## Section2.3Total & Marginal Revenue

After the previous two Decision Reports you now have a full understanding of the demand for your product --- that is, you've modeled the relationship between the price point you charge $p$ and the total number of sales quantity in the national market at that price $q\text{,}$ and that relationship is captured in the demand function $p = D(q)\text{.}$

The next step is to go beyond demand and construct function models for sales revenue, defined as the total dollar amount that would be collected from selling $q$ units at a price of $p$ dollars each:

\begin{equation*} R(q) = q\cdot p = q\cdot D(q) \end{equation*}

We'll also ultimately want a function model for the total cost of production $C(q)\text{,}$ so that at any given production quantity we can compare revenue and cost to determine the net profit, which is defined to be their difference:

\begin{equation*} P(q) = R(q) - C(q) \end{equation*}

(In other words, profit is the amount of money your firm can "keep" after paying a total cost of $C(q)$ dollars to produce $q$ units and then collecting $R(q)$ dollars of revenue from selling those units.)

The key tool in the next decision is "marginal analysis:" the study not only of the functions' values themselves, but of how those values change in response to changes in quantity. At a certain level of production $q\text{,}$ would increasing production have a positive effect on sales revenue (if so, we would say that "marginal revenue is positive")? Or a negative effect ("marginal revenue is negative")?

### Subsection2.3.1Motivation 3

#### Subsubsection2.3.1.1Derivatives: The Why

We could begin this discussion by focusing again on the demand function from our previous Decision Reports. What can we learn from how this function changes from one point to the next, that is, how a change in the independent variable quantity ($q$) effects a change in the dependent variable price ($p = D(q)$)?

In this interactive's default state it shows that, if $q=400$ thousand units of my product are being produced, and if I were to increase my production quantity, I would need to decrease my unit price in order to sell them all. Specifically,

\begin{gather} \text{For each additional thousand units I produce, I'd have to drop the unit price by \\$0.37.}\label{marginal-demand}\tag{2.3.1} \end{gather}

We call this a "marginal" analysis, or a computation of the marginal demand function.

###### Definition2.3.2.Marginal function.

Let $f(q)$ be a function whose independent variable is $q\text{,}$ the quantity of a product or service produced for sale.

The marginal function $Mf(q)$ measures the amount this function's value would change as a result of increasing the production $q$ by a single unit.

This is typically defined for $f$ being demand, revenue, cost, or profit. For example, if $f = R$ is a revenue function, then $MR(60) =$ "at a production of $q=60$ thousand units, how much revenue would be earned by producing 'one more' unit?"

Referring back to the marginal demand computation in (2.3.1), where we found in our example that $MD(400) \approx -0.37\text{,}$ we'll see next that this quantity can be expressed and communicated in four important ways:

1. Numerically, as a rate of change: "$-\0.37$ per thousand units of increase."
2. Graphically, as the slope of the demand graph at $q=400\text{,}$ or more specifically the slope of a line drawn tangent to the grpah at this point, as in the purple line below: 3. In writing, as in (2.3.1), and lastly,
4. In a formula, by deriving an expression for $MD(q)$ and then evaluating it at $q=400\text{.}$ Just how we get that formula is the subject of Subsection 2.3.2.

#### Subsubsection2.3.1.2Derivatives: The What, and the Rule of Four

Numerical: Derivative as a Rate of Change

Graphical: Derivative as a Slope

Verbal: Derivative as a Measure of Marginal Change

Algebraic: Derivative as a Function. If we follow the graphical approach of measuring the slope of a function's graph at each point, we can record those slopes at each point and thereby make a new function, called the derivative function

\begin{align*} f'(x) \amp\amp \text{or sometimes denoted }\frac{df}{dx} \end{align*}

read as "the derivative of $f$ with respect to $x\text{.}$"

Because the derivative $f'$ measures the slope of the graph of $f\text{,}$ there will always be a close relationship between the values of the derivative $f'$ and the direction of the function $f\text{,}$ as you can see in the following interactive from PhET Sims:

### Subsection2.3.3Apply It 3

Let's turn our attention to the revenue function, to prepare for the upcoming decision report. What can we learn from marginal analysis of the revenue function, using each of the Rule of Four (Subsubsection 2.3.1.2)?

Click to launch interactive.

### Subsection2.3.4Decision Report 3

In your previous decision report (Subsection 2.2.3), your team determined a quadratic function $p = D(q)$ which best fits the demand relationship between quantity of items sold $q$ (in thousands) and sale price $p$ (in dollars per unit) on the national market for your product. This function looked something like the example below:

\begin{equation*} p = D(q) = -0.0000535\, q^2 - 0.034403\, q + 414.535. \end{equation*}

This relationship between price and quantity permits your team to also determine the sales revenue --- that is, the total amount of income your company would take in from sales --- at each quantity. This is also a function $R(q)\text{,}$ defined as the amount of money earned by selling $q$ thousand units at a price of $p$ dollars per unit. In other words, revenue is quantity multiplied by price, but price is given by your demand function: in this example,

\begin{equation*} R(q) = q\, p = q \bigl( -0.0000535\, q^2 - 0.034403\, q + 414.535 \bigr) \end{equation*}

(Because $q$ is measured in thousands of units, $R(q)$ is measured in thousands of dollars.)

Meanwhile, the marginal revenue at a given quantity refers to the additional amount of revenue that would be earned by producing one additional unit of your product. More precisely, marginal revenue is the (instantaneous) rate of change of revenue with respect to a change in quantity, which is the derivative:

\begin{equation*} MR(q) \quad {\it or} \quad R'(q) = -0.0001605\, q^2 - 0.068806\, q + 414.535. \end{equation*}

Marginal revenue, being a slope on the graph of $q$ in thousands of units vs. $R$ in thousands of dollars, has units of dollars per unit. Because it is its derivative, $MR$ can tell you whether and how much an increase in quantity will increase revenue (if $MR$ is positive), or decrease revenue (if $MR$ is negative).

Now that you've done some analysis, the project lead who was responsible for the test marketing data wants a status report. What does each of the test markets predict about total revenue on the national market?

#### Exercises2.3.4.1Exercises

###### 1.Decision Report 3: Where will sales revenue top out?

Decide: Which one of the test market price points predicts the highest amount of national sales revenue? Between which two prices is the actual maximum possible revenue likely to be found? Explain your findings in a memo to the test marketing project lead, complete with a table of data and a graph.

Deliver: Determine formulas for revenue $R(q)$ and marginal revenue $MR(q)$ as a function of quantity in the national market (in thousands), as well as a table of values showing $q\text{,}$ $p\text{,}$ $R\text{,}$ and $MR$ at at least four different price points from your test markets.