<p>Observing the highest possible iso-profit curve passing through the feasible set, we obtain the supply of <span class="math-container">$y_2$</span> and demand for <span class="math-container">$y_1$</span> as follows. Here iso-profit curve is set of <span class="math-container">$(y_1,y_2)$</span> pairs that yields the same level of profit given <span class="math-container">$p$</span>.</p>
<p><a href="https://i.sstatic.net/iSQuy.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/iSQuy.png" alt="enter image description here" /></a></p>
<p>So,
<span class="math-container">\begin{eqnarray*} (y_1^d,y_2^s)(p)\in\begin{cases}\{(0,0),(1+a,1)\} &amp; \text{if } p = 1+a \\ \{(1+a,1)\} &amp; \text{if } p &gt; 1+a \\ \{(0,0)\} &amp; \text{if } p &lt; 1+a \end{cases} \end{eqnarray*}</span></p>
<p>Corresponding optimal profit is</p>
<p><span class="math-container">\begin{eqnarray*} \pi(p)=\max(0,p-(1+a)) \end{eqnarray*}</span></p>
<p>Given that <span class="math-container">$a\sim\text{Unif}(0,1)$</span>, aggregate input demand is
<span class="math-container">\begin{eqnarray*}Y_1^d(p) = \int_0^1y_1^d(p)da = \begin{cases} \int_0^{1}(1+a)da=\frac{3}{2} &amp; \text{if } p \geq 2 \\ 0 &amp; \text{if } p \leq 1 \\ \int_0^{p-1}(1+a)da=\dfrac{(p-1)(p+1)}{2} &amp; \text{if } 1 &lt; p &lt; 2\end{cases} \end{eqnarray*}</span></p>
<p>Aggregate output supply is
<span class="math-container">\begin{eqnarray*}Y_2^s(p) = \int_0^1y_2^s(p)da = \begin{cases} \int_0^{1}da=1 &amp; \text{if } p \geq 2 \\ 0 &amp; \text{if } p \leq 1 \\ \int_0^{p-1}da=p-1 &amp; \text{if } 1 &lt; p &lt; 2\end{cases} \end{eqnarray*}</span></p>

CC BY-SA 4.0
2023-12-08T15:27:30.607Z
Hi - What is the average production set in this case?
Ludwig Gershwin
CC BY-SA 4.0
2023-12-11T02:34:43.190Z
Boundary of the aggregate Production Set is given by $g(y_1)=\begin{cases} 1 & \text{for } y_1<-2\\ \int_0^1 \min(-y_1-a,1)da & \text{for } -2\leq y_1\leq -1 \\ \int_0^1 \max(-y_1-a,0)da & \text{for } -1<y_1\leq 0 \end{cases}$ and aggregate production set is simply $\{(y_1,y_2)|0\leq y_2\leq g(y_1)\}$
Amit
CC BY-SA 4.0
2023-10-31T13:38:43.363Z
2023-10-31T13:38:43.363Z
Amit
<p>Let's first look at the profit of a single firm with parameter <span class="math-container">$a$</span>. Let us denote by <span class="math-container">$x = -y_1$</span> the input and <span class="math-container">$f(x) = y_2$</span> the output.</p>
<p>The production function is the following:
<span class="math-container">$$
f(x) = \begin{cases} 0 &amp;\text{ if } x &lt; a,\\ x - a &amp;\text{ if } a \le x \le 1+a,\\ 1 &amp;\text{ if } x &gt; 1+a\end{cases}
$$</span></p>
<p>The firm optimizes profits:
<span class="math-container">$$
x^\ast = \arg\max_x p f(x) - x.
$$</span></p>
<p>You can solve this by looking at various scenarios and taking the best (profit maximising) one.</p>
<ol>
<li>if <span class="math-container">$x^\ast &lt; a$</span> then profits are <span class="math-container">$p 0 - x^\ast$</span>, so here it is best to put <span class="math-container">$x^\ast = 0$</span>.</li>
<li>if <span class="math-container">$a \le x^\ast \le 1+a$</span> then profits are <span class="math-container">$p(x^\ast - a) - x^\ast= (p-1)x^\ast - pa$</span>. If <span class="math-container">$p &gt; 1$</span> then profits increase in <span class="math-container">$x^\ast$</span> so it is best to put <span class="math-container">$x^\ast = 1+a$</span> and profits will be <span class="math-container">$p - 1 - a$</span>. if <span class="math-container">$p &lt; 1$</span>, then it is best to set <span class="math-container">$x^\ast = a$</span> and profits will be <span class="math-container">$-a$</span> (which is negative). In the latter case, it will be best to produce nothing.</li>
<li>if <span class="math-container">$x^\ast \ge 1+a$</span> then profits are <span class="math-container">$p - x^\ast$</span> so here it is optimal to put <span class="math-container">$x = 1+a$</span> and profits are given by <span class="math-container">$p - 1 - a$</span>.</li>
</ol>
<p>As such, we see that profits are either equal to <span class="math-container">$p-1-a$</span> with optimal demand <span class="math-container">$x^\ast = 1+a$</span> or equal to zero with optimal demand <span class="math-container">$x^\ast = 0$</span>.</p>
<p>The first will be optimal if <span class="math-container">$a \le p-1$</span>, while the second is optimal if <span class="math-container">$a &gt; p-1$</span>. This gives the following optimal demand for a firm with parameter <span class="math-container">$a$</span>:
<span class="math-container">$$
x^\ast = \begin{cases} 1+a &amp;\text{ if } a\le p-1 \\ 0 &amp;\text{ if } a &gt; p-1\end{cases}
$$</span></p>
<p>Now let us derive optimal aggregate demand. If <span class="math-container">$p &lt; 1$</span> then <span class="math-container">$a \ge 0 &gt; p-1$</span> so no firm will demand anything: aggregate demand equals zero.</p>
<p>If <span class="math-container">$p &gt; 2$</span> then <span class="math-container">$a \le 1 \le p - 1$</span> so the firm of type <span class="math-container">$a$</span> will demand <span class="math-container">$1+a$</span>. Aggregate demand is then given by:
<span class="math-container">$$
\int_0^1 (1+a) da = [a + a^2/2]^1_0 = \frac{3}{2}.
$$</span></p>
<p>If <span class="math-container">$1 \le p \le 2$</span> then firms with <span class="math-container">$a \le p-1$</span> will demand <span class="math-container">$1+a$</span> and firms with <span class="math-container">$a &gt; p-1$</span> will not demand anything. As such, aggreagate demand is:
<span class="math-container">$$
\begin{align*}
\int_0^{p-1} (1+a) da &amp;= [a + a^2/2]^{(p-1)}_0 \\
&amp;= (p-1) + \frac{(p-1)^2}{2} \\
&amp;= p-1 + \frac{p^2}{2} - \frac{2p}{2} + \frac{1}{2},\\
&amp; = \frac{p^2}{2} - \frac{1}{2}.
\end{align*}
$$</span></p>
<p>We get
<span class="math-container">$$
\text{aggregate demand} = \begin{cases} 0 &amp;\text{ if } p &lt; 1\\ \frac{p^2}{2} - \frac{1}{2} &amp;\text{ if } 1 \le p \le 2\\ 
\frac{3}{2} &amp;\text{ if } p &gt; 2.\end{cases}
$$</span></p>
<p>Aggregate demand is given below.
<a href="https://i.sstatic.net/Yr5xn.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/Yr5xn.png" alt="aggregate supply" /></a></p>

CC BY-SA 4.0
2023-11-03T09:37:06.193Z
According to the "production function" (<<a production set Yn as depicted in the figure below>>), the firm cannot produce more than 1 unit of output.
escaiguolquer
CC BY-SA 4.0
2023-11-03T12:48:32.570Z
@escaiguolquer thanks. I made some corrections (supply->demand)
tdm
CC BY-SA 4.0
2023-12-08T15:17:30.677Z
Hi - What is the average production set in this case?
Ludwig Gershwin
CC BY-SA 4.0
2023-10-31T11:38:02.707Z
2023-11-03T12:47:38.220Z
2023-11-03T12:47:38.220Z
tdm
<p>Consider a continuum of firms indexed by <span class="math-container">$a\in [0,1]$</span> and that <span class="math-container">$a$</span> is uniformaly distributed. a firm of type <span class="math-container">$a$</span> has a production set <span class="math-container">$Y^n$</span> as depicted in the figure below. note that good 1 is used as an input to produce good 2. suppose that the price of the input (good 1) is fixed at <span class="math-container">$w=1$</span> while the price of output is <span class="math-container">$p&gt;0$</span>.</p>
<p>How should I get the input demand and output supply of a firm of type <span class="math-container">$a$</span> when the price of output is <span class="math-container">$p$</span>? And what is the aggregate or average input demand and output supply when the output price is <span class="math-container">$p$</span>? And what is the aggregate (or average) production set?</p>
<p><a href="https://i.sstatic.net/VqAeK.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/VqAeK.png" alt="enter image description here" /></a></p>

<aggregate-production>
How to solve this aggregate production problem?
user45481
# How to solve this aggregate production problem?

(Asked by: username_0 on 2023-10-30)

Consider a continuum of firms indexed by $a\in [0,1]$ and that $a$ is uniformaly distributed. a firm of type $a$ has a production set $Y^n$ as depicted in the figure below. note that good 1 is used as an input to produce good 2. suppose that the price of the input (good 1) is fixed at $w=1$ while the price of output is $p>0$.

How should I get the input demand and output supply of a firm of type $a$ when the price of output is $p$? And what is the aggregate or average input demand and output supply when the output price is $p$? And what is the aggregate (or average) production set?

[![enter image description here](https://i.sstatic.net/VqAeK.png)](https://i.sstatic.net/VqAeK.png)

**Tags:** aggregate-production


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## 2 answers

### Answer by username_1 on 2023-10-31. Score: 4

Observing the highest possible iso-profit curve passing through the feasible set, we obtain the supply of $y\_2$ and demand for $y\_1$ as follows. Here iso-profit curve is set of $(y\_1,y\_2)$ pairs that yields the same level of profit given $p$.

[![enter image description here](https://i.sstatic.net/iSQuy.png)](https://i.sstatic.net/iSQuy.png)

So,
\begin{eqnarray\*} (y\_1^d,y\_2^s)(p)\in\begin{cases}\{(0,0),(1+a,1)\} & \text{if } p = 1+a \\ \{(1+a,1)\} & \text{if } p > 1+a \\ \{(0,0)\} & \text{if } p < 1+a \end{cases} \end{eqnarray\*}

Corresponding optimal profit is

\begin{eqnarray\*} \pi(p)=\max(0,p-(1+a)) \end{eqnarray\*}

Given that $a\sim\text{Unif}(0,1)$, aggregate input demand is
\begin{eqnarray\*}Y\_1^d(p) = \int\_0^1y\_1^d(p)da = \begin{cases} \int\_0^{1}(1+a)da=\frac{3}{2} & \text{if } p \geq 2 \\ 0 & \text{if } p \leq 1 \\ \int\_0^{p-1}(1+a)da=\dfrac{(p-1)(p+1)}{2} & \text{if } 1 < p < 2\end{cases} \end{eqnarray\*}

Aggregate output supply is
\begin{eqnarray\*}Y\_2^s(p) = \int\_0^1y\_2^s(p)da = \begin{cases} \int\_0^{1}da=1 & \text{if } p \geq 2 \\ 0 & \text{if } p \leq 1 \\ \int\_0^{p-1}da=p-1 & \text{if } 1 < p < 2\end{cases} \end{eqnarray\*}

> **Comments:**
> **username_2**: Hi - What is the average production set in this case?  
> **username_1**: Boundary of the aggregate Production Set is given by $g(y_1)=\begin{cases} 1 & \text{for } y_1<-2\\ \int_0^1 \min(-y_1-a,1)da & \text{for } -2\leq y_1\leq -1 \\ \int_0^1 \max(-y_1-a,0)da & \text{for } -1<y_1\leq 0 \end{cases}$ and aggregate production set is simply $\{(y_1,y_2)|0\leq y_2\leq g(y_1)\}$  

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### Answer by username_3 on 2023-10-31. Score: 4

Let's first look at the profit of a single firm with parameter $a$. Let us denote by $x = -y\_1$ the input and $f(x) = y\_2$ the output.

The production function is the following:
$$
f(x) = \begin{cases} 0 &\text{ if } x < a,\\ x - a &\text{ if } a \le x \le 1+a,\\ 1 &\text{ if } x > 1+a\end{cases}
$$

The firm optimizes profits:
$$
x^\ast = \arg\max\_x p f(x) - x.
$$

You can solve this by looking at various scenarios and taking the best (profit maximising) one.

1. if $x^\ast < a$ then profits are $p 0 - x^\ast$, so here it is best to put $x^\ast = 0$.
2. if $a \le x^\ast \le 1+a$ then profits are $p(x^\ast - a) - x^\ast= (p-1)x^\ast - pa$. If $p > 1$ then profits increase in $x^\ast$ so it is best to put $x^\ast = 1+a$ and profits will be $p - 1 - a$. if $p < 1$, then it is best to set $x^\ast = a$ and profits will be $-a$ (which is negative). In the latter case, it will be best to produce nothing.
3. if $x^\ast \ge 1+a$ then profits are $p - x^\ast$ so here it is optimal to put $x = 1+a$ and profits are given by $p - 1 - a$.

As such, we see that profits are either equal to $p-1-a$ with optimal demand $x^\ast = 1+a$ or equal to zero with optimal demand $x^\ast = 0$.

The first will be optimal if $a \le p-1$, while the second is optimal if $a > p-1$. This gives the following optimal demand for a firm with parameter $a$:
$$
x^\ast = \begin{cases} 1+a &\text{ if } a\le p-1 \\ 0 &\text{ if } a > p-1\end{cases}
$$

Now let us derive optimal aggregate demand. If $p < 1$ then $a \ge 0 > p-1$ so no firm will demand anything: aggregate demand equals zero.

If $p > 2$ then $a \le 1 \le p - 1$ so the firm of type $a$ will demand $1+a$. Aggregate demand is then given by:
$$
\int\_0^1 (1+a) da = [a + a^2/2]^1\_0 = \frac{3}{2}.
$$

If $1 \le p \le 2$ then firms with $a \le p-1$ will demand $1+a$ and firms with $a > p-1$ will not demand anything. As such, aggreagate demand is:
$$
\begin{align\*}
\int\_0^{p-1} (1+a) da &= [a + a^2/2]^{(p-1)}\_0 \\
&= (p-1) + \frac{(p-1)^2}{2} \\
&= p-1 + \frac{p^2}{2} - \frac{2p}{2} + \frac{1}{2},\\
& = \frac{p^2}{2} - \frac{1}{2}.
\end{align\*}
$$

We get
$$
\text{aggregate demand} = \begin{cases} 0 &\text{ if } p < 1\\ \frac{p^2}{2} - \frac{1}{2} &\text{ if } 1 \le p \le 2\\
\frac{3}{2} &\text{ if } p > 2.\end{cases}
$$

Aggregate demand is given below.
[![aggregate supply](https://i.sstatic.net/Yr5xn.png)](https://i.sstatic.net/Yr5xn.png)

> **Comments:**
> **username_4**: According to the "production function" (<<a production set Yn as depicted in the figure below>>), the firm cannot produce more than 1 unit of output.  
> **username_3**: @username_4 thanks. I made some corrections (supply->demand)  
> **username_2**: Hi - What is the average production set in this case?  

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