Question

A monopolistic firm produces three products Q1, Q2, and Q3, The average revenue functions are given as: AR 63-40 AR, 105-50, AR 75-62, and the total cost function is given as C-20+150 where Q-2 22+Q. Find the profit-maximizing quantities and the maximum profit that the firm can earn. 3*

Answer 1

Since Average Revenue = Price (P),

AR1 = P1

AR2 = P2 and AR3 = P3.

Total cost (C) = 20 + 15Q = 20 + 15(Q1 + Q2 + Q3) = 20 + 15Q1 + 15Q2 + 15Q3

MC1 = $\partial$ C/$\partial$Q1 = 15

MC2 = $\partial$ C/$\partial$Q2 = 15

MC3 = $\partial$ C/$\partial$Q3 = 15

Profit is maximized when MR1 = MC1, MR2 = MC2 and MR3 = MC3

For product 1,

Total revenue (TR1) = P1 x Q1 = 63Q1 - 4Q1²

Marginal revenue (MR1) = dTR1/dQ1 = 63 - 8Q1

Equating MR1 and MC1,

63 - 8Q1 = 15

8Q1 = 48

Q1 = 6

P1 = 63 - (4 x 6) = 63 - 24 = 39

For product 2,

Total revenue (TR2) = P2 x Q2 = 105Q2 - 5Q2²

Marginal revenue (MR2) = dTR2/dQ2 = 105 - 10Q2

Equating MR2 and MC2,

105 - 10Q2 = 15

10Q2 = 90

Q2 = 9

P2 = 105 - (5 x 9) = 105 - 45 = 60

For product 3,

Total revenue (TR3) = P3 x Q3 = 75 - 6Q3²

Marginal revenue (MR3) = dTR3/dQ3 = 75 - 12Q3

Equating MR3 and MC3,

75 - 12Q3 = 15

12Q3 = 60

Q3 = 5

P3 = 75 - (6 x 5) = 75 - 30 = 45

Market quantity (Q) = 6 + 9 + 5 = 20

Total profit = TR1 + TR2 + TR3 - C

= (P1 x Q1) + (P2 x Q2) + (P3 x Q3) - (20 + 15Q)

= (6 x 39) + (60 x 9) + (45 x 5) - [20 + (15 x 20)]

= 234 + 540 + 225 - (20 + 300)

= 999 - 320

= 679