HOW TO SOLVE THIRD-DEGREE PRICE DISCRIMINATION

Third-degree price discrimination, as you probably know, is the act of charging different prices to different categories of customers based on their elasticity of demand. It is also called multi-market price discrimination

I have explained "price discrimination" in this blog post, so, today we are just going to be solving three questions on multi-market price discrimination. 

Ok, let's get started

Example 1

Suppose a monopoly is practicing price discrimination in the sale of a product by charging different prices in two different markets

Let the demand curve be

$Q_1=300-P_1$

$Q_2=180-P_2$

$MC=20$

A) How much product should be sold in the two markets to maximize profit

B) What price should be charged to maximize profit?

Solution: 

First, you will find the inverse demand curve

$Q_1=300-P_1$

Making P the subject of the formula

$P_1=300-Q_1$

For the other market

$Q_2=180-P_2$

Making P the subject of the formula

$P_2=180-Q_2$

Then, you solve for total revenue.

Recalled that:

$TR=P\times Q$

Since

$P_1=300-Q_1$

$P_2=180-Q_2$

Therefore, 

$TR_1=(300-Q_1)Q$

    $=300Q-(Q_1)^2$

While,

$TR_2=(180-Q_2)Q$

    $=180Q-(Q_2)^2$

Marginal revenues are obtained by taking the derivative of the total revenue curve concerning output.

Hence,

$MR_1=\frac{dTR_1}{dQ_1}=300-2Q_1$

And 

$MR_2=\frac{dTR_2}{dQ_2}=180-2Q_2$

To Maximize profit, $MR_1=MR_2=MC$

in words, the marginal revenues from the first market calculated are equal to the marginal revenues from the second market and both marginal revenues are equal to the marginal cost computed.

For the first market

$MR_1=MC$ will translate into

$300-2Q_1=20$

$300-20=2Q_1$

$280=2Q_1$

$\frac{280}{2}=\frac{2Q_1}{2}$

$140=Q_1$

$Q_1=140$

While $MR_2=MC$ will become

$180-2Q_2=20$

$180-20=2Q_2$

$160=2Q_2$

$\frac{160}{2}=\frac{2Q_2}{2}$

$80=Q_2$

$Q_2=80$

The price is easily obtained by substituting the value of $Q$ into our demand curve

So for the first market

$Q_1=300-P_1$

$140=300-P_1$

$P_1=300-140$

           $=160$

and for the second market,

$Q_2=180-P_2$

$80=180-P_2$

$P_2=180-80=100$

Therefore to maximize profit in the first market, the firm produces where

Price=160 and Quantity=140

And for the second market, the firm will produce where Price=100 and Quantity=80 

Example 2

A firm is selling its product in two markets. In the first market, the demand is given by $Q_1=100-P_1$ and in the second market, the demand is $Q_2=120-0.5P_2$. The firm's total cost is $TC=2000+20(Q_1+Q_2)$

1. Find its profit-maximizing quantity and price.

2. Find the firm profit

Solution:

This is almost identical to the first question, the only difference is that Marginal cost is not given in this question.

Ok, let's play with this question

The inverse demand function will be

$P_1=100-Q_1$

$P_2=240-2Q_2$

$TR=P\times$ Q

 $TR_1=(100-Q_1)Q_1$

            $=100Q_1-Q_1^2$

$TR_2=(240-2Q_2)Q_2$

             $=240Q_2-2Q_2^2$

$MR_1=\frac{dTR_1}{dQ_1}=100-2Q_1$

$MR_2=\frac{dTR_2}{dQ_2}=240-4Q_2$

Marginal costs are obtained by taking the derivative of the total costs curve concerning output.

$MC=\frac{dTC}{dQ}=20$

To maximize profit

$MR=MC$

For the First market

$100-2Q_1=20$

$80=2Q_1$

$2Q_1=80$

Divide both sides by 2

$\frac{2Q_1}{2}=\frac{80}{2}$

$Q_1=40$

For the second market

$240-4Q_2=20$

$240-20=4Q_2$

$220=4Q_2$

$\frac{220}{4}=\frac{4Q_2}{4}$

$55=Q_2$

$Q_2=55$

The price is easily obtained by substituting the value of Q into our demand curve

$Q_1=100-P_1$

Substituting the value of $Q_1$

$40=100-P_1$

$P_1=100-40$

$P_1=60$

For the other market,

$Q_2=120-0.5P_2$

$55=120-0.5P_2$

$0.5P_2=120-55$

$0.5P_2=65$

Divide both sides by $0.5$

$\frac{0.5P_2}{0.5}=\frac{65}{0.5}$

$P_2=130$

Profit can be calculated using the profit-maximizing quantity and price.

$P_1\times Q_1+P_2\times Q_2-TC(Q_1+Q_2 )$

($60\times40)+(130\times55-(2000+20(50+45))=5650$

Example 3

A monopoly is facing two groups of consumers. The inverse demand of market 1 is $P_1=50-\frac{Q_1}{2}$ and in market 2, the inverse demand is $P_2=20-\frac{Q_2}{4}$. The firm's total cost is $C=\frac{(Q_1+Q_2)²}{12}$

1) Find the profit-maximizing quantity and price.

2) calculate its profit if it can successfully price discriminate.

Solution:

First, let's find the total revenue for each market

$TR=P\times Q$

$TR_1=(50-\frac{Q_1}{2})Q$

            $=50Q_1-\frac{Q_1^2}{2}$

$TR_2=(20-\frac{Q_2}{4})Q$

            $=20Q_2-\frac{Q_2^2}{4}$ 

solve for the marginal revenue

$MR_1=\frac{dTR_1}{dQ_1}=50-Q_1$

$MR_2=\frac{dTR_2}{dQ_2}=20-\frac{Q_2}{2}$

Marginal costs are obtained by taking the derivative of the total costs curve concerning output.

$TC_1=\frac{(Q_1+Q_2)²}{12}$

$MC=\frac{dTC_1}{dQ_1}=\frac{Q_1+Q_2}{6}$

To maximize profit, firms will choose sales where

$MR_1=MR_2=MC_2$

in words, the marginal revenues from the first market calculated at profit-maximizing $Q_1$  is equal to the marginal revenues from the second marker calculated at the profit-maximizing $Q_2$  and both marginal revenues are equal to the marginal cost computed at ($Q_1+Q_2$)

For the first market

$50-Q_1= \frac{Q_1+Q_2}{6}$

cross multiply

$6(50-Q_1)=Q_1+Q_2$

$300-6Q_1=Q_1+Q_2$

$300-Q_2=Q_1+6Q_1$

$7Q_1=300-Q_2$.............eqn 1

For second market

$\frac{Q_1+Q-2}{6}=20-\frac{Q_2}{2}$

$\frac{Q_1+Q_2}{6}=\frac{40-Q_2}{2}$

Cross multiply

$2(Q_1+Q_2)= 6(40-Q_2)$

$2Q_1+2Q_2=240-6Q_2$

$8Q_2=240-2Q_1$

$\frac{8Q_2}{8}=\frac{240}{8}-\frac{2Q_1}{8}$

$Q_2=30-\frac{Q_1}{4}$......eqn 2

Substituting eqn 1 into eqn 2

$300-(30-\frac{Q_1}{4})=7Q_1$

$300-\frac{(120-Q_1)}{4}=7Q_1$

Opening bracket

$300-\frac{120+Q_1}{4}=7Q_1$

$\frac{1200-120+Q_1}{4}=7Q_1$

Again cross multiply

$1200-120+Q_1=28Q_1$

$1200-120=28Q_1-Q_1$

$1080=27Q_1$

$\frac{1080}{27}=\frac{27Q_1}{27}$

$40=Q_1$

$Q_1=40$

substitute the value of $Q_1$ in eqn 1

$7(40)=300-Q_2$

$280=300-Q_2$

$Q_2=300-280$

         $=20$

Related post

Now,  let's substitute the value of the Q in the demand equation

$P_1= 50-\frac{Q_1}{2}= 50-\frac{40}{2}$

          $=50-20$

          $=30$

$P_2=20 -\frac{20}{4}$         

           $=20-5$

           $=15$

Profit can be computed using the profit-maximizing quantity and price.

($P_1\timesQ_1)+(P_2\timesQ_2)-TC(Q_1+Q_2$ ) 

$=(30\times40)+(15\times 20)-(\frac{(40+20)²}{12}$)=1200$

I hope with this, you will be able to answer any question on this topic. But, if you have got questions, suggestions? Pls, feel free to tell me in the comment box. Likewise, if you have any educational questions? Join our Facebook community to get expert answers.

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