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1

Preface

This is a supplement to Mathematical Methods for Economics, 2nd edition,by Michael W. Klein.

This manual provides answers to all the end-of-section exercises in thetext. Solving the end-of-section exercises is an integral part of the learn-ing process for a course using this textbook. The exercises in Mathemati-cal Methods for Economics are designed to develop students' economic andmathematical intuition as well as to hone their skills at practical problemsolving. The end-of-section exercises also oer students some additionaleconomic applications. Deirdre M. Savarese developed most of the end-of-section exercises in Mathematical Methods for Economics, and her assistanceis gratefully acknowledged.

This manual also oers concordances to aid instructors who would liketo supplement their microeconomic or macroeconomic courses with Mathe-matical Methods for Economics. The concordances list, for seven leadingmicroeconomic and macroeconomic textbooks, the mathematical tools re-quired in specic chapters of those books and the place where those tools arepresented in Mathematical Methods for Economics.1 The concordances alsolist applications in Mathematical Methods for Economics that instructors canuse when teaching material from specic chapters in these leading macroeco-

nomic and microeconomic textbooks. A conversion chart is also presented toaid instructors who want to switch from another mathematics for economicstextbook to Mathematical Methods for Economics. Deirdre M. Savareseput together the concordances and the conversion chart for the 1rst editionof Mathematical Methods for Economics and Kieran Brenner updated thismaterial for the 2nd edition.

This manual also oers reproductions of gures from Mathematical Meth-ods for Economics and these reproductions can serve as transparency mas-ters. Graphs are important pedagogic tools in a course like this one. In-structors may nd transparencies more accurate, easier to use, and less time-

consuming to present than gures drawn on a chalkboard in the midst ofclass.

1There are concordances for microeconomic texts by Perlo, Pindyck and Rubinfeld,

and Varian, and for macroeconomic texts by Abel and Bernanke, Blanchard, Gordon, and

Mankiw.

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2

Chapter 2

Section 2.1

1. The intervals are

(a) (5; 0)(b) [5; 0)(c) (1; 100)

(d) (1; 100](e) (0; 1)(f) (1; 1)

2. Do the following represent a function?

(a) Yes

(b) No

(c) Yes

(d) Yes(e) No

(f) The function y is not dened at x = 3:

(g) No

3. Can the function be dened according to the mapping?

(a) Yes

(b) Yes

(c) No(d) No

4. It would not be a function since there would be a mapping betweenone argument and more than one value.

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5. With this cost function

(a) When Q = 10; T C = 125: When Q = 25; T C = 200: When thereis no production (Q = 0) ; T C = 75:

(b) Linear graph.

(c) Domain [0:50] Range [75; 325]

6. Four representative ordered pairs are

(a) (2; 140) ; (1; 120) ; (0; 100) ; (1; 80)(b) (1; 2) ; (0; 0) ; (1; 2) ; (2; 10)

(c) (100; 101) ; (0; 1) ; (1; 2) ; (100; 101)7. The limits are

(a) limx!1

= 2

(b) limx!7+

= 1(c) lim

x!7+= 7

(d) limx!1

= 0

8. Are the functions continuous?

(a) No

(b) Yes

(c) Yes

(d) Yes

9. The function presented in 8(c)

y = 3 + 1x + 7

is not continuous over the domain (1; 0] since the function is unde-ned at the point x = 7:

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4

Section 2.2

1. The respective functions are

(a) strictly monotonic

(b) nonmonotonic

(c) strictly monotonic

(d) monotonic

2. Are these one-to-one?

(a) no. An individual may be a citizen of more than one country

(b) no. While a street address has one zip code, a given zip codeapplies to more than one street address.

(c) yes

(d) no. While one identication number is linked with one coursegrade, a given course grade may correspond to numerous students'identication numbers.

3. The inverse functions, if they exist, are

(a) x = f1

(y) =

y147

(b) This function does not have an inverse unless we restrict the do-main to y 6 so that the inverse is x = f1(y) = py 6

(c) This function does not have an inverse unless we restrict the do-main to y 0 so that the inverse is x = f1(y) = y2

(d) x = f1(y) = y13

4. The inverse is x = y10

+ 12

. To check this, note that

10x 510

+1

2= x and 10

y

10+

1

2 5 = y:5. For continuous functions with extreme points, the answers are

(a) No

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(b) Yes

(c) Either two minima and one maximum or two maxima and oneminimum

6. This is a linear function so the global minimum and maximums arethe respective endpoints. The global minimum is (0; 50) and the globalmaximum is (100; 100)

7. The average rates of return are

(a) Average rate of return = 12

(b) Average rate of return= 16

(c) Average rate of return = 12(d) Average rate of return = 0

8. For a strictly increasing function, f(xB) > f(xA) and xB > xA, so theaverage rate of change will always be positive since both the numeratorand the denominator yield a positive answer. For a strictly decreasingfunction, the numerator will be negative and the denominator will bepositive.

9. The answers are

(a) y0 = 5

(b) The slope of the secant line f(xB)f(xA)xBxA = 4: The value of the

slope of the secant line represents the average rate of change ofa function over the interval dened by the two endpoints of thesecant line.

(c) The function is strictly convex over the given interval since thesecant line lies wholly above the function.

10. The answers are

(a) x0 = 2:8; f(x0) = 28:16

(b) y0 = 26

(c) The function is strictly concave since f(x0) > y 0; 28:16 > 26

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11. The answers are

(a) A ( B(b) A ( B(c) A ) B(d) A , B This set satises the necessary and sucient condition

Section 2.3

1. These expressions can be written as

(a) x1

(b) (xy)4

(c) x20

(d) x3y2 (no further simplication possible)

(e)

1xy

62. These expressions can be condensed as

(a) x(a+b+cd)

(b) x53

(c) x3112

(d) x2y2

3. Simplifying the expressions, we get

(a) 8

(b) 212

= 4096(c) 1

2

(d) x+3x+2

(e) (x + 1)25

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4. The quadrants through in which the graphs of the functions appear are

(a) Quadrants I and II

(b) Quadrants I and III

(c) Quadrants I and II

(d) Quadrants I and III

5. The roots are

(a) x = 65

(b) x = 1; 6(c) x = 3 (two equal roots)(d) f(x) = (x + 1) (x2 3x + 2) = (x + 1)(x 1)(x 2) = 0 so

x = (1; 1; 2)

6. The root is x =q

pq2pr

p

7. Some points of these functions are

x 0 14

12

34

12x2 0 1

812

98

22x 1 1.19 1.41 1.68 2

The two functions do not share a y-intercept but do have a commonvalue when x = 1:

8. The average rate of change for the function y = 2x2 is 10. The averagerate of change for the function y = 2x is 14

3: Both functions are strictly

convex such that f(x0) < y0.

9. Some points of these functions are

x 0 1 2 3 4

12 2x 12 1 2 4 812 4x 12 2 8 32 128The two curves share a y-intercept since in both cases when x = 0,y = 1

2. When A = 1

4, the curves no longer share a common y value

since the graph of the rst function shifts down.

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10. When the domain ofx is restricted, some points of these functions are

x -2 -1 012

2x 1

814

12

12

4x 1

3218

12

The functions are graphed in Quadrant II. The y-intercept is still at 12

but the curve slopes down and to the left, asymptotically approachingthe x-axis.

11. Matching the functions to the economic relations, we have

(a) a = iii

(b) b = v

(c) c = i

(d) d = ii

(e) e = iv

12. a. There are two tax rates consistent with raising $60 billion, t = 0:3,t = 0:4. These are found by using the quadratic formula.

t1; t2 = 350

q(350)2

4

(

60)

(

500)

2 (500)b. There is one tax rate consistent with raising $61:25 billion, t = 0:35,since, using quadratic formula.

t =350

q(350)2 4 (61:25) (500)

2 (500) =350

2 (500) = 0:35

This is a case of a single root to a quadratic equation.

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Chapter 3

Section 3.1

1. The values are

(a) Xt = 100

(b) Xt+5 = 115:93

(c) Xt+1 = 200

(d) Xt+50 = 710:67

2. The values are

(a) Xt4 = 36:75

(b) Xt4 = 39:6

(c) Xt1 = 47:17

(d) Xt10 = 41:01

3. The values are

(a) Xt = 22:22

(b) Xt = 20:4

(c) Xt = 17:82

(d) Xt = 25

(e) Xt = 27:6

4. The value of Xt+n approaches zero as n approaches innity but neverequals zero. Rather the function is a horizontal asymptote, with nplotted on the horizontal axis and Xt+n on the vertical.

5. Your salary will be

(a) $52; 500

(b) $56; 784

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(c) $80; 549

6. Let Pt stand for prots in year _t. With P2000 = $50 million, and agrowth rate of 6%, we have

P2001 = $50 million (1:06) = $53 million

P2003 = $50 million (1:06)3 = $59:55 million

P1998 = $50 million (1:06)2 = $44:50 million

7. The levels of national income in 2010, Y2010, under dierent assumptionsabout the rate of growth beginning in 1994, and given Y1994, the levelof income in 1994, are

Growth Rate of 2:5% : Y2010 =

e160:025

Y1994 = 1:49Y1994


e160:030

Y1994 = 1:62Y1994


e160:035

Y1994 = 1:75Y1994

The ratios of national income in 2010 under dierent scenarios con-cerning the rate of growth are

Growth Rate of 3:0% compared to 2:5% :

1:62Y19941:49Y1994 = 1:087

Growth Rate of 3:5% compared to 2:5% :1:75Y19941:49Y1994

= 1:174

8. For Indonesia,

207 (1:0175)11 = 250:5

and for China

1; 250 (1:0104)11 = 1; 400:7

9. 100(1:02)5 = 110:41; 100(1:03)5 = 86:26

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Section 3.2

1. The values are

(a) Xt+1 = 21:65

(b) Xt+1 = 20:1

(c) Xt+1 = 22:1

2. The value ofXt is impacted more by a change in r than by a change in k.Doubling the frequency of compounding, holding r constant, reducedthe value of Xt by 0.03%. Doubling the rate of decay, r, increased thevalue of Xt by 5.2%.

(a) Xt = 105:19

(b) Xt = 105:16

(c) Xt = 110:66

3. Choose Bank B which will oer a 0.25% better return ($3,916).

4. rE = erA 1 so rE = 8:3%

5. The values are

(a) Xt+3 = 98:25

(b) Xt+0:5 = 75:94

(c) Xt2 = 60:19

(d) Xt+0:25 = 76:13

(e) Xt+0:75 = 73:33

6. If there is no capital appreciation, the return on the stock is simply thevalue of the dividend, or 3%. With capital appreciation and a dividendpayment, the following returns apply:

Q1 Q2 Q3 Q4

Stock 46.14 47.31 48.50 49.72Dividend 00.35 00.35 00.36 00.37Total Value 46.49 47.66 48.86 50.09

The total return on the original investment is thus $5.09, or 11.31%.

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7. Using the data given, we nd, for Indonesia

207 e110:0175 = 250:9and for China

1; 250

e110:0104

= 1; 401:5

The ratio of population in 2010 is 5:59, whereas the ratio in 1999 is6:04. The percentage dierence between the continuous compound-ing calculation and the discrete compounding calculation is 0:16% forIndonesia and 0:06% for China.

8. At 7%, PV= 13; 986. At 5%, PV= 14; 268. At 9.5%, P V = 13; 641

9. P V = 5; 578

10. Yt = Y0 exp(:01 + (0:03 0:7) + (0:02 0:3)) = Y0 exp(0:037) whereY0 = A0L0K0

11. The answers are

(a) L (0) = 1

(b) limt!1

L (t) = 2

(c) L (2) L (1) = 1:181 1:095 = 0:086: L (11) L (10) = 1:667 1:632 = 0:035: This represents diminishing marginal advances inproductivity over time.

12. The answers are

(a) $1320.00

(b) $1322.50

(c) In each case, with continuously compounded interest, the bondpays $1349.86.

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Section 3.3

1. The answers are

(a) 100

(b) 0

(c) 5log10 x(d) log2(a + b)

(e) a + bx + cz

(f) 3ln4x = 3(ln 4 + ln x)

(g) 5 + 2( ln x ln y)2. The answers are

(a) log2 256 = 8

(b) log3 27 = 3

(c) log3 1 = 0

3. Assuming that x > 1 and since e > 2, ln32 < log2 32: Note thatln 32 = 3:47 and log2 32 = 5.

4. In the stock market you would get $2,585.71. Keeping the collectionyou'll get $3,000 but you will have to pay storage fees equivalent to$329.75, which leaves you with $2,670.25 so you should still keep thecollection.

5. U =nP

i=1

qii

6. The various means are

(a) Arithmetic mean =420+1070+3470

3 = 1653:33(b) Geometric mean= (420 1070 3470)13 = 1159:6(c) Log Transformation= 1

3(ln 420 + ln 1; 070 + ln3; 470) = 7:056 so

Xg = exp(ln Xg) = exp(7:056) = 1; 159:6

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7. The ratio of health care in 1999 (H1999) to GDP in 1999 (Y1999) is

H1999=Y1999 = 0:10. With Health Care growing at a continuouslycompounded annual rate of 4:8% and GDP growing at a continuouslycompounded rate of 2%, we would have, after 30 years,

H2029Y2029

=H1999e

300:048

Y1999e300:02= 0:10e300:028 = 0:23

and health care would command almost a quarter of GDP. It is likely,however, that economic forces rein in the rise in health care costs beforethey reach this proportion of national income.

8. Bank Betterrates eective rate of 8.25% is equivalent to an annual

interest rate of 7.9% through the equation rE = erA 1 and taking thenatural logarithm of both sides. Bank Highrates is oering its investorsa better deal.

9. When n is bigger, the Rule of 70 appears to hold better.

(a) n = 69:66 Rule of 70= 70

(b) n = 14:21 Rule of 70 = 14

(c) n = 7:27 Rule of 70 = 7

(d) n = 3:10 Rule of 70 = 2.8

(e) n = 1 Rule of 70 = 0.7

10. ln Q = ln 15 + 45

ln L + 15

ln K. When L = 10 and K = 5, then ln Q =4:872 and Q = exp(ln Q) = 130:6

11. n = ln1:50:05

= 8:1 years

12. The average ination rate is calculated by solving for in

144:8

e9

= 192:9

Rearranging this, we have

=ln (192:9=144:8)

9 100% = 3:2%:

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Chapter 4

Section 4.1

1. The equilibrium values are

(a) (w;x;y) =14

; 4; 4

(b) (w;x;y) = (19; 232; 116)(c) (w;x;y) = (3; 6; 14)

(d) (w;x;y) = 64gb2h5 ; (8 5g b 4h) ; (8 5g b 4h)2. For this system of equations

(a) (x;y;z) = [(8a 9h 2) ; (8a 9h 2) ; (2a 2h 2)](b) (x; y; z) = (8a; 8a; 2a) I

(c) (x; y; z) = (16; 16; 4)

3. With h = 3, the answers are

(a) (x;y;z) = [(8a 9h 2) ; (8a 9h 2) ; (2a 2h 2)](b) (x; y; z) = (9h; 9h; 2h)(c) (x; y; z) = (27; 27; 6)

4. P = a+cb+d

; Q = adbcb+d

The parameters b and d are slope parameters.The linear demand curve must have a negative slope reecting theinverse relationship between the quantity demanded and price. Thelinear supply curve has a positive slope since producers are willing tosupply more at a higher price. The parameters a and c are y-interceptparameters, and are positive to reect the notion that prices are alwayspositive.

5. In this Keynesian model, the equilibrium values are

(a) Y = 12; 000; C = 11; 000

(b) Y = 10; 500; C = 10; 000

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6. Y = 3I

7. In this IS/LM model, the answers are

(a) Y = 20; 500 R = 0:05

(b) For M = 400, Y = 5400 = 2000 and R = 0:0005400 =0:2.

8. The equilibrium quantity in this model is

Q = + N T G

+ :

Therefore the change in the equilibrium quantity with a given increase

T in the tax isQ =

+ T:

A given increase in the tax has a bigger eect on the equilibrium quan-tity with an increase in the sensitivity either demand or supply to pricesince the fraction

+increases with an increase in either or .

9. In solving the model, we found the common wage to be

W =BY LB

and therefore

W = BY LB

:

Using the solution for LB, we obtain, after some manipulation,

W =A + B

2Y:

10. Solving these equations, we nd

LA =

B AA + B

Y

LB = A BA + BYFirm A experiences job creation and Firm B experiences job destruc-tion in the face of Y > 0 if B > A, that is, if labor demand is lessresponsive to the increase in wages in Firm A than in Firm B.

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Section 4.2

1. The answers are

(a)5P

i=1

axi = ax1 + ax2 + ax3 + ax4 + ax5

(b)nP

i=1

= bixi = b1x

1 + b2x2 + :::bnx

n

(c) 3

Pi=2xi+13

Pi=2yi1 = (x3 + x4) (y1 + y2) = x3y1 + x3y2 + x4y1 +x4y2(d)

3Pi=1

ixi(xi + 2) = x1 (x1 + 2) + 2x2 (x2 + 2) + 3x3 (x3 + 2)

(e)4P

i=1xi = 1

x+ 1

x2+ 1

x3+ 1

x4

2. The dimensions are

(a) 2 x 2 square matrix

(b) 1 x 3 row vector(c) 3 x 1 column vector

(d) 2 x 3 matrix

3. The values of the elements are

(a) a22 = 100(b) b13 = m

(c) c21 = 1

(d) d23 =

4. For this matrix

(a) Diagonal: z;b; 3; 1: O-diagonal: y;x;c;w;d; 4; a; 1; 4; 2; 3; 2

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(b) B0 = 2664z y x wa b c d1 2 3 44 3 2 1

3775

(c) (B0)0 =

2664z a 1 4y b 2 3x c 3 2w d 4 1

3775(d) b21 = y (in B

0 it becomes b12); b32 = c (in B0 it becomes b23);b44 =(in B

0 it remains b44). The diagonal and o-diagonal elementsin matrix B are the same diagonal and o-diagonal elements in

matrix B0:

5. These matrices are

(a) not conformable

(b) conformable; solution dimension is 1 1(c) not conformable

(d) conformable; solution dimension is 1 l

6. For this system of equations

(a) y1 = b11cj + b12cj + b13cj; y2 = b21cj + b22cj + b23cj

(b) y =

y1y2

B =

b11 b12 b13b21 b22 b23

c =

24 c1c2c3

35(c) The system is conformable

(d) yi =3P

j=1

bijcj

7. The products of the matrices are

(a) S =

24 7a 7b 7c 7d9a 9b 9c 9d4a 4b 4c 4d

35

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(b) S = 1 2 43 7 16

(c) S =

7 5 4 2

(d) S =

24 a g + 7h 4g + 2hx w + 7p 4w + 2pb c + 7d 4c + 2d

358. For the matrices A and B

(a) AB is dened. BA is not.

(b) AB is 3

1

(c) A0 = 2 1 7

4 5 6

(d) Not conformable

9. The four equations which comprise the model are

AD = C+ I

C = 2000 +3

4Y

I = 500

1000r

AD = Y

10. For this model

(a) The matrix system is

24 1 0 200 1 101 1 0

3524 XM

E

35 =24 1000 + 0:20YF450 + 0:15YD

0

35(b) The real exchange rate E = 0:2YF0:15YD+550

30. The equilibrium level

of imports (and exports) is M = X = 633:3 + 0:67YF + 0:1YD

(c) E = 0:230

YF

and E =

0:15

30Y

D

(d) Given the changes in YF and YD, E = 0:18

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Section 4.3

1. The answers are

(a)

1 00 1

(b)

2 a1 b

(c)

3 9 25 5 2. For the matrices A and B

(a) A + B = B + A =24 5 8 141 2

30

7 8 0

35(b) A B =

24 25 8 1013 23

23 2 8

35(c) B =

10 0 26 2

31

5 5 4B + A =

24 25 8 1013 23

23 2 8

35 = A B3. For the matrices K, L, and M

(a) M L K(b) (M L) K = M (L K)

4. For A as given

(a) Identity matrix in AI is 4 4: Identity matrix in IA is 3 3:

(b) AI = IA = 242 4 5 70 0

1 6

5 8 9 2 35(c) See (b).

5. The transposes are

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(a) X 0 = 4 16 2 Y 0 = 26643 75 20 41 3

3775

(b) (XY)0

= Y 0X 0 =

266430 1132 9

24 822 6

3775

Section 4.4

1. The respective matrices are

(a) nonsingular

(b) singular

(c) nonsingular

(d) nonsingular

2. The matrix is nonsingular if

(a) 6= cbad

(b) 6= adcd

(c) The matrix is singular if a = bcd

.

3. The only system which can be solved is 2(a).

4. The determinants are

(a) jAj = 44

(b) jAj = 1(c) jAj = 0(d) jAj = 2 nonsingular as long as 6= 0(e) jAj = 0

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(f) jAj = a(d ) (d ) cb

5. The adjoints are

(a) adjA =

7 3

4 2

(b) adjA =

4 19 3

(c) adjA =

1 22 0

(d) adjA = 4 63 12

6. A1A =

536

236

336

636

6 23 5

=

1 00 1

= I

7. The inverses (if they exist) are

(a) A1 =

1044

4446

44244

(b) A1 =

0 11 1 (c) Singular so no inverse

(d) A1 =

"

22

12

12

#(e) Singular so no inverse

(f) A1 =

"d

a(d)(d)cbb

a(d)(d)cbc

a(d)(d)cba

a(d)(d)cb

#

(g) A1 =

72

32

2 1

(h) A1 = 4

23323

923

123

(i) A1 =

14

12

12

0

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(j) A1 = 420

620

3

20

1

40 8. The system, with its solution, is

(a) A =

1 1

b 1

x =

YC

y =

I+ G

a

(b) x = A1y = 1

1b

I+ G + ab(I+ G)a

9. The answer is

(a) 1 1212

1 AB = FG (b) A = B = 37:50

(c) A = 0; B = 112:50

10. The system is

(a)

12 42 6

wr

=

PSPW

(b) jAj = 64

(c) wr = 332PS 116PW 1

32PS +

316

PW

(d) w = 9; r = 5

(e) r = 316

PW = 3 w = 116PW = 1. Thereturn to capital rises because wheat production uses capital moreintensively.

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Chapter 5

Section 5.1


(a) jBj = 54(b) jBj = 0(c) jBj =

2. The minors and cofactors are

(a)

jM12j = 6 jC12j = 6jM22j = 12 jC22j = 12

jM32j = 6 jC32j = 6

(b)

jM12

j= 7

jC12

j=

7

jM22j = 28 jC22j = 28jM32j = 28 jC32j = 28

(c)

jM12j = b2 ac jC12j =

b2 acjM22j = ab c2 jC22j = ab c2

jM32j = a2 bc jC32j =

a2 bc3. The determinants are as in question 1, namely

(a) jBj = 54(b) jBj = 0(c) jBj =

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4. jAj = 6


(a) jAj = 10:50(b) jAj = 441(c) jAj = 4

6. jHj = kd j (1 b)7. jTj = t11 t22 t33 t44 t55

8. jAj = 0. This matrix is redundant in that each element of the thirdrow consists of k times each element in the second row plus the corre-sponding element in the rst row.

Section 5.2

1. The adjoints are

(a) adj(A) =24 25 35 1720 15 5

40 30 3335

(b) adj(A) =

24 bc a2 (bc ac) ab bc (c2 ab) ac bc (a2 c2)ac b2 (a2 b2) ab bc

35

(c) adj(A) =

2664

6 92 4 1630 82 20 8078 64 11 44

192 298 128

134

3775

(d) adj(A) =

266418 4 32 8

18 0 36 0216 56 412 40

162 42 300 30

3775

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26

2. The inverses of the above matrices are

(a) A1 =24 25215 35215 17215 20

21515215

5215

40215

30215

33215

35(b) A1 = 1

abca3bc2+ab2+ac2cb2

24 bc a2 (bc ac) ab bc (c2 ab) ac bc (a2 c2)ac b2 (a2 b2) ab bc

35

(c) A1 = 1422

2664

6 92 4 1630 82 20 8078 64 11 44

192 298 128

134

3775

(d) A1 = 1236

266418 4 32 8

18 0 36 0216 56 412 40

162 42 300 30

37753. For this macroeconomic model

(a)

24 1 1 10:8 1 00 0 1

35 24 YC

I

35 =24 G200

1000 2000R

35(b) A1 = 24 5 5 54 5 4

0 0 1

35(c) Y = 250. The change in income due to a change in government

spending is more in this scenario than in Chapter 4.1 since moneydemand is not included in this model and there is no crowdingout.

4. For this export/import model

(a) 24 1 0 200 1 101 1 0 35 24

X

ME35 = 24 1000 + 0:2YF450 + 0:15YD

035 . The inverse of

the matrix of parameters, A1 =

24 13 23 2313

23

13

130

130

130

35. Given the

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solution x = A1y, the following are the values of the endogenous

variables: X = 650 +

0:2YF+0:3YD

3 ; M = 650 +

0:2YF+0:3YD

3 ; andE = 55030

+ 0:2YF0:15YD30

:

(b) The impact on exports of a $100 increase in foreign income isX = 62

3:

5. The solutions are

(a) (x; y) = (4; 0)

(b) (x; y) =32017

; 2817

(c) (x;y;z) =

539

; 449

; 1529

(d) (x;y;z) = 3828 ; 9628 ; 72286. The Keynesian style model is set up as follows:

(a)

24 1 1 10:8 1 00 0 1

35 24 YC

I

35 =24 G200 1000R

1000 2000R

35(b) Y = 5G 5000R + 4000

7. If the matrix A is nonsingular, a homogeneous equation system yieldsthe solution that x1 = x2 = = xn = 0: Using Cramer's rule, which, inthis case, replaces the rst column of matrix A with a column of zeroes

suggests that jAjj = 0. The value of jAj must therefore be non-zero inorder for there to be a dened solution since xj =

jAj jjAj =

0jAj = 0. If

matrix A were singular then its determinant would be zero making thesolution ofxj undened.

8. Using Cramer's rule, we have the change in the output of manufacturedgoods as

M =

0 aKF

L aLF

jAj =

aKFaKMaLF aLMaKF L < 0

and the change in the output of food as

F =

aKM 0aLM L

jAj =

aKMaKMaLF aLMaKF L > 0:

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Section 5.3

1. The characteristic equations and characteristic roots are as follows;

(a) 2 6 + 5 = 0; 1; 2 = 1; 5; jAj = 8 3 = 5(b) 2 + 4 + 3 = 0; 1; 2 = 1; 3; jAj = 3 0 = 3(c) 2 3 4 = 0; 1; 2 = 1; 4; jAj = 10 (6) = 4

2. For the matrix

4 1

3 2

with characteristic roots 1 = 5 and 2 = 1.

The characteristic vector associated with 1 solves4 1

3 2

p1p2

=

5p15p2

which gives us p1 = p2 and the characteristic vector associated with2 solves

4 13 2

p1

p2

=

p1

p2

which gives us 3p1 = p2. Thus we have

P = 1 11 3 and P1 = 34

34

14 14 and

P1AP =

34

34

14

14

4 1

3 2

1 11 3

=

5 00 1

:

For the matrix

1 02 3

with characteristic roots 1 = 1 and

2 = 3. The characteristic vector associated with 1 solves

1 0

2 3 p1

p2 = p1p2

which gives us p1 = p2 and the characteristic vector associated with 2solves 1 0

2 3

p1p2

=

3p13p2

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which gives us p1 = 0. Thus we have

P = 1 0

1 1

and P1 = 1 0

1 1

and

P1AP =

1 01 1

1 02 3

1 01 1

=

1 00 3

:

For the matrix

5 6

1 2

with characteristic roots 1 = 1 and2 = 4. The characteristic vector associated with 1 solves

5 61 2

p1

p2

=

p1p2

which gives us p1 = p2 and the characteristic vector associated with2 solves

5 61 2

p1

p2

=

4p14p2

which gives us p1 = 6p2. Thus we have

P = 1 61 1 and P1 = 15 6515

15

and

P1AP =

15

65

15

15

5 6

1 2 1 6

1 1

=

1 00 4

:

3. For this system

(a) In matrix format,

2 61 3

x1x2

= 20

4

(b) The characteristic equation is 2 + 12 = 0; 1; 2 = 3; 4

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(c) Associated with the characteristic root 3,

2 61 3

p1p2

= 3p13p2

which gives us p1 = 6p2. Associated with the characteristic root4,

2 61 3

p1

p2

=

4p14p2

which gives us p1 = p2. So the matrix P is (up to a multiplicativeconstant)

P = 6 11

1

(d) We have

P1 =

17

17

17

67

and by matrix multiplication we nd P1AP = where is adiagonal matrix with the elements 3 and 4.

4. For this system, with P and P1 as dened in exercise 3;

(a) The vectors u = P1x and v = P1y are

u = 17x + 17y17

x + 67

y v = 2474

7

:(b) We have

AP u =

2x + 6yx 3y

P v =

204

which is just the original system.

(c) We have

P1AP = =

3 00 4

and therefore

u =

3 00 4

u1u2

= v =

24747

:

Solving this we nd u1 =87

and u2 = 17 :

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31

(d) Transforming back to the original system using P u = x we have

x1 = 7 and x2 = 1:

(e) These values solve the original system as well, as they must.

5. The characteristic equations and characteristic roots (found by usingthe quadratic formula) are as follows;

(a) 2 0:5 + 0:0225 = 0; 1; 2 = 0:05; 0:45. This is stable.

(b) 2 0:3125 = 0; 1; 2 = 0:25; 1:25. This is not stable.

(c) 2 1:5 + 0:3125 = 0; 1; 2 = 0:25; 1:25. This is not stable.

The characteristic polynomial is found by evaluating the determinant

a 0

0 b

and setting it equal to zero. This gives us 2

(a + b) +(ab) = (a ) (b ) =0. The two roots of this equation are a and b. More generally, as discussedin Section 5.1, the determinant of a diagonal matrix D with diagonal ele-ments d11; d22;:::dnn is (d11 ) (d22 ) ::: (dnn ) and, therefore, the nvalues of for which this determinant equals zero are d11; d22;:::dnn.

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Chapter 6

Section 6.2

1. The dierence quotients are

(a) yx

= 5

(b) yx

= 15(c) y

x= 2 + 12x0 + 6x

(d) yx

= 2x0 x

2. When x = 2, the dierence quotients for questions a and b do not

change. The dierence quotients for questions c and d are 38 and -6,respectively, where for each the variance is equal to cx:

(a) The average rate of change = 5:

(b) The average rate of change = 15.(c) The average rate of change = 50:

(d) The average rate of change = 8:


(a) zx = 3gx20 + 3gx0x + gx2

(b) wx

= b + 2cx0 + cx + 3gx20 + 3gx0x + gx

2


(a) yx

= b + 2cx0 + cx

(b) hx

= b + 2cx0 + cx + 3gx20 + 3gx0x + gx

2 + 4hx30 + 6hx20x +

4hx0x2 + hx3

5. With the given functions

x0 = 1; x = 1 x0 = 2; x = 1 x0 = 2; x = 2y = 10x 4 y

x= 10 10 10

y = 3x2 + 6x 5 yx

= 15 21 24y = x3 + 4x2 6x + 12 y

x= 13 33 46

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33

Changing the values of x0 and x has no impact on (a) since it is

linear. Doubling the value of x0 has a greater impact on the quadraticand cubic functions than does doubling the value of x:

6. The change in wages are: w(20) = 49:75; w(30) = 44:75; w(40) =39:75; w(50) = 34:75. As this employee ages, his wages increase at adecreasing rate as indicated by the negative coecient on the squaredterm.

7. The revenue function and its properties are as follows

(a) T R = 10Q 0:5Q2

(b)

TR

Q = 10 Q 0:5Q. When Q = 1 and Q0 = 5;the impacton total revenue of a one unit change in Q is 4.5.(c) The impact on total revenue is 6.5 when Q0 = 3 and Q = 1:

(d) The impact on total revenue is 6 when Q0 = 3 and Q = 2:

8. When x0 = 3 and x = 3,yx

= 34

(a) When x = 1:5; yx

= 28: When x = 0:5; yx

= 24:

(b) In the limit, as x approaches zero, yx

= 22:

Section 6.3

1. The derivatives are

(a) f0(x) = 30

(b) f0(x) = 16x 6(c) f0(x) = 0

(d) f0(x) = 4x2. The derivatives evaluated at 3 and at 6 are

(a) f0(3) = 30; f0(6) = 30

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34

(b) f0(3) = 42; f0(6) = 90

(c) f0(3) = 0; f0(6) = 0(d) f0(3) = 12; f0(6) = 24

3. R0(t) = 25 150t; tm = 0:167: An increase in the tax rate beyond tmwould decrease tax revenue since the eect of lower sales on tax revenuemore than osets the eect of the higher tax rate so higher taxes wouldreduce consumption.

4. Are the functions dierentiable?

(a) Yes

(b) Yes(c) No; Not smooth at x = 2

(d) Yes

(e) No; Not continuous at x = 1

5. Equal average tax rates: at a given level of income, the tax systemsshould have secant lines drawn from the origin with the same slope.Equal marginal tax rates: the line tangent to each of the three taxfunctions at a given level of income should have the same slope.

6. In all cases, the average rate of change of a function is equal to the

average of the function's marginal changes over a given interval. Theanswers for these functions are

Average Change Marginal Changey = x2 + 2x 3 7 f0(2) = 6; f0(3) = 8y = 1

4x2 2 f0(2) = 1; f0(6) = 3

y = 4 3x x2 3 f0(1) = 1; f0(5) = 7y = x

12

15

f0(4) = 14

; f0(9) = 16

7. C0(q) = 4q; AC(q) = 10q

+ 2q. The total cost curve, C(q) = 10 + 2q2;where q 0; is a convex function. Therefore, the marginal cost, whichis measured as the slope of a line tangent to the total cost function at agiven point, is greater than the slope of the secant line from the originto that same point. An increase in output will increase the slope ofthe secant line, but the slope of the marginal cost curve will always begreater than that of the secant line.

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8. For this cost function

(a) The total revenue function is a concave function with a y-interceptof (0,0). It is increasing for Q = 0 to Q = 8.

(b) The derivative of the function, which is marginal revenue, is dTRdQ

=8 Q

(c) Average revenue is TR(Q)Q

= 8 12

Q.

(d) In the range where the function is increasing, the slope of themarginal revenue function is less than that of the average revenuefunction.

(e) The slope of the AR curve decreases with increasing values ofQ:

9. The roots of the equation y = 4x2 8x + 3 are 128

; 12

, which means thatthe value of the function at these values of x equals zero: The followingordered pairs can be plotted for this function: (0 ; 3) ; (1; 1) ; (2; 3)

(a) f0(x) = 8x 8. When f0(x) = 0; x = 1: The extreme value is aglobal minimum.

(b) The graph is a convex curve with y-intercept (0; 3). It reachesa global minimum at (1; 1) and passes through the x-axis at

12; 0and 128 ; 0 :10. The roots of this function are (3; 3; 1), which are all values where

y = 0: Other points on the function are (0; 9) ; (2; 5) ;and (1; 16)

(a) f0(x) = 3x2 2x 9(b) When f0(x) = 0; x = 2:08 and 1:4 which are the two roots of

the derivative equation and indicate the values of x where thederivative function passes through the x- axis.

(c) The derivative function is convex.

11. The derivative of this function is dPd lnY

= 58:2 + 7:8 ln Y:

(a) This derivative equals zero for ln Y = 58:2=7:8 = 7:46, whichcorresponds to a level of income ofe7:46 = $1737.

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(b) Evaluating this derivative at ln Y = 7:04 we obtain 3:29. Eval-

uating this derivative at ln Y = 8:60 we obtain 8:88. A sketch ofthis function, then, would show that it is \atter" than the func-tion presented in the application (and, because the value of ln Ywhere the derivative is equal to zero is smaller in this function,the minimum point of the function is to the left of the minimumpoint of the function in the application. The association of theminimum point of a function with the point where the derivativeequals zero will be discussed in detail in Chapter 9).

Section 6.4

1. The dierentials are

(a) dy = (14x 3) dx(b) dy =

10 1

2x

dx

(c) dy = (2x)dx(d) dy = (3x2 + 3)dx

2. Using the dierential to approximate the changes in the value of thefunction, we get

(a) y = 13:70

(b) y = 0:5

(c) y = 2(d) y = 30:3

3. The dierentials and approximate changes are

(a) dy = (6x + 13

)dx When x = 0:5; y = 616

. When x = 2; y =

2423(b) dy = (x) dx When x = 1

4; y = 1

2: When x = 10; y = 20

(c) dy = (3x2 4) dx When x = 8; y = 64: When x = 0:2; y =1:6

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4. The percentage dierence results should be read as how much smaller

the estimate is than the actual result.

(a)Actual Estimated % Dierence

x = 0:5 15.58 14.833 4.8%x = 2 45.333 33.333 26.5%

(b)Actual Estimated % Dierence

x = 14

-11.47 -11.50 0.26%x = 10 58 8 86%

(c)Actual Estimated % Dierence

x = 8 976 80 92%

x = 0:2 17.84 17.6 1.4%

5. With this consumption function

(a) The derivative of the consumption function is dCdY

= 0:8, whichin economic terms is the marginal propensity to consume, or theportion of current income which is consumed in a given period.

(b) Using the dierential approximation, C = f0(y)y = 800 whichin this case equals the actual dierence in consumption due to theY since the consumption function is linear.

6. The answers are in the following table

r I (actual) I (estimated) % dierence0.02 0.0 0.0 0%0.025 -0.66 -0.67 1.5%0.030 -1.30 -1.34 3.1%

7. The roots of the function are x1; x2 = 1; 11. The dierential equationis dy = (10 2x) dx

Actual Estimate % Dierencex = 0:5 22.75 24 5.5%x = 1 27 28 3.7%x = 1:75 30.94 34 9.9%x = 3 35 44 26%

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The percentage dierence should be read as how much bigger the dif-

ferential estimate is than the actual value ofy. The dierential approx-imation becomes a less accurate estimate for the actual value of y asthe x gets larger.

8. The dierential can be expressed as

dP = (58:2 + 7:8 ln Y0) d ln Y

where ln Y0 is the initial level of income.

(a) ln Y0 = 6:64 and d ln Y = 0:1 so dP = 0:64.

(b) ln Y0 = 6:64 and d ln Y = 0:25 so dP = 1:60.(c) ln Y0 = 9:08 and d ln Y = 0:1 so dP = 1:26.

(d) ln Y0 = 9:08 and d ln Y = 0:25 so dP = 3:16.

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Chapter 7

Section 7.1


(a) f0(x) = 32

x12

(b) f0(x) = 0

(c) f0(x) = 7x5

(d) f0(x) = 16x +3

2px

(e) f0(x) = 23

x3 2

2. Evaluated at 1 and 4, the derivatives are

(a) g0(1) = 24; g0(4) = 96

(b) g0(1) = 2a; g0(4) = a32

(c) g0(1) = 2; g0(4) = 1

(d) g0(1) = 2e2 = 14:8; g0(4) = 2e8 = 5; 962

3. f0(x) = 0 is the derivative of a constant function such as f(x) = 5for any x. f0(x0) = 0 is the derivative of a function evaluated at aparticular x, which we may call x0, equals 0. For example, for f(x) =2x2 + 4x with the derivative f0(x) = 4x + 4, we have at x0 = 1;f0 (x0) = f0 (1) = 0.


(a) f0(x) = 24x 1(b) f0(x) = 45x2 + 100x + 58

(c) f0(x) = 20x3 + 66x2 + 16x

(d) f0(x) = 5x4 20x(e) f0(x) = 6x2 24

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5. T R = AR Q = f(Q) Q so the derivative of T R, marginal revenue

(MR) is MR = Qf0(Q) + f(Q)


(a) f0(x) = ex + xex

(b) f0(x) = 2e(x5 )(c) f0(x) = a(e(bx) + xbe(bx))

(d) f0(x) = 2e(x2+4) + 4x2e(x

2+4)

(e) f0(x) = (9x2

2x + 5)e(3x3

x2+5x

6)

7. The derivatives evaluated at particular points are as follows

(a) f0(x) = 2 0 8x; f0(1) = 28; f0(1) = 12 so the derivative isdecreasing

(b) f0(x) = 2e2x; f0(0) = 2; f0(2) = 109 so the derivative is increasing

(c) f0(x) = 3x

; f0(4) = 34

; f0(6) = 12

so the derivative is decreasing

(d) f0(x) = 4(x + 1); f0(5) = 16; f0(1) = 8 so the derivative isincreasing

8. MR(Q) = 70 20q; MC(Q) = 40q 50; Q = 2

9.dMP

di= e(i);When the interest rate rises, people want to hold fewer

real balances and more of other assets which earn interest so MP

falls asi rises.

10. The instantaneous growth rates are

(a) 5 percent

(b) 4.5 percent

(c) 100 percent

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Section 7.2


(a) f0(x) = 4x3 9x2 + 14x + 3(b) f0(x) = 198(2x + 4)98

(c) f0(x) = (200x + 200)(5x2 + 10x + 3)19

(d) f0(x) = abeabx

(e) f0(x) = b

exab1 axa1exa

(f) f0(x) = ea+bx+cx2

10

10 (b + 2cx)


(a) f0(x) = 2x214x2

(x21)2

(b) f0(x) = 2bx3+cx2+4x2

(c) f0(x) = 2xe(2x)2e(2x)

x3

(d) f0(x) = 9x24x2

3. For this cost function

(a) The derivative of average cost is A0(x) = (f0(x)x)(1(f(x))

x2= 1

x

hf0(x) f(x)

x

i=

1x

[f0(x) A(x)] where the last term in brackets is marginal costminus average cost times 1

x:

(b) When the derivative of average cost is less than zero then it isdecreasing. Marginal cost is less than average cost.

(c) When the derivative of average cost is more than zero then it isincreasing. Marginal cost is greater than average cost.

(d) If marginal cost equals average cost, then A0(x) = 0, which is a

point of horizontal tangency


(a) f0(x) = 4x5 + x1

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(b) f0(x) = 8x2 + 24x2(ln x)

(c) f0(x) =1

x(1+x)

(d) f0(x) = 1x

(e) f0(x) = 1x(ln2)

5.dYfdDef(+)

=dYf

dTBB(+)

dTBAed(+)

deddKf(+)

dKfdid(+)

diddDef(+)

. A (+) indicates a positive

relationship between the two variables and a () indicates a negativerelationship between the variables. In the case of dTB

ed, where the ex-

change rate is measured in units of domestic currency, a strengtheningcurrency is measured as a decrease in the exchange rate since it takes

fewer units of domestic currency to purchase one unit of foreign cur-rency. A stronger exchange rate has a dampening eect on country A'strade balance, since it encourages imports, but it has a strengtheningeect on country's B's trade balance by stimulating that country's ex-ports. The ultimate eect of an increase in country A's budget decit,therefore, is to increase country B's income.

6. ddt

logb f(x) =f0(x)f(x)

1ln b


(a) f0(x) =2

ln2(2x+3)

(b) f0(x) = 2x ln 4

(c) f0(x) = 3x2 log2 x +x2

ln2

(d) f0(x) =1ln3

log3 x2x2

8. d(lnQ)dt

= c + f + (1 ) g9. These elasticities are

(a) " =

4 and therefore elastic. Elasticity will change because linear

function.

(b) " = 1:2 and therefore elastic. Elasticity will change with changesin x and y:

(c) " = 0:5 which is inelastic and constant.

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(d) " = 2 which is elastic and constant

10. The elasticities are

Quadratic LinearSri Lanka 2.0017 1.037Venezuela 2.0000 1.007Sweden 2.0000 1.001

11. The answers are

(a) QM = 10; PM = 20. If QM = 8; PM = 24 and demand is elastic.If Q

Mrises to 11, this part of the demand curve is inelastic.

(b) "q;p = 0:5 which is inelastic. Because it is a constant elasticity,it will not change when quantity demanded changes.

12. The answers are

(a) P = 1; Q = 21

(b) "D = 17

(c) "S = 27

(d) Response ofPP to a change in T =

13

. Response ofPC to change

in T = 23

(e) The price facing consumers rises by more than the price facingproducers for a given increase in the tax becasue the elasticityof demand is less than the elasticity of supply in absolute valueterms. With T = 0:3, the new PC = 1:20 and the new PP = 0:9:The new Q = 20:40:

13. The answers are

(a) P = 7:389; Q = 403:4

(b) "D = 2(c) "S = 0:5

(d) Response of PP to a change in T = 0:8. Response of PC tochange in T = 0:2

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(e) The price facing consumers rises by more than the price facing

producers for a given increase in the tax because the elasticityof demand is less than the elasticity of supply in absolute valueterms. With T = 1, the new PC = 7:589 and the new PP = 6:589:The new Q is approximately 381:

14. The marginal change in the eective labor input with respect to timeis

dL

dt= e0:1t

which is positive. Thus, this specication shows assumes that workersbecome more eective the longer their tenure in a position.

Section 7.3

1. The second derivatives are

(a) f00(x) = 14 6x(b) f00(x) = 10x3

(c) f00(x) =

x2

(d) f00(x) = 2ex + 4xex + x2ex

(e) f00(x) = 600x26

2. f0(x) = 8x53 ; f0(0) = 0; f00(x) = 40

3x

23 ; f00(0) = 0; f000(x) = 80

9x

13 ; f000(0)

is undened

3. Determining the second derivatives we nd

(a) f00(x) = 2 Convex. The second derivative is constant.

(b) f00(x) = 32px3

Concave. The second derivative is always negative.

(c) f00(x) = 6x12 Concave and convex portions. The second deriva-tive changes sign.

(d) f00(x) = 3x2 Concave. The second derivative is always nega-tive.

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4. The second derivative is f00(x) = 2x 8 . The function is concave over

the interval [0; 4] since f"(x) 0: Using the denition for concavitywhich appears in Chapter 2

f(2:4) 0:4(8) + 0:6(1023

)

5. f00(x) = 30x4 so the second derivative is unambiguously positive, andthe function is strictly convex since a secant line connecting two pointslies wholly below the function itself. When evaluated at f(x0), thesecond derivative equals zero but the function remains strictly convex.

6. T P(Q) = 150L23 ; T P0(Q) = MP(Q) = 100L1

3; T P00(Q) = 100

3L43 .

The negative second derivative indicates diminishing marginal returns.

7. d lnUdqi

= iqii > 0 A positive marginal utility implies that the total util-

ity is a monotonically increasing function, as consumer demand the-ory requires. d

2 lnUdq2i

= i(qii)2

< 0 The function exhibits diminishing

marginal utility and is concave.

8. The necessary restrictions are found by considering the derivatives andthe second deivatives.

(a) u0(c) = aec > 0 if a > 0: u00(c) = a2ec < 0 if a < 0 therefore < 0

(b) u0(c) = c1 if > 0: u00(c) = ( 1) c2 < 0 if 0 < < 1

9. Coecient of Relative Risk Aversion is x. Coecient of AbsoluteRisk Aversion is

10. The coecient of absolute risk aversion for the square root utility func-tion is = 1

2c. It is = 1

cfor the logarithmic function.

11. The relationship between the actual function and its approximationsare

(a)

x = 2 x = 2:1Actual f(x) = 3 f(x) = 3:73Linear h(x) = 3 + 7(x 2) h(x) = 3:7Quadratic j(x) = 3 + 7(x 2) + 3(x 2)2 j(x) = 3:73

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(b)

x = 2 x = 2:1Actual f(x) = 1:3863 f(x) = 1:4351Linear h(x) = ln 4 + 1

2(x 2) h(x) = 1:4363

Quadratic j(x) = ln 4 + 12

(x 2) 14

(x 2)2 j(x) = 1:4338

(c)

x = 2 x = 2:1Actual f(x) = 403:4 f(x) = 544:6Linear h(x) = e6 + 3e6(x 2) h(x) = 566:8Quadratic j(x) = e6 + 3e6(x 2) + 9

2e6(x 2)2 j(x) = 591:3

12. f(x0) = e0 = 1: The derivative of ex is simply ex for each successively

higher order derivative. Thus the Taylor series expansion for ex evalu-ated around the point x = 0 is ex = 1 + x + x

2

2!+ x

3

3!+ ::: x

n1

(n1)!

where

each x in the Taylor series term represents a value around the originalpoint x = 0:

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Chapter 8

Section 8.2

1. The partial derivatives are

(a) @y@x1

= 48x31 12x1x2; @y@x2 = 6x21 + 12x22(b) @y

@x1= 6x1x2 + 24x1 + 5x2 + 20;

@y@x2

= 3x21 + 5x1 + 1

(c) @y@x1

=2x2214(x12)2 ;

@y@x2

= 2x1x2x12

(d) @y@x1

= 6x22e(3x1); @y

@x2= 4x2e(3x1)

(e) @y@x1

= 2x1

; @y@x2

= 4x2

(f) @y@x1

= 2x1 + 2p

x2;@y@x2

= x1px2

4

2. The second partial derivatives and cross partial derivates are

(a) @2y

@x21= 144x21 12x2; @

2y@x22

= 6x21 + 12x22; @2y

@x1@x2= 12x1

(b) @2y

@x21= 6x2 + 24;

@2y@x22

= 0; @2y

@x1@x2= 6x1 + 5

(c) @2y@x21= 4(x1x222x227x1+14)

((x12)2) ;@2y@x22

= 4x2(x12)2 ;

@2y@x1@x2

= 4x2(x12)2

(d) @2y

@x21= 18x22e

3x1; @2y

@x22= 4e3x1; @

2y@x1@x2

= 12e3x1x2

(e) @2y

@x21= 2

x21; @

2y@x22

= 4x22

; @2y

@x1@x2= 0

(f) @2y

@x21= 2; @

2y@x22

= 12

x1x 3

22 ;

@2y@x1@x2

= x1

22

3. The partial derivatives at x1 = 1 and x2 = 4 are

(a) @y@x1

= 0; @y@x2

= 186

(b) @y@x1

= 88; @y@x2

= 9

(c) @y@x1

= 18; @y@x2

= 8

(d) @y@x1

= 1928; @y@x2

= 321

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(e) @y@x1

= 2; @y@x2

= 1

(f) @y@x1 = 6; @y@x2 = 724. The partial derivatives are

@u@v

= 4v34xyz

v4+x4+y4+z44vxyz@u@x

= 4x34vyz

v4+x4+y4+z44vxyz@u@y

= 4y34vxz

v4+x4+y4+z44vxyz@u@z

= 4z34vxy

v4+x4+y4+z44vxyz

Adding together the above partial derivatives and simplifying results

in4(v4+x4+y4+z44vxyz)v4+x4+y4+z44vxyz

= 4:

5. The solution to this problem is as follows

(a)

fw(w;x;y;z) = w1; fww(w;x;y;z) = ( 1) w2

fx(w;x;y;z) =y

ln(z); fxx(w;x;y;z) = 0fy(w;x;y;z) = x(y)2 ln (z); fyy(w;x;y;z) = 22

x

(y)3 ln (z)

fz(w;x;y;z) =xy

1z

; fzz(w;x;y;z) = xy 1z2(b) fxz(w;x;y;z) =

y

1z

; fwy(w;x;y;z) = 0

(c) According to Young's Theorem, this function has six cross partialderivatives.

6. In this demand and supply model

(a) @Q@Y

= :05; @Q@PS

= 10Ps; @Q@PC = 10(b) "QD ;Y = 1:12:

(c) "QD ;PS = 0:56(d) "QD ;PC = 0:16

7. The partial derivative is

fx (x;y;z) =

1

z+ x y + x

(z+ x)2

=z y

(z+ x)2

which is positive if z > y , negative if y > z, and 0 ify = z.

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8. The properties of the earnings relationship are as follows

(a) @lnE@x

= 0:004S+ 0:15 0:004X(b) 5 percent

(c) Earnings increase by $495

(d) Earnings increase by 0.2 percent.

(e) This specication exhibits diminishing marginal returns with re-spect to experience since the second partial derivative @

2 lnE@X2

isnegative.

9. The relative return relationship has the following properties.

(a) rIndiarUS

=23

2:5 115

1:5 21:08(b)

@

rIrUS

@

AIAUS

= 11

AIAUS

1 Q

L

1

10. For this production function

(a) When w = 10, L = 16: When the wage falls to 8, the demand forlabor, L, increases to 25. The labor demand schedule is downwardsloping reecting the inverse relationship between labor demandand real wages.

(b) When capital is allowed to vary and the real wage, w = 8, thelabor demanded will be L = 39:06: An increase in the amount ofcapital is represented by an outward shift in the labor demandschedule, which reects an increase in labor demand at any realwage due to an increase in the productivity of labor.

(c) The cross partial derivative, @2y

@L@K

= 5

2pLK: The cross partial is

positive since the marginal product of either input rises as moreof the other input is used.

11. The demand for hamburgers

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(a) varies with price in according to the relationship

@QD@P

= 0:25e4P1:25 in September@QD@P

= 0:25e3:5P1:25 in January@QD@P

= 0:25e3P1:25 in May

(b) varies with the number of months since September according tothe relationship

dQD

dt=

0:125e4P0:25e0:125t

12. The partial derivative of the weighted arithmetic mean with respect tothe iith argument is

@xA@xi

= !i

and the second derivative is zero. The partial derivative of the weightedarithmetic mean with respect to the iith argument is

@xG@xi

= !ix!i1i

i6=j!jx

!ji

and the second partial derivative is negative (since !i 1 < 0) andequal to

@xG@xi

= (!i 1) !ix!i2i

i6=j!jx!ji

:

These results are consistent since they show that the eect of an outlierdiminshes for the geometric mean, but not the arithmetic mean (whichhas a second partial derivative of zero).

Section 8.3


(a) dydz

= 40x 8w

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(b) dydz

= (12x2 + 14

z2) (2z3) +

12

xz 2

(c) dydz = @y@v dvdz + @y@w dwdz + @y@z(d) dy

dz= 4xt

2+10tz

+ 2x2t+5xp

z


(a) @z@u

= 48xu + 12yu + 2x + 2y; @z@v

= 4x 4y(b) @z

@u= (3ax22bxy)+ (bx2 + c); @z

@v= (3ax22bxy) + (bx2 +

c) (2v)(c) @z

@u= (2ex + xy)1

4+

12

x2 4y

2u; @z@v

= 3x2 24y

(d) @z@u

= (6x2 3y2) 12pu+v

+ 0:75y 10u; @z@v

= (6x2 3y2) 12pu+v

+

(6xy + 0:75u) 2v

3. The partial derivatives of the trade balance dTBdM

are

(a) dTBdM

= fEdEdM

gE dEdM gYUS dYUSdM(b)

dT B

dM= fE

dE

dM gE dE

dM gYUS

dYUSdM

R 0

dT BdYJ

= fYJ > 0

The total impact on the trade balance due to dM is ambiguouswhile the eect due to dYJ is positive.

4. The homogeneity properties are as follows

(a) homogeneous of degree 0

(b) homogeneous of degree 1

(c) homogeneous of degree 3(d) homogeneous of degree 1

(e) not homogeneous

(f) not homogeneous

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5. The homogeneity properties are

(a) Partial derivatives are homogeneous of degree -1. For example,

fx(sx;sy;sw) = sw1 = s1(w1)

(b) Partial derivatives are homogeneous of degree 0. For example,

fx(sx;sy;sw) = 2sx1sw1 = s0(2x1w1)

(c) Partial derivatives are homogeneous of degree 2. For example,

fx(sx;sy;sw) = 3s2x2s1y1s1w1+2s1y1s1w1 = s2(3x2yw1)+s2(2yw)

(d) Partial derivatives are homogeneous of degree 0. For example,

fx(sx;sy) =s1p

s2x2 + s2y2=

s1x1

s1 p

x2 + y2

6. The production function properties are as follows

(a) Consider the case of a doubling of inputs: A (2K) (2L) =A 2K2L = AKL2(+) = 2+Q. Since + = 1, 2Qresults which represents a doubling of output given a doubling ofinputs (constant returns to scale).

(b) For increasing returns to scale, Q increases more than proportion-ately so 2+ > 2. This requires that + > 1:

(c) + < 1

7. The homogeneity properties are

(a) The function is homogeneous of degree k = 712

since

f(sx1; sx2) = (sx1)14 (sx2)

13 = s

712 f(x1; x2)

(b) The partial derivatives of the production function are

@y

@x1=

1

4x 3

41 x

132 and

@y

@x2=

1

3x

141 x

23

2

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where the fi are homogeneous of degree k 1 = 512 as follows:

f1 (sx1; sx2) =14

(sx1)3

4 (sx2)13 = s

512 f1 (x1; x2)

f2 (sx1; sx2) =1

3(sx1)

14 (sx2)

23 = s

512 f1 (x1; x2)

(c) Euler's Theorem shows that

x1(1

4(sx1)

34 (sx2)

13 ) + x2(

1

3(sx1)

14 (sx2)

23 ) =

7

12s

512 f(x1; x2)

where f(x1; x2) = y = x141 x

132

8. wL = 30K0:4L0:6; rK = 20K0:4L0:6 The sum of wL + rK = 50K0:4L0:6

9. The answers are

(a) Homogeneous of degree 2

(b) Homothetic but not homogeneous

(c) Homothetic and homogeneous of degree 2

(d) Homothetic and homogeneous of degree 4

(e) Homothetic but not homogeneous

10. The answers are

(a) ey = xz

(b) ey = L0:3K0:7

(c) ey = x2yw

(d) ln(y) = xz

Exercise 8.4

1. The total dierentials are

(a) dw =

4x + 12

y

dx +12

x 9y2 dy(b) dy = (12x21 1x1 )dx1 +

6 1

x2

dx2

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(c) dz =

2xy3+x2y

(y3+xy)2 dx

3x2y2+x3

(y3+xy)2 dy

(d) dy = 4x1e(3x2) dx1 + 6x21e(3x2) dx22. The actual and approximate changes in the values of the functions are

(a) y = x2 + 4x z2 2xz y =24 dy =16(b) y = ex

2+3z y =7007 dy =-2194

(c) y = ln x3 4z+ 2xz y =19.3 dy =6(d) dy = 2xdx + 1

2ez=2dz = 4 1

2e = 2:64 while the actual change,

y = 3:93.

3. The function y = 3x22x+z31:5z2xz, evaluated at x; z = (2; 1) =5:5

Actual Change (y) Dierential Approx. (dy) Percent Diex; z = 1 11.5 7 40%x; z = 0:5 4.5 3.5 22%x; z = 0:1 0.74 0.7 5.4%

4. dI = (afa1) df +

p

dp + () dc. If protability increases by 1%,

salary increases by 100p

.

5. The answers are

(a) .05

(b) -50

(c) 5

(d) -2.50

6. The deriviatives are

(a) 6x+x2

3 y

1

3

6y+y23 x

13

(b) yx

2x+y+12y+x+1

(c) 3x1+2xy2

2x2y4

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(d) 3x2+wy3w3y2+wx

(e) yx(2+y)7. With this production function

(a) 73KL

(b) Slope = 7; Slope = 12

(c) MRTS is 7 and 12

, respectively.

8. With this production function

(a) Slope of isoquant = 13x2

x1. The function is homogeneous of

degree 1.

(b) The partial derivatives are homogeneous of degree 0 so

dx2dx1

=f1 (sx1; sx2)f2 (sx1; sx2)

Therefore,sk1f1(x1; x2)sk1f2 (x1; x2)

=f1(x1; x2)f2 (x1; x2)

Regardless of the scaling factor, s, the slopes of the isoquants areequal and unchanged.

9. The marginal rates of substitution are

(a) MRSA;B =BA

(b) MRSA;B =B3A

10. This utility function

(a) Is homogeneous of degree 2 since (sL) (sT) = s2LT. The slope ofthe indierence curve is dL

dT= L

T.

(b) The function z = ln L+ ln T is homothetic because ln U is strictlyincreasing and U = LT is homogeneous.

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(c) The total dierential is

dz = ssL

dL + ssT

dT = 1L

dL + 1T

dT

and therefore dLdT

= LT

and s does not appear.

11. For a given level of M,dY

dH= 0:4 Y

H

12. To maintain a constant value ofM, (that is, dM = 0), for a given levelof r, we have

dH

dc =1

r

(c + r) (1 + c) > 0:

To maintain a constant value of M for a given level of H, we have

dc

dr= 1 + c

1 r < 0:

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Chapter 9

Section 9.1

1. The stationary points are

(a) x = 5 which is a minimum

(b) x = (1; 2) which are maximum and minimum, respectively

(c) x = 0 which is a maximum

(d) x = 12

which is a maximum

(e) x = (1:80; 0:46) which are maximum and minimum, respec-tively

(f) x = exp = 2:718 which is a maximum

2. Using the second-order conditions we nd

(a) f00(x) = 2 which is sucient for a minimum

(b) f00(x) = 2x 3 At x = 1, sucient for a maximum. At x = 2,sucient for minimum.

(c) f00(x) = 15x4 which is sucient but not necessary for maximum(d) f00(x) = 2x2 which is sucient for a maximum(e) f00(x) = 12x + 8 At x = 1:80, sucient for a maximum. At

x = 0:46, sucient for minimum.

(f) f00(x) = 2ln(x)3x3

which is sucient for a maximum.

3. The maximum number of extreme points for the function is n 1:4. The maximum number of inection points for a polynomial of degree

n is n 2.5. The general formula for determining the critical points of a quadratic

function is: + 2x = 0. The general formula for the critical points ofa cubic function is:

2q

(2)2 126

= 0:

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6. A quadratic function has only one bend and is either concave or convex.

The second derivative is constant because f0(x) is linear, therefore,there is no inection point since the second derivative never changessign.

7. Substituting the functions, we get

V (R) = 9R 2

R2

4

= 9R R

2

2:

The vote-maximizing use of resources is found by setting the derivativeequal to zero,

V0 (R) = 9

R = 0

so R = 9 and then V (9) = 81 812

= 40:5 percent of the vote; sheloses anyway. This is a maximum since V00 (R) = 1.

8. The critical points are Y1Y2 = 13:72 (local minimum), 3:95 (local max-imum)

9. The prot function is also a cubic function with two bends. The rstcurvature is convex and crosses the x-axis at the quantity where theconcave portion of the total cost curve intersects the total revenuecurve. The prot function reaches a maximum at Q the point wherethere is a maximum positive dierence between the total revenue and

total cost curves. The marginal revenue curve is a horizontal line. Themarginal cost curve is a u-shaped convex curve, and the marginal protcurve is a concave curve.

10. FOC: dTCdn

= c+(2) iY(2n)2

= 0 so solving for n =q

iY2c

. Using the SOC

to check that this is a minimum gives us d2TC(dn)2

= 8( iY(2n)3

) > 0. Assume

that average cash balances = M = Y2n

then M =q

cY2i

.

Section 9.2

1. d2Sd2 = (2 + 2)exp(( + r) + ay) = 0 Since the exponential termcannot equal zero, the rst set of terms on the left must equal zero forthe product to equal zero. The inection point is = 2

:

2. For the rm described in this question

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(a) The rm's prot function is = 12Q (Q3 4:5Q2 + 18Q 7).

The extreme values of this function are Q = 1 and Q = 2: Toobtain maximum prots, the rm should produce 2 units. Thesecond order condition is: 9 6Q < 0 so Q > 3

2

(b) Marginal cost = 3Q2 9Q + 18 = 12, which is price. The twovalues which obtain are Q = 1 and Q = 2: The slope of themarginal revenue curve is zero, and at Q = 2, the slope of themarginal cost curve = 3, which is greater than the slope of themarginal revenue curve. At Q = 1; the slope of the marginal costcurve is -3 which is less than the slope of the marginal revenuecurve.

3. In order to maximize prots, the monopolist should produce at Q = 5which corresponds to a price of P = 2:

4. For the monopolist Babydrink

(a) Q = ac2b

; P = a ac2

(b) Q = act2b

; P = a act2

(c) t = ac2

5. For these two duopolists, rms A and B,

(a) A = (pA c) QA = (pA c) (a pA + bpB) ; B = (pB c) QB =(pB c) (a pB + bpA)

(b) pA =a+c2

+ b2pB = RA; pB =

a+c2

+ b2pA = RB

(c) pA = pB =

a+c2b

6. In the Solow growth model

(a) Long-run consumption per capita = c = y i = ka(+ n)k: Thelevel of k that maximizes consumption is dc

dk= aka1 (+ n)

(b) The Golden Rule level of capital accumulation sets the marginalproduct of capital equal to the sum of depreciation and populationgrowth. The economy with the higher population growth, holdingeverything else equal, will have the higher marginal product ofcapital.

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(c) An increase in the number of workers causes capital per worker to

decline because capital is spread over a larger population since inthe steady state k = 0: The Solow model also predicts that coun-tries with higher population growth will have lower income/outputper capita.

7. The intensive production function is y = ka which incorporates labor-augmenting technological progress and i = (+ n + g) k. Accordingto the Solow growth model, c = y i = ka (+ n + g) k so dc

dk=

12pk0:145 = 0 and k = 11:91 which is the Golden Rule level of capital

accumulation that maximizes consumption per eciency unit of labor.Labor-augmenting technological progress and population growth, or an

increase in the labor force, have the same directional impact on k:8. For this monopsonist who is the only purchaser of labor

(a) The First order condition is F OC = 9L23 8L13 = 0: The optimal

amount of labor to employ is L = 98

(b) The second order condition is SOC = 6L53 83

L23 < 0. The

negative coecient suggests diminishing marginal returns to la-bor.

9. The cost-minimizing quantity is Q = 14:47. The original function

has a y-intercept at (0; 500), a local maximum at Q = 5:53, a localminimum at Q = 14:47, and an inection point at Q = 10. The protmaximizing quantity is Q = 12, however, and if the rm producedthe cost-minimizing quantity it would be earning less than maximumprot.

10. a. The derivative of the pro t function is

d

dL= pL1 (w + b) :

Setting this equal to zero we have the optimal amount of labor de-

manded, L, equal to

L =

w + b

p

1=(1)=

p

w + b

1=(1):

This shows that L decreases as b rises.

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b. The second derivative is

d2dL2

= p( 1) L2 < 0

which shows that the critical point derived in part(a) represents a maximum.

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Chapter 10

Section 10.1


(a) (x; z) =2; 1

4

(b) (x; z) = (4; 1)

(c) (x; z) =

164

; 9

(d) (x; z) = (0; 0)2. The stationary points are

(a) (u; v) = (2; 6)

(b) (u; v) =11

8; 2

(c) u = v = (1; 2)(d) (u; v) = (128; 256)


(a) (a; b) = (5; 3)(b) (a; b) = (0; 0)

(c) (a; b) =1

4; 94

(d) (a; b) = (1; 1)

4. The function in (d) should be f(x; z) = 6x+2z2x2+4z. The answersare

Original Function x + a z + a x a z aa. (3; 2) = 25 24 22 24 21 Maximum

b. (18; 8) = 196 195:5 194 195:5 194 Maximumc. 12

; 2

= 12 4 11 4 8 Minimumd. (3; 1) = 7 6 9 6 9 Saddle Point

5. For this government

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(a) The level of ination that maximizes seignorage revenue is =b(X

ct)

2(abg)(b) The tax rate that maximizes income tax revenues is t = X+g

2c

(c) They choose dierent outcome. = b(Xct)+gt2(abg) which is

gt2(abg)

bigger than answer 4a. t = X+gbc2c

which is b2

smaller thananswer 4b. R is higher in the cooperation case.


@C

@E= 4E+ 78 2X

@C@X

= 4X+ 66 2E:

Setting these equal to zero, we nd the stationary point is (E; X) =(15; 9) :

7. (E; U) = (70; 30)

8. (L; K) = (2; 4); fLL = 4; fKK = 1 and fLL fKK > (fLK)2

9. For these two cigarette rms

(a) AC = 500 and AM = 250;C = 187; 500 and M = 62; 500;C + M = 250; 000

(b) AC =30007

and AM =10007

; C + M = 306; 122:45

(c) A = 250 ; C + M = 250; 000

10. For this publishing rm

(a) (QT; QM) =

ac2

; c2

. Both of the second partial derivatives

are negative so these stationary points represent a maximum.

(b) (QT; QM) = a2cQM2(+c) ; 2cQT2(+c) (c) With linear costs, Q = +c(+)

2:

11. The critical points of this function are (QB; QS) = (18526

; 4013

).

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64

12. For this multi-plant monopolist

(a) = 40(Q1+Q2+Q3)(Q1 + Q2 + Q3)2(Q1 + Q21)(Q22 2Q2)(2Q23 Q3)

(b) (Q1; Q2; Q3) = (1; 16:5; 1)

Section 10.2


(a) Saddle Point

(b) Maximum

(c) Minimum

(d) Maximum


(a) Maximum

(b) Minimum

(c) Maximum

(d) Maximum


(a) Saddle Point

(b) Saddle Point

(c) Saddle Point

(d) Minimum

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4. This leads to maximum government revenue under the following con-

dition. We have@2R

@2= 2a 2bg < 0

@2R

@t2= 2c < 0

@2R

@@t= g bc

so the condition needed for a maximum is

4ac + 6bcg > g2 + b2c2:

5. For this problem

@2C

@E2= 4 < 0

@2C

@X2= 4 < 0

@2C

@E@X= 2

and (

4) (

4) > (

2)2.

6. 1 4 > 12, so maximum7. Conrming maximum prots in part (a)

2 2 > 02

Conrming maximum prots in part (b)

(2B 2C) (2 2C) = 4 + 4C + 4C + 4C2 > (2C)2

8. Conrming maximum prots:

fQBQB = 6fQSQS = 7

fQBQB fQSQS > (fQBQS)2

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9. The stationary point is r = s = 0. The second derivatives are @2z

@r2=

2 and@2z

@s2 = 2:

(a) > 0, > 0 so minimum

(b) < 0, < 0 so maximum

(c) and are dierent, so saddle point

10. The second order conditions are

@2

@P2= 2b

1 e SN

N = 2bQ

q< 0

@2

@S2

=

1

N

eSN aP bP2 C(a bP) =

1

N

eSNq(P

C) < 0

@2

@P@S= e

SN [(a 2bP) + bC] = e SN [q b (P C)]

where the signs assigned to the second partial derivatives assume thatP C > 0, S > 0 and Q > 0. Some algebra will show that thecondition

@2

@P2@2

@S2>

@2

@P@S

2is also met.

11. The rst order conditions are

12. The rst order conditions are

@y

@x= 4x 4 + 4z = 0

@y

@z= z+ 4 + 4x = 0:

(a) The stationary point is x = 1 and z = 0.(b) The second partial derivatives and the cross partial derivative are

@2y

@x2 = 4@2y

@z2= 1

@2y

@x@z= 4:

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67

Thus, we have @2y

@x2< 0 and @

2y@z2

< 0, two of the conditions needed

for the stationary point to be a maximum, but the other conditionis not met since @

2y@x2

@2y@z2

< @2y

@x@z

2and, therefore, this stationary

point is neither a maximum nor a minimum.

Exercises 10.3

1. For this function, all three second partial derivatives @2y

@x2

i

= 2 for i =

1; 2; 3 and all cross partial derivatives are zero. Therefore the stationarypoint, (0; 0; 0), is a minimum.

2. If all cross partial derivatives are zero, then the stationary point is aminimum if fii > 0 for all i and it is a maximum if fii < 0 for all i:

3. For a function with all cross partial derivatives equal to zero, a sta-tionary point is a minimum if all second derivatives are positive whenevaluated at the stationary point and a stationary point is a maximumif all second partial derivatives are negative when evaluated at the sta-tionary point. If some second partial derivatives are zero, or if they are

of dierent signs, then the stationary point is a saddle point.

4. (x1; x2; x

3) = (12; 9; 9) The sucient condition shows that these repre-

sent a local minimum

f11 = 4 > 0

f11f22 = 4 6 > (f12)2 = 9f11f22f33 + 2f12f23f13 > f11 (f23)

2 + f22 (f13)2 + f33(f12)

2

5. (x1; x2; x

3) = (4; 2; 4). The sucient condition shows that these repre-

sent a local maximum

f11 = 1 < 0f11f22 = 6 > (f12)

2 = 4

f11f22f33 + 2f12f23f13 < f11 (f23)2 + f22 (f13)

2 + f33(f12)2

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6. Conrming point of maximum prots

f11 = 4 < 0f11f22 > (f12)

2 = 4 2 > 22f11f22f33 + 2f12f23f13 < f11 (f23)

2 + f22 (f13)2 + f33(f12)

2 = 64 < 48

7. In the matrix

A =

a11 a12a21 a22

the two 1rst order principal minors are a11 and a22 and the single 2

nd

order principal minor is the determinant of the matrix, jAj = a11a22 a12a21. This matrix is positive denite if the two leading principleminors, a11 and jAj, are positive. This matrix is positive semideniteif a11 0, a22 0, and jAj 0. This matrix is negative denite ifa11 < 0 and jAj > 0. This matrix is negative semidenite if a11 0,a22 0, and jAj 0. Therefore, using these criteria we nd thatthe matrix in (a) is negative denite, the matrix in (b) is positivesemidenite, the matrix in (c) is negative semidenite, and the matrixin (d) is positive denite.

8. A 4 4 matrix has 4 leading principal minors and 15 principal minors.

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Chapter 11

Section 11.1

1. The extreme points are

(a) (x; z) = (8; 0) Minimum

(b) (x; z) =13

; 23

Minimum

(c) (x; z) = (4; 1)Minimum

(d) (x; z) = 43 ; 323 Maximum(e) (x; z) =

12

; 12

Maximum

2. (x; z) = (8; 4) :

3. R = 16; L = 8 (R = 0 yields minimum)

4. U(S) = 12

ln(S) + ln(12 S2

); @U@S

= 0 , S = 12; @2U@S2

= 2(2S)2

1

2(24S)2 < 0:

5. For the consumption problem with three items

(a) S4

+ V2

+ J12

= 6

(b) U(V; J) = 13

ln(24 2V J3

) + 13

ln(V) + 13

ln(J). The ratio is6V = J

(c) U(S; J) = 13

ln(S)+ 13

ln(12 J6 S

2)+ 1

3ln(J). The ratio is 3S = J

(d) The optimal amounts of soup, salad, and juice are S = 8, V = 4,and J = 24:

(e) The solution provides a maximum level of utility since each of thesecond order conditions are negative.

6. This problem requires the minimization of 50C+ 100P subject to theconstraint 5 = C1=3P2=3. Solving for C from the constraint we get

C = 125P2

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which then gives us the internalized function

f(P) =6250

P2+ 100P:

The rst order condition is

f0 (P) = 12500P3

+ 100 = 0

and the solution is P3 = 125, that is, P = 5. Substituting this intothe constraint, we have C = 5 as the optimal amount of cappuccino.The second derivative, evaluated at P = 5, is

f00 (P) =37500

P4=

37500

625= 60 > 0

which shows that the stationary point is a minimum.

7. (x;y;z) = (1; 4; 9) It is a saddle point.

8. V = 6; S = 12; J = 36 Optimal relative proportion is same as the caseof $6 budget.

9. The constraint is T + P = 12. Expressing this as T = 12 P,substituting it into the revenue function and taking the deriviative, wend the rst order condition to be

R0 (P) = 12

(12 P) 12 + 14

P

2

12

= 0:

Solving, we obtain the optimal level P = 4 and, using the constraint,T = 8. The second derivative of the internalized objective function is

U00 (M) = 14

(12 P)32 116P

2 32 < 0

for P = 4. Therefore, this choice ofP and T represents a maximumincrease in revenues.

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Section 11.2

1. The solutions are

(a) (x;z;) = (12; 4; 15).

(b) (a;b;c;) = (12; 6; 1; 16)

(c) (x;z;) = ( 51007

; 1367

; 4(5100)3(136)376

).

(d) (x;y;) = (4; 4; 132

).

2. For this school lunch problem

(a) L = 2C12 + A 12 (C+ 12A 24); (A; C) = (16; 16)(b) = 1

4; This is the marginal utility of money available for cookies

and apples. 0

= 14p2

Marginal utility is diminishing according tothe increase in budget.

3. X = ln2+43

; M = 8ln 23

, is the marginal utility regarding the increasein time you devote to these activities.

4. For these two pastries, the optimal amounts are

(a) R = 2; T = 4

(b) R = 4; T = 16

(c) Answer: (ii), exactly twice as much

5. (M; L) = (4; 8); GPA = 4:0

6. The optimal amounts are

(a) K = Cr

; L = C(1)w

(b) K =QA 11 wr1; L = QA 1 wr

7. (w;x;y) = (4; 4; 4)

8. The bordered Hessians are

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72

(a) For part (a)

H = 24 0 2 22 1 12 1 6

35which has the determinant 24 so the stationary point is a mini-mum.

(b) For part (b)

H =

2664

0 12

3 112

1 0 03 0 8 01 0 0 16

3775The determinant is 184. The determinant of the leading princi-

pal minor, which is 24 0 12 312

1 03 0 8

35is 11 and therefore this is a minimum.

(c) For part (d), the bordered Hessian is24

0 4 44 1

64164

4

164

164

35

and with = 132

, the determinant is 1 so the critical point is aminimum.

9. The Lagrangean function is

L = G13 U

23 (WGG + WUU B)

where B is the total wage-bill. The rst order conditions are

@L

@G =

1

3G23

U

23

WG = 0@L

@U=

2

3G

13 U

13 WU = 0

@L

@= WGG + WUU B = 0

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73

a. Solving the rst order conditions, we get

GU

= 12

WUWG

b. If WU rises thenGU

rises.(c) With B = 1500, WG = 50 andWU = 25, we have G = 10 and U = 40.

Section 11.3

1. The student's utility is maximized with

(a) (x1; x2) = (15; 30)

(b) (x1; x

2) = (15; 30)

(c) PART(a) =

r1

32= 0:1777; PART(b) =

r1

120= 0:00833

2. Expenditure minimization requires

(a) (x1; x2) = (12:5; 25)

(b) The results are the same because the utility constraint in Question2 was set at exactly the level of utility achieved in Question 1through the maximization of x1 and x2

(c) If the utility constraint rises to u = 20, x1 rises to 14.14 and x2increases to 28.28.

3. Solve the Lagrangian function F(K;L;) = rK+wL(K L1Q)and get

=r

L1 K1 =w

(1 ) L K

, KL

=

1 w

r

, ln

K

L= ln

1

+ ln

w

rtherefore, the elasticity of substitution is = d ln

KL

d ln

wr

= 1

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4. Max U(X1; X2; X3) = (X1 +X

2 +X

3 )

1 s.t. P1X1+P2X2+P3X3 = I.

Set Lagrangian functionL = (X1 + X

2 + X

3 )

1 (P1 X1 + P2 X2 + P3 X3 I):

The elasticity of substitution is

= d logX2X1

d log

P2P1

= d log X3X2d log

P3P2

= d log X3X1d log

P3P1

= = 11

5. The Euler equations are

(a) 1C1

= (1 + R) C2

(b) 11 = (1 + R) 1 , 1 = (1 + R) (c) exp(C1) = (1 + R) exp(C2)

6. For this problem,

I1 = C1 + S1 (1); I2 + (1 + R) S1 = C2 + S2 (2);I3 + (1 + R) S2 = C3 (3); , (1 + R) I1 + I2 + I3

1 + R= (1 + R) C1 + C2

If we put (1 + R) I1 + I2 + I31+R = I, the Lagrangian function will be

L = u(C1) + u(C2) + u(C3) (1 + R) C1 + C2 + C31 + R

I :

Solve this for .

=u0(C1)1 + R

= u0(C2) = (1 + R) u0(C3):

7. The savings problem with period 2 income is as follows

(a) Put Y1; C1 as income and consumption at the rst period.

A = Y1

C1

(1 + r)A + Y2 = C2 + R + (E S)A consolidated intertemporal budget constraint is

(1 + r )Y1 + Y2 = (1 + r )C1 + C2 + R + (1 )E

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(b) Put (1 + r )Y1+ Y2 = Y : The Lagrangian function is written

asL = 2

pC1 + 2

pC2 + 2

pE+ 22

pR

(1 + r )C1 + C2 + R + (1 )E Y

Solve the Lagrangian function and we get

C1 =(1 ) (Y + )

(1 + r ) (1 ) + (1 + 2) 2 (1 ) (1 + r ) + 2 (1 + r C2 =

(1 + r ) 2 (1 ) (Y + )(1

) + (1 + 2)

2

(1

)

(1 + r

) + 2

(1 + r

)

R =(1 + r ) 4 (1 ) (Y + )

(1 ) + (1 + 2) 2 (1 ) (1 + r ) + 2 (1 + r )E =

(1 + r ) 2 (Y + )(1 ) (1 ) + (1 + 2) 2 (1 ) (1 + r ) + 2 (1 + r )

(Hint: Express all variables in terms of C2 and plug them into the

constraint. You will get C2

h1

2(1+r) + 1 + 2 + 1

1

i= Y )

8. I will be W0 + I1 +I2

(1+R), rather than I1 +

I2(1+R)

. However, the Euler

equation will be the same as u0(C1) = (1 + R) u0(C2). A non-zeroinitial stock of wealth does NOT alter the results of the model exceptfor the fact that the initial income dier. (W0 + I1 , rather than I1)

9. This time-allocation problem has the following solution.

(a) The optimal allocations are

A1 = 3:75

A2 = 2:50

w = 10:5

t1 = 7:5

t2 = 6:0

(b) G1 = 7:50; G2 = 15

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76

(c) This technological change would reduce n1 thereby increasing the

relative amount of activity A1(d) This improvement in \physical productivity" will reduce n2 thereby

increasing the relative amount of activity A2

(e) The increase in wages will decrease the consumption of the moretime-intensive activity, A2.

Section 11.4

1. The Lagrangian function is

L = 2x2 y2 1

x2 + y2 1

+ 2x + 3y

and the solution that provides the maximum value must satisfy@L

@x= 4x 21x + 2 = 0

@L

@y= 2y 21y + 3 = 0

1

x2 + y2 1 = 02x = 0

3y = 0

2 0; 3 0;

x

2

+ y

2

= 1The optimal solution is x = 1, y = 0, which means that 2 = 0:


L = 2a2 + b2 (2a + b 9) a2 + b2 16and the solution that provides the maximum value must satisfy

@L

@a= 4a 2 2a = 0

@L

@b= 2b 2b = 0

(2a + b 9) = 0

a2 + b2 16 = 0 0; 0;

16 a2 + b2; 2a + b 9

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The optimal solution has 6= 0, = 0, with a = b = 3:


L =1

3ln (S)+

1

3ln (V)+

1

3ln (J)

S

4+

V

2+

J

12 6

(S+ J 40)

and the solution that provides the maximum value must satisfy

@L

@S=

1

3S

4 = 0

@L

@V=

1

3V

2= 0

@L@J = 13J 12 = 0

S

4+

V

2+

J

12 6

= 0

(S+ J 40) = 0 0; 0;

S

4+

V

2+

J

12 6; S+ J 40

The optimal solution has 6= 0, = 0, with S = 8, V = 4, and J = 24.

4. For this problem;

(a) The Lagrangian function is

L = AKN1 (wN + rK c) + 1N + 2Kwhere labor is N and capital is K. The rst order conditons are

@L

@K= AK1N1 r + 2 = 0

@L

@N= (1 ) AKN w + 1 = 0

(wN + rK c) = 01N = 0; 2K = 0

0; 1 0; 2 0wN + rK c

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The optimal solution involves 1 = 0 and 2 = 0 while wN+rK =

c and 6= 0which gives us rK = c and wN = (1 ) c foroptimal amounts spent on capital and labor services, respectively.(b) The Lagrangian function is

L = (wN + rK) AKN1 Q+ 1N + 2K:The rst order conditons are

@L

@K= r AK1N1 + 2 = 0

@L@L

= w (1 ) AKN + 1 = 0

AKN1 Q = 01N = 0; 2K = 0

0; 1 0; 2 0AKN1 Q

The optimal solution involves 1 = 0 and 2 = 0 while AKN1 =

Q and 6= 0 which gives us

KN

1

= wr

(c) These solutions are the same as the respective solutions in exercise11.2.6 since the optimal solution requires using some of each fac-tor of production and either using the full budget, in the outputmaximization problem, or producing no more than the requiredamount, in the cost minimization problem. The optimization func-tions with inequality constraints thus are the same as those withequality constraints.


L = u (x1; x2) (p1x1 +p2x2 m) (t1x1 + t2x2 T)

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and the solution that provides the maximum value must satisfy

L1 = u1 p1 t1 = 0L2 = u2 p2 t2 = 0

(p1x1 +p2x2 m) = 0 (t1x1 + t2x2 T) = 0

0; 0;m p1x1 +p2x2; T t1x1 + t2x2:

It is will not be the case that p1x1 + p2x2 = m and t1x1 + t2x2 = Tsince then more of one or both goods could be consumed, which would

raise utility. So it is unlikely that the solution is one where = 0 and = 0. It is also unlikely that both 6= 0 and 6= 0 since this wouldmean that both constraints would be binding. It is more likely thateither 6= 0 and = 0, in which case the budget constraint is bindingand the time constraint is not binding, or = 0 and 6= 0, in whichcase the time constraint is binding and the budget constraint is notbinding.

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Chapter 12

Exercises 12.1

1. For this asset;

(a) 1; 000

(b) 941:76

(c) Overestimate equals 985; Underestimate equals 955

(d) Overestimate equals 977:60; Underestimate equals 963(e) Overestimate equals 973:04; Underestimate equals 968:19

(f) Continuous stream equals 970:61

2. The price of each of the following bonds is

(a) 975:41

(b) 1812:69

(c) 3625:39

(d) 2673:76

3. The antiderivatives can be checked by taking the derivatives.

4. The antiderivatives are

(a) F(x) = 5x22

+ C

(b) G(x) = 13

x3 + C

(c) H(x) = 13

x3 5x22

+ C

(d) J(x) = 2ex2 + C

(e) K(x) = ex2

+ C

5. The values of the antiderivatives evaluated over the interval 1 to 5 are

(a) 60

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