Question

THIRD-DEGREE PRICE DISCRIMINATION A seller faces two groups of buyers: Pi-16- Q1 and P2-24-Q2. Marginal cost is constant at $4 and fixed costs are zero a. Assuming that resale of the good by consumers is impossible, find profit-maximizing quantities and prices under 3rd-degree price discrimination. No need to calculate profit. Show graphs and math. Suppose someone argues "Under this outcome Pi and P2 differ, so this cannot be profit maximizing since a seller could always transfer a unit from the low price market to the high price market and make a profit in doing so." What is wrong with this argument? BE BRIEF (ONE sentence) b. Suppose price discrimination is not possible, so all buyers face the same price. Find profit-maximizing total quantity, price, and profit. Note: make sure you show ALL work. Show this result on a graph. Your graph should include the quantity at which kink in the market demand occurs along with all relevant information about the MR curve (such as values at Qkink) c. Suppose that consumers can resell the good from one market to the next at a cost of S2 per unit. Using the Lagrangian method, find the profit maximizing quantities and prices. Show work.

Answer 1

a)  

A p_1 = 16 - Q_1 p_2 = 24 - Q_2 Me = $ 4 Fe = 0 Profit Maximization problem Max P_1 Q_1 + P_2 Q_2 - TC OPTIMAL SOLUTION MUST HAVE TR_1 = (16 - Q_1)Q_1 therefore TR_1 = 16Q_1 - Q^2_1 MR_1 = 16 - 2Q_1 MR_1 = Me therefore 16 - 2Q_1 - 4 implies 12 = 2Q_1 implies Q^+_1 = 6 TR_2 = (24 - Q_2) Q_2 TR_2 = 24Q_2 Q^2_2 MR_2 = 24 - 2Q_2 MR_2 = Me 24 - 2Q_2 = 4 implies 20 = 2Q_2 implies Q^+_2 = 10 graph not drawn to scale If seller transfers a unit from lower priced market to higher priced market, then demand equations changes and sum of MR OF 1 AND MR OF 2 are not equal to MC. So, not possible. b) Following from (a), the seller cannot discriminate now. So, he must charge equal price in both markets. Total demand (Q) + Q_1 + Q_2 = 16 - P + 24 - P = 40 - 2P

If seller transfers a unit from lower priced market to higher priced market, then demand equations changes and sum of MR OF 1 AND MR OF 2 are not equal to MC. So, not possible.

b) Following from (a), the seller cannot discriminate now. So, he must charge equal price in both markets.

Total demand (Q) + Q1 + Q2 = 16 - P + 24 - P = 40 - 2P

INVERSE DEMAND (P) = (40 - Q ) /2 = 20 - 0.5Q

MC = 4 (GIVEN)

TOTAL REVENUE (P*Q) = 20Q - 0.5 Q² .  

So, MR = 20 - Q

At equlibrium, MC = MR

or, 4 = 20 - Q

or, Q* = 16 Now, P = 20 -0.5(16) = 16 - 8 = 8 or, P*=8

PROFIT = TR - TC = 20(16) - 0.5(16²) - 4(16) = 320 - 128 - 64 = 128.

c) max R(q)

subject to $\pi$ = R(q) - C(q) >= $\pi$* ---------------------------------------------------- (1)

Lagrangian function : L = R(q) + $\lambda$ [ R(q) - C(q) - $\pi$*] ;  $\pi$* is min acceptable profit.

Kuhn- tucker conditions are : (1) dL/dq = R'(q) + $\lambda$ [ R'(q) - C'(q) - $\pi$ ] <= 0

(2) q >= 0 q(dL/dq) = 0

(3)  $\lambda$ >= 0 $\lambda$(dL/d$\lambda$) = 0

(4) dl/d$\lambda$ = R(q) - C(q) - $\pi$* >= 0  

Assuming $\pi$* = 85.75 < profit of 128

Unrestricted Max R'(q) = 200 at Q* = 20 and $\pi$ = 120

Equality 1 is

20q - 0.5q² - (4q+2q) = 85.75

or, 14q - 0.5q² - 85.75 = 0

or, q² -28q +171.5 = 0

Using sridhar acharya formula to solve the quadatic equation, we get

q = 9,19

therefore,p = 15.5, 10.5