Lecture 16: Profit Maximization

Demand and Inverse Demand

In the last lecture, we derived the cost function for a firm: for any quantity of output $q$ we determined the total cost $c(q)$ of producing that quantity. From that function, in turn, we determined the firm’s average cost $AC(q) = c(q)/q$ and marginal cost $MC(q) = c’(q)$.

In this chapter, we’ll determine the firm’s optimal quantity $q^\star$ to produce. We’re going to assume that the firm is trying to maximize its profit, which we’ll write as $\pi(q)$, which is the revenue it receives from selling $q$ units, $r(q)$, minus the cost of producing those $q$ units, $c(q)$: \(\pi(q) = r(q) - c(q)\) We will assume that the firm sets a single price $p$, at which it sells all of its units. In particular, this means that we’re not thinking about a firm like an airline or movie theater that might sell the same product (a seat) to different customers for a different price. (That’s called “price discrimination,” and we’ll get to it much later.)

Furthermore, we will assume that the price the firm can charge may depend on the quantity of output it wants to sell: that is, if it produces $q$ units, the most it can charge is given by some function $p(q)$. Note that this is related to the notion of a “demand function,” but a demand function describes consumers’ behavior as a response to price: that is, $D(p)$ gives the quantity demanded by consumers when the price is $p$. Because we’re thinking of this from the firm’s perspective, we reverse the logic: we think of the price $p$ the firm could charge as a function of the number of units it wants to sell, $q$. For this reason we call it an “inverse demand function,” or, when plotted, an “inverse demand curve:”

We’re only going to be dealing in the short run for this lecture; we’ll come back to the long run in the next lecture when we talk about competitive firms.

Revenue

Because we’re assuming that the firm sells every unit at the same price $p$, the total revenue for a firm facing inverse demand curve $p(q)$ is just \(\text{Total revenue from selling }q\text{ units at price }p(q) = r(q) = p(q) \times q\) Just as the average cost is the total cost divided by the quantity, average revenue is the total revenue divided by the quantity: \(\text{Average revenue }AR(q) = {r(q) \over q} = p(q)\) Note that since every unit is being sold at price $p$, the average revenue is just the price.

For example, if a firm faces the inverse demand curve \(p(q) = 20 - q\) then its revenue function is \(r(q) = p(q)q = 20q - q^2\) and its average revenue is \(AR(q) = {r(q) \over q} = 20 - q\) which is the same as the inverse demand function.

Visually, this means the average revenue curve is the same as the inverse demand curve $p(q)$, and the total revenue is the area represented by price times quantity:

Marginal revenue

We can define marginal revenue as the increase in revenue from increasing output by a bit. Using calculus and the product rule, we have that \(\text{Marginal revenue =}MR(q) = {dr \over dq} = {dp \over dq} \times q + p\) Let’s see what this means. The ratio $dp/dq$ is the slope of the inverse demand curve. If the inverse demand curve is downward sloping, this is negative: that is, if the firm wants to sell $dq$ more units, the price it could charge would drop by $dp$. Note that if we multiply both sides by $dq$, we get \(dr = dp \times q + dq \times p\) This says that if the firm wants to sell $dq$ more units, the resulting change in revenue $dr$ will be composed of two changes:

• Price effect: Revenue will decrease by $dp \times q$, because the firm is dropping its price by $dp$ on the $q$ units it was previously selling; and
• Output effect: Revenue will increase by $dq \times p$, because the firm will sell $dq$ additional units for price $p$.

Visually, we can see these two effects as the red area (lost revenue due to the price effect) and green area (new revenue due to the output effect) in the diagram below:

Relationship between total and marginal revenue

Mathematically, marginal revenue is just the derivative of total revenue; so if, for example, we have the total revenue function \(r(q) = 20q - q^2\) then the marginal revenue will be \(MR(q) = r'(q) = 20 - 2q\)

Visually, we can see the relationship between total and marginal revenue by plotting them together. Since MR is the derivative of TR, the height of the MR curve is the slope of the TR curve. Where TR is increasing, MR is positive; and where TR is decreasing, MR is negative:

Having analyzed the firm’s revenue as a function of $q$, let’s now combine that with our cost function to analyze its profit.

Finding the profit-maximizing quantity to produce

Having defined a firm’s costs in lecture 13, and a firm’s revenues in the previous section, we can now combine these to define its profit function, $\pi(q)$: \(\overbrace{\pi(q)}^\text{PROFIT} = \overbrace{r(q)}^\text{REVENUE} - \overbrace{c(q)}^\text{COST}\)

For example, let’s combine the cost function from Chapter 13 \(c(q) = 64 + \tfrac{1}{4}q^2\) with the revenue function from the last section \(r(q) = 20q - q^2\) For such a firm, their total profit would be \(\pi(q) = r(q) - c(q) = 20q - q^2 - [64 + \tfrac{1}{4}q^2] = -64 + 20q - \tfrac{5}{4}q^2\) If we plot the firm’s total revenue, total cost, and total profit curves, we can see that the total profit is the vertical distance between the total revenue and total cost curves. Where $TR > TC$, the firm is running a profit; where $TR < TC$, the firm is running a loss:

Marginal revenue, cost, and profit

The marginal profit from the $q^\text{th}$ unit produced is the derivative of this function with respect to $q$: \(\pi^\prime(q) = \overbrace{r^\prime(q)}^{MR} - \overbrace{c^\prime(q)}^{MC}\) The firm’s version of a “gravitational pull” argument is by now familiar to us: if producing another unit increases a firm’s profit ($\pi^\prime(q) > 0$, or $MR > MC$) then the firm will expand its output, while if its marginal profit is negative ($\pi^\prime(q) < 0$, or $MR < MC$) then a firm will cut back on production; and as long as things are “well behaved,” the firm’s profit-maximizing quantity will occur at the point where $MR = MC$.

For the specific example above, the marginal profit of the firm would be \(\pi^\prime(q) = r^\prime(q) - c^\prime(q) = \overbrace{20 - 2q}^{MR} - \overbrace{\tfrac{1}{2}q}^{MC}\) Setting this equal to zero gives us \(\begin{aligned} MR &= MC\\ 20 - 2q &= \tfrac{1}{2}q\\ q^\star &= 8 \end{aligned}\)

Visually, we can see that this occurs at the point where the slopes of the TR and TC curves are the same:

To see why this is the profit-maximizing point, try dragging $q$ so it’s a little less than 8 — say, $q = 6$. Notice that in this range, $MR > MC$, so the distance representing profit is increasing as you move to the right. Conversely, if you move $q$ to be a little more than 8 (for example $q = 10$), you can see the opposite is true: increasing production in this range narrows the distance between revenues and cost, decreasing the firm’s profits.

Unit cost and revenue analysis

Let’s now turn to analyzing the profit maximization problem in a unit cost and revenue diagram. The following diagram shows the firm’s average revenue and cost, and its marginal revenue and cost:

Notice that the area between AR and AC, up to the quantity $q$, is shaded in. This represents the profit or loss of the firm. Why? Look what happens if we multiply the profit function for the firm by $q/q$: \(\text{Profit} = TR - TC = \left[{TR \over q} - {TC \over q}\right] \times q = [AR - AC] \times q\) In other words, the firm’s total profit is its average profit (that is, $AR - AC$) times the number of units it sells. Visually, this is the area with height $AR - AC$ and width $q$.

Finally, notice that (although we haven’t said it explicitly) the price the firm charges is found by plugging the optimal quantity back into the inverse demand curve. In this case, the firm maximizes its profit by producing $q^\star = 8$, so the price it charges is $p(8) = 20 - 8 = 12$.

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