<p>The idea that the long run average cost curve (LRAC) must pass through the minimum points of the short run average cost curves (SRAC) is a fallacy, but it seems to be a remarkably plausible one.  It was the source of a famous error by the economist Jacob Viner, referred to in <a href="https://www.jstor.org/stable/1183036?seq=1#page_scan_tab_contents" rel="nofollow noreferrer">this paper by Silberberg</a>.  Underlying the fallacy is perhaps an assumption that the points of tangency with the LRAC must be the minimum points of the SRAC’s.  These points are coincident in the special case of a SRAC tangent to the LRAC <em>at the minimum point of the latter</em>.  But usually they are distinct, as in the numerical example below.</p>

<p>Suppose a firm has a Cobb-Douglas production function with increasing returns  <span class="math-container">$y = x_1^{0.6}x_2^{0.6}$</span>.  The inputs are bought in markets in which supply is not perfectly elastic, so that costs are increasing functions of quantities:</p>

<p><span class="math-container">$\quad c_1(x_1)  =  20x_1 + x_1^2$</span></p>

<p><span class="math-container">$\quad c_2(x_2)  =  20x_2 + x_2^2$</span></p>

<p>The symmetry between the two inputs in respect of both technology and costs is not necessary to obtain a suitable example, but it is convenient because it implies that every point on the LRAC must satisfy <span class="math-container">$x_1=x_2$</span> (see Appendix below for proof).  This simplifies the derivation of  the equation of the LRAC.  </p>

<p><strong>LRAC</strong></p>

<p>Writing <span class="math-container">$c(a,b)$</span> for the total cost function with inputs <span class="math-container">$a, b$</span> and given <span class="math-container">$x_1=x_2$</span> we have:</p>

<p><span class="math-container">$\quad c(x_1,x_2) = 40x_1 + 2x_1^2\qquad(1)$</span></p>

<p><span class="math-container">$\quad y = x_1^{1.2}\qquad(2)$</span></p>

<p>Hence:</p>

<p><span class="math-container">$\quad x_1 = y^{5/6}\qquad(3)$</span></p>

<p><span class="math-container">$\quad c(x_1,x_2)  =  40y^{5/6} + y^{5/3}\qquad(4)$</span></p>

<p>and so:</p>

<p><span class="math-container">$\quad LRAC = \frac{40y^{5/6} + 2y^{5/3}}{y}  =  40y^{-1/6}+2y^{2/3}\qquad(5)$</span></p>

<p>To find the minimum point:</p>

<p><span class="math-container">$\quad \frac{dLRAC}{dy}  =  (-40/6)y^{-7/6} + (4/3)y^{-1/3} = 0\qquad(6)$</span></p>

<p><span class="math-container">$\quad -40/6 + (4/3)y^{5/6} = 0\qquad(7)$</span></p>

<p><span class="math-container">$\quad y^{5/6} = 5\qquad(8)$</span></p>

<p><span class="math-container">$\quad y=6.90\qquad(9)$</span></p>

<p>To confirm this is a minimum:</p>

<p><span class="math-container">$\quad \frac{d^2LRAC}{dy^2} = (280/36)y^{-13/6} + (-4/9)y^{-4/3}$</span>
<span class="math-container">$\quad = (280/36)0.0152 – (4/9)0.0761 = 0.118 – 0.034 = 0.084 \boldsymbol{&gt; 0}\qquad(10)$</span></p>

<p>The inputs at this minimum, using (3), are:</p>

<p><span class="math-container">$\quad x_1 = x_2 = 6.90^{5/6} = 5.00\qquad(11)$</span></p>

<p><strong>SRAC</strong></p>

<p>Suppose now that <span class="math-container">$x_1$</span> is freely variable but <span class="math-container">$x_2$</span> is fixed in the short run at a value other than <span class="math-container">$5.00$</span>, say <span class="math-container">$2$</span>.  Then:</p>

<p><span class="math-container">$\quad y = x_1^{0.6}(2^{0.6})\qquad(12)$</span></p>

<p><span class="math-container">$\quad c(x_1,x_2)  =  20x_1 + x_1^2 + 44\qquad(13)$</span></p>

<p>Hence:</p>

<p><span class="math-container">$\quad x_1 = (2^{-0.6}y)^{5/3} =  (1/2)y^{5/3}\qquad(14)$</span></p>

<p><span class="math-container">$\quad c(x_1,x_2) = 10y^{5/3} + (1/4)y^{10/3} + 44\qquad(15)$</span></p>

<p>and so:</p>

<p><span class="math-container">$\quad SRAC(x_2 = 2) = \frac{10y^{5/3} + (1/4)y^{10/3} + 44}{y} = 10y^{2/3} + (1/4)y^{7/3} + 44y^{-1}\qquad(16)$</span></p>

<p>The first derivative is:</p>

<p><span class="math-container">$\quad \frac{dSRAC}{dy}  =  (20/3)y^{-1/3} + (7/12)y^{4/3} – 44y^{-2}\qquad(17)$</span></p>

<p><strong>Relation between LRAC and SRAC</strong></p>

<p>The two curves meet when <span class="math-container">$x_1=x_2=2$</span> implying <span class="math-container">$y = 2^{1.2} =  2.2974$</span> since at that point, using (5) and (16):</p>

<p><span class="math-container">$\quad LRAC = 40(2.2974^{-1/6}) + 2(2.2974^{2/3}) = 34.822 + 3.482 = \boldsymbol{38.30}\qquad(18)$</span></p>

<p><span class="math-container">$\quad SRAC = 10(2.2974^{2/3}) + (1/4)(2.2974^{7/3}) + 44(2.2974^{-1})$</span>
<span class="math-container">$\quad = 17.411 + 1.741 + 19.152 = \boldsymbol{38.30}\qquad(19)$</span></p>

<p>Moreover they are tangential at that point since using (6) and (17) the respective slopes are:</p>

<p><span class="math-container">$\quad \frac{dLRAC}{dy} = (-40/6)(2.2974^{-7/6}) + (4/3)(2.2974^{-1/3})$</span>
<span class="math-container">$\quad = -2.526 + 1.010 = \boldsymbol{-1.52}\qquad(20)$</span></p>

<p><span class="math-container">$\quad \frac{dSRAC}{dy}  =  (20/3)2.2974^{-1/3} + (7/12)2.2974^{4/3} + (-44)2.2974^{-2}$</span>
<span class="math-container">$\quad = 5.052 + 1.768 – 8.336 = \boldsymbol{-1.52}\qquad(21)$</span></p>

<p>However, this point of tangency is <strong>not</strong> the minimum point of the SRAC.  Using (17) to find the minimum:</p>

<p><span class="math-container">$\quad \frac{dSRAC}{dy}  =  (20/3)y^{-1/3} + (7/12)y^{4/3} – 44y^{-2}  =  0\qquad(22)$</span></p>

<p><span class="math-container">$\quad (20/3)y^{5/3} + (7/12)y^{10/3} – 44 = 0\qquad(23)$</span></p>

<p>Treating this as a quadratic equation in <span class="math-container">$y^{5/3}$</span>, or by trial and error, it can be found that <span class="math-container">$y$</span> is approximately <span class="math-container">$2.525$</span>.  To confirm this is a minimum:</p>

<p><span class="math-container">$\quad \frac{d^2SRAC}{dy^2}  =  (-20/9)2.525^{-4/3} + (28/36)2.525^{1/3} + (88)2.525^{-1}  =  -0.646 + 1.059 + 34.851 = 35.26 \boldsymbol{&gt; 0}\qquad(24)$</span></p>

<p>At this minimum point:</p>

<p><span class="math-container">$\quad SRAC = 10(2.525^{2/3}) + (1/4)2.525^{7/3} + 44(2.525^{-1})$</span>
<span class="math-container">$\quad = 18.543 + 2.170 + 17.426 = \boldsymbol{38.14}\qquad(25)$</span></p>

<p>This is lower than the point of tangency with the LRAC (<span class="math-container">$\boldsymbol{38.30}$</span>), but above the LRAC at <span class="math-container">$y = 2.525$</span> which using (5) is:</p>

<p><span class="math-container">$\quad LRAC = 40(2.525^{-1/6}) + 2(2.525^{2/3}) = 34.278 + 3.709 = \boldsymbol{37.99}\qquad(26)$</span></p>

<p><strong>Appendix</strong></p>

<p>Suppose <span class="math-container">$x_1\neq x_2$</span> and let <span class="math-container">$x* = \sqrt{x_1x_2}$</span>.  Then:</p>

<p><span class="math-container">$\quad y(x_1,x_2) = (x_1x_2)^{0.6} = (x*^2)^{0.6} = y(x*,x*)\qquad(27)$</span></p>

<p><span class="math-container">$\quad c(x_1,x_2) = 20(x_1 + x_2) + x_1^2 + x_2^2$</span>
<span class="math-container">$\quad = 20[(\sqrt{x_1} - \sqrt{x_2})^2 + 2\sqrt{x_1x_2}] + (x_1 – x_2)^2 + 2x_1x_2$</span>
<span class="math-container">$\quad \boldsymbol{&gt;} 2[20\sqrt{x_1x_2}) + (\sqrt{x_1x_2})^2] = c(x*,x*)\qquad(28)$</span></p>

<p>Thus the input combination <span class="math-container">$(x*,x*)$</span> yields the same output at lower cost than <span class="math-container">$(x_1,x_2)$</span>, and so the latter does not correspond to a point on the LRAC.</p>

CC BY-SA 4.0
2019-11-09T20:55:31.787Z
It is a little confusing that you use the function $c()$ for three different things: total costs spent on input 1, total costs spent on input 2 and total costs.
Giskard
CC BY-SA 4.0
2019-11-10T12:59:01.467Z
@Giskard  I've edited to add subscripts to $c$ for total costs spent on input 1 and total costs spent on input 2. $c$ without subscripts is thus reserved for total costs spent on the combined inputs.
Adam Bailey
CC BY-SA 4.0
2019-11-09T19:47:19.770Z
2019-11-10T12:55:04.030Z
2019-11-10T12:55:04.030Z
Adam Bailey
<p>Adam Bailey is correct.</p>

<p>Consider the production function <span class="math-container">$f(x_1,x_2) = x_1 + x_2/2$</span> where <span class="math-container">$(x_1,x_2)$</span> are inputs.</p>

<p>If the input costs are <span class="math-container">$w_1=w_2=1$</span> and all inputs are freely chosen, the solution to the cost minimization problem is
<span class="math-container">\begin{align*}
x_1 &amp; = y \\
\\
x_2 &amp; = 0.
\end{align*}</span>
However, in the short run one or more of the input quantities may be fixed. If <span class="math-container">$x_2 = \bar{x}_2 &gt; 0$</span>, this is never optimal, the short term cost is always higher than the long term cost. The long and short run cost functions in this case would be
<span class="math-container">\begin{align*}
C(y) &amp; = y \\
\\
C_s(y,\bar{x}_2) &amp;= \bar{x}_2 + \max(y - \bar{x}_2/2;0) \geq y + \bar{x}_2/2 &gt; C(y).
\end{align*}</span></p>

<p>Adam's other point (mentioned in his comment under this question) is that
<span class="math-container">$$
C(y) = \min_{\bar{x}_2} C_s(y,\bar{x}_2).
$$</span></p>

CC BY-SA 4.0
2019-10-31T21:37:35.990Z
What is y and how did you get the solution to the cost minimization problem? Legrangian?
Aqqqq
CC BY-SA 4.0
2019-11-01T07:50:56.983Z
The notation $y$ is from the question you yourself linked to. You could solve the cost minimization with a Lagrangian. There are also many other ways, but that is a separate question, one that is dealt with in most textbook.
Giskard
CC BY-SA 4.0
2019-11-01T08:10:10.967Z
I thought of a simpler example.
Giskard
CC BY-SA 4.0
2019-10-31T11:08:13.820Z
2019-11-01T08:17:23.567Z
2019-11-01T08:17:23.567Z
Giskard
<p>Looking at the context of the link posted, seems like you have the correct idea and the answerer might have misspoken. The minimum of SR cost curve should be <em>on</em> the LR cost curve. Suppose (for contradiction) that SR cost curve is above the LR cost curve at point <span class="math-container">$x$</span> on LR cost curve. This point <span class="math-container">$x$</span> involves a set of fixed variable (which is adjustable in the long run). You could set the fixed variables to this level and obtain another, cheaper SR cost curve.</p>

CC BY-SA 4.0
2019-10-31T12:20:06.780Z
Please see my comment on the question above.
Adam Bailey
CC BY-SA 4.0
2019-10-31T13:16:29.100Z
Thank you for your answer. Would you mind if you give me an interpretation that  "SR cost curve is above the LR cost curve at point x on LR cost curve"? How is it possible?
Aqqqq
CC BY-SA 4.0
2021-05-06T12:00:35.340Z
(-1) This answer is incorrect, or we mean something different under "minimum of SR cost curve".
Giskard
CC BY-SA 4.0
2019-10-31T10:30:05.853Z
2019-10-31T10:30:05.853Z
Art
<p>In the answer to <a href="https://economics.stackexchange.com/questions/14900/why-is-the-short-run-average-cost-curve-not-a-tangent-to-the-long-run-average-co">this question</a>, the answerer said "the minimum point of a short run cost curve will be above the long run cost curve". Is it true? If so, how would it be so?</p>

<p>I thought that if e.g. the short-run capacity is too small or too big, it would just be represented as different short-run cost curve tangent with the long-run cost curve.</p>

CC BY-SA 4.0
2019-10-31T08:51:06.097Z
You mean "in the answer to this question"?
Art
CC BY-SA 4.0
2019-10-31T08:56:16.213Z
Sorry about that. Just corrected.
Aqqqq
CC BY-SA 4.0
2019-10-31T12:18:39.453Z
Let me try to clarify.  There is a difference between: 1) If you take the _particular_ SR cost curve corresponding to particular values of the fixed inputs, then its minimum will in most cases be above the LR cost curve; and 2) The minimum of the minima of all possible SR cost curves will be above the LR cost curve.  In my answer to the linked question I was asserting (1).
Adam Bailey
<microeconomics><cost><long-run>
Is it possible that the minimum point of a short run cost curve does not touch the long run cost curve?
Aqqqq
# Is it possible that the minimum point of a short run cost curve does not touch the long run cost curve?

(Asked by: username_0 on 2019-10-31)

In the answer to [this question](https://economics.stackexchange.com/questions/14900/why-is-the-short-run-average-cost-curve-not-a-tangent-to-the-long-run-average-co), the answerer said "the minimum point of a short run cost curve will be above the long run cost curve". Is it true? If so, how would it be so?

I thought that if e.g. the short-run capacity is too small or too big, it would just be represented as different short-run cost curve tangent with the long-run cost curve.

**Tags:** microeconomics, cost, long-run

> **Comments:**
> **username_1**: You mean "in the answer to this question"?  
> **username_0**: Sorry about that. Just corrected.  
> **username_2**: Let me try to clarify.  There is a difference between: 1) If you take the _particular_ SR cost curve corresponding to particular values of the fixed inputs, then its minimum will in most cases be above the LR cost curve; and 2) The minimum of the minima of all possible SR cost curves will be above the LR cost curve.  In my answer to the linked question I was asserting (1).  

---

## 3 answers

### Answer by username_2 on 2019-11-09. Score: 2

The idea that the long run average cost curve (LRAC) must pass through the minimum points of the short run average cost curves (SRAC) is a fallacy, but it seems to be a remarkably plausible one. It was the source of a famous error by the economist Jacob Viner, referred to in [this paper by Silberberg](https://www.jstor.org/stable/1183036?seq=1#page_scan_tab_contents). Underlying the fallacy is perhaps an assumption that the points of tangency with the LRAC must be the minimum points of the SRAC’s. These points are coincident in the special case of a SRAC tangent to the LRAC *at the minimum point of the latter*. But usually they are distinct, as in the numerical example below.

Suppose a firm has a Cobb-Douglas production function with increasing returns $y = x\_1^{0.6}x\_2^{0.6}$. The inputs are bought in markets in which supply is not perfectly elastic, so that costs are increasing functions of quantities:

$\quad c\_1(x\_1) = 20x\_1 + x\_1^2$

$\quad c\_2(x\_2) = 20x\_2 + x\_2^2$

The symmetry between the two inputs in respect of both technology and costs is not necessary to obtain a suitable example, but it is convenient because it implies that every point on the LRAC must satisfy $x\_1=x\_2$ (see Appendix below for proof). This simplifies the derivation of the equation of the LRAC.

**LRAC**

Writing $c(a,b)$ for the total cost function with inputs $a, b$ and given $x\_1=x\_2$ we have:

$\quad c(x\_1,x\_2) = 40x\_1 + 2x\_1^2\qquad(1)$

$\quad y = x\_1^{1.2}\qquad(2)$

Hence:

$\quad x\_1 = y^{5/6}\qquad(3)$

$\quad c(x\_1,x\_2) = 40y^{5/6} + y^{5/3}\qquad(4)$

and so:

$\quad LRAC = \frac{40y^{5/6} + 2y^{5/3}}{y} = 40y^{-1/6}+2y^{2/3}\qquad(5)$

To find the minimum point:

$\quad \frac{dLRAC}{dy} = (-40/6)y^{-7/6} + (4/3)y^{-1/3} = 0\qquad(6)$

$\quad -40/6 + (4/3)y^{5/6} = 0\qquad(7)$

$\quad y^{5/6} = 5\qquad(8)$

$\quad y=6.90\qquad(9)$

To confirm this is a minimum:

$\quad \frac{d^2LRAC}{dy^2} = (280/36)y^{-13/6} + (-4/9)y^{-4/3}$
$\quad = (280/36)0.0152 – (4/9)0.0761 = 0.118 – 0.034 = 0.084 \boldsymbol{> 0}\qquad(10)$

The inputs at this minimum, using (3), are:

$\quad x\_1 = x\_2 = 6.90^{5/6} = 5.00\qquad(11)$

**SRAC**

Suppose now that $x\_1$ is freely variable but $x\_2$ is fixed in the short run at a value other than $5.00$, say $2$. Then:

$\quad y = x\_1^{0.6}(2^{0.6})\qquad(12)$

$\quad c(x\_1,x\_2) = 20x\_1 + x\_1^2 + 44\qquad(13)$

Hence:

$\quad x\_1 = (2^{-0.6}y)^{5/3} = (1/2)y^{5/3}\qquad(14)$

$\quad c(x\_1,x\_2) = 10y^{5/3} + (1/4)y^{10/3} + 44\qquad(15)$

and so:

$\quad SRAC(x\_2 = 2) = \frac{10y^{5/3} + (1/4)y^{10/3} + 44}{y} = 10y^{2/3} + (1/4)y^{7/3} + 44y^{-1}\qquad(16)$

The first derivative is:

$\quad \frac{dSRAC}{dy} = (20/3)y^{-1/3} + (7/12)y^{4/3} – 44y^{-2}\qquad(17)$

**Relation between LRAC and SRAC**

The two curves meet when $x\_1=x\_2=2$ implying $y = 2^{1.2} = 2.2974$ since at that point, using (5) and (16):

$\quad LRAC = 40(2.2974^{-1/6}) + 2(2.2974^{2/3}) = 34.822 + 3.482 = \boldsymbol{38.30}\qquad(18)$

$\quad SRAC = 10(2.2974^{2/3}) + (1/4)(2.2974^{7/3}) + 44(2.2974^{-1})$
$\quad = 17.411 + 1.741 + 19.152 = \boldsymbol{38.30}\qquad(19)$

Moreover they are tangential at that point since using (6) and (17) the respective slopes are:

$\quad \frac{dLRAC}{dy} = (-40/6)(2.2974^{-7/6}) + (4/3)(2.2974^{-1/3})$
$\quad = -2.526 + 1.010 = \boldsymbol{-1.52}\qquad(20)$

$\quad \frac{dSRAC}{dy} = (20/3)2.2974^{-1/3} + (7/12)2.2974^{4/3} + (-44)2.2974^{-2}$
$\quad = 5.052 + 1.768 – 8.336 = \boldsymbol{-1.52}\qquad(21)$

However, this point of tangency is **not** the minimum point of the SRAC. Using (17) to find the minimum:

$\quad \frac{dSRAC}{dy} = (20/3)y^{-1/3} + (7/12)y^{4/3} – 44y^{-2} = 0\qquad(22)$

$\quad (20/3)y^{5/3} + (7/12)y^{10/3} – 44 = 0\qquad(23)$

Treating this as a quadratic equation in $y^{5/3}$, or by trial and error, it can be found that $y$ is approximately $2.525$. To confirm this is a minimum:

$\quad \frac{d^2SRAC}{dy^2} = (-20/9)2.525^{-4/3} + (28/36)2.525^{1/3} + (88)2.525^{-1} = -0.646 + 1.059 + 34.851 = 35.26 \boldsymbol{> 0}\qquad(24)$

At this minimum point:

$\quad SRAC = 10(2.525^{2/3}) + (1/4)2.525^{7/3} + 44(2.525^{-1})$
$\quad = 18.543 + 2.170 + 17.426 = \boldsymbol{38.14}\qquad(25)$

This is lower than the point of tangency with the LRAC ($\boldsymbol{38.30}$), but above the LRAC at $y = 2.525$ which using (5) is:

$\quad LRAC = 40(2.525^{-1/6}) + 2(2.525^{2/3}) = 34.278 + 3.709 = \boldsymbol{37.99}\qquad(26)$

**Appendix**

Suppose $x\_1\neq x\_2$ and let $x\* = \sqrt{x\_1x\_2}$. Then:

$\quad y(x\_1,x\_2) = (x\_1x\_2)^{0.6} = (x\*^2)^{0.6} = y(x\*,x\*)\qquad(27)$

$\quad c(x\_1,x\_2) = 20(x\_1 + x\_2) + x\_1^2 + x\_2^2$
$\quad = 20[(\sqrt{x\_1} - \sqrt{x\_2})^2 + 2\sqrt{x\_1x\_2}] + (x\_1 – x\_2)^2 + 2x\_1x\_2$
$\quad \boldsymbol{>} 2[20\sqrt{x\_1x\_2}) + (\sqrt{x\_1x\_2})^2] = c(x\*,x\*)\qquad(28)$

Thus the input combination $(x\*,x\*)$ yields the same output at lower cost than $(x\_1,x\_2)$, and so the latter does not correspond to a point on the LRAC.

> **Comments:**
> **username_3**: It is a little confusing that you use the function $c()$ for three different things: total costs spent on input 1, total costs spent on input 2 and total costs.  
> **username_2**: @username_3  I've edited to add subscripts to $c$ for total costs spent on input 1 and total costs spent on input 2. $c$ without subscripts is thus reserved for total costs spent on the combined inputs.  

---

### Answer by username_3 on 2019-10-31. Score: 1

username_2 is correct.

Consider the production function $f(x\_1,x\_2) = x\_1 + x\_2/2$ where $(x\_1,x\_2)$ are inputs.

If the input costs are $w\_1=w\_2=1$ and all inputs are freely chosen, the solution to the cost minimization problem is
\begin{align\*}
x\_1 & = y \\
\\
x\_2 & = 0.
\end{align\*}
However, in the short run one or more of the input quantities may be fixed. If $x\_2 = \bar{x}\_2 > 0$, this is never optimal, the short term cost is always higher than the long term cost. The long and short run cost functions in this case would be
\begin{align\*}
C(y) & = y \\
\\
C\_s(y,\bar{x}\_2) &= \bar{x}\_2 + \max(y - \bar{x}\_2/2;0) \geq y + \bar{x}\_2/2 > C(y).
\end{align\*}

Adam's other point (mentioned in his comment under this question) is that
$$
C(y) = \min\_{\bar{x}\_2} C\_s(y,\bar{x}\_2).
$$

> **Comments:**
> **username_0**: What is y and how did you get the solution to the cost minimization problem? Legrangian?  
> **username_3**: The notation $y$ is from the question you yourself linked to. You could solve the cost minimization with a Lagrangian. There are also many other ways, but that is a separate question, one that is dealt with in most textbook.  
> **username_3**: I thought of a simpler example.  

---

### Answer by username_1 on 2019-10-31. Score: 0

Looking at the context of the link posted, seems like you have the correct idea and the answerer might have misspoken. The minimum of SR cost curve should be *on* the LR cost curve. Suppose (for contradiction) that SR cost curve is above the LR cost curve at point $x$ on LR cost curve. This point $x$ involves a set of fixed variable (which is adjustable in the long run). You could set the fixed variables to this level and obtain another, cheaper SR cost curve.

> **Comments:**
> **username_2**: Please see my comment on the question above.  
> **username_0**: Thank you for your answer. Would you mind if you give me an interpretation that  "SR cost curve is above the LR cost curve at point x on LR cost curve"? How is it possible?  
> **username_3**: (-1) This answer is incorrect, or we mean something different under "minimum of SR cost curve".  

---

