In the answer to this question, 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.

  • $\begingroup$ You mean "in the answer to this question"? $\endgroup$ – Art Oct 31 '19 at 8:51
  • $\begingroup$ Sorry about that. Just corrected. $\endgroup$ – Aqqqq Oct 31 '19 at 8:56
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    $\begingroup$ 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). $\endgroup$ – Adam Bailey Oct 31 '19 at 12:18

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.

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  • $\begingroup$ Please see my comment on the question above. $\endgroup$ – Adam Bailey Oct 31 '19 at 12:20
  • $\begingroup$ 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? $\endgroup$ – Aqqqq Oct 31 '19 at 13:16

Adam Bailey 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 $$ C(y) = \min_{\bar{x}_2} C_s(y,\bar{x}_2). $$

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  • $\begingroup$ What is y and how did you get the solution to the cost minimization problem? Legrangian? $\endgroup$ – Aqqqq Oct 31 '19 at 21:37
  • $\begingroup$ 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. $\endgroup$ – Giskard Nov 1 '19 at 7:50
  • $\begingroup$ I thought of a simpler example. $\endgroup$ – Giskard Nov 1 '19 at 8:10

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. 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.


Writing $c(a,b)$ for the total cost function with inputs $a, b$ and given $x_1=x_2$ we have:

$\quad c(x_1,x_2) = 40x_1 + 2x_1^2\qquad(1)$

$\quad y = x_1^{1.2}\qquad(2)$


$\quad x_1 = y^{5/6}\qquad(3)$

$\quad c(x_1,x_2) = 40y^{5/6} + y^{5/3}\qquad(4)$

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:

$\quad x_1 = x_2 = 6.90^{5/6} = 5.00\qquad(11)$


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:

$\quad y = x_1^{0.6}(2^{0.6})\qquad(12)$

$\quad c(x_1,x_2) = 20x_1 + x_1^2 + 44\qquad(13)$


$\quad x_1 = (2^{-0.6}y)^{5/3} = (1/2)y^{5/3}\qquad(14)$

$\quad c(x_1,x_2) = 10y^{5/3} + (1/4)y^{10/3} + 44\qquad(15)$

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:

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


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.

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  • $\begingroup$ 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. $\endgroup$ – Giskard Nov 9 '19 at 20:55
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    $\begingroup$ @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. $\endgroup$ – Adam Bailey Nov 10 '19 at 12:59

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