<p>The cost minimization problem (in general) is given by:
<span class="math-container">$$
\min_{L_i} \int_0^A p(i) L(i) di \text{ s.t. } \left(\int_0^A L(i)^{\phi}\right)^{1/\phi} = y.
$$</span>
where <span class="math-container">$y$</span> is the output level.</p>
<p>The Lagrangian is given by:
<span class="math-container">$$
L = \int_0^A p(i) L(i) - \lambda \left( \left(\int_0^A L(i)^\phi\right)^{1/\phi} - y\right).
$$</span>
The first order conditions give:
<span class="math-container">$$
\begin{align*}
&amp;p(i) = \lambda L(i)^{\phi - 1},\\
&amp;\left(\int_0^A L(i)^\phi\right)^{1/\phi} = y.
\end{align*}
$$</span></p>
<p>The first condition gives:
<span class="math-container">$$
L(i) = \left(\frac{p(i)}{\lambda}\right)^{1/(\phi - 1)}
$$</span>
Substituting into the second condition produces:
<span class="math-container">$$
\begin{align*}
&amp;\lambda^{-1/(\phi - 1)} \left(\int_0^A p(i)^{\phi/(\phi - 1)}\right)^{1/\phi} = y.\\
\leftrightarrow&amp; \lambda = y^{(1-\phi)} \left(\int_0^A p(i)^{\phi/(\phi - 1)}\right)^{(\phi - 1)/\phi}
\end{align*}
$$</span></p>
<p>This gives:
<span class="math-container">$$
L(i) = y p(i)^{1/(\phi - 1)} \left(\int_0^A p(j)^{\phi/(\phi - 1)}\right)^{-1/\phi}. 
$$</span></p>
<p>So the cost function equals:
<span class="math-container">$$
c(p,y) = \int_0^A p(i) L(i) = y \left(\int_0^A p(i)^{\phi/(\phi - 1)}\right)^{(\phi - 1)/\phi}
$$</span></p>
<p>Average cost is:
<span class="math-container">$$
\frac{c(p,y)}{y} = \left(\int_0^A p(i)^{\phi/(\phi - 1)}\right)^{(\phi - 1)/\phi}
$$</span>
Marginal cost is:
<span class="math-container">$$
\frac{\partial c(p,y)}{\partial y} = \left(\int_0^A p(i)^{\phi/(\phi - 1)}\right)^{(\phi - 1)/\phi}
$$</span>
So the two are the same.</p>
<p>In order to show that profits are zero (under constant returns to scale) the issue is a bit more subtle as it relies on market equilibrium effects. Note that profits can be written as:
<span class="math-container">$$
Py - \text{ cost} = (P - AC)y.
$$</span>
Where <span class="math-container">$P$</span> is the output price and <span class="math-container">$AC$</span> are the average costs. Assume that <span class="math-container">$AC$</span> is constant in the sense that they do not depend on <span class="math-container">$y$</span>, (which is the case under constant returns to scale). Then if <span class="math-container">$P &lt; AC$</span>, profits are negative unless <span class="math-container">$y = 0$</span>. In this case, the optimal thing to do for the firm is to set <span class="math-container">$y = 0$</span>, which basically means that the firm exists the market.</p>
<p>If <span class="math-container">$P &gt; AC$</span> then the profits are strictly positive, and the optimal thing to do for the firm is to set <span class="math-container">$y$</span> very large (<span class="math-container">$= \infty$</span>). These profits will also attract other firms into the market. This increasing output will lead to a drop in the price level <span class="math-container">$P$</span> (equilibrium effects) until there are not more profits. So in equilibrium, profits will be zero and equilibrium price will be such that <span class="math-container">$P = AC$</span>. In other words, with constant returns to scale, it is impossible to determine the output level of each firm (and hence the number of firms in the market). However total market output is determined by the condition that sets profits equal to zero.</p>

CC BY-SA 4.0
2024-02-06T13:03:30.767Z
This answer is everything that I could've hoped for and more. I can't even begin to thank you enough for your clear and comprehensive answer to my question, which I've struggled with considerably before deciding to post here. The derivation of the proof (both) are not trivial and required rich insights, which you've shared here. Even with the provided answer, it still took me a while to derive all the results, which further highlights the non-triviality of the proof. I am immensely grateful. Thank you so much.
Tan2525
CC BY-SA 4.0
2024-02-06T12:14:40.260Z
2024-02-06T12:14:40.260Z
tdm
<p>On page 124 of David Romer's Advanced Macroeconomics (5th Ed), he mentioned that for the Romer model,</p>
<blockquote>
<p>Because firms producing final output face constant costs for each input and the production function exhibits constant returns, marginal cost equals average cost. As a result, these firms earn zero profits.</p>
</blockquote>
<p>I'm struggling to mathematically prove this result.</p>
<p>The Ethier production function is <span class="math-container">$Y=\left[\int_{i=0}^{A}L(i)^{\phi}di\right]^{\frac{1}{\phi}}$</span>, where <span class="math-container">$0&lt;\phi&lt;1$</span> and <span class="math-container">$L(i)$</span> is the quantity of labor devoted to producing input <span class="math-container">$i$</span> and the quantity of input <span class="math-container">$i$</span> that goes into final-goods production. Let <span class="math-container">$L_{Y}$</span> be the total number of workers producing inputs and assume that the number producing each available input is the same. Then, <span class="math-container">$L(i)=\frac{L_Y}{A}$</span> for all <span class="math-container">$i$</span>. We also assume that the patent-holder is a monopolist who hires workers in a competitive labor market to produce the input and then sell the input to producers of the final output, where the monopolist charges a constant price for each unit of the output. Competitive firms producing the output take the prices of inputs as given. For the cost-minimization problem of a representative output producer, the Lagrangian for the problem of producing one unit of output at minimum cost is given by
<span class="math-container">$$
\mathcal{L}=\int_{i=0}^{A} p(i) L(i) di-\lambda\left\{\left[\int_{i=0}^{A}L(i)^{\phi}di\right]^{\frac{1}{\phi}}-1\right\},
$$</span>
where <span class="math-container">$p(i)$</span> is the price charged by the holder of the patent on idea <span class="math-container">$i$</span> for each unit of the input embodying that idea.</p>
<p>The text derives the result that the first-order condition for an individual <span class="math-container">$L(i)$</span> is
<span class="math-container">$$
p(i) = \lambda L(i)^{\phi -1},
$$</span>
and then gives us the clue that</p>
<blockquote>
<p>One could use the condition that <span class="math-container">$\left[\int_{i=0}^{A}L(i)^{\phi}di\right]^{\frac{1}{\phi}} = 1$</span> to solve for <span class="math-container">$\lambda$</span>, and then solve for the cost-minimizing levels of the <span class="math-container">$L(i)$</span>’s and the level of marginal cost.</p>
</blockquote>
<p>However, I'm unsure of how to go about this and can't derive an expression for the (1) marginal cost and (2) average cost, which should be equal here. Also, how do we go about showing that when marginal cost equal to average cost, the profit is zero?</p>

<macroeconomics><economic-growth><technology><endogenous-growth>
Marginal and Average Cost in the Romer Model of Endogenous Growth
Tan2525
# Marginal and Average Cost in the Romer Model of Endogenous Growth

(Asked by: username_0 on 2024-02-06)

On page 124 of David Romer's Advanced Macroeconomics (5th Ed), he mentioned that for the Romer model,

> Because firms producing final output face constant costs for each input and the production function exhibits constant returns, marginal cost equals average cost. As a result, these firms earn zero profits.

I'm struggling to mathematically prove this result.

The Ethier production function is $Y=\left[\int\_{i=0}^{A}L(i)^{\phi}di\right]^{\frac{1}{\phi}}$, where $0<\phi<1$ and $L(i)$ is the quantity of labor devoted to producing input $i$ and the quantity of input $i$ that goes into final-goods production. Let $L\_{Y}$ be the total number of workers producing inputs and assume that the number producing each available input is the same. Then, $L(i)=\frac{L\_Y}{A}$ for all $i$. We also assume that the patent-holder is a monopolist who hires workers in a competitive labor market to produce the input and then sell the input to producers of the final output, where the monopolist charges a constant price for each unit of the output. Competitive firms producing the output take the prices of inputs as given. For the cost-minimization problem of a representative output producer, the Lagrangian for the problem of producing one unit of output at minimum cost is given by
$$
\mathcal{L}=\int\_{i=0}^{A} p(i) L(i) di-\lambda\left\{\left[\int\_{i=0}^{A}L(i)^{\phi}di\right]^{\frac{1}{\phi}}-1\right\},
$$
where $p(i)$ is the price charged by the holder of the patent on idea $i$ for each unit of the input embodying that idea.

The text derives the result that the first-order condition for an individual $L(i)$ is
$$
p(i) = \lambda L(i)^{\phi -1},
$$
and then gives us the clue that

> One could use the condition that $\left[\int\_{i=0}^{A}L(i)^{\phi}di\right]^{\frac{1}{\phi}} = 1$ to solve for $\lambda$, and then solve for the cost-minimizing levels of the $L(i)$’s and the level of marginal cost.

However, I'm unsure of how to go about this and can't derive an expression for the (1) marginal cost and (2) average cost, which should be equal here. Also, how do we go about showing that when marginal cost equal to average cost, the profit is zero?

**Tags:** macroeconomics, economic-growth, technology, endogenous-growth


---

## 1 answers

### Answer by username_1 on 2024-02-06. Score: 5 | Accepted answer

The cost minimization problem (in general) is given by:
$$
\min\_{L\_i} \int\_0^A p(i) L(i) di \text{ s.t. } \left(\int\_0^A L(i)^{\phi}\right)^{1/\phi} = y.
$$
where $y$ is the output level.

The Lagrangian is given by:
$$
L = \int\_0^A p(i) L(i) - \lambda \left( \left(\int\_0^A L(i)^\phi\right)^{1/\phi} - y\right).
$$
The first order conditions give:
$$
\begin{align\*}
&p(i) = \lambda L(i)^{\phi - 1},\\
&\left(\int\_0^A L(i)^\phi\right)^{1/\phi} = y.
\end{align\*}
$$

The first condition gives:
$$
L(i) = \left(\frac{p(i)}{\lambda}\right)^{1/(\phi - 1)}
$$
Substituting into the second condition produces:
$$
\begin{align\*}
&\lambda^{-1/(\phi - 1)} \left(\int\_0^A p(i)^{\phi/(\phi - 1)}\right)^{1/\phi} = y.\\
\leftrightarrow& \lambda = y^{(1-\phi)} \left(\int\_0^A p(i)^{\phi/(\phi - 1)}\right)^{(\phi - 1)/\phi}
\end{align\*}
$$

This gives:
$$
L(i) = y p(i)^{1/(\phi - 1)} \left(\int\_0^A p(j)^{\phi/(\phi - 1)}\right)^{-1/\phi}.
$$

So the cost function equals:
$$
c(p,y) = \int\_0^A p(i) L(i) = y \left(\int\_0^A p(i)^{\phi/(\phi - 1)}\right)^{(\phi - 1)/\phi}
$$

Average cost is:
$$
\frac{c(p,y)}{y} = \left(\int\_0^A p(i)^{\phi/(\phi - 1)}\right)^{(\phi - 1)/\phi}
$$
Marginal cost is:
$$
\frac{\partial c(p,y)}{\partial y} = \left(\int\_0^A p(i)^{\phi/(\phi - 1)}\right)^{(\phi - 1)/\phi}
$$
So the two are the same.

In order to show that profits are zero (under constant returns to scale) the issue is a bit more subtle as it relies on market equilibrium effects. Note that profits can be written as:
$$
Py - \text{ cost} = (P - AC)y.
$$
Where $P$ is the output price and $AC$ are the average costs. Assume that $AC$ is constant in the sense that they do not depend on $y$, (which is the case under constant returns to scale). Then if $P < AC$, profits are negative unless $y = 0$. In this case, the optimal thing to do for the firm is to set $y = 0$, which basically means that the firm exists the market.

If $P > AC$ then the profits are strictly positive, and the optimal thing to do for the firm is to set $y$ very large ($= \infty$). These profits will also attract other firms into the market. This increasing output will lead to a drop in the price level $P$ (equilibrium effects) until there are not more profits. So in equilibrium, profits will be zero and equilibrium price will be such that $P = AC$. In other words, with constant returns to scale, it is impossible to determine the output level of each firm (and hence the number of firms in the market). However total market output is determined by the condition that sets profits equal to zero.

> **Comments:**
> **username_0**: This answer is everything that I could've hoped for and more. I can't even begin to thank you enough for your clear and comprehensive answer to my question, which I've struggled with considerably before deciding to post here. The derivation of the proof (both) are not trivial and required rich insights, which you've shared here. Even with the provided answer, it still took me a while to derive all the results, which further highlights the non-triviality of the proof. I am immensely grateful. Thank you so much.  

---

