You are on page 1of 4

Solutions for Homework 2

Chapter 3: Problem 2

Q x Px 140
a. QXd = 1,200 – 3(140) – 0.1(300) = 750 => E P|Px 140   3  0.56 .
Px Q x 750
Since |EP|Px=154| < 1, demand is inelastic. If the firm charged a lower price, TR would decrease.

Q x Px 240
b. QXd = 1,200 – 3(240) – 0.1(300) = 450 => E P| Px  240   3  1.6 .
Px Q x 450
Since |EP|Px=354| > 1, demand is elastic. If the firm charged a higher price, TR would decrease.

c. QXd = 1,200 – 3(140) – 0.1(300) = 750 =>
Q x Pz 300
E P|Pz 300   .01  0.04  0 , goods X and Z are complements.
Pz Q x 750

Chapter 3: Problem 4

%Q x
a. EP   %Q x  E P * %Px   3 * (5%)  15%
%Px
% Q x
b. E xy   % Q x  E xy * % Py   4 * 8%  32%
% Py
% Q x
c. EA   %Q x  E P * %A  2 * (4%)  8%
%A
% Q x
d. E I   %Q x  E I * %I  1 * (4%)  4%
%I

Chapter 3: Problem 15

Toyota should charge the price that makes |EP| = 1 <=> MR = 0 => Max TR.
Q P P
EQ , P   1.5 1 => revenue maximizing P =
P Q 150,000  1.5 P
$50,000.

Problem Set: Problem 3 P Q AR TR MR EP $50 1 $50 $50 $50 40 2 40 80 30 -3. Since max TR where MR = 0. dP dP dQ -9 dP Q 9 4. P = 81 – 9Q => MR = 81 – 18Q b.5 dP Q 8 Problem Set: Problem 6 dQ P 4 a. MR = 81 – 18Q = 0 => Q = 4. Q|P=3 = 16 + 9P – 2P2 = 25.38 Problem Set: Problem 4 a.1.3 = .71 13 5 13 65 -15 -0. (dQ/dP)|P=4 = -7 => p = = -7 = . dP Q 25 .36. dP Q 20 dQ P 3 b. (dQ/dP)|P=3 = -3 => p = =-3 = . TR is maximized when Q = 4. dQ 1 1 dQ P 1 40.52 8 6 8 48 -17 -0.5.1.0.40 20 4 20 80 -10 -0.0.5.5 = 40.5. Q|P=4 = 16 + 9P – 2P2 = 20. Q|P=4 = 20 – 3P = 8 => p =  .50 d. = = => p = . P|Q=4.18 dP Q 17 dQ P 4 b. = . c.0.00 30 3 30 90 10 -1.1 . Q|P=1 = 20 – 3P = 17.5 Problem Set: Problem 5 dQ P 1 a.4. dQ/dP = -3 => p =  -3 = .

000 + 3Q – 4Q2 = 630 => 1 P 1 630 p = . TR = 1.0.0285 Q 2 = 4.7 = 0.0.6 = .000) = .057Q  PT PT Q 5. Problem Set: Problem 9 Price elasticity of demand for U. A and B are substitutes (positive cross elasticity). Problem Set: Problem 10 Q Q Q Q Q Q a. dPB Q A 750 b.0285( Q 2 + 5.000) 1.14 Q 2 = 4.82.S.0.5PB2 = 750. EZ = = 0.0.6 = .750 Q Refreshment price rise = .2 Z Z PT PT PR PR Q 0. Steel should be higher.S.Problem Set: Problem 7 a.55 Pt Q TR from tickets TR from tickets + refreshment .S.7% Q Z 60.7 EPT = = .857. Steel charges lower prices.5 = .000 + 6Q – 12Q2 Problem Set: Problem 8 dQ A PB 10 a.057 Q = . Q Q Q 0. since buyers can relatively easy substitute the products of U.5. Price elasticity is high when good substitutes are readily available. = . QA|Pb=10 = 80PB – 0.722.0. dP dQ Q 77 10 b. Assume current sale Q1 = 5000 tickets.25 ( Q 2 .93% change in the sales of A in the same direction.Z 1.7 = 0. P|Q=10 = 1.8% => Decline of 8% Q b.500 Population increase = 0.0.0.93. Steel for the products of other steel companies if U. as is the case with many steel products.0.000Q + 3Q2 – 4Q3 c.2 * 40% = . but the relationship is not particularly strong: 1% change in the price of B will cause a 0. MR = dTR/dQ = 1.017 => Rise of 1.0. (dQA/dPB)|P=10 = 70 =>  AB = = 70 = 0.6 EPR = = .

5 5.5*0.0.000 5.55 25.3906 %TRticket      0.932.1953Q1 TR2  TR1 2(5.29% (CTR2  CTR1 ) 2 (8. = .1953Q1 10.057(Q2  Q1 )  2(Q2  Q1 )  2.3% Q2  Q1 2(Q2  Q1 ) Using general numbers: %Q    0.038  TRtotal 135.000 + 5. whereas elasticity refers to relative change as they are expressed percentage wise.943Q1  Q1  Q2 2.057 TR1 = P1*Q1 = 5Q1 and TR2 = P2*Q2 = 5.864.057Q2  1.55*4 = 44.487 =.000*4 = 45.8% .8 25.5Q2 = 9.1953Q1  5Q1 ) .9737Q1 Problem Set: Problem 10 The statement is incorrect since it refers to absolute changes in physical sales volume and prices.0.9737Q1  9Q1 )  0.722.0383 or 3.974 + 4.943  0.1953Q1  5Q1 ) 10.0526Q1 %TRcombined      0.9446Q1 = 5.722.00 25. .057  (Q2  Q1 ) / 2 Q2  Q1 1.83% (TR2  TR1 ) 2 (5.3906Q1 .0029 or  0.1 TRtotal + 3.5Q2 = 5.5 + 4)Q2 = 9.2 TR = 974 = 0.9737Q1 CTR2  CTR1 2(8.000 25.5 4.9446Q1 = 8.000.974 25.9737Q1  9Q1 ) 17.1953 Combined TR1 = (P1 + R1)Q1 = (5 + 4)Q1 = 9Q1 Combined TR2 = (P2 + R1)Q2 = (5.5*0.003 TR 44.