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```									BUSS 230-                         Instructor: Pierre Abou Ezze.

Solution to:             Chapter 8, Pro. No. 12&15.
Chapter 9, Pro. No. 12 & appendix problems.

Chapter 8
12. (a) See Figure 7.

(b) At the best short-run level of output of 80 haircuts, MR=SMC (point E in the figure) and P=\$8 (point
A). Since ATC=\$10 the firm incurs a loss of A'A=\$2 per unit and \$160 in total. However, since P=\$8 and
AVC=\$6, the firm minimizes losses by remaining in business in the short run.

Specifically, if the firm stopped producing it would incur a total loss of \$320 (equal to the \$4 average fixed
costs at Q = 80 times the output of 80 units) as compared to the total loss of only \$160 by continuing to
produce.

15. (a)     QD = QS
60,000 - 25,000P = 25,000P
50,000P = 60,000
P = \$1.20 (equilibrium price)
Q = 60,000 - 25,000 (\$1.20) = 30,000 or Q = 25,000(\$1.20) = 30,000

(b) See Figure 9.

(c) In Figure 9, the market price of gasoline is \$1.20 per gallon and the equilibrium quantity in the market is
30,000 gallons per day (point E in the left panel). Since the market is nearly perfectly competitive, each of
the 100 identical firms will face a horizontal demand curve at P = \$1.20. Each firm will supply 1/100 of the
market or 300 gallons per day (point E’, at which the
MC = S curve crosses the d curve).

Assuming that the cost of the gasoline and the cost of the other inputs remain constant, the horizontal
summation of the MC = S curves above P = \$0.60 give the market supply curve (S in the left panel). The
intersection of the D and S curves in the left panel then determine the equilibrium price which is given to
each firm. Thus, the circle is complete-from the market to the firm and back to the market.

(d) See Figure 10.

Figure 10 shows that the level of output of the monopolist is 20,000 per day (given by point E at the
intersection of the MR and the S = ÓMC curve).

(e) The monopolist would operate 66.5 or 67 gasoline stations and close the other 33. If ATC = \$1.20, the
monopolist would earn \$0.40 per gallon and \$8,000 in total per day. This compares to zero profits for the
perfect competitors.

(f). The monopolist supplies less (20,000 gallons per day as compared with 30,000
under perfect competition) and charges a higher price (\$1.60 per gallon as            compared with \$1.20
under perfect competition). This results in a less efficient use of society’s resources than under perfect
competition.
The reason is that consumers are willing to pay \$1.60 for the last gallon of gasoline supplied while it costs the
monopolist only \$0.80 to supply it. The true or deadweight loss to society from moving from perfect competition
to monopoly in this market is given by area E*EE' = \$4,000 in Figure 10.
Chapter 9
12.    (a) See figure 5 on the next page.

(b) If the cartel wants to minimize costs of production, it will set a quota of 8 units of production
for each firm (given by the condition MC1 = MC2 = ... = MC10 = MR = \$8, where the
subscripts refer to the firms in the cartel). This is the same as for the multiplant monopolist.
Using Figure 5, we see that the best level of output for this cartel is 80 units and is given by the point
where MR=Ó MC. The cartel will set the price of \$16. This is the monopoly solution.
(b) If the cartel wants to minimize costs of production, it will set a quota of 8 units of production for each firm
(given by the condition MC1 = MC2 = ... = MC10 = MR = \$8, where the
subscripts refer to the firms in the cartel). This is the same as for the multiplant monopolist.

(c) If the ATC=\$12 at Q=8 for each firm, each firm will earn a profit of \$4 per unit and \$32 in total. The cartel
as a whole will earn a profit of \$320. In this case, each firm will very likely share equally in the cartel’s
profits. In other more complicated and realistic cases, it may not be so easy to decide on how the cartel’s
profits should be shared. The bargaining strength of each firm then becomes important.

Problem Appendix: The Centralized Cartel Model

With Q = 150 - 10P or P = 15 - 0.lQ

TR = PQ = (15 - 0.lQ)Q = 15Q - 0.1Q2

MR = d(TR) = 15 - 0.2Q
dQ

Setting ∑MC = MR, we have
3 + 0.1Q = 15 - 0.2Q

so that 12 = 0.3Q and Q = 40

Then P = 15 - 0.1(40) = 15 - 8 = \$7
At Q=40, MR = 15 - 0.2(40) = 15 - 8 = \$7
Setting MC1 and MC2 equal to MR, we have

4 + 0.2Q1 = \$7 so that Q1 = 15
and
2 + 0.2Q2 = \$7 so that Q2 = 25

Thus, Q = Q1 + Q2 = 15 + 25 = 40

Therefore,
1 = TR1 - TC1 = PQ1 - 4Q1 - 0.lQ1 2

= 11(15) - 4(15) -0.1(15)2 = 165 - 60 -22.50 = \$82.50

2 = TR2 - TC2 = PQ2 -2Q2 -0. 1Q2   2

= 11(25) - 4(25) - 0.1(25)2 = 275 - 100-62.50 = \$112.50

and = 1 + 2 = \$82.50 + \$112.50 = \$195

Problem Appendix: The Market Sharing Cartel

The monopolist will face the market demand function

Q = 120- 10P or P = 12 - 0.1Q

Therefore, TR = PQ = (12 - 0.1Q)Q = 12Q - 0.1Q2

and MR =            d(TR) = 12-0.2Q
dQ

The MC function of the monopolist has one half the slope of the MC curve for each duopolist, so
that
∑MC = 1/2(MC') = 0.2Q/2 = 0.1Q

Setting ∑MC equal to MR, we get

0. 1Q = 12 - 0.2Q

so that Q = 40 and P = 12 - 0.1(40) = \$8 (the same as for each duopolist)

181
The total profits of the monopolist are

=TR-TC

TC =∑MC = 0.05Q2

Therefore,

= 12Q - 0.1Q2 - 0.05Q2

= 12(40) - 0.1(40)2 - 0.05(40)2

= 480 - 160 - 80

= \$240

This is the same as the sum of the total profits of the two duopolists under the equal market-sharing
cartel.

Problem Appendix Sales Maximization Model

(a) Q = 120 - 10P so that P = 12 - 0.1Q and TR = 12Q - 0.1Q2

Since TC = 90 + 2Q, = TR - TC = 12Q - 0.1Q2 - 90 - 2Q = -90 + 10Q - 0.1Q2

The firm maximizes profits where

d= 10 - 0.2Q = 0
dQ

so that Q = 50 and P = 12 - 0.1(50) = \$7

TR = 12(50) - 0.1(50)2 = 600 - 250 = \$350

= -90 + 10(50) - 0.1(50)2= -90 + 500 - 250 = \$160
(b) The firm maximizes TR where

d(TR) = 12 - 0.2Q = 0; so that Q = 60 and P = 12 - 0.1(60) = \$6
dQ

TR = 12(60) - 0.1(60)2 = 720 - 360 =
\$360
= -90 + 10(60) - 0.1(60)2= -90 + 600 - 360 = \$150

(c) When the minimum profit constraint of the firm is \$155

= -90 + 10Q-0.1Q2 = \$155

-245 + 10Q-0.1Q2 = 0

0.1Q2 - 10Q + 245 = 0

Taking the larger of these outputs, we have

P = 12-0.1(57.07) = \$6.29

and

TR = 12(57.07) - 0.1(57.07)2 = 684.84-325.70 = \$359.14

so that

= -90 + 10(57.07) - 0.1(57.07)2= -90+570.70-325.70 = \$155

This is the minimum required.

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