# Obtain profit-maximizing supply curve (an exercise)

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?

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

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