Obtain profit-maximizing supply curve (an exercise)

Question

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?

Answer 1

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

Answer 2

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, $$Y^s(\alpha,p,w)\in \begin{cases}\phi & \text{if} \ \ \alpha p>w \\ \mathbb{R}^+ & \text{if} \ \ \alpha p = w\\ \{0\} & \text{if} \ \ \alpha p < w\\ \end{cases}$$