<p>Given the assumption that <span class="math-container">$C(q)$</span> is continuously differentiable we have for all <span class="math-container">$q$</span>:
<span class="math-container">$$
qAC(q) = C( q ) = FC + \int_0^{q} MC(x) \text{d}x. 
$$</span>
Taking the difference for any pair <span class="math-container">$q,\hat{q}$</span>:
<span class="math-container">$$
qAC(q) - \hat{q}AC(\hat{q}) = \int_{\hat{q}}^{q} MC(x) \text{d}x. \tag{1}
$$</span>
The left hand side may be reformulated:
<span class="math-container">$$
q\left(AC(q) - AC(\hat{q}) \right) + \left(q - \hat{q}\right)AC(\hat{q}) = \int_{\hat{q}}^{q} MC(x) \text{d}x 
$$</span>
and the form we will use is
<span class="math-container">$$
q\left(AC(q) - AC(\hat{q}) \right) = -\left(q - \hat{q}\right)AC(\hat{q}) + \int_{\hat{q}}^{q} MC(x) \text{d}x \tag{2}
$$</span>
Assume <span class="math-container">$C(q)$</span> is strictly convex and <span class="math-container">$MC'(q)$</span> switches sign somewhere; furthermore assume that <span class="math-container">$q_c$</span> exists where <span class="math-container">$AC(q_c) = MC(q_c)$</span>, and <span class="math-container">$AC(q)$</span> is minimized at this point.</p>
<hr />
<p><strong>We will first show that <span class="math-container">$q_c$</span> is not in the range where <span class="math-container">$MC(q)$</span> is decreasing.</strong></p>
<p><strong>Proof by contradiction:</strong>
<br> Assume <span class="math-container">$MC(q)$</span> is strictly decreasing at <span class="math-container">$q_c$</span>, thus there exists a small environment <span class="math-container">$(q_c, q_c + \epsilon)$</span> such that for <span class="math-container">$q \in (q_c, q_c + \epsilon)$</span> we have
<span class="math-container">$$
\int_{q_c}^{q} MC(x) \text{d}x &lt; \left(q - q_c\right)MC(q_c).
$$</span>
Combining this with (2), we get
<span class="math-container">$$
q\left(AC(q) - AC(q_c) \right) &lt; -\left(q - q_c\right)AC(q_c) + \left(q - q_c\right)MC(q_c) = 0.
$$</span>
From this it follows that
<span class="math-container">$$AC(q) &lt; AC(q_c),$$</span>
contradicting the assumption that <span class="math-container">$q_c$</span> minimizes <span class="math-container">$AC(q)$</span>.</p>
<hr />
<p><strong>We will now show that there is no <span class="math-container">$q_b &gt; q_c$</span> for which <span class="math-container">$AC(q_b) = MC(q_b)$</span>.</strong></p>
<p><strong>Proof:</strong>
<br> At <span class="math-container">$q_b$</span>  <span class="math-container">$MC(q)$</span> is strictly increasing, thus
<span class="math-container">$$
\int_{q_c}^{q_b} MC(x) \text{d}x &lt; \left(q_b - q_c\right)MC(q_b).
$$</span>
As <span class="math-container">$AC(q)$</span> is minimized at <span class="math-container">$q_c$</span>, we also have <span class="math-container">$-AC(q_c) &gt; -AC(q_b)$</span>, and thus
<span class="math-container">$$
q_bAC(q_b) - q_cAC(q_c) &gt; q_bAC(q_b) - q_cAC(q_b) = \left(q_b - q_c\right)AC(q_b) 
$$</span>
Combining these inequalities with
<span class="math-container">$$
qAC(q) - \hat{q}AC(\hat{q}) = \int_{\hat{q}}^{q} MC(x) \text{d}x. \tag{1}
$$</span>
we get
<span class="math-container">$$
\left(q_b - q_c\right)AC(q_b) &lt; q_bAC(q_b) - q_cAC(q_c) = \int_{q_c}^{q_b} MC(x) \text{d}x &lt; \left(q_b - q_c\right)MC(q_b),
$$</span>
thus <span class="math-container">$AC(q_b) &lt; MC(q_b)$</span>.</p>

CC BY-SA 4.0
2021-09-03T07:13:10.677Z
I feel that this proof could be simplified and made more elegant, but I have spent too much time on it as it is.
Giskard
CC BY-SA 4.0
2021-09-04T08:44:17.263Z
thank you so much for the detailed proof! that was exactly what helped me to understand!
user37865
CC BY-SA 4.0
2021-09-03T07:12:39.660Z
2021-09-04T12:11:40.937Z
2021-09-04T12:11:40.937Z
Giskard
<p>You claim that</p>
<blockquote>
<p>AC is minimised at MC=AC</p>
</blockquote>
<p>thus at this quantity <span class="math-container">$q_c$</span> we have <span class="math-container">$AC'(q_c) = 0$</span>.</p>
<p>We will show that given the assumptions <span class="math-container">$MC'(q_c) \geq 0$</span>, that is <span class="math-container">$MC$</span> cannot be decreasing in <span class="math-container">$q$</span> at this location.</p>
<p>For all <span class="math-container">$q$</span> it is true that
<span class="math-container">$$
AC'(q) = \frac{MC(q)-AC(q)}{q}.
$$</span>
For all <span class="math-container">$q &gt; q_c$</span> we have
<span class="math-container">$$
0 = AC'(q_c) = \frac{MC(q_c) - AC(q_c)}{q_c} = \frac{MC(q_c) - AC(q_c)}{q}  
$$</span>
(as the numerator is <span class="math-container">$0$</span>),
and if <span class="math-container">$MC$</span> is decreasing at <span class="math-container">$q_c$</span>, then for all <span class="math-container">$q$</span> close enough to <span class="math-container">$q_c$</span> we have
<span class="math-container">$MC(q_c) &gt; MC(q)$</span>. Combine this with the above and we get
<span class="math-container">$$
0 = \frac{MC(q_c) - AC(q_c)}{q} &gt; \frac{MC(q) - AC(q)}{q} = AC'(q), 
$$</span>
where <span class="math-container">$-AC(q_c)&gt;-AC(q)$</span> holds because <span class="math-container">$q_c$</span> is supposed to be where <span class="math-container">$AC$</span> is minimized.</p>
<p>We have shown that for all <span class="math-container">$q &gt; q_c$</span> the inequality <span class="math-container">$0 &gt; AC'(q)$</span>, which (together with the assumption that second derivatives are continuous) contradicts <span class="math-container">$AC'(q_c) = 0$</span>.</p>

CC BY-SA 4.0
2021-09-02T14:15:45.547Z
2021-09-02T14:15:45.547Z
Giskard
<p>Question as in title, does anyone know how to show it? with mean value theorem?</p>
<p>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</p>
<p>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</p>
<p>I am thinking how to show using mean value theorem that AC &amp; MC can only cross at the upward sloping MC curve (where MC' &gt;0); I know graphically how and why but not how to start off to show?</p>
<p><a href="https://i.sstatic.net/rmJaY.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/rmJaY.png" alt="Second-order of AC''" /></a></p>

CC BY-SA 4.0
2021-08-30T14:31:17.013Z
When you proved AC is minimised at MC=AC, did you look at the second order conditions? If yes, can you please [edit](https://economics.stackexchange.com/posts/47374/edit) your work into your question?
Giskard
CC BY-SA 4.0
2021-08-30T15:45:06.673Z
You need to make some assumptions on your cost function for this to be true
Michael Greinecker
CC BY-SA 4.0
2021-09-02T11:09:44.683Z
@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
user37865
CC BY-SA 4.0
2021-09-02T11:12:27.603Z
@Giskard added the AC'',
user37865
<microeconomics><mathematical-economics><cost>
Mathematically show that MC cross AC can only occur on upward sloping part of MC curve
user37865
# Mathematically show that MC cross AC can only occur on upward sloping part of MC curve

(Asked by: username_0 on 2021-08-30)

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''](https://i.sstatic.net/rmJaY.png)](https://i.sstatic.net/rmJaY.png)

**Tags:** microeconomics, mathematical-economics, cost

> **Comments:**
> **username_1**: When you proved AC is minimised at MC=AC, did you look at the second order conditions? If yes, can you please [edit](https://economics.stackexchange.com/posts/47374/edit) your work into your question?  
> **username_2**: You need to make some assumptions on your cost function for this to be true  
> **username_0**: @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  
> **username_0**: @username_1 added the AC'',  

---

## 2 answers

### Answer by username_1 on 2021-09-03. Score: 1 | Accepted answer

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 $AC(q\_b) = MC(q\_b)$.**

**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 $AC(q\_b) < MC(q\_b)$.

> **Comments:**
> **username_1**: I feel that this proof could be simplified and made more elegant, but I have spent too much time on it as it is.  
> **username_0**: thank you so much for the detailed proof! that was exactly what helped me to understand!  

---

### Answer by username_1 on 2021-09-02. Score: 2

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 $AC'(q\_c) = 0$.


---

