker yes; the assumptions are total cost function is increasing and smooth (first and second derivatives are continuous) with , and MC function is first strictly decreasing and then strictly increasing in q Sep 2 '21 at 11:09
• @Giskard added the AC'', Sep 2 '21 at 11:12

Given the assumption that $$C(q)$$ is continuously differentiable we have for all $$q$$: $$qAC(q) = C( q ) = FC + \int_0^{q} MC(x) \text{d}x.$$ Taking the difference for any pair $$q,\hat{q}$$: $$qAC(q) - \hat{q}AC(\hat{q}) = \int_{\hat{q}}^{q} MC(x) \text{d}x. \tag{1}$$ The left hand side may be reformulated: $$q\left(AC(q) - AC(\hat{q}) \right) + \left(q - \hat{q}\right)AC(\hat{q}) = \int_{\hat{q}}^{q} MC(x) \text{d}x$$ and the form we will use is $$q\left(AC(q) - AC(\hat{q}) \right) = -\left(q - \hat{q}\right)AC(\hat{q}) + \int_{\hat{q}}^{q} MC(x) \text{d}x \tag{2}$$ Assume $$C(q)$$ is strictly convex and $$MC'(q)$$ switches sign somewhere; furthermore assume that $$q_c$$ exists where $$AC(q_c) = MC(q_c)$$, and $$AC(q)$$ is minimized at this point.

We will first show that $$q_c$$ is not in the range where $$MC(q)$$ is decreasing.

Assume $$MC(q)$$ is strictly decreasing at $$q_c$$, thus there exists a small environment $$(q_c, q_c + \epsilon)$$ such that for $$q \in (q_c, q_c + \epsilon)$$ we have $$\int_{q_c}^{q} MC(x) \text{d}x < \left(q - q_c\right)MC(q_c).$$ Combining this with (2), we get $$q\left(AC(q) - AC(q_c) \right) < -\left(q - q_c\right)AC(q_c) + \left(q - q_c\right)MC(q_c) = 0.$$ From this it follows that $$AC(q) < AC(q_c),$$ contradicting the assumption that $$q_c$$ minimizes $$AC(q)$$.

We will now show that there is no $$q_b > q_c$$ for which $$AC(q_b) = MC(q_b)$$.

Proof:
At $$q_b$$ $$MC(q)$$ is strictly increasing, thus $$\int_{q_c}^{q_b} MC(x) \text{d}x < \left(q_b - q_c\right)MC(q_b).$$ As $$AC(q)$$ is minimized at $$q_c$$, we also have $$-AC(q_c) > -AC(q_b)$$, and thus $$q_bAC(q_b) - q_cAC(q_c) > q_bAC(q_b) - q_cAC(q_b) = \left(q_b - q_c\right)AC(q_b)$$ Combining these inequalities with $$qAC(q) - \hat{q}AC(\hat{q}) = \int_{\hat{q}}^{q} MC(x) \text{d}x. \tag{1}$$ we get $$\left(q_b - q_c\right)AC(q_b) < q_bAC(q_b) - q_cAC(q_c) = \int_{q_c}^{q_b} MC(x) \text{d}x < \left(q_b - q_c\right)MC(q_b),$$ thus $$AC(q_b) < MC(q_b)$$.

• I feel that this proof could be simplified and made more elegant, but I have spent too much time on it as it is. Sep 3 '21 at 7:13
• thank you so much for the detailed proof! that was exactly what helped me to understand! Sep 4 '21 at 8:44

You claim that

AC is minimised at MC=AC

thus at this quantity $$q_c$$ we have $$AC'(q_c) = 0$$.

We will show that given the assumptions $$MC'(q_c) \geq 0$$, that is $$MC$$ cannot be decreasing in $$q$$ at this location.

For all $$q$$ it is true that $$AC'(q) = \frac{MC(q)-AC(q)}{q}.$$ For all $$q > q_c$$ we have $$0 = AC'(q_c) = \frac{MC(q_c) - AC(q_c)}{q_c} = \frac{MC(q_c) - AC(q_c)}{q}$$ (as the numerator is $$0$$), and if $$MC$$ is decreasing at $$q_c$$, then for all $$q$$ close enough to $$q_c$$ we have $$MC(q_c) > MC(q)$$. Combine this with the above and we get $$0 = \frac{MC(q_c) - AC(q_c)}{q} > \frac{MC(q) - AC(q)}{q} = AC'(q),$$ where $$-AC(q_c)>-AC(q)$$ holds because $$q_c$$ is supposed to be where $$AC$$ is minimized.

We have shown that for all $$q > q_c$$ the inequality $$0 > AC'(q)$$, which (together with the assumption that second derivatives are continuous) contradicts $$AC'(q_c) = 0$$.