Question

32. (a)    A firm wishes to third degree price discriminate to meet competition in two markets A and B. The firm’s total cost schedule is TC = $10*QA + $25*QB. Demands in the two markets are QA = 4,000*PA^-1.5 and QB = 100,000*PA^-2. The profit-maximizing prices to charge in markets A and B are:

A         $20, $30.                     B.             $30, $50.                     C.             $40, $50                      D.             $60, $80                      E. None of the above.

(b) Refer back to the previous problem. The amounts the firm will sell in each market are:

A         20, 30.                        B.             24, 40.                        C.             40, 100                        D.             60, 80             

Answer 1

1)

QA = 4000PA- 1.5

QB = 100,000PB-2

A demand function of the form

Q = aP- b

dQ/dP = a(-b)P- b - 1  

Ed = (dQ/dP)(P/Q)

= (-abP- b - 1  )(P/Q)  

= -( abP- b - 1 + 1)/Q

= - abP - b/Q

= - b(aP - b/Q)

= - b(Q/Q) = - b  

Now

QA = 4000PA- 1.5

therefore EdA = - 1.5  

likewiwe QB = 100,000PB-2   

EdB = - 2

Profit maximization condition

MR = MC

P(1 - 1/| Ed |) = MC

TC = 10QA + 25QB  

MCA = 10

MCB = 25

Now PA(1 - 1/|EdA |) = MCA

PA(1 - 1/| - 1.5 | ) = 10

PA(1 - 1/1.5) = 10

PA{1 - 1/(3/2)} = 10  

PA(1 - 2/3) = 10

PA/3 = 10  

PA = 30  

MRB = MCB  

PB(1 - 1/| - 2 |) = 25

PB(1 - 1/2) = 25

PB/2   = 25

PB = 50

2)

QA = 4000PA- 1.5

= 4000(30)- 1.5

= 24.34

= 24 (Taking nearest whole number)

QB = 100,000PB-2

  = 100,000(50)- 2

= 40