# BM2.4 – Monopoly

Chapter 2, Lesson 4

In this lesson students will:

• understand the difference between a linear and a quadratic total profit function
• understand and give an example of an economy under monopoly
• create a quadratic total profit function given some information
• use the intercept method or the power law of derivative calculus to determine the maximal production level and maximal profit level for a company

## A Non-Calculus Approach to Monopoly

In our last lesson, our total profit function was linear in nature; that is, the highest power of the variable $x$ was one. While linear functions do occur quite readily in nature, it is often not the most ideal model to use when modelling the total profit function. In this lesson, we will allow the selling price $S$ to fluctuate according to market demands, creating a total profit function that is quadratic, or parabolic, in nature. This means the highest power of our variable $x$ will be two. These kind of total profit functions often model economies that are known as monopolies. A monopoly is an economy in which one company holds all of the production and selling power. For example, the Insurance Corporation of British Columbia (ICBC) is a monopoly – they are the only insurance company in which drivers can insure their vehicles. As such, they have a lot of control over pricing and legislation regarding insurance of vehicles.

Allowing the selling price to fluctuate will open us up to an interesting set of questions: what is the maximal profit that a company can make, and how many items should be produced in order to do so? With a linear model, we cannot ask such a question since the more items produced, the more profit we obtain. It doesn’t really make sense to discuss maximal profit output for a linear model – we simply produce as many items as we possibly can. Yet, for a quadratic model, we know from our previous studies that a parabola must attain a maximal or minimal value – this is sometimes called the vertex of the parabola. The focus of this lesson is to determine a maximal value for a given total profit function.

#### Using Intercepts to Determine Maximal Profit

The first method that we can use to help us determine the maximal profit is to find the $x$-intercepts of the total profit function. Given a quadratic total profit function, we can try to factor the right hand side of the equation. Once we obtain the two intercepts, we can find the optimal production level by using symmetry. That is, since we are working with a parabola, the $x$-coordinate of the vertex must be halfway between the two intercepts. Substitution will then give us the maximal profit as well. In brief:

1. Factor the right hand side to obtain the two intercepts, if possible. If this seems challenging, you might consider using the quadratic formula.
2. Add the two intercepts together, then divide by two. This will give the $x$-intercept of the vertex, or the number of items to produce to make maximal profit.
3. Substitute the value obtained in 2. into the total profit function. This will give the $y$-intercept of the vertex, or the maximal profit.

## A Calculus Approach to Monopoly

#### The Power Law of Derivative Calculus

For those of you that have taken calculus, we will explore a second option that is a bit quicker than the previous method, and has further connections to economics. Let’s first recall the power law of derivative calculus. The power law states that the rate of change of a polynomial expression of the form $x^{m}$ can be found by bringing the power down and multiplying and then subtracting one from the power. That is

\begin{aligned} x^{m} &\longrightarrow mx^{m-1}. \end{aligned}

For example, the expression $x^{3}$ changes to $3x^{2}$ when applying the power law to find the rate of change. Now this law by itself doesn’t seem like it would be helpful to answer our big questions from the beginning of this lesson; however, it turns out that the rate of change at a maximal value (like the vertex of a parabola) happens to be zero. This means, if we could determine the rate of change of our total profit function, and set this rate of change equal to zero, we could solve for the $x$-value that would lead us to the optimal production level. Before we explore this concept, let’s go over a few more examples of how to find the rate of change of various polynomial expressions.

#### Marginal Revenue and Marginal Cost

Now that we know how to determine the rate of change of a polynomial expression, let’s connect this idea back to business economics. Recall that the total profit function is given by

\begin{aligned} TP &= TR - TC. \end{aligned}

Let’s denote the rate of change of the total profit function using a dash: $TP'$. The rate of change of the total profit function is known in economics as the marginal profit. Similarly, we can denote $TR'$ and $TC'$ as the marginal revenue and marginal cost; these are the rates of change of the revenue and cost functions, respectively. We know from calculus that the maximal profit will be found when the marginal profit is equal to zero; that is $TP' = 0$. Applying the rate of change to our three functions, then making this substitution allows us to arrive at an interesting conclusion:

\begin{aligned} TP &= TR - TC\\ TP' &= TR' - TC' \\ 0 &= TR' - TC' \\ TC' &= TR'. \end{aligned}

In other words, the maximal profit can be found by setting the marginal revenue and the marginal cost equal to each other, and solving for the variable $x$. This gives us two methods for determining maximal profit, given some information:

1. If given $TP$ explicitly, find $TP'$ by using the power law. Set $TP' = 0$ and solve for $x$ to determine the production level required for maximal profit. If needed, substitute this $x$-value back into the total profit function (not the marginal profit) to obtain the maximal profit level.
2. If given $TR$ and $TC$ explicitly, find $TR'$ and $TC'$ by using the power law. Set $TR' = TC'$ and solve for $x$ to determine the production level required for maximal profit. If needed, create the total profit function and substitute this $x$-value to obtain the maximal profit level.

Let’s work through a few examples of these techniques to end the lesson.