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