CHAPTER 1epstein/FMText/Ch1SM.pdfSince we have 8 hours per day, 100 days become 100(8) = 800 hours. Renting will cost less. Notice, this is true either by calculating the cost of rental

CHAPTER 1

EXERCISE SET 1.1

1.

Cost =(Variable Cost)+(Fixed Cost)=(Cost per item) × (number produced)+ (Fixed Cost)= 3x + 10, 000

3. Revenue=(Price per item)×(number of items sold)= 5x

5. Profit=(Revenue)−(Cost)=R−CIn view of Problems 1 and 3, we have,

P = 5x − (3x + 10000) = 2x − 10, 000

7. We have (x, p) = (10, 8) and(x, p) = (15, 6). The slope is6 − 8

15 − 10= −0.40. The demand equation

becomes

p − 8 = −0.40(x − 10)

= −0.4x + 6

p = −0.4x + 14


120 − 100= −0.25. The demand equation

becomes

p − 35 = −0.25(x − 100)

= −0.25x + 25

p = −0.25x + 60


100 − 50= 1.6. The supply equation

becomes

p − 95 = 1.6(x − 50)

= 1.6x − 80

p = 1.6x + 15

13. We have (x, p) = (10, 60) and

(x, p) = (14, 80). The slope is80 − 60

14 − 10= 5.

The supply equation becomes

p − 60 = 5(x − 10)

= 5x − 50

p = 5x + 10

15. We assume the linear depreciationequation to be V = mt + b, with t given inyears. Initially, when t = 0, the value is$130. The value decreases by $15 per year,so m = −15. The linear appreciationequation is then V = −15t + 135.

17. We assume the linear depreciationequation to be V = mt + b, with t given inyears. Initially, when t = 0, the value is$15,000. So, b = 15, 000. Six years later thecar is sold for $6000. So,6000 = V (6) = m(6) + 15, 000. Solving form gives m = −1500. The linearappreciation equation is thenV = −1500t + 15, 000.

19. Fixed costs are $150 and variable costsare $10. The cost equation is thenC = 10x + 150, where x is in units of hoursand C in dollars.

1

21. Fixed costs are $91,000 and variablecosts are $2. The cost equation is thenC = 2x + 91, 000, where x is in numbers ofshirts and C in dollars.

23.

C = mx + b

= mx + 1000

5000 = m(800) + 1000

m = 5

C = 5x + 1000

25. We have (x, p) = (50, 1.25) and(x, p) = (100, 1). The slope is then1 − 1.25

100 − 50= −0.005. The equation becomes

p − 1 = −0.005(x − 100)

= −0.005x + 0.5

p = −0.005x + 1.5

27. If x is number of items and R is indollars, we have

R = xp

7200 = 600p

p = 12

R = 12x

29. Let x be the number of items and letR, C, and P be in dollars. Since an itemsells for $10, we have R = 10x. Fixed costsare $2000, so C = mx + 2000. Then

P = R − C

= 10x − mx − 2000

= (10 − m)x − 2000

7000 = (10 − m)(1000) − 2000

= −1000m + 8000

m = 1

So, C = −x + 2000.

31. Let x be the number of items and letR, C, and P be in dollars. Since fixes costsar $1000 and it takes $3 to make each item,the cost equation is C = 3x + 1000. Then

P = R − C

= px − (3x + 1000)

= (p − 3)x − 1000

7000 = (p − 3)(1000) − 1000

p − 3 = 8

p = 11

So, R = 11x.

33. Let x be the number of packages andlet p be in dollars. The slope is−0.25

5= −0.05. Then the demand equation

is

p − 3.18 = −0.05(x − 27)

= −0.05x + 1.35

p = −0.05x + 4.53

35. Let x be the number of packages andlet p be in dollars. The slope is

m =1

20= 0.05. Then the supply equation

is

p − 2 = 0.05(x − 15)

= 0.05x − 0.75

p = −0.05x + 1.25

37. Let V = mt + b, where t is in yearsand V is in dollars. The new machine is$50,000, so b = 50, 000. To find m we have

V = mt + 50, 000

2

5000 = m(9) + 50, 000

m = −5000

V = −5000t + 50, 000

After one year the value in dollars isV (1) = −5000(1) + 50, 000 = 45, 000After five years the value in dollars isV (5) = −5000(5) + 50, 000 = 25, 000

39. The fixed cost are $674,000 and thevariable costs are $21.

41. Let x be the number of pairs offenders manufactured. Then the costsfunctions in dollars are as follows:Steel: C(x) = 260, 000 + 5.26xAluminum: C(x) = 385, 000 + 12.67xRMP: C(x) = 95, 000 + 13.19xNPN: C(x) = 95, 000 + 9.53x

43. Let x be in tons and R in dollars.Then R = 2274x.

45. The revenue function is R(x) = 40x.Set 40x = 100, 000 and obtain x = 2500.

45. The revenue function is R = 40x. Set40x = 100, 000 and obtain x = 2500.

47. We have R = px. Given the units, wemust have 105.6 = 150p. We can calculatep = 0.704. Thus, R = 0.0704x.

49. The total cost in dollars isC = 75 + 371 = 446. Profits, in dollars isP = R − C = 573 − 446 = 127.

51. Let x be the number of tons. Fixedcosts are $447,917 and variable costs are209.03x. Therefore, costs areC(x) = 209.03x + 447, 917. Revenue is$266.67 per ton, so we haveR(x) = 266.67x. Finally, profits are

P = R−C = 266.67x−(209.03x+447, 917) =57.64x − 447, 917.

53. We have (x, p) = (10, 99) and(x, p) = (20, 43). The slope is

m =43 − 99

20 − 10= −5.6. The ehr demand

equation is

p − 99 = −5.6(x − 10)

= −5.6x + 56

p = −5.6x + 155

55. (a) p ≈ 300, 000 (b) 0.1 yen per ton

57. The revenue function is R(x) = 2700x.Set 2700x = 31, 750 and obtain x ≈ 11.76.

59. Since no sales result in no revenue, wemust have R(0) = 0.

61. The area is the base times the height.But the height is the price and the base isthe number of items sold. Therefore, thearea of the rectangle is the price times thenumber sold, which is the revenue.

63. Here (x, c) = (0, 12) and (x, c) = (2, 0).

The slope is m =0 − 12

2 − 0= −6. The

equation then is

c − 0 = −c(x − 2)

c = −6x + 12

65. Let x be the number of cows, then thecost C(x) in dollars isC(x) = 13, 386 + 393x. The revenuefunction in dollars is R(x) = 470x. Theprofit function in dollars isP (x) = 77x − 13, 386. The profit for anaverage of 97 cows is P (97) = −5917, that

3

is a loss of $5917. For many such farms theproperty and buildings have already beenpaid off. Thus, the fixed costs for thesefarms are lower than stated in the table.

67. Let the cost revenue, and profits begivin in thousands of dollars. We haveC(x) = 3.84x + 300, 000R(x) = 4.8xP (x) = R(x) − C(x) = 0.96x − 300, 000If x = 200, 000, then

C(200, 000) = 3.84(200, 000) + 300, 000

= 1, 068, 000

R(200, 000) = 4.8(200, 000)

= 960, 000

P (200, 000) = 960, 000 − 1, 068, 000

= −108, 000

Thus, if 200,000 tons are produced andsold, the cost is 1,068,000 thousand dollarsor $1068 million. The revenue is $960million, and there is a loss of $108 million.Doing the same for some other values of x,we have the results shown in the table,given im millions of dollars.

Number 200,000 300,000 400,000

Cost 1068 1452 1836Revenue 960 1440 1920Profit -108 -12 84

We can see from the table, that for smallervalues of x, P (x) is negative; that is, theplant has losses. For larger values of x,P (x) turns positive and the plant has(positive) profits.

4

EXERCISE SET 1.2

1. The break-even quantity is the numberof items for which the revenue equals thecost. Setting revenue equal to costs gives

4x = 2x + 4

2x = 4

Thus x = 2 is the break-even quantity.

3. Setting revenue equal to cost gives

0.2x = (0.1x + 2)

0.1x = 2

Hence x = 20 is the break-even quantity.

5. The equilibrium point is the point atwhich supply equals to demand.

x + 3 = −x + 6

2x = 3

The equilibrium quantity is x = 1.5 and theequilibrium price is p = 1.5 + 3 = 4.5. Thus,the equilibrium point is (1.5, 4.5).

7. The equilibrium point is the pointwhere supply equals demand.

5x + 10 = −10x + 25

15x = 15

x = 1

The equilibrium price is thenp = 5(1) + 10 = 15. The equilibrium pointis then (1,15).9. We have

R(x) = 35x

C(x) = 15x + 40, 000

P = R − C

= 20x − 40, 000

Setting P = 0 or R = C gives 20x = 40, 000or x = 2000 as the break-even quantity.

11. Let d be the number of days. (a) Thecost to rental is CR = 320d, (b) while thecost to buy is CB = 28, 000 + 40d. SetCR = CB and obtain

320d = 28, 000 + 40d

280d = 28, 000

d = 100

Since we have 8 hours per day, 100 daysbecome 100(8) = 800 hours. Renting willcost less. Notice, this is true either bycalculating the cost of rental or buy or justnotice that the graph of the rental costfunction is below the graph of the buy costfuntion for days less than 100. (They-intercept is smaller.)

13. (a) We have CU = 5000x, (b) whileCW = 40, 000 + 1000x. (c) Setting thesetwo cost functions equal gives

5000x = 1000x + 40, 000

4000x = 40, 000

x = 10

or 10 years.

15. Let x be the miles. (a) The cost fromAcme is CA = 75 + 0.40x and the cost fromBell is CB = 105 + 0.25x.(b) First set the two cost function equaland obtain

75 + 0.40x = 105 + 0.25x

0.15x = 30

x = 200

At this milage, the two cost equations areequal. Acme will cost less. Notice, this is

1

true either by calculating the cost of Acmeor Bell or just notice that the graph of theAcme cost function is below the graph ofthe Bell cost funtion for hours less than 200.(The y-intercept is smaller.)

17. The cost of the outside supplier isCO = 0.75x and for 50,000 manuals the costis CO = 0.75(50, 000) = 37, 500 dollars.The cost for making them in-house isCI = 10, 000 + 0.50x and for 50,000manuals the cost isCO = 10, 0000 + 0.50(50, 000) = 35, 000dollars.So contracting in-house is cheaper.

19. Let x be the number of bunches and p

the price per bunch in dollars. The slope of

the supply equation is m =.25

2= 0.125.

The equation is

p − 2.50 = 0.125(x − 8)

p = 0.125x + 1.5

Setting this equal to the demand equationp = −0.1x + 6 yields

0.125x + 1.5 = −0.1x + 6

0.225x = 4.5

x = 20

The equilibrium quantity is 20 bunches andthe equilibrium price is −0.1(20) + 6 = 4dollars per bunch.

21. The total cost is 75 + 371 = 446dollars. The least price per hectare is the$446.

23. We have R = 266.67x andC = 235, 487 + 201.68x. Setting these equal

to each other, yields

266.67x = 206.68 + 235, 487

x ≈ 3925

25. We have R = 470x andC = 13, 386 + 393x. Setting these equal toeach other yields

470x = 13, 386 + 393x

x ≈ 174

27. We have

225 + 0.012x = 550 + 0.007x

0.005 = 225

x = 45, 000

We can either calculate the cost for each ornotice that the graph of the steel equationstarts out below the other. Since 5000 milesis less than the 45,000 found above, steel isthe lower cost.

29. We have

21 + 0.00085x = 23 + 0.00080x

0.00005 = 2

x = 40, 000

We can either calculate the cost for each ornotice that the graph of the steel equationstarts out below the other. Since 30,000miles is less than the 40,000 found above,steel is the lower cost.

31. Let x be the number of items producedper year.(a) The annual Cost in Houston isCH = 4, 000, 000 + 0.25xThe annual Cost in Boston is

2

CB = 4, 200, 000 + 0.22xWhen x = 10, 000, 000,

CH = 4, 000, 000 + 2.5(10, 000, 000)

= 6, 500, 000

CB = 4, 200, 000 + 2.2(10, 000, 000)

= 6, 400, 000

Therefore Boston has lower annual cost, notcounting initial investment.(c) Counting initial investment, the totalcost over five years is:

C̃H = 16, 000, 000 + 5(CH)

= 16, 000, 000 + 5(6, 500, 000)

= 48, 500, 000

C̃B = 20, 000, 000 + 5(CB)

= 20, 000, 000 + 5(6, 400, 000)

= 52, 000, 000

So Houston has the lower cost over the fiveyears if the initial investment is counted.

33 . We have

13.19 + 95, 000 = 12.67 + 385, 000

0.52x = 290, 000

x ≈ 557, 692

35 . We have

12.55 + 95, 000 = 5.26 + 260, 000

7.29x = 165, 000

x ≈ 22, 634

37. The total cost for a manual machine isCm = 1000 + 16t, where t is the processingtime.For an automatic machine the total cost is

Ca = 8000 +2n

100.

The total cost for a manual machine is

Cm = 1000 +16x

10, where x is the number of

items. For an automatic machine the total

cost is Ca = 8000 +2x

100.

Set the two costs equal to each other andobtain

1000 +16x

10= 8000 +

2x

100158x

100= 7000

x ≈ 4430

3

EXERCISE SET 1.3

1. We have{

x + 2y = 122x + 3y = 19

Proceeding with Gauss elimination we must remove x from the second equation. We thenperform E2 − 2E1 → E2 and obtain

{

x + 2y = 12y = 5

We have y = 5. Back substituting gives

x + 2y = 12

x + 2(5) = 12

x = 2

So the solution is (x, y) = (2, 5).

3. We have{

4x − 8y = 20−x + 3y = −7

We first divide the first row by 4,1

4E1 → E1. We then have

{

x − 2y = 5−x + 3y = −7

Proceeding with Gauss elimination we must remove x from the second equation. We thenperform E2 + E1 → E2 and obtain

{

x − 2y = 5y = −2

We have y = −2. Back substituting gives

x − 2y = 5

x − 2(−2) = 5

x = 1

So the solution is (x, y) = (1,−2).

5. We have{

−2x + 8y = −6−2x + 3y = −1

1

We first divide the first row by −2,1

−2E1 → E1. We then have

{

x − 4y = 3−2x + 3y = −1

Proceeding with Gauss elimination we must remove x from the second equation. We thenperform E2 + 2E1 → E2 and obtain

{

x − 4y = 3−5y = 5

We first divide the second row by −5,1

−5E1 → E1.

{

x − 4y = 3y = −1


x − 4y = 3

x − 4(−1) = 3

x = −1

So the solution is (x, y) = (−1,−1).

7. We have{

3x + 2y = 0x − y = −3



{

x + 2y = 0x − y = −3

Proceeding with Gauss elimination we must remove x from the second equation. We thenperform E2 − E1 → E2 and obtain

{

x + 2y = 0−3y = −3


−3E1 → E1.

{

x + 2y = 0y = 1

2


x + 2y = 0

x + 2(1) = 0

x = −2

So the solution is (x, y) = (−2, 1).

9. We have{

x + 2y = 52x − 3y = −4

Proceeding with Gauss elimination we must remove x from the second equation. We thenperform E2 − 2E1 → E2 and obtain

{

x + 2y = 5−7y = −14


−7E1 → E1.

{

x + 2y = 5y = 2


x + 2y = 5

x + 2(2) = 5

x = 1

So the solution is (x, y) = (1, 2).

11. The augmented matrix is

3 −3 6 −32 1 2 42 −2 5 −2

Proceeding with Gauss elimination we have

3 −3 6 −32 1 2 42 −2 5 −2

1

3R1 → R1

−→

1 −1 2 −22 1 2 42 −2 5 −2

R2 − 2R1 → R2

R3 − 2R1 → R3

−→

1 −1 2 −10 3 −2 60 0 1 0

1

3R2 → R2

−→

1 −1 2 −10 1 −2

32

0 0 1 0

3

This corresponds to the system of equations

E1 : x − y + 2z = −1E2 : y −

2

3z = 2

E3 : 0 z = 0

This gives z = 0. Now backward substitute and obtain

E2 : y −2

3z = 2

y −2

3(0) = 2

y = 2E1 : x − y + 2z = −1

x − (2) + 2(0) = −1x = 1

The solution is (x, y, z) = (1, 2, 0).


1 1 1 101 −1 1 101 1 −1 0


1 1 1 101 −1 1 101 1 −1 0

R2 − R1 → R2

R3 − R1 → R3

−→

1 1 1 100 −2 0 00 0 −2 −10

1

−2R2 → R2

1

−2R3 → R3

−→

1 1 1 100 1 0 00 0 1 5


E1 : x + y + z = 10E2 : y − = 0E3 : 0 z = 5

This gives y = 0 and z = 5. Now backward substitute and obtain

E1 : x + y + z = 10x + (0) + (5) = 10

x = 5



1 1 1 62 1 2 103 2 1 10

4


1 1 1 62 1 2 103 2 1 10

R2 − 2R1 → R2

R3 − 3R1 → R3

−→

1 1 1 60 −1 0 −20 −1 −2 −8

−R2 → R2

−→

1 1 1 60 1 0 20 −1 −2 −8

R3 + R2 → R3

−→

1 1 1 60 1 0 20 0 −2 −6

−1

3R3 → R3

−→

1 1 1 60 1 0 20 0 1 3


E1 : x + y + z = 6E2 : y = 2E3 : z = 3

This gives y = 2 and z = 3. Now backward substitute and obtain

E1 : x + y + z = 6x + (2) + (3) = 6

x = 1



1 −1 2 −12 1 −3 60 1 −1 2


1 −1 2 −12 1 −3 60 1 −1 2

R2 − 2R1 → R2 −→

1 −1 2 −10 3 −7 80 1 −1 2

1

3R2 → R2

−→

1 −1 2 −10 1 −7/3 8/30 1 −1 2

R3 − R2 → R3

−→

1 −1 2 −10 1 −7/3 8/30 0 4/3 −2/3

3

4R3 → R3

5

−→

1 −1 2 −10 1 −7/3 8/30 0 1 −0.5


E1 : x − y + 2z = −1E2 : y − −

7

3z = 8

3

E3 : z = −0.5

This gives z = −0.5. Now backward substitute and obtain

E2 : y −7

3z = 8

3

y −7

3(−1

2) = 8

3

y = 1.5E1 : x − y + 2z = −1

x − (1.5) + 2(−0.5) = −1x = 1.5

The solution is (x, y, z) = (1.5, 1.5,−0.5).


2 −4 8 22 3 2 32 −3 5 0


2 −4 8 22 3 2 32 −3 5 0

1

2R2 → R2

−→

1 −2 4 12 3 2 32 −3 5 0

R2 − 2R1 → R2

R3 − 2R1 → R3

−→

1 −2 4 10 7 −6 10 1 −3 −2

1

7R2 → R2

−→

1 −2 4 10 1 −6/7 1/70 1 −3 −2

R3 − R2 → R3

−→

1 −2 4 10 1 −6/7 1/70 0 −15/7 −15/7

−7

15R3 → R3

−→

1 −2 4 10 1 −6/7 1/70 0 1 1


E1 : x − 2y + 4z = 1E2 : y − −

6

7z = 1

7

E3 : z = 1

6


E2 : y −6

7z = 1

7

y −6

7(1) = 1

7

y = 1E1 : x − 2y + 4z = 1

x − 2(1) + 4(1) = 1x = −1



1 1 1 1 60 1 −1 2 40 0 1 1 31 2 3 −1 5


1 1 1 1 60 1 −1 2 40 0 1 1 31 2 3 −1 5

R4 − R1 → R4

−→

1 1 1 1 60 1 −1 2 40 0 1 1 30 1 2 −2 −1

R4 − R2 → R4

−→

1 1 1 1 60 1 −1 2 40 0 1 1 30 0 3 −4 −5

R4 − R3 → R4

−→

1 1 1 1 60 1 −1 2 40 0 1 1 30 0 0 −7 −14

−1

7R4 → R4

−→

1 1 1 1 60 1 −1 2 40 0 1 1 30 0 0 1 2


E1 : x + y + z + u = 6E2 : y − z + 2u = 4E3 : z + u = 3E4 : u = 2

7

This gives u = 2. Now backward substitute and obtain

E3 : z + u = 3z + (2) = 3

z = 1E2 : y − z + 2u = 4

y − (1) + 2(2) = 4y = 1

E1 : x + y + z + u = 6x + (1) + (1) + (2) = 6

x = 2

The solution is (x, y, z, u) = (2, 1, 1, 2).


1 2 1 −1 −21 2 2 2 90 1 1 −1 −20 1 −2 3 4


1 2 1 −1 −21 2 2 2 90 1 1 −1 −20 1 −2 3 4

R2 − R1 → R2−→

1 2 1 −1 −20 0 1 3 110 1 1 −1 −20 1 −2 3 4

R2 ↔ R3

−→

1 2 1 −1 −20 1 1 −1 −20 0 1 3 110 1 −2 3 4

R4 − R2 → R4

−→

1 2 1 −1 −20 1 1 −1 −20 0 1 3 110 0 −3 4 6

R4 + 3R3 → R4

−→

1 2 1 −1 −20 1 1 −1 −20 0 1 3 110 0 0 13 39

1

13R4 → R4

−→

1 2 1 −1 −20 1 1 −1 −20 0 1 3 110 0 0 1 3

8


E1 : x + 2y + z − u = −2E2 : y + z − u = −2E3 : z + 3u = 11E4 : u = 3

This gives u = 3. Now backward substitute and obtain

E3 : z + 3u = 11z + 3(3) = 11

z = 2E2 : y + z − u = −2

y + (2) − (3) = −2y = −1

E1 : x + 2y + z − u = −2x + 2(−1) + (2) − (3) = −2

x = 1

The solution is (x, y, z, u) = (1,−1, 2, 3).

25. Let x be the number of quarters and y the number of dimes. Since there are 25 cins, wehave x + y = 25. The face value of quarters is $0.25x and that of dimes is $0.10y. Since thetotal face value is $3.25, we have 0.25x + 0.10y = 3.25. We then have

{

x + y = 250.25x + 0.10y = 3.25

Proceeding with Gauss elimination we must remove x from the second equation. We thenperform E2 − 0.25E1 → E2 and obtain

{

x + y = 25−0.15y = −3

We first divide the second row by −0.15,1

−0.15E1 → E1.

{

x + y = 25y = 20


x + y = 25

x + (20) = 25

x = 5

So the solution is (x, y) = (5, 20). Thus, the person has 5 quarters and 20 dimes.

9

27. Let x be the number of nickels, y the number of dimes, and z the number of quarters.since the total number of coins is 36, we have x + y + z = 36. Since there are twice as manydimes as nickels, y = 2x. The face values of nickels, dimes, and quarters are $0.05x, $0.10y, and$0.25z, respectively. Since the total value of the money is $4, we have 0.05x + 0.10y + 0.25z.Thus,

E1 : x + y + z = 36E2 : 2x − y = 0E3 : 0.05x + 0.10y + 0.25z = 4

The augmented matrix is

1 1 1 362 −1 0 0

0.05 0.10 0.25 4


1 1 1 362 −1 0 0

0.05 0.10 0.25 4

R2 − 2R1 → R2

R3 − 0.05R1 → R3

−→

1 1 1 360 3 2 720 0.05 0.2 2.2

1

3R2 → R2

−→

1 1 1 360 1 2/3 240 0.05 0.2 2.2

R3 − 0.05R2 → R3

−→

1 1 1 360 1 2/3 240 0 1/6 1

6R3 → R3

−→

1 1 1 360 1 2/3 240 0 1 6


E1 : x + y + z = 36E2 : y + 2

3z = 24

E3 : 0 z = 6


E2 : y + 2

3z = 24

y + 2

3(6) = 24

y = 20E1 : x + y + z = 36

x + (20) + (6) = 36x = 10

The solution is (x, y, z) = (10, 20, 6). Thus, the person has 10 nickels, 20 dimes, and 6 quarters.

10

29. Let x be the number of type A documents processed and let y be the number of type B.The accountant spends 2x hours on type A documents and 4y on type B. Since the accountantspends 40 hours each week on these documents, 2x + 4y = 40. The attorney spends 3x hourson type A documents and y on type B. Since the attorney spends 36 hours each week on thesedocuments, 3x + y = 30. We have

{

2x + 4y = 403x + y = 30



{

x + 2y = 203x + y = 30

We then have

[

1 2 203 1 30

]

R2 − 3R1 → R2

−→

[

1 2 200 −5 −30

]

−1

5R2 → R2

−→

[

1 2 200 1 6

]


x + 2y = 20

x + 2(6) = 20

x = 8

So the solution is (x, y) = (8, 6). Thus, they can process 8 of type A documents and 6 of typeB each week.

31. Let x be the number of dollars deposited in the first bank and y the number of dollars inthe second. Since a total of $1000 was deposited in the two banks, we have x + y = 1000. Theinterest received from the first bank was 0.08x and from the second was 0.1y. Since the totalinterest was $86, we have 0.08x + 0.1y = 86. We have

{

x + y = 10000.08x + 0.1y = 86

Then

[

1 1 10000.08 0.1 86

]

R2 − 0.08R1 → R2

−→

[

1 1 10000 0.02 6

]

50R2 → R2

−→

[

1 1 10000 1 300

]

11


x + y = 1000

x + (300) = 1000

x = 700

So the solution is (x, y) = (700, 300). Thus, the person deposits $700 in the first bank and $300in the second.

33. Let x be the number of style A sweaters produced and y the number of style B. Thecutting department spends 0.5x labor-hours on style A sweaters and 0.4y on style B. Since thisdepartment has 200 labor-hours each day, we have 0.5x + 0.4y = 200. The sewing departmentspends 0.6x labor-hours on style A sweaters and 0.3y on style B. Since this department has 186labor-hours each day, we have 0.6x + 0.3y = 186.

{

0.5x + 0.4y = 2000.6x + 0.03y = 186

We first divide the first row by 0.5 (or multiply by 2). We then have

{

x + 0.8y = 4000.6x + 0.3y = 186


[

1 0.8 4000.06 0.3 186

]

R2 − 0.06R1 → R2

−→

[

1 0.8 4000 −0.18 −54

]

−

1

0.18R2 → R2

−→

[

1 0.8 4000 1 300

]


x + 0.8y = 400

x + 0.8(300) = 400

x = 160

So the solution is (x, y) = (160, 300). Thus, 160 style A and 300 of style B can be producedeach day.

35. Let x be the number od style A documents processed, y the number of style B, and z thenumber of style C. The accountants spends 2x hours on tyle A documents, 4y on the type B,and 2z on tyle C. Since the accountant has 34 hours, we have 2x + 4y + 2z = 34. The attorneyspends 3x hours on tyle A documents, 2y on the type B, and 4z on tyle C. Since the attorney

12

has 35 hours, we have 3x + 2y + 4z = 35. Since the4 secretary needs 3 hours to type eachdocument and has 36 hours available, 3x + 3y + 3z = 36. Thus,

2x + 4y + 2z = 343x + 2y + 4z = 353x + 3y + 3z = 36

We then have

2 4 2 343 2 4 353 3 3 36

1

2R1 → R1

−→

1 2 1 173 2 4 353 3 3 36

R2 − 3R1 → R2

R3 − 3R1 → R3

−→

1 2 1 170 −4 1 −160 −3 0 −15

−

1

4R2 → R2

−→

1 2 1 170 1 −0.25 40 −3 0 −15

We may just divide the last equation by −3 and have

1 2 1 170 1 −0.25 40 1 0 5

We then have y = 5. Back substituting we have

y − 0.25z = 4

(5) − 0.25z = 4

z = 4

x + 2y + z = 17

x + 2(5) + (4) = 17

x = 3

We have (x, y, z) = (3, 5, 4). Thus we can process 3 of type A, 5 of type B, and 4 of type C.

37. Let x be the number of shares of MathOne bought, y the number of shares of NewModulebought, and z the number of shares of JavaTime bought. The stocks cost $20, $60, and $20,respectively. She wants to invest $4200. So, we have 20x + 60y + 20z = 4200. The stocks paydividents of $1, $2, and $3, respectively. Jennifer wants $290 in dividends, so x+2y+3z = 290.Finally, she wants twice as much money in MathOne as in JavaTime. So, x = 2z. We mustsolve the system

x − 2z = 0x + 2y + 3z = 290

20x + 60y + 20z = 4200

13

We then have

1 0 −2 01 2 3 290

20 60 20 4200

R2 − R1 → R2

R3 − 20R1 → R3

−→

1 0 −2 00 2 5 2900 60 60 4200

1

2R2 → R2

−→

1 0 −2 00 1 2.5 1450 60 60 4200

R3 − 60R2 → R3

−→

1 0 −2 00 1 2.5 1450 0 −90 −4500

−1

90R3 → R3

−→

1 0 −2 00 1 2.5 1450 0 1 50

We have z = 50. Back sustituting we have

y + 2.5z = 145

y + 2.5(50) = 145

y = 20

x − 2z = 0

x − 2(50) = 0

x = 100

The solution is (x, y, z) = (100, 20, 50). She then buys 100 shares of MathOne, 20 shares ofNewModule, and 50 shares of JavaTime. This represents a dollar amount of 100(20) = 2000,20(60) = 1200, and 50(20) = 1000, respectively.

39. Let x be the number of units of wood, y the number of units of fabric, and z the numberof units of stuffing. Then we have the system

x + 2y + 3z = 54

2x + 2y + z = 63

2x + y + z = 43

Then we have

1 2 3 542 2 1 632 1 1 43

R2 − 2R1 → R2

R3 − 2R1 → R3

−→

1 2 −3 00 −2 −5 −450 −3 −5 −65

1

2R2 → R2

−→

1 2 3 540 1 2.5 22.50 −3 −5 −65

R3 + 3R2 → R3

14

−→

1 2 3 540 1 2.5 22.50 0 2.5 2.5

1

2.5R3 → R3

−→

1 2 3 540 1 2.5 22.50 0 1 1


y + 2.5z = 22.5

y + 2.5(1) = 22.5

y = 20

x + 2y + 3z = 54

x + 2(20) + 3(1) = 54

x = 11

The solution is (x, y, z) = (11, 20, 1). So, they use 11 units of wood, 20 units of fabric, and 1unit of stuffing.

41. Let x be the number of batches of muffins, y the number of batches of scones, and z thenumber of batches of croissants. We then have

2x + 2y + z = 17

3x + 5y + 4z = 37

6(2.50)x + 10(2.00)y + 12(1.50) = 169

Dividing the first equation by 2 and then

1 1 0.5 8.53 5 4 37

15 20 18 169

R2 − 3R1 → R2

R3 − 15R1 → R3

−→

1 1 0.5 8.50 2 2.5 11.50 5 10.5 41.5

1

2R2 → R2

−→

1 1 0.5 8.50 1 1.25 5.750 5 10.5 41.5

R3 − 5R2 → R3

−→

1 1 0.5 8.50 1 1.25 5.750 0 4.25 12.75

1

4.25R3 → R3

−→

1 1 0.5 8.50 1 1.25 5.750 0 1 3


y + 1.25z = 5.75

15

y + 1.25(3) = 5.75

y = 2

x + y + 0.5z = 8.5

x + (2) + 0.5(3) = 8.5

x = 5

The solution is (x, y, z) = (5, 2, 3). So, they sell 5 batches of muffins, 2 batches of scones, and3 batches of croissants.

43. Let x, y, and z be the Foods I, II, and III, respectively. Then we have

5x + 3y + 2z = 27

20x + 60y + 40z = 540

8x + 4y + 10z = 128

To simplify, we divide the second equation by 20 and switch it with the first equation. Wethen have

1 3 2 275 1 3 518 4 10 128

R2 − 5R1 → R2

R3 − 8R1 → R3

−→

1 3 2 270 −14 −7 −840 −20 −6 −88

−

1

14R2 → R2

−→

1 3 2 270 1 0.5 60 −20 −6 −88

R3 + 20R2 → R3

−→

1 3 2 270 1 0.5 60 0 4 32

1

4R3 → R3

−→

1 3 2 270 1 0.5 60 0 1 8


y + 0.5z = 6

y + 0.5(8) = 6

y = 2

x + 3y + 2z = 27

x + 3(2) + 2(8) = 27

x = 5

The solution is (x, y, z) = (5, 2, 8). So, she has 5 units of iron, 2 units of calcium, and 8 unitsof folic acid

45. Let x be the number of oranges used, y the number of cups of strawberries, and z thenumber of cups of blackberries. The oranges contain x grams of fiber, the strawberries contain

16

2y grams, and the blackberries contain 6z grams. Since a total of 13 grams of fiber is needed,x + 2y + 6z = 13. The oranges contain 75x mg of vitamin C, the strawberries contain 60ymg, and the blackberries contain 30z mg. Since a total of 374 mg of vitamin is needed, 75x +60y + 30z = 375. The oranges contain 50x mg of phosphorus, the strawberries contain 50ymg, and the blackberries contain 40z mg. Since a total of 290 mg of phosphorus is needed,50x + 50y + 40z = 290. Thus, we wish to solve the system

x + 2y + 6z = 13

75x + 60y + 30z = 375

50x + 50y + 40z = 290

Use the augmented matrix and obtain

1 2 6 1375 60 30 37550 50 40 290

R2 − 75R1 → R2

R3 − 50R1 → R3

−→

1 2 6 130 −90 −420 −6000 −50 −260 −300

−

1

90R2 → R2

−→

1 2 6 130 1 14/3 20/30 −50 −260 −360

R3 + 50R2 → R3

−→

1 2 6 130 1 14/3 20/30 0 80/3 80/3

3

80R3 → R3

−→

1 2 6 130 1 14/3 20/30 0 1 1

We have z = 1. Backward substituting we have

y + (14/3)z = 20/3

y + (14/3)(1) = 20/3

y = 2

x + 2y + 6z = 13

x + 2(2) + 6(1) = 13

x = 3

So the unique solution is (3, 2, 1). Thus, 3 oranges, 2 cups of strawberries, and 1 cup ofblackberries are needed.

47. Let x , y, and z be the three foods I, II, and III, respectively.

1 4 2 372 6 2 36

12 3 21 177

R2 − 2R1 → R2

R3 − 12R1 → R3

−→

1 4 2 370 −2 −− 2 −380 −45 −3 −267

−

1

2R2 → R2

17

−→

1 4 2 370 1 1 190 −45 −3 −267

R3 + 45R2 → R3

−→

1 4 2 370 1 1 190 0 42 588

1

42R3 → R3

−→

1 4 2 370 1 1 190 0 1 14


y + z = 19

y + (14) = 19

y = 5

x + 2y + 2z = 27

x + 4(5) + 2(14) = 37

x = −11

The solution is (x, y, z) = (−11, 5, 14). But since x = −11, we have no solution for the dietitian.We can not have a negative amount of Food I.

49. (a) We have

x + 2y + 3u = 1000 + 1000 + 1200 = 3200 < 3500

x + 2z + 2u = 1000 + 700 + 800 = 2500 < 2700

z + u = 350 + 400 = 750 < 900

3z + 2u = 1050 + 800 = 1850 < 3000

Yes, there is sufficient food.(b) Looking above, we could have

x = 3200 − 2200 = 1000

x = 2700 − 1500 = 1200

So, an additional 1000 of x-species could be supported.

18

EXERCISE SET 1.4

1. Yes

3. Yes

5. No, the second row should be at the bottom.

7. x = 2 and y = 3

9. y = t and we have

x + 2y = 4

x + 2(t) = 4

x = 4 − 2t

So, the general solution is (x, y) = (4 − 2t, t), where t is any number.

11. We set z = s and u = t. Backward substituting, we have

y + 2z + 3u = 5

y + 2(s) + 3(t) = 5

y = 5 − 2s − 3t

x + 2z + 3u = 4

x + 2(s) + 3(t) = 4

x = 4 − 2s − 3t

The general solution is (x, y, z, u) = (4− 2s− 3t, 5− 2s− 3t, s, t), where s and t are anynumbers.

13. We set z = s and u = t. Backward substituting, we have

y + 2z + 3u = 1

y + 2(s) + 3(t) = 1

y = 1 − 2s − 3t

x + 2z + 4u = 6

x + 2(s) + 4(t) = 6

x = 6 − 2s − 4t

The general solution is (x, y, z, u) = (6− 2s− 4t, 1− 2s− 3t, s, t), where s and t are anynumbers.

1

15. We set x = r, z = s, and v = t. Backward substituting, we obtain

w = 3

u + v = 2

u + (t) = 2

u = 2 − t

y + 2z + 2v = 1

y + 2(s) + 2(t) = 1

y = 1 − 2s − 2t

The general solution is (x, y, z, u, v, w) = (r, 1 − 2s − 2t, s, 2 − t, t, 3), where r, s, and t,are any numbers.

17. The augmented matrix is[

2 −4 8−1 2 4

]


[

2 −4 8−1 2 4

]

1

2R1 → R1

−→

[

1 −2 4−1 2 4

]

R2 + R1 → R2

−→

[

1 −2 40 0 8

]

The last row implies that 0 = 8. Since this is a contradiction, the system has no solution.


3 −6 12−1 2 −4

]


[

3 −6 12−1 2 −4

]

1

3R1 → R1

−→

[

1 −2 4−1 2 −4

]

R2 + R1 → R2

−→

[

1 −2 40 0 0

]

There are an infinite number of solutions. We set y = t, and have

x − 2y = 4

x − 2(t) = 4

x = 2t + 4

2

The general solution is (2t + 4, t) where t is any number.


−1 3 72 −6 −14

]


[

−1 3 72 −6 −14

]

−R1 → R1−→

[

1 −3 −72 −6 −14

]

R2 − 2R1 → R2

−→

[

1 −3 −70 0 0

]

There are an infinite number of solutions. We set y = t, and have

x − 3y = −7

x − 3(t) = −7

x = 3t − 7

The general solution is (3t − 7, t) where t is any number.


[

0.1 −0.3 0.4−2 6 4

]


[

0.1 −0.3 0.4−2 6 4

]

10R1 → R1−→

[

1 −3 4−2 6 4

]

R2 + 2R1 → R2

−→

[

1 −3 40 0 12

]

The last row implies that 0 = 12. Since this is a contradiction, the system has no solution.


[

−1 2 3 142 −1 2 2

]


3

[

−1 2 3 142 −1 2 2

]

−R1 → R1−→

[

1 −2 −3 −142 −1 2 2

]

R2 − 2R1 → R2

−→

[

1 −2 −3 −140 3 8 30

]

1

3R2 → R2

−→

[

1 −2 −3 −140 1 8/3 10

]

There are an infinite number of solutions. We set z = t, and have

y +8

3z = 10

y +8

3t = 10

y = 10 −

8

3t

x − 2y − 3z = −14

x − 2(10 −

8

3t) − 3t = −14

x = 6 −

7

3t

The general solution is (5 −7

3t, 10 − 8

3t, t) where t is any number.


[

−2 6 4 123 −9 −6 −18

]


[

−2 6 4 123 −9 −6 −18

]

−1

2R1 → R1

−→

[

1 −3 −2 −63 −9 −6 −18

]

R2 − 3R1 → R2

−→

[

1 −3 −2 −60 0 0 0

]

There are an infinite number of solutions. We set y = s and z = t, and have

x − 3y − 2z = −6

x − 3(s) − 2(t) = −6

x = 3s + 2t − 6

4

The general solution is (3s + 2t − 6, s, t) where s and t are any numbers.


[

−1 5 −3 72 −10 6 −14

]


[

−1 5 −3 72 −10 6 −14

]

−R1 → R1−→

[

1 −5 3 −72 −10 6 −14

]

R2 − 2R1 → R2

−→

[

1 −5 3 −70 0 0 0

]

There are an infinite number of solutions. We set y = s and z = t, and have

x − 5y + 3z = −7

x − 5(s) + 3(t) = −7

x = 5s − 3t − 7

The general solution is (5s − 3t − 7, s, t) where s and t are any numbers.


1 1 1 11 −1 −1 23 1 1 4


1 1 1 11 −1 −1 23 1 1 4

R2 − R1 → R2

R3 − 3R1 → R3

−→

1 1 1 10 −2 −2 10 −2 −2 1

−

1

2R2 → R2

−→

1 1 1 10 1 1 −0.50 −2 −2 1

R3 + 2R2 → R3

−→

1 1 1 10 1 1 −0.50 0 0 0


y + z = −0.5

5

y + (t) = −0.5

y = −t − 0.5

x + y + z = 1

x + (−t − 0.5) + (t) = 1

x = 1.5

The general solution is (1.5,−t − 0.5, t) where t is any number.


2 1 −1 03 −1 2 11 −2 3 2


2 1 −1 03 −1 2 11 −2 3 2

0.5R1 → R1

−→

1 0.5 −0.5 03 −1 2 11 −2 3 2

R2 − 3R1 → R2

R3 − R3 → R3

−→

1 0.5 −0.5 00 −2.5 3.5 10 −2.5 3.5 2

−0.4R2 → R2

−→

1 0.5 −0.5 00 1 −1.4 −0.40 −2.5 3.5 2

R3 + 2.5R2 → R3

−→

1 0.5 −0.5 00 1 −1.4 −0.40 0 0 1

The last row implies that 0 = 1. since this is a contradiction, there is no solution to thissystems.


1 −2 2 12 1 −1 23 −1 1 3


1 −2 2 12 1 −1 23 −1 1 3

R2 − 2R1 → R2

R3 − 3R1 → R3

−→

1 −2 2 10 5 −5 00 5 −5 0

0.2R2 → R2

6

−→

1 −2 2 10 1 −1 00 5 −5 0

R3 − 5R2 → R3

−→

1 −2 2 10 1 −1 00 0 0 0


y − z = 0

y − (t) = 0

y = t

x − 2y + 2z = 1

x − 2(t) + 2(t) = 1

x = 1

The general solution is (1, t, t) where t is any number.


1 1 42 −3 −73 −4 −9


1 1 42 −3 −73 −4 −9

R2 − 2R1 → R2

R3 − 3R1 → R3

−→

1 1 40 −5 −150 −7 −21

−

1

5R2 → R2

−→

1 1 40 1 30 −7 −21

R3 + 7R2 → R3

−→

1 1 40 1 30 0 0


E1 : x + y = 4E2 : y = 3

This gives y = 3. Now backward substitute and obtain

E1 : x + y = 4x + (3) = 4

x = 1

7

The solution is (x, y) = (1, 3).


1 2 42 −3 51 −5 2


1 2 42 −3 51 −5 2

R2 − 2R1 → R2

R3 − R1 → R3

−→

1 2 40 −7 −30 −7 −2

−

1

7R2 → R2

−→

1 2 40 1 3/70 −7 −2

R3 + 7R2 → R3

−→

1 2 40 1 3/70 0 1

The last inequation indicates a contradiction, so the system has no solution.


1 1 1 11 −1 1 23 1 3 1


1 1 1 11 −1 1 23 1 3 1

R2 − R1 → R2

R3 − 3R1 → R3

−→

1 1 1 10 −2 0 10 −2 0 −2

1

−2R2 → R2

−→

1 1 1 10 1 0 −

1

2

0 −2 0 −2

R3 + 2R2 → R3

−→

1 1 1 10 1 0 −

1

2

0 0 0 3

Since the last equation is a contradiction, the system has no solution.

43. Using a graphing calculator or computer gives

45. Using the graphing calculator or the computer gives the following graph. There isno solution.

8

47. Using the graphing calculator or the computer gives the following graph. There isno solution.

49. Let x be the number of nickels, y the number of dimes, and z the number ofquarters. Since there are 36 coins, x + y + z = 36. Since the face value of the coins is $4,0.05x + 0.10y + 0.25z = 4. Thus, we wish to solve the system

x + y + z = 360.05x + 0.10y + 0.25z = 4

Using the augmented matrix we have

[

1 1 1 360.05 0.10 0.25 4

]


[

1 1 1 360.05 0.10 0.25 4

]

R2 − 0.05R1 → R2

−→

[

1 1 1 360 0.05 0.02 2.2

]

20R2 → R2

−→

[

1 1 1 360 1 4 44

]


y + 4z = 44

y + 4t = 44

y = 44 − 4t

x + y + z = 4

x + (44 − 4t) + t = 36

x = 3t − 8

The general solution is (3t − 8, 44 − 4t, t) where t is a parameter, but does have somerestrictions. First, since the number of quarters must certainly be a nonnegative intergerand z = t, t must be a nonnegative integer. We must have 0 ≤ x = 3t − 8 and0 ≤ y = 44 − 4t. The first inequality implies that t ≥ 8/3 and the second inequalityimplies that t ≤ 11. We conclude that the parameter t can be any natural number biggerthan 2 and less than 12.

51. Let x be the amount invested in bonds, y the amount invested in the low-risk stocks,and z the amount invested in the medium risk stocks. Since the client invests a total of$900,000 among the three groups, x+ y + z = 900, 000. The annual return from bonds is0.07x, from the low-risk stocks is 0.09y, and from the medium risk stocks is 0.11z. Sincethe total annual return is (0.9)(900, 000) = 81, 000, we have 0.07x + 0.09y + 0.11z =81, 000. Thus, we wish to solve the system

9

x + y + z = 900, 000

0.07x + 0.09y + 0.11z = 81, 000


[

1 1 1 900, 0000.07 0.09 0.11 81, 000

]

R2 − 0.07R1 → R2

−→

[

1 1 1 900, 0000 0.02 0.04 18, 000

]

50R2 → R2

−→

[

1 1 1 900, 0000 1 2 900, 000

]

We set z = t as the parameter. Then

y + 2z = 900, 000

y + 2(t) = 900, 000

y = 900, 000− 2t

x + y + z = 900, 000

x + (900, 000 − 2t) + (t) = 900, 000

x = t

The general solution is (t, 900000 − 2t, t). The only constraint on the parameter t is0 ≤ t ≤ 450, 000. Thus, the pension fund manager can allocate $t to bonds and the sameto the medium risk stocks, and 900, 000−2t to the low-risk stocks, where 0 ≤ t ≤ 450, 000.

1.4.53. Let x be the number of Tiderunner sandals sold, y the number of Sport sandals,and z the number of Harbor sandals. Since the store sold a total of 300 sandals, x+y+z =300. Since the sales of Sport is the sum of the other two, y = x+ z or x− y + z = 0. Thestore sold 30x worth of Tiderunners, 40x worth of Sports, and 50x worth of Harbors. Sincethe store sold $11,500 worth of these ssandals in that month, 30x + 40y + 50z + 11, 500.Thus, we wish to solve the system

x + y + z = 300

x − y + z = 0

30x + 40y + 50z = 11, 500


10

1 1 1 3001 −1 1 0

30 40 50 11, 500

R2 − R1 → R2

R3 − 30R1 → R3

−→

1 1 1 3000 −2 0 −3000 10 20 2500

−0.5R2 → R2

−→

1 1 1 3000 1 0 1500 10 20 2500

R3 − 10R2 → R3

−→

1 1 1 3000 1 0 1500 0 20 1000

1

20R3 → R3

−→

1 1 1 3000 1 0 1500 0 1 50

We have z = 50 and y = 150. Solving for x we obtain

x + y + z = 300

x + 150 + 50 = 300

x = 100

So the unique solution is (100, 150, 50). Thus, the store sold 100 of Tiderunner sandals,150 of Sports, and 50 of Harbors.

55. Let x, y, z, and u be the number of units of X, Y , Z, and U , respectively, that ismanufactured each day. The firm needs 0.1x, 0.3y, 0.6z, and 0.2u units of the perishableingredient I to manufacture X, Y , Z, and U , respectively. Since the firm can produce 100units of I, we have 0.1x + 0.3y + 0.6z + 0.2u = 100. The firm needs 0.4x, 0.2y, 0.4z, and0.3u units of the perishable ingredient II to manufacture X, Y , Z, and U , respectively.Since the firm can produce 200 units of I, we have 0.4x+0.2y +0.4z +0.3u = 200. Sincean order for 500 units of U has been received, u = 500. Thus, after subsituting u = 500,we wish to solve the system

0.1x + 0.3y + 0.6z = 0

0.4x + 0.2y + 0.4u = 50

Using the augmented matrix we obtain

[

0.1 0.3 0.6 00.4 0.2 0.4 50

]

10R1 → R1−→

[

1 3 6 00.4 0.2 0.4 50

]

R2 − 0.4R1 → R2

11

−→

[

1 3 6 00 −1 −2 50

]

−R2 → R2

−→

[

1 3 6 00 1 2 −50

]

Since 0 ≤ y and 0 ≤ z and x + 2z = −50, the latter equation can not be satisfied.Thus, the production manager can not meet the demands of his boss.

57 Let x be the number of type A documents to process, y the number of type B, and zthe number of type C. The accountant spends 2x hours on type A, 4y hours on type B,and 2z hours on type C. Since the accountant has 34 hours available, 2x + 4y + 2z = 34.The attorney spends 3x hours on type A, 2y hours on type B, and 4z hours on type C.Since the attorney has 35 hours available, 3x + 2y + 4z = 35. Thus, we wish to solve thesystem

2x + 4y + 4z = 343x + 2y + 4z = 35

Using the augmented matrix we have

[

2 4 2 343 2 4 35

]

0.5R1 → R1−→

[

1 2 1 173 2 4 35

]

R2 − 3R1 → R2

−→

[

1 2 1 170 −4 1 −16

]

−0.25R2 → R2

−→

[

1 2 1 170 1 −0.25 4

]


y − 0.25z = 4

y − 0.25(t) = 4

y = 0.25t + 4

x + 2y + z = 17

x + 2(0.25t + 4) + t = 17

x = 9 − 1.5t

The general solution is (9 − 1.5t, 0.25t + 4, t), where t is a parameter that we needto check the constraints. Since z must be a nonnegative interger and z = t, t must bea nonnegative integer. Since 0 ≤ x = 9 − 1.5t, t ≤ 6. Since y must be a nonnegativeinteger and y = 0.25t+4, then t must be a multiple of 4. We conclude that t can only be0 or 4. This means that there are exactly two solutions, (9, 4, 0) and (3, 5, 4). So, they

12

can process 9 of type A documents and 4 of type B, or process 3 of type A, 5 of type B,and 4 of type C.


1 2 a b2 5 c d

]


[

1 2 a b2 5 c d

]

R2 − 2R1 → R2

−→

[

1 2 a b0 1 c − 2a d − 2b

]


y + (c − 2a)z = d − 2b

y + (c − 2a)(t) = d − 2b

y = (2a − c)t + d − 2b

x + 2y + az = b

x + 2[(2a − c)t + d − 2b] + a(t) = b

x = −2[(2a − c)t + d − 2b] − at + b

x = (2c − 5a)t − 2d + 5b

The general solution is ([2c− 5a]t− 2d + 5b, [2a− c]t + d− 2b, t) where t is any number.Notice that this works for any of the numbers a, b, c, and d.

61. From the flow diagram we obtain the following system of equations.

x1 + x4 = 700x1 + x2 = 600

x2 + x3 = 800x3 + x4 = 900

Use the augment matrix and obtain

1 0 0 1 7001 1 0 0 6000 1 1 0 8000 0 1 1 900

R2 − R1 → R2−→

1 0 0 1 7000 1 0 −1 −1000 1 1 0 8000 0 1 1 900

R3 − R2 → R3

13

−→

1 0 0 1 7000 1 0 −1 −1000 0 1 1 9000 0 1 1 900

R4 − R3 → R4

Set x4 = t. Then x3 = 900 − t and x2 = t − 100, and x1 = 700 − t. The generalsolution is then (700 − t, t − 100, 900 − t, t). The 4 variables must be nonnegative. Thisyields 100 ≤ t ≤ 700.

63 We wish to solve the system

xA + 2pA − pB = 6xB − pA + 3pB = 16

xA + xB − pA = −1xA + xB − pB = −1

Use the augment matrix and obtain

1 0 2 −1 60 1 −1 3 16

0.5 1 −1 0 −11 1 0 −1 −1

R3 − 0.5R1 → R3

R4 − R1 → R4

−→

1 0 2 −1 60 1 −1 3 160 1 −2 0.5 −40 1 −2 0 −7

R3 − R2 → R3

R4 − R2 → R4

−→

1 0 2 −1 60 1 −1 3 160 0 −1 −2.5 −200 0 −1 −3 −23

−R3 → R3

−→

1 0 2 −1 60 1 −1 3 160 0 1 2.5 200 0 −1 −3 −23

−R4 + R3 → R4

−→

1 0 2 −1 60 1 −1 3 160 0 1 2.5 200 0 0 −0.5 −3

−2R4 → R4

14

−→

1 0 2 −1 60 1 −1 3 160 0 1 2.5 200 0 0 1 6

We have pB = 6. Subsitute this into the third equation gives pA = 20 − 2.5(6) = 5.Substituting into the second equation gives xB = 16 + 5 − 3(6) = 3. Finally, xA =6 − 2(5) + 6 = 2. Thus, the point of equilibrium is (2, 3, 5, 6).

65. Let x be the amount of money in the maney market fund, y the amount of moneyin a bond fund, z the amount in a conservative stock fund, and u the amount in aspeculative fund. The amount in the money market is the same as in speculative fund.So x = u. The amount in in the bond fund is to be the sum of the amounts in in thetwo stock funds, so y = z + u. The investor has $100,000, so x + y + z + u = 100, 000The investor wants to yield $9000, so 0.06x + 0.08y + 0.10z + 0.13u = 9000. We wish tosolve the system

x − u = 0

y − z − u = 0

x + y + z + u = 100, 000

0.06x + 0.08y + 0.10z + 0.13u = 9000

We have

1 0 0 −1 00 1 −1 −1 01 1 1 1 100, 000

0.06 0.08 0.10 0.13 10, 000

R3 − R1 → R3

R4 − 0.06R1 → R4

−→

1 0 0 −1 00 1 −1 −1 00 1 1 2 100, 0000 0.08 0.10 0.19 10, 000

R3 − R2 → R3

R4 − 0.08R1 → R4

−→

1 0 0 −1 00 1 −1 −1 00 0 2 3 100, 0000 0 0.18 0.27 10, 000

1

2R3 → R3

−→

1 0 0 −1 00 1 −1 −1 00 0 1 1.5 50, 0000 0 0.18 0.27 10, 000

R4 − 0.18R3 → R3

15

−→

1 0 0 −1 00 1 −1 −1 00 0 1 1.5 50, 0000 0 0 0 1000

But the last equation gives 0 = 1000. Thus, there is no solution.

67. We know that our row operations will automatic place a zero or a one where wewant them. So we do not count these as a multiplication or a division.

We start with doing the problem by the suggested way.

1 3 2 275 1 3 518 4 10 128

R2 − 5R1 → R2 3 multiplicationsR3 − 8R1 → R3 3 multiplications

−→

1 3 2 270 −14 −7 −840 −20 −6 −88

−

1

14R2 → R2 2 divisions

−→

1 3 2 270 1 0.5 60 −20 −6 −88

R3 + 20R2 → R3 2 multiplications

−→

1 3 2 270 1 0.5 60 0 4 32

1

4R3 → R3 1 division

−→

1 3 2 270 1 0.5 60 0 1 8

R1 − 2R3 → R1 1 multiplicationR2 − 0.5R3 → R2 1 multiplication

−→

1 3 0 110 1 0 20 0 1 8

R1 − 3R2 → R1 1 multiplication

−→

1 0 0 50 1 0 20 0 1 8

There is a total of 11 multiplications and 3 divisions.

We now look at another way of doing this problem.

16

1 3 2 275 1 3 518 4 10 128

R2 − 5R1 → R2 3 multiplicationsR3 − 8R1 → R3 3 multiplications

−→

1 3 2 270 −14 −7 −840 −20 −6 −88

−

1


−→

1 3 2 270 1 0.5 60 −20 −6 −88

R1 − 3R2 → R1 2 multiplications

R3 + 20R2 → R3 2 multiplications

−→

1 0 0.5 90 1 0.5 60 0 4 32

1


−→

1 0 0.5 90 1 0.5 60 0 1 8

R1 − 0.5R3 → R3 1 multiplicationR2 − 0.5R3 → R2 1 multiplication

−→

1 0 0 50 1 0 20 0 1 8

Here we have 12 multiplications and 3 divisions.

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CHAPTER 1epstein/FMText/Ch1SM.pdfSince we have 8 hours per day, 100 days become 100(8) = 800 hours. Renting will cost less. Notice, this is true either by calculating the cost of rental