Solutions to Homework Assignment 2 (Chapter 2)
6.a.
Ep,ivm = (Q2 - 1,000)/(-200) x (600 + 800)/(Q2 + 1,000) = - 2.
-7Q2 + 7,000 = -2Q2 - 2,000.
5Q2 = 9,000.
Q2 = 1,800.
6.b.
TR2 = Q2 x P2 = 1,800 x 600 = 1,080,000.
6.c.
The corresponding increase in cost.
6.d.
Ep,v = (Q2 - 1,800)/(800 - 900) x (1,700)/(Q2 + 1,800) = 0.5.
-17Q2 + 30,600 = 0.5Q2 + 900.
17.5Q2 = 29,700.
Q2 = 1,697.
10.a.
Ep,c=(Q2 - 220,000)/(1.65 - 1.89) x (1.65 + 1.89)/(Q2 + 220,000)=-1.8.
14.75Q2 - 3,245,000 = 1.8Q2 + 396,000.
12.95Q2 = 3,641,000.
Q2 = 281,158.
10.b.
New TR = 821,158 x 1.65 = 463,911.
This compares with a previous TR = 220,000 x 1.89 = 415,800.
10.c.
Ep,a=(Qm2 - 160,000)/(1.65 - 1.89) x (1.65 + 1.89)/(Qm2 + 160,000)=2.2.
14.75Qm2 - 2,360,000 = -2.2Qm2 - 352,000.
116.95Qm2 = 2,008,000.
Qm2 = 118,466.
10.d.
For Mindy's the change in TR is (118,466 - 160,000) x 1.79 = -74,346.
C3.a.
Given estimated demand curve Q = 2,840 - 20P:
AR = P = 142 - 0.05Q (inverse demand curve).
TR = Q x P(Q) = Q X (142 - 0.05Q) = 142Q - 0.05Q^2.
MR = dTR/dQ = 142 - 0.1Q.
C3.b.
FONC: dTR/dQ = MR = 142 - 0.1Q = 0, solve for maximizing Q = 1,420.
Substitute Q into P(Q), for corresponding P(Q) = 142 -0.05x1,420 = 71.
Maximum TR = 1,420 x 71 = 100,820.
C3.c.
Differentiate the Q equation: dQ/dP = -20.
For Q = 1,600, P(Q) = 142 - 0.05x1,600 = 62.
Ep = dQ/dP x P/Q = -20 x (62/1,600) = -0.78, inelastic since |Ep| < 1.
C3.d.
For Q1 = 1,000, P1 = 142 - 0.05x1,000 = 92.
For Q2 = 1,100, P2 = 142 - 0.05x1,100 = 87.
Ep = (1,100 - 1,000)/(87 - 92) x (87 + 92)/(1,100 + 1,000) = -1.70.
In the range from Q1 = 1,000 to Q2 = 1,100, demand is elastic, and a 1%
price cut will lead to an increase of 1.7% in quantity demanded. If
price is cut from $92 to $87, TR will increase from 1,000 x 92 = 92,000
to 87 x 1,100 = 95,700.
C6.a.
Qs=89830-40x9000+20x9500+15x10000+2x15000+0.001x170000+10x160 = 101,600
Ep,s = dQs/dPs x Ps/Qs = -40 x 9,000/101,600 = -3.54.
C6.b.
Elastic, since |Ep| > 1.
C6.c.
If price is lowered, TR will increase, because 1% decrease in price will
result in 3.54% increase in quantity demanded. For effect on profit,
we need to calculate the increase in cost from higher levels of sale.
C6.d.
Qs=89830-40x9000+20x10000+15x10000+2x15000+0.001x170000+10x160 = 111,600
Ep,x = (Q2 - Q1)/(P2 - P1) x (P2 + P1)/(Q2 + Q1)
Ep,x=(111600-101600)/(10000-9500) x (10000+9500)/(111600+101600) = 1.83
Substitutes, since the cross price elasticity of demand is positive.
C6.e.
Ei = dQs/dI x I/Qs = 2 x 15,000/101,600 = 0.295
Normal good since the income elasticity of demand is positive.
C7.a.
Qx = 1,420 - 20Px - 10x40 + 0.02x8,000 + 0.04x1,200 = 1,228 - 20Px
C7.b.
Y is complement of X, since Py = -10 is negative, and hence the cross
price elasticity of demand for X with respect to Y will be negative.
C7.c.
For Px = 50, Qx = 1,228 - 20x50 = 228.
Ep = dQ/dP x P/Q = -20 x 50/228 = -4.39.
C7.d.
Inverse demand function is P(Q) = -0.05Q + 61.4.
TR = QxP(Q) = Q x (-0.05Q + 61.4) = -0.05Q^2 + 61.4Q
FONC: dTR/dQ = -0.1Q + 61.4 = 0, solve for maximizing Q = 614.
Corresponding P(Q) = -0.05x614 +61.4 = 30.70
Maximum TR = MR = Q x P = 614 x 30.70 = 18,849.80.