Solutions to Homework Assignment 7 (Chapter 9)
3.a.
Inverse demand function P(Q) = 0.2 - 0.00004Q. TR = 0.2Q - 0.00004Q^2
FONC: dTR/dQ = 0.2 - 0.00008Q = 0, solve for maximizing Q = 2,500.
P(Q) = 0.2 - 0.00004x2,500 = 0.1.
3.b.
Under the Cournot assumption, the two firms together produce (2/3)Q*,
with P(Q*) = 0.2 - 0.00004Q* = 0. Solving for Q* = 5,000.
Total quantity produced = (2/3)x5,000 = 3,333. Each firm will produce
(1/2)x3,333 = 1,667 units.
The price will be P = 0.2 - 0.00004x3,333 = 0.07.
In each firm TR = 0.07x1,667 = 116.69.
3.c.
Yes, if the firms were to cooperate and each produce one half of the
monopoly output (where TR is maximized at Q = 2,500). Each firm should
produce Q = 1,250 at P = 0.10. In each firm TR = 0.10x1,250 = 125, an
increase of 8.31 for each firm.
5.a.
Large firm's demand function Ql=Qm-Qs=81,000-200P-1,000-50P=80,000-250Pl
Inverse demand function Pl(Ql) = 320 - 0.004Ql. TRl = 320Ql - 0.004Ql^2.
To maximize profit MRl - MCl = 320-0.008Ql-100-0.014Ql = 220-0.022Ql = 0
Solve for maximizing Ql = 100,000. Pl(Ql) = 320 - 0.004x100,000 = 280.
5.b.
Ql = 80,000 - 250x280 = 10,000.
5.c.
Qs = 1,000 + 50x280 = 15,000.
8.a.
MR
115
105
95
85
75
65
55
45
35
8.b.
Q = 50, because for Q > 50 SMC > MR.
8.c.
P = 95. PROF = TR - TFC - TVC = 4750 - 600 - 1500 = 2,650.
Q
0
10
20
30
40
50
60
70
80
90
P
120
115
110
105
100
95
90
85
80
75
TR
0
1150
2200
3150
4000
4750
5400
5950
6400
6750
TVC
0
600
800
900
1120
1500
2400
3500
4800
7200
AVC
60
40
30
28
30
40
50
60
80
AFC
60
30
20
15
12
10
8.57
7.50
6.67
SMC
60
20
10
22
38
90
110
130
240
C1.a. Since AVC is constant AVC = MC = 5.
Inverse demand function P(Q) = 12.5 - 0.0005Q. TR = 12.5Q - 0.0005Q^2.
To maximize profit MR - MC = 12.5 - 0.001Q - 5 = 7.5 - 0.001Q = 0.
Solve for maximizing Q = 7,500. P(Q) = 12.5 - 0.0005x7,500 = 8.75.
C1.b. PROF = TR - FC - AVCxQ = 8.75x7,500 - 12,000 - 5x7,500 = 16,125.
C1.c. Inverse demand function P(Q') = 11 - 0.0005Q'. TR = 11Q - 0.0005Q^2.
To maximize profit MR - MC = 11 - 0.001Q' - 5 = 6 - 0.001Q' = 0.
Solve for maximizing Q' = 6,000. P(Q') = 11 - 0.0005x6,000 = 8.
C1.d. PROF' = TR' - FC - AVCxQ' = 8x6,000 - 12,000 - 5x6,000 = 6,000.
PROF with 5,000 additional advertising expenditures =16,125-5,000=11,125
Thus, profit will decrease by 11,125 - 6,000 = 5,125 when advertising
expenditures are cut by 5,000.
C5.a. Large, dominant firm's demand function Ql = Qm-Qs = 7,520-75P-120-25P.
Inverse demand function P(Ql) = 74 - 0.01Ql. TRl = 74Ql - 0.01Ql^2.
To maximize profit MRl - MCl = 74-0.02Ql - 8-0.002Ql = 62 - 0.022Ql = 0.
Solve for maximizing Ql = 3,000. P(Ql) = 74 - 0.01x3,000 = 44.
C5.b. Qs = 120 + 25x44 = 1,220.
C5.c. PROFl = TRl - TCl = 44x3,000 - 85,000-8x3,000-0.001x3,000^2 = 14,000.
C6.a. Use Starcom's kinked demand function (Q=3,000-20P for P increases and
Q=1,800-10P for price decreases) to find P from 3,000-20P = 1,800-10P.
P = 120. By substitution into either demand curve Q = 600.
C6.b. Inverse demand curve for price increases P(Q) = 150-0.05Q. MR = 150-0.1Q
Inverse demand curve for price decreases P(Q) = 180-0.1Q. MR = 180-0.2Q
Substituting Q = 600 we get upper limit of gap in MR = 150-0.1x600 = 90
and lower limit of gap in MR = 180 - 0.2x600 = 60.
Since, for Q = 600, MC = 3 + 0.1Q = 3 + 0.1x600 = 63, lies within the
gap in MR, Stracom is maximizing profit.
C6.c. For new cost function, when Q = 600, MC' = 7+0.16Q = 7+0.16x600 = 103.
Since MC > upper limit of MR gap, use MR(rising P)-MC=150-0.1Q-7-0.16Q=0
Solve for Q = 550. Using P(Q) for rising P = 150 - 0.05x550 = 122.50.
PROF = TR - STC' = 122.50x550 - 20,000 - 7x550 - 0.08x550^2 = 19,325.