<p>The producer's problem is</p>
<p><span class="math-container">$
\begin{align*}
\max_{l \in \mathbb{R_+}}              &amp;\; pq - wL \\
\text{subject to} &amp;\;
\begin{alignedat}[t]{1}
   q &amp; \leq f(L) \\
f(L) &amp; = L\alpha
\end{alignedat}
\end{align*}
$</span></p>
<p>Simply put, we want to maximise profit
<span class="math-container">$\pi(L) = pL\alpha - wL$</span></p>
<p>Note that <span class="math-container">$\pi(L)$</span> is a linear function of <span class="math-container">$L$</span>, so this will require a little more care than just setting the derivative to 0!</p>
<p><span class="math-container">$
\pi(L) = pL\alpha - wL \\
\pi'(L) = p\alpha - w \\
\pi''(L) = 0
$</span></p>
<p>Since the second derivative is 0, the first derivative test for maximisation is inconclusive. (This generalises across linear functions - their rate of change is constant).</p>
<p>In this case, let's look more carefully at <span class="math-container">$\pi'(L)$</span>, which is a constant.</p>
<p><strong>Case 1:</strong> <span class="math-container">$\pi'(L) &gt; 0 $</span><br />
Profit is an increasing function of labour. Without any constraints on <span class="math-container">$L$</span>, the entrepreneur can always increase profit by employing an additional unit of <span class="math-container">$L$</span>. So no maximum exists in this case.</p>
<p><strong>Case 2:</strong> <span class="math-container">$\pi'(L) &lt; 0 $</span><br />
Profit is a decreasing function of labour. So every unit of labour employed actually reduces profit (or increases losses). The maximum profit therefore is where we employ the least amount of <span class="math-container">$L$</span>, i.e., <span class="math-container">$L=0$</span>.</p>
<p><strong>Case 3:</strong> <span class="math-container">$\pi'(L) = 0 $</span><br />
Profit is constant at all points, independent of the units of labour employed. So the optimum amount of labour is any <span class="math-container">$L \in \mathbb{R_+}$</span></p>
<p>Summarizing, we get the labour demand <span class="math-container">$L^d$</span> as a function of <span class="math-container">$(p, w, \alpha)$</span>:</p>
<p><span class="math-container">$L^d(p, w, \alpha) \in 
\begin{cases}
\phi, \ \ \ \textrm{if } p\alpha &gt; w \\
\mathbb{R_+}, \ \ \ \textrm{if } p\alpha = w \\
\{0\}, \ \ \ \textrm{if } p\alpha &lt; w \\
\end{cases}$</span></p>
<p>The supply curve is obtained by feeding <span class="math-container">$L^d$</span> back into our production function: <span class="math-container">$q^s(p, w, \alpha) = f(L^d(p, w, \alpha)) = \alpha L^d(p, w, \alpha)$</span></p>
<p><strong>EDIT:</strong> Here's a visual to build intuition: <a href="https://www.desmos.com/calculator/hfbj67ulf0" rel="nofollow noreferrer">https://www.desmos.com/calculator/hfbj67ulf0</a></p>

CC BY-SA 4.0
2024-08-31T06:40:25.530Z
2024-08-31T06:48:28.477Z
2024-08-31T06:48:28.477Z
shgwani2804
<p>Note that our production function is  <span class="math-container">$Y=\alpha L$</span> where Y is the number of widgets produced, then
our objective is to maximise profits of the of the entrepreneur so,
The Profit Maximisation problem of the entrepreneur is,
<span class="math-container">$$\max_{Y\geq 0}Y-wL$$</span>
after substituting for Y,
<span class="math-container">$$\max_{L\geq 0}\alpha p L-wL$$</span>
For solving the above problem note that our objective is linear in <span class="math-container">$L$</span> so solving the problem will give us the unconditional labor demand function of the entrepreneur,
<span class="math-container">$$L^d(\alpha,p,w)\in \begin{cases}\phi &amp; \text{if} \ \ \alpha p &gt; w\\\mathbb{R}^+ &amp; \text{if} \ \ \alpha p = w\\ \{0\} &amp; \text{if} \ \ \alpha p &lt; w\\
\end{cases}$$</span>
Substituting it back into the production function <span class="math-container">$Y=\alpha L$</span> will give us the optimal supply function of the entrepreneur,
<span class="math-container">$$Y^s(\alpha,p,w)\in \begin{cases}\phi &amp; \text{if} \ \ \alpha p&gt;w \\
\mathbb{R}^+ &amp; \text{if} \ \ \alpha p = w\\ \{0\} &amp; \text{if} \ \ \alpha p &lt; w\\
\end{cases}$$</span></p>

CC BY-SA 4.0
2024-08-31T06:34:37.143Z
2024-08-31T07:43:00.283Z
2024-08-31T07:43:00.283Z
econhead
<p>Exercise 2.2.3 from <em>Introduction to economic analysis</em> by McAfee:</p>
<blockquote>
<p>An entrepreneur has a factory that produces <span class="math-container">$L\alpha$</span> widgets, where
<span class="math-container">$\alpha&lt;1$</span>, when <span class="math-container">$L$</span> hours of labor is used. The cost of labor (wage
and benefits) is <span class="math-container">$w$</span> per hour. If the entrepreneur maximizes profit,
what is the supply curve for widgets?</p>
<p>Hint: The entrepreneur’s profit, as a function of the price, is
<span class="math-container">$pL\alpha – wL$</span>. The entrepreneur chooses the amount of labor to
maximize profit. Find the amount of labor that maximizes, which is a
function of <span class="math-container">$p$</span>, <span class="math-container">$w$</span> and <span class="math-container">$\alpha$</span>. The supply is the amount of output
produced, which is <span class="math-container">$L\alpha$</span>.</p>
</blockquote>
<p>How to approach this at all? I do understand that given a price(per unit) <span class="math-container">$p$</span>, since the amount of supply produced is <span class="math-container">$L\alpha$</span>, the expected profit without net is <span class="math-container">$pL\alpha$</span> and the net profit is therefore
<span class="math-container">$$pL\alpha - wL$$</span>
<span class="math-container">$wL$</span> is the amount you pay to the labors for <span class="math-container">$L$</span> hours.</p>
<p>What does the author mean by <strong>maximizing profit by choosing the amount of labor</strong> here? Is it just <span class="math-container">$L$</span>?</p>
<p>But then equating the derivative to <span class="math-container">$0$</span> would result in
<span class="math-container">$$p\alpha = w \Leftrightarrow p = \frac{w}{\alpha} = \frac{wL}{q},$$</span>
here <span class="math-container">$q = L\alpha$</span> being the amount of quantity produced as in the hint. This can't be the supply curve since it is decreasing as <span class="math-container">$q \rightarrow \infty$</span>. Am I wrong somehow?</p>
<p>My questions:</p>
<ol>
<li>What is the amount of labor?</li>
<li>How to approach this question at all?</li>
</ol>

<microeconomics><mathematical-economics>
Obtain profit-maximizing supply curve (an exercise)
Duck Gia
# Obtain profit-maximizing supply curve (an exercise)

(Asked by: username_0 on 2024-08-31)

Exercise 2.2.3 from *Introduction to economic analysis* by McAfee:

> An entrepreneur has a factory that produces $L\alpha$ widgets, where
> $\alpha<1$, when $L$ hours of labor is used. The cost of labor (wage
> and benefits) is $w$ per hour. If the entrepreneur maximizes profit,
> what is the supply curve for widgets?
>
> Hint: The entrepreneur’s profit, as a function of the price, is
> $pL\alpha – wL$. The entrepreneur chooses the amount of labor to
> maximize profit. Find the amount of labor that maximizes, which is a
> function of $p$, $w$ and $\alpha$. The supply is the amount of output
> produced, which is $L\alpha$.

How to approach this at all? I do understand that given a price(per unit) $p$, since the amount of supply produced is $L\alpha$, the expected profit without net is $pL\alpha$ and the net profit is therefore
$$pL\alpha - wL$$
$wL$ is the amount you pay to the labors for $L$ hours.

What does the author mean by **maximizing profit by choosing the amount of labor** here? Is it just $L$?

But then equating the derivative to $0$ would result in
$$p\alpha = w \Leftrightarrow p = \frac{w}{\alpha} = \frac{wL}{q},$$
here $q = L\alpha$ being the amount of quantity produced as in the hint. This can't be the supply curve since it is decreasing as $q \rightarrow \infty$. Am I wrong somehow?

My questions:

1. What is the amount of labor?
2. How to approach this question at all?

**Tags:** microeconomics, mathematical-economics


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## 2 answers

### Answer by username_1 on 2024-08-31. Score: 1 | Accepted answer

The producer's problem is

$
\begin{align\*}
\max\_{l \in \mathbb{R\_+}} &\; pq - wL \\
\text{subject to} &\;
\begin{alignedat}[t]{1}
q & \leq f(L) \\
f(L) & = L\alpha
\end{alignedat}
\end{align\*}
$

Simply put, we want to maximise profit
$\pi(L) = pL\alpha - wL$

Note that $\pi(L)$ is a linear function of $L$, so this will require a little more care than just setting the derivative to 0!

$
\pi(L) = pL\alpha - wL \\
\pi'(L) = p\alpha - w \\
\pi''(L) = 0
$

Since the second derivative is 0, the first derivative test for maximisation is inconclusive. (This generalises across linear functions - their rate of change is constant).

In this case, let's look more carefully at $\pi'(L)$, which is a constant.

**Case 1:** $\pi'(L) > 0 $  
Profit is an increasing function of labour. Without any constraints on $L$, the entrepreneur can always increase profit by employing an additional unit of $L$. So no maximum exists in this case.

**Case 2:** $\pi'(L) < 0 $  
Profit is a decreasing function of labour. So every unit of labour employed actually reduces profit (or increases losses). The maximum profit therefore is where we employ the least amount of $L$, i.e., $L=0$.

**Case 3:** $\pi'(L) = 0 $  
Profit is constant at all points, independent of the units of labour employed. So the optimum amount of labour is any $L \in \mathbb{R\_+}$

Summarizing, we get the labour demand $L^d$ as a function of $(p, w, \alpha)$:

$L^d(p, w, \alpha) \in
\begin{cases}
\phi, \ \ \ \textrm{if } p\alpha > w \\
\mathbb{R\_+}, \ \ \ \textrm{if } p\alpha = w \\
\{0\}, \ \ \ \textrm{if } p\alpha < w \\
\end{cases}$

The supply curve is obtained by feeding $L^d$ back into our production function: $q^s(p, w, \alpha) = f(L^d(p, w, \alpha)) = \alpha L^d(p, w, \alpha)$

**EDIT:** Here's a visual to build intuition: <https://www.desmos.com/calculator/hfbj67ulf0>


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### Answer by username_2 on 2024-08-31. Score: 1

Note that our production function is $Y=\alpha L$ where Y is the number of widgets produced, then
our objective is to maximise profits of the of the entrepreneur so,
The Profit Maximisation problem of the entrepreneur is,
$$\max\_{Y\geq 0}Y-wL$$
after substituting for Y,
$$\max\_{L\geq 0}\alpha p L-wL$$
For solving the above problem note that our objective is linear in $L$ so solving the problem will give us the unconditional labor demand function of the entrepreneur,
$$L^d(\alpha,p,w)\in \begin{cases}\phi & \text{if} \ \ \alpha p > w\\\mathbb{R}^+ & \text{if} \ \ \alpha p = w\\ \{0\} & \text{if} \ \ \alpha p < w\\
\end{cases}$$
Substituting it back into the production function $Y=\alpha L$ will give us the optimal supply function of the entrepreneur,
$$Y^s(\alpha,p,w)\in \begin{cases}\phi & \text{if} \ \ \alpha p>w \\
\mathbb{R}^+ & \text{if} \ \ \alpha p = w\\ \{0\} & \text{if} \ \ \alpha p < w\\
\end{cases}$$


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