# Monopoly

Christopher Makler

Stanford University Department of Economics

Econ 50: Lecture 16

# Profit

The profit from $$q$$ units of output

\pi(q) = r(q) - c(q)

PROFIT

REVENUE

COST

is the revenue from selling them

minus the cost of producing them.

# Revenue

We will assume that the firm sells all units of the good for the same price, $$p$$. (No "price discrimination")

r(q) = p(q) \times q

The revenue from $$q$$ units of output

REVENUE

PRICE

QUANTITY

is the price at which each unit it sold

times the quantity (# of units sold).

The price the firm can charge may depend on the number of units it wants to sell: inverse demand $$p(q)$$

• Usually downward-sloping: to sell more output, they need to drop their price
• Special case: a price taker faces a horizontal inverse demand curve;
can sell as much output as they like at some constant price $$p(q) = p$$

# Today's Agenda

• Overview of market structures
• Quick review of revenue, marginal revenue, and elasticity
• Profit maximization with market power
• Monopolistic competition

# Market Structures

What were economists modeling when they came up with all these models?

Farmers producing commodities: price takers, no market power.

Railroads transporting goods:
price setters, lots of market power.

# "Market Power" doesn't actually require a monopoly

## Competition

• Lots of "small" firms selling basically the same thing

## Market Power

• One or a few "medium" or "large" firms selling differentiated products
• Firms face essentially horizontal demand curve
• Firms face downward sloping demand curve

## Monopoly as Metaphor

• True "monopolies" are rare

• Firms with some (at least local) market power are common

• We'll use "monopoly" as a metaphor
to analyze any firm that doesn't take prices as given,
and therefore faces a downward-sloping demand curve.

# Demand and Inverse Demand

Demand curve:

quantity as a function of price

Inverse demand curve:
price as a function of quantity

QUANTITY

PRICE

\text{revenue} = r(q) = p(q) \times q
\text{marginal revenue} = {dr \over dq} = {dp \over dq} \times q + p(q)
\displaystyle{\text{average revenue} = \frac{r(q)}{q} = p(q)}

If the firm wants to sell $$q$$ units, it sells all units at the same price $$p(q)$$

Since all units are sold for $$p$$, the average revenue per unit is just $$p$$.

By the product rule...
let's delve into this...

# Total, Average, and Marginal Revenue

\text{marginal revenue} = {dr \over dq} = {dp \over dq} \times q + p
dr = dp \times q + dq \times p
p
p(q)
q

The total revenue is the price times quantity (area of the rectangle)

\text{marginal revenue} = {dr \over dq} = {dp \over dq} \times q + p
dr = dp \times q + dq \times p
p
p(q)
q
dp
dq

Note: $$MR < 0$$ if

dq \times p
{dq \over q} < {dp \over p}
\% \Delta q < \% \Delta p
|\epsilon| < 1
dp \times q
<

The total revenue is the price times quantity (area of the rectangle)

If the firm wants to sell $$dq$$ more units, it needs to drop its price by $$dp$$

Revenue loss from lower price on existing sales of $$q$$: $$dp \times q$$

Revenue gain from additional sales at $$p$$: $$dq \times p$$

\text{marginal revenue} = {dr \over dq} = {dp \over dq} \times q + p

## Marginal Revenue and Elasticity

= \left[{dp \over dq} \times {q \over p}\right] \times p + p
= {p \over {dq \over dp} \times {p \over q}} + p
= - {p \over |\epsilon_{q,p}|} + p

(multiply first term by $$p/p$$)

(simplify)

(since $$\epsilon < 0$$)

Notes

Elastic demand: $$MR > 0$$

Inelastic demand: $$MR < 0$$

In general: the more elastic demand is, the less one needs to lower ones price to sell more goods, so the closer $$MR$$ is to $$p$$.

# Profit Maximization with Market Power

\pi(q) = r(q) - c(q)

Optimize by taking derivative and setting equal to zero:

\pi'(q) = r'(q) - c'(q) = 0
\Rightarrow r'(q) = c'(q)

Profit is total revenue minus total costs:

"Marginal revenue equals marginal cost"

Example:

p(q) = 20 - q \Rightarrow r(q) = 20q - q^2
c(q) = 64 + {1 \over 4}q^2

What is the profit-maximizing value of $$q$$?

p(q) = 20 - q \Rightarrow r(q) = 20q - q^2
c(q) = 64 + {1 \over 4}q^2

What is the profit-maximizing value of $$q$$?

r'(q)=20-2q
c'(q)={q^2 \over 2}
20-2q={q \over 2}
40=5q
q^*=8
40-4q = q
\pi(q) = r(q) - c(q)

Multiply right-hand side by $$q/q$$:

\pi(q) = \left[{r(q) \over q} - {c(q) \over q}\right] \times q
= (AR - AC) \times q

Profit is total revenue minus total costs:

"Profit per unit times number of units"

AVERAGE PROFIT

CHECK YOUR UNDERSTANDING

p(q)=20-2q
c(q)=10+5q+{1 \over 2}q^2

Find the profit-maximizing quantity.

r(q)=20q-2q^2
r'(q)=20-4q
c'(q)=5+q
20-4q=5+q
15=5q
q^*=3

# Elasticity and Profit Maximization

MR = p\left[1 - \frac{1}{|\epsilon_{q,p}|}\right]

Recall our elasticity representation of marginal revenue:

MR = MC

Let's combine it with this
profit maximization condition:

p\left[1 - \frac{1}{|\epsilon_{q,p}|}\right] = MC
p = \frac{MC}{1 - \frac{1}{|\epsilon_{q,p}|}}

Really useful if MC and elasticity are both constant!

Inverse elasticity pricing rule:

p = \frac{MC}{1 - \frac{1}{|\epsilon_{q,p}|}}

Find the optimal price and quantity if a firm has the cost function $$c(q) = 200 + 4q$$ and faces the demand curve $$D(p) = 6400p^{-2}$$

Inverse elasticity pricing rule:

## One more way of slicing it...

P = \frac{MC}{1 - \frac{1}{|\epsilon|}}
1 - \frac{1}{|\epsilon|} = \frac{MC}{P}
\frac{P - MC}{P} = \frac{1}{|\epsilon|}

Fraction of price that's markup over marginal cost
(Lerner Index)

What if $$|\epsilon| \rightarrow \infty$$?

By Chris Makler

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