Instructor’s Manual to accompany Fundamental Methods of Mathematical Economics Fourth Edition Alpha C. Chiang University of Connecticut Kevin Wainwright British Columbia Institute of Technology Title of Supplement to accompany FUNDAMENTAL METHODS OF MATHEMATICAL ECONOMICS Alpha C. Chiang, Kevin Wainwright Published by McGraw-Hill, an imprint of The McGraw-Hill Companies, Inc., 1221 Avenue of the Americas, New York, NY 10020. Copyright 2005, 1984, 1974, 1967 by The McGraw-Hill Companies, Inc.
All rights reserved. The contents, or parts thereof, may be reproduced in print form solely for classroom use with FUNDAMENTAL METHODS OF MATHEMATICAL ECONOMICS provided such reproductions bear copyright notice, but may not be reproduced in any other form or for any other purpose without the prior written consent of The McGraw-Hill Companies, Inc., including, but not limited to, in any network or other electronic storage or transmission, or broadcast for distance learning.com Contents CONTENTS 1 CHAPTER 2 6 Exercise 2.com Chiang/Wainwright: Fundamental Methods of Mathematical Economics Instructor’s Manual CHAPTER 2 Exercise 2. All are valid. First part: A ∪(B ∩ C) = {4, 5, 6} ∪{3, 6} = {3, 4, 5, 6} ; and (A ∪B)∩ (A∪ C) = {3, 4, 5, 6, 7}∩ {2, 3, 4, 5, 6} = {3, 4, 5, 6} too.
Second part: A ∩ (B ∪ C) = {4, 5, 6} ∩ {2, 3, 4, 6, 7} = {4, 6} ; and (A ∩ B) ∪ (A ∩ C) = {4, 6} ∪ {6} = {4, 6} too. There are 24 = 16 subsets: ∅, {a}, {b}, {c}, {d}, {a,b}, {a,c}, {a,d}, {b,c}, {b,d}, {c,d}, {a,b,c}, {a,b,d}, {a,c,d}, {b,c,d}, and {a,b,c,d}. The complement of U is Ũ = {x | x ∈ / U }. Here the notation of ”not in U ” is expressed via the ∈ / symbol which relates an element (x) to a set (U ).
In contrast, when we say ”∅ is a subset of U,” the notion of ”in U” is expressed via the ⊂ symbol which relates a subset(∅) to a set (U ). Hence, we have two different contexts, and there exists no paradox at all.com Chiang/Wainwright: Fundamental Methods of Mathematical Economics Instructor’s Manual 3. Only (d) represents a function. The range is the set of all nonpositive numbers.
For each level of output, we should discard all the inefficient cost figures, and take the lowest cost figure as the total cost for that output level. This would establish the uniqueness as required by the definition of a function. (a) and (c) differ in the constant terms; a larger constant means a higher vertical intercept. A negative coefficient (say, -1) for the x2 term is associated with a hill.
as the value of x is steadily increased or reduced, the −x2 term will exert a more dominant influence in determining the value of y. Being negative, this term serves to pull down the y values at the two extreme ends of the curve. If negative values can occur there will appear in quadrant III a curve which is the mirror image of the one in quadrant I. By Rules VI and V, we can successively write xm/n = (xm )1/n = n xm ; by the same two rules, √ we also have xm/n = (x1/n )m = ( n x)m 8.
Rule VI: (xm )n = |xm × xm{z ×. × x} = x mn n term s mn term s 7 www.com Chiang/Wainwright: Fundamental Methods of Mathematical Economics Instructor’s Manual Rule VII: xm × y m = x × x × .com Chiang/Wainwright: Fundamental Methods of Mathematical Economics Instructor’s Manual CHAPTER 3 Exercise 3. (a) By substitution, we get 21 − 3P = −4 + 8P or 11P = 25. Substituting 2 P ∗ into the second equation or the third equation, we find Q∗ = 14 11.
(b) With a = 21, b = 3, c = 4, d = 8, the formula yields P ∗ = 25 3 11 = 2 11 Q∗ = 156 2 11 = 14 11 2. If b+d = 0 then P ∗ and Q∗ in (3.5) would involve division by zero, which is undefined. If b + d = 0 then d = −b and the demand and supply curves would have the same slope (though different vertical intercepts). The two curves would be parallel, with no equilibrium intersection point in Fig.
By Theorem III, we find: (a) Yes.com Chiang/Wainwright: Fundamental Methods of Mathematical Economics Instructor’s Manual 5. (a) By Theorem I, any integer root must be a divisor of 6; thus there are six candidates: ±1, ±2, and ±3. Among these, −1, 12 , and − 14 (b) By Theorem II, any rational root r/s must have r being a divisor of −1 and s being a divisor of 8. The r set is {1, −1}, and the s set is {1, −1, 2, −2, 4, −4, 8, −8}; these give us eight root candidates: ±1, ± 12 , ± 14 , and ± 18.
Among these, −1, 2, and 3 satisfy the equation, and they constitute the three roots. (c) To get rid of the fractional coefficients, we multiply every term by 8. The resulting equation is the same as the one in (b) above. (d) To get rid of the fractional coefficients, we multiply every term by 4 to obtain 4x4 − 24x3 + 31x2 − 6x − 8 = 0 By Theorem II, any rational root r/s must have r being a divisor of −8 and s being a divisor of 4.
The r set is {±1, ±2, ±4, ±8}, and the s set is {±1, ±2, ±4}; these give us the root candidates ±1, ± 12 , ± 14 , ±2, ±4, ±8. Among these, 12 , − 12 , 2, and 4 constitute the four roots. (a) The model reduces to P 2 + 6P − 7 = 0. By the quadratic formula, we have P1∗ = 1 and P2∗ = −7, but only the first root is acceptable.
Substituting that root into the second or the third equation, we find Q∗ = 2. √ √ (b) The model reduces to 2P 2 −10 = 0 or P 2 = 5 with the two roots P1∗ = 5 and P2∗ = − 5. Only the first root is admissible, and it yields Q∗ = 3.7) is the equilibrium stated in the form of ”the excess supply be zero.com Chiang/Wainwright: Fundamental Methods of Mathematical Economics Instructor’s Manual 2. Since we have c0 = 18 + 2 = 20 c1 = −3 − 4 = −7 c2 = 1 γ 0 = 12 + 2 = 14 γ1 = 1 γ 2 = −2 − 3 = −5 it follows that P1∗ = 14+100 57 6 35−1 = 17 = 3 17 and P2∗ = 20+98 59 8 35−1 = 17 = 3 17 Substitution into the given demand or supply function yields Q∗1 = 194 7 17 = 11 17 and Q∗2 = 143 7 17 = 8 17 Exercise 3.
(a) Three variables are endogenous: Y, C, and T.com Chiang/Wainwright: Fundamental Methods of Mathematical Economics Instructor’s Manual 2. (a) The endogenous variables are Y, C, and G. (b) g = G/Y = proportion of national income spent as government expenditure. (c) Substituting the last two equations into the first, we get Y = a + b(Y − T0 ) + I0 + gY Thus a − bT0 + I0 Y∗ = 1−b−g (d) The restriction b + g 6= 1 is needed to avoid division by zero.
Upon substitution, the first equation can be reduced to the form Y − 6Y 1/2 − 55 = 0 or w2 − 6w − 55 = 0 (where w = Y 1/2 ) The latter is a quadratic equation, with roots ∙ ¸ 1 w1∗ , w2∗ = 6 ± (36 + 220)1/2 = 11, −5 2 From the first root, we can get Y ∗ = w1∗2 = 121 and C ∗ = 25 + 6(11) = 91 On the other hand, the second root is inadmissible because it leads to a negative value for C: C ∗ = 25 + 6(−5) = −5 12 www.com Chiang/Wainwright: Fundamental Methods of Mathematical Economics Instructor’s Manual CHAPTER 4 Exercise 4. ⎡Coeffi cient M atrix:⎤ ⎡ of Constants: Vector ⎤ Qd −Qs =0 1 −1 0 0 ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ Qd +bP =a ⎢ 1 0 b ⎥ ⎢ a ⎥ ⎣ ⎦ ⎣ ⎦ Qs −dP = −c 0 1 −d −c 2. Qd1 −Qs1 =0 Qd1 −a1 P1 −a2 P2 = a0 Qs1 −b1 P1 −b2 P2 = b0 Qd2 −Qs2 =0 Qd2 −α1 P1 −α2 P2 = α0 Qs2 −β 1 P1 −β 2 P2 = β0 ⎡ Coeffi cient m atrix: ⎤ ⎡ Variable ⎤ vector: ⎡ Constant ⎤ vector: 1 −1 0 0 0 0 Qd1 0 ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ 1 0 0 0 −a1 −a2 ⎥ ⎢ Qs1 ⎥ ⎢ a0 ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ 0 1 0 0 −b1 −b2 ⎥ ⎢ Qd2 ⎥ ⎢ b0 ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ 0 0 1 −1 0 0 ⎥ ⎢ Qs2 ⎥ ⎢ 0 ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ 0 0 1 0 −α1 −α2 ⎥ ⎢ P1 ⎥ ⎢ α0 ⎥ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦ 0 0 0 1 −β 1 −β 2 P2 β0 3. No, because the equation system is nonlinear 4.
Y −C = I0 + G0 −bY + C = a The coefficient matrix and constant vector are ⎡ ⎤ ⎡ ⎤ 1 −1 I0 + G0 ⎣ ⎦ ⎣ ⎦ −b 1 a 13 www.com Chiang/Wainwright: Fundamental Methods of Mathematical Economics Instructor’s Manual 5. First expand the multiplicative expression (b(Y − T ) into the additive expression bY − bT so that bY and −bT can be placed in separate columns. Then we can write the system as Y −C = I0 + G0 −bY +bT +C =a −tY +T =d Exercise 4. No, not conformable.
⎣ ⎦ 13 8 ⎡ ⎤ ⎡ ⎤ 14 4 20 16 (b) Both are defined, but BC = ⎣ ⎦ 6= CB = ⎣ ⎦ 69 30 21 24 ⎡ ⎤ ⎡ ⎤ − 15 + 12 0 − 35 + 10 6 1 0 0 ⎢ 10 ⎥ ⎢ ⎥ ⎢ 14 ⎥ ⎢ ⎥ 3. BA = ⎢ −3 + 15 + 28 1 −2 + 3 + ⎥ = ⎢ 0 1 0 ⎥ ⎣ 10 5 10 ⎦ ⎣ ⎦ 2 4 6 2 5 − 10 0 5 − 10 0 0 1 Thus we happen to have AB = BA in this particular case. (a) x2 + x3 + x4 + x5 (b) a5 + a6 x6 + a7 x7 + a8 x8 (c) b(x1 + x2 + x3 + x4 ) (d) a1 x0 + a2 x1 + · · · + an xn−1 = a1 + a2 x + a3 x2 + · · · + an xn−1 14 www.com Chiang/Wainwright: Fundamental Methods of Mathematical Economics Instructor’s Manual (e) x2 + (x + 1)2 + (x + 2)2 + (x + 3)2 P 3 P 4 P n 1 P n 1 7.