Suppose the short-run cost function is written as $SC(\bar{w}, \bar{r}, y, k)$ and the long-run cost function as $C(\bar{w},\bar{r},y)$ where the rates $w$ and $r$ are fixed. $y$ determines the quantity of output and $k$ is the capital which can be fixed in the short-run.

For further definition: $SC(w,r,y,\bar{k}) = \underset{l}{\min}(wl + r\bar{k})$ subject to $f(l,\bar{k}) = y$ and $C(w,r,y) = \underset{l,k}{\min}(wl + rk)$ subject to $f(l,k) = y$.

Suppose the capital is fixed at $k = \bar{k}$. What's the quantity $y$ for which the short-run cost curve is tangent to the long-run cost curve?

enter image description here

In other words, let's say you're given the $SC(w,r,y,k)$ and $C(w,r,y)$ functions. And you're given $k = k_0$ (as in the picture). How will find $q_0$ (or the point where the $SAC(k_0)$ curve is tangent to the $LAC$ curve) in terms of $\bar{w}, \bar{r}, k_0$?

Edit: I think the way is to do one of the two:

  1. Extract the equation $k(y)$ from $\frac{\partial SC(\bar{w}, \bar{r}, k, y)}{\partial k} = 0$. Substitute $k = \bar{k}$ to find $y$.
  2. Extract the value of $y$ from $\frac{\partial C(w,r,k)}{\partial w} = \bar{k}$. This is the desired $y$.
  • 1
    $\begingroup$ What is $y$? Labour? And how can it be that there's no capital in the long run? $\endgroup$ Oct 18, 2022 at 7:44
  • $\begingroup$ What do you mean what is the quantitiy of $y$? Do you expect an answer like $\overline{k}/\overline{r}$? This is impossible without the functional forms. $\endgroup$
    – Giskard
    Oct 18, 2022 at 7:59
  • $\begingroup$ What is your contribution to the question? What is the relationship between SC and C? How can you be sure that an intersection exists? $\endgroup$
    – Bertrand
    Oct 18, 2022 at 8:20
  • $\begingroup$ @MatteoBulgarelli $y$ is the output (units). Labour is assumed to not be fixed in any of the runs. $k$ is fixed in the short-run, hence the inclusion of variable $k$ and non-inclusion of $l$ in $SC(\cdot)$. As for "no capital" in the long-run, that's because $SC(\cdot)$ and $C(\cdot)$ give the optimal values of all $l$ in both runs (and also over all $k$ in the long-run). $\endgroup$
    – Rick_Morty
    Oct 18, 2022 at 12:59
  • $\begingroup$ @Giskard I have added the definitions and also a motivation. $\endgroup$
    – Rick_Morty
    Oct 18, 2022 at 13:00

1 Answer 1


In order to compute the level of output $y_0$ such that short-run and long-run average cost functions are equal, let us start with the relationship between $C$ and $SC$ (with abuse of notations to keep them short): $$ C(y) = \min_k CS(k,y) \leq CS(k,y) $$ which defines the optimal long-run capital level $k^*$ for which $$ C(y) = CS(k^*(y),y). $$ In terms of average costs, for any $y>0$: $$ \frac{C}{y}(y) = \frac{CS}{y}(k^*(y),y) \leq \frac{CS}{y}(k,y). $$ This weak inequality as well as the tangency (when the inequality is binding) is illustrated on your figure. At the tangency point, we have the equality between optimal and restricted capital level: $$k^*(y)=k.$$ If you want to know which output level, if it exists, is compatible with this optimality of the capital stock, you can find it by inverting this last relationship (if it is possible), and find: $$y_0=(k^*)^{-1}(k).$$ As an exercise, we could code and represent $C/y$, $CS/y$ and $y_0$ for the Cobb-Douglas case (for instance), and reproduce the above Figure.


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