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?

Second-order of AC''

  • 2
    $\begingroup$ 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? $\endgroup$
    – Giskard
    Aug 30, 2021 at 14:31
  • 4
    $\begingroup$ You need to make some assumptions on your cost function for this to be true $\endgroup$ Aug 30, 2021 at 15:45
  • $\begingroup$ @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 $\endgroup$
    – user37865
    Sep 2, 2021 at 11:09
  • $\begingroup$ @Giskard added the AC'', $\endgroup$
    – user37865
    Sep 2, 2021 at 11:12

2 Answers 2


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.

Proof by contradiction:
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 $AC(q_b) = MC(q_b)$.

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 $AC(q_b) < MC(q_b)$.

  • $\begingroup$ I feel that this proof could be simplified and made more elegant, but I have spent too much time on it as it is. $\endgroup$
    – Giskard
    Sep 3, 2021 at 7:13
  • 1
    $\begingroup$ thank you so much for the detailed proof! that was exactly what helped me to understand! $\endgroup$
    – user37865
    Sep 4, 2021 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 $AC'(q_c) = 0$.


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