<p>In order to compute the level of output <span class="math-container">$y_0$</span> such that short-run and long-run average cost functions are equal, let us start with the relationship between <span class="math-container">$C$</span> and <span class="math-container">$SC$</span> (with abuse of notations to keep them short):
<span class="math-container">$$ C(y) = \min_k CS(k,y) \leq CS(k,y) $$</span>
which defines the optimal long-run capital level <span class="math-container">$k^*$</span> for which
<span class="math-container">$$ C(y) = CS(k^*(y),y). $$</span>
In terms of average costs, for any <span class="math-container">$y&gt;0$</span>:
<span class="math-container">$$ \frac{C}{y}(y) = \frac{CS}{y}(k^*(y),y) \leq \frac{CS}{y}(k,y). $$</span>
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: <span class="math-container">$$k^*(y)=k.$$</span>
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:
<span class="math-container">$$y_0=(k^*)^{-1}(k).$$</span>
As an exercise, we could code and represent <span class="math-container">$C/y$</span>, <span class="math-container">$CS/y$</span> and <span class="math-container">$y_0$</span> for the Cobb-Douglas case (for instance), and reproduce the above Figure.</p>

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2022-10-19T11:04:38.260Z
2022-10-19T14:30:24.033Z
2022-10-19T14:30:24.033Z
Bertrand
<p>Suppose the short-run cost function is written as <span class="math-container">$SC(\bar{w}, \bar{r}, y, k)$</span> and the long-run cost function as <span class="math-container">$C(\bar{w},\bar{r},y)$</span> where the rates <span class="math-container">$w$</span> and <span class="math-container">$r$</span> are fixed. <span class="math-container">$y$</span> determines the quantity of output and <span class="math-container">$k$</span> is the capital which can be fixed in the short-run.</p>
<p>For further definition: <span class="math-container">$SC(w,r,y,\bar{k}) = \underset{l}{\min}(wl + r\bar{k})$</span> subject to <span class="math-container">$f(l,\bar{k}) = y$</span> and <span class="math-container">$C(w,r,y) = \underset{l,k}{\min}(wl + rk)$</span> subject to <span class="math-container">$f(l,k) = y$</span>.</p>
<p><strong>Suppose the capital is fixed at <span class="math-container">$k = \bar{k}$</span>. What's the quantity <span class="math-container">$y$</span> for which the short-run cost curve is tangent to the long-run cost curve</strong>?</p>
<p><a href="https://i.sstatic.net/68V5u.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/68V5u.png" alt="enter image description here" /></a></p>
<p><strong>In other words, let's say you're given the <span class="math-container">$SC(w,r,y,k)$</span> and <span class="math-container">$C(w,r,y)$</span> functions. And you're given <span class="math-container">$k = k_0$</span> (as in the picture). How will find <span class="math-container">$q_0$</span> (or the point where the <span class="math-container">$SAC(k_0)$</span> curve is tangent to the <span class="math-container">$LAC$</span> curve) in terms of <span class="math-container">$\bar{w}, \bar{r}, k_0$</span>?</strong></p>
<p>Edit: I think the way is to do one of the two:</p>
<ol>
<li>Extract the equation <span class="math-container">$k(y)$</span> from <span class="math-container">$\frac{\partial SC(\bar{w}, \bar{r}, k, y)}{\partial k} = 0$</span>. Substitute <span class="math-container">$k = \bar{k}$</span> to find <span class="math-container">$y$</span>.</li>
<li>Extract the value of <span class="math-container">$y$</span> from <span class="math-container">$\frac{\partial C(w,r,k)}{\partial w} = \bar{k}$</span>. This is the desired <span class="math-container">$y$</span>.</li>
</ol>

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2022-10-18T07:44:17.920Z
What is $y$? Labour? And how can it be that there's no capital in the long run?
Matteo Bulgarelli
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2022-10-18T07:59:52.963Z
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.
Giskard
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2022-10-18T08:20:43.813Z
What is your contribution to the question? What is the relationship between SC and C? How can you be sure that an intersection exists?
Bertrand
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2022-10-18T12:59:42.007Z
@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).
Rick_Morty
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2022-10-18T13:00:16.383Z
@Giskard I have added the definitions and also a motivation.
Rick_Morty
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2022-10-18T13:11:54.220Z
@Bertrand I am considering a continuous number of plants, so by fixing $k$, we get one of the SRAC curves. That is bound to intersect with the LRAC at one point. $\frac{\partial SC}{\partial k} = 0$ gives an equation of the form $k(y)$. Assuming continuous number of plants (or $\text{range}(k) = \mathbb{R}_{0}$), I think we can tell there's always one such intersection.
Rick_Morty
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2022-10-18T16:14:25.420Z
@Citrus I am afraid I still don't understand what sort of answer you are looking for. A formula? A number? An algorithm? A verbal characterization?
Giskard
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2022-10-18T19:33:17.573Z
@Giskard Alright, an SRAC curve is given by $\frac{SC(\bar{w},\bar{r},y,\bar{k})}{y}$ by fixing the $k$. It intersects with the LRAC at the cost-minimizing level of production. I am trying to find this $y^{*}$ that gives us the cost-minimizing level of production.
Rick_Morty
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2022-10-19T00:57:14.753Z
@Giskard I have shortened the post and updated it with a more direct and hopefully, less confusing, question. Please see the last sentence (in bold). Thank you!
Rick_Morty
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2022-10-19T00:58:09.407Z
@MatteoBulgarelli I have updated the post. Please see the last sentence (in bold).
Rick_Morty
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2022-10-19T03:03:09.350Z
$q_0$ is where $SAC(k_0)$ is tangent to $AC$. Since you're given $SC(\cdot)$ and $C(\cdot)$, simply compute $SAC$ and $AC$ from them, differentiate with respect to $y$, equate the derivatives, and voila you're done.
Herr K.
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2022-10-19T03:05:18.203Z
@HerrK. What a coincidence! I made the last edit and right then, you added this comment. Also, no, I wanted a way to do it without having to equate them.
Rick_Morty
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2022-10-19T04:05:00.277Z
Why, what is the problem with equating them?
Giskard
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2022-10-19T04:33:20.070Z
@Giskard I was trying to find out if the point of tangency has any significance, so I wanted a way that avoided equating one to the other. I was also trying to find out how the two cost curves are related to one another, especially at the intersection (tangency) points. I still don't know if there's a significance, but that was the intent.
Rick_Morty
CC BY-SA 4.0
2022-10-19T04:36:34.970Z
None of this is mentioned in the question?
Giskard
<microeconomics><cost-functions>
Where does the short-run and long-run costs intersect if $k$ is fixed?
Rick_Morty
# Where does the short-run and long-run costs intersect if $k$ is fixed?

(Asked by: username_0 on 2022-10-18)

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](https://i.sstatic.net/68V5u.png)](https://i.sstatic.net/68V5u.png)

**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 $y$.

**Tags:** microeconomics, cost-functions
> **Comments:**
> **username_1**: What is $y$? Labour? And how can it be that there's no capital in the long run?  
> **username_2**: 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.  
> **username_3**: What is your contribution to the question? What is the relationship between SC and C? How can you be sure that an intersection exists?  
> **username_0**: @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).  
> **username_0**: @Giskard I have added the definitions and also a motivation.  
> **username_0**: @Bertrand I am considering a continuous number of plants, so by fixing $k$, we get one of the SRAC curves. That is bound to intersect with the LRAC at one point. $\frac{\partial SC}{\partial k} = 0$ gives an equation of the form $k(y)$. Assuming continuous number of plants (or $\text{range}(k) = \mathbb{R}_{0}$), I think we can tell there's always one such intersection.  
> **username_2**: @Citrus I am afraid I still don't understand what sort of answer you are looking for. A formula? A number? An algorithm? A verbal characterization?  
> **username_0**: @Giskard Alright, an SRAC curve is given by $\frac{SC(\bar{w},\bar{r},y,\bar{k})}{y}$ by fixing the $k$. It intersects with the LRAC at the cost-minimizing level of production. I am trying to find this $y^{*}$ that gives us the cost-minimizing level of production.  
> **username_0**: @Giskard I have shortened the post and updated it with a more direct and hopefully, less confusing, question. Please see the last sentence (in bold). Thank you!  
> **username_0**: @MatteoBulgarelli I have updated the post. Please see the last sentence (in bold).  
> **username_4**: $q_0$ is where $SAC(k_0)$ is tangent to $AC$. Since you're given $SC(\cdot)$ and $C(\cdot)$, simply compute $SAC$ and $AC$ from them, differentiate with respect to $y$, equate the derivatives, and voila you're done.  
> **username_0**: @HerrK. What a coincidence! I made the last edit and right then, you added this comment. Also, no, I wanted a way to do it without having to equate them.  
> **username_2**: Why, what is the problem with equating them?  
> **username_0**: @Giskard I was trying to find out if the point of tangency has any significance, so I wanted a way that avoided equating one to the other. I was also trying to find out how the two cost curves are related to one another, especially at the intersection (tangency) points. I still don't know if there's a significance, but that was the intent.  
> **username_2**: None of this is mentioned in the question?  

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## 1 answers

### Answer by username_3 on 2022-10-19. Score: 2 | Accepted answer

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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