a coffee manufacturer uses colombian and brazilian coffee ... · resource allocation.a coffee...

432 6 Systems of Equations and Inequalities

of packages weighing 1 pound or less and a surcharge foreach additional pound (or fraction thereof). A customer isbilled $27.75 for shipping a 5-pound package and $64.50for shipping a 20-pound package. Find the base price andthe surcharge for each additional pound.

70. Delivery Charges. Refer to Problem 69. Federated Ship-ping, a competing overnight delivery service, informs thecustomer in Problem 69 that it would ship the 5-poundpackage for $29.95 and the 20-pound package for $59.20.(A) If Federated Shipping computes its cost in the same

manner as United Express, find the base price and thesurcharge for Federated Shipping.

(B) Devise a simple rule that the customer can use tochoose the cheaper of the two services for each pack-age shipped. Justify your answer.

71. Resource Allocation. A coffee manufacturer usesColombian and Brazilian coffee beans to produce two

blends, robust and mild. A pound of the robust blend re-quires 12 ounces of Colombian beans and 4 ounces ofBrazilian beans. A pound of the mild blend requires 6ounces of Colombian beans and 10 ounces of Brazilianbeans. Coffee is shipped in 132-pound burlap bags. Thecompany has 50 bags of Colombian beans and 40 bags ofBrazilian beans on hand. How many pounds of each blendshould it produce in order to use all the available beans?

72. Resource Allocation. Refer to Problem 71.(A) If the company decides to discontinue production of

the robust blend and only produce the mild blend, howmany pounds of the mild blend can it produce and howmany beans of each type will it use? Are there anybeans that are not used?

(B) Repeat part A if the company decides to discontinueproduction of the mild blend and only produce the ro-bust blend.

SECTION 6-2 Gauss––Jordan Elimination

• Reduced Matrices• Solving Systems by Gauss–Jordan Elimination• Application

Now that you have had some experience with row operations on simple augmentedmatrices, we will consider systems involving more than two variables. In addition,we will not require that a system have the same number of equations as variables. Itturns out that the results for two-variable–two-equation linear systems, stated in The-orem 1 in Section 6-1, actually hold for linear systems of any size.

In the last section we used row operations to transform the augmented coefficientmatrix for a system of two equations in two variables

a11x1 � a12x2 � k1

a21x1 � a22x2 � k2�a11

a21

a12

a22

� k1

k2�

• Reduced Matrices

Possible Solutions to a Linear System

It can be shown that any linear system must have exactly one solution, no solu-tion, or an infinite number of solutions, regardless of the number of equationsor the number of variables in the system. The terms unique, consistent, incon-sistent, dependent, and independent are used to describe these solutions, just asthey are for systems with two variables.

6-2 Gauss–Jordan Elimination 433

into one of the following simplified forms:

Form 1 Form 2 Form 3

(1)

where m, n, and p are real numbers, p � 0. Each of these reduced forms representsa system that has a different type of solution set, and no two of these forms are row-equivalent. Thus, we consider each of these to be a different simplified form. Nowwe want to consider larger systems with more variables and more equations.

EXPLORE-DISCUSS 1 Forms 1, 2, and 3 above represent systems that have, respectively, a unique solu-tion, an infinite number of solutions, and no solution. Discuss the number of solu-tions for the systems of three equations in three variables represented by the fol-lowing augmented coefficient matrices.

(A) (B) (C)

Since there is no upper limit on the number of variables or the number of equa-tions in a linear system, it is not feasible to explicitly list all possible “simplifiedforms” for larger systems, as we did for systems of two equations in two variables.Instead, we state a general definition of a simplified form called a reduced matrix thatcan be applied to all matrices and systems, regardless of size.

DEFINITION 1 Reduced Matrix

A matrix is in reduced form if:

1. Each row consisting entirely of 0’s is below any row having at least onenonzero element.

2. The leftmost nonzero element in each row is 1.

3. The column containing the leftmost 1 of a given row has 0’s above andbelow the 1.

4. The leftmost 1 in any row is to the right of the leftmost 1 in the precedingrow.

EXAMPLE 1 Reduced Forms

The matrices below are not in reduced form. Indicate which condition in the defini-tion is violated for each matrix. State the row operation(s) required to transform thematrix to reduced form, and find the reduced form.

�1

0

0

0

1

0

0

0

1

� 234��1

0

0

1

0

0

1

0

0

� 200��1

0

0

1

0

0

1

0

0

� 230�

�1

0

m

0 � np��1

0

m

0 � n0��1

0

0

1 � mn�


(A) (B)

(C) (D)

Solutions (A) Condition 4 is violated: The leftmost 1 in row 2 is not to the right of the left-most 1 in row 1. Perform the row operation R1 ↔ R2 to obtain the reduced form:

(B) Condition 3 is violated: The column containing the leftmost 1 in row 2 does nothave a zero above the 1. Perform the row operation 2R2 � R1 → R1 to obtainthe reduced form:

(C) Condition 1 is violated: The second row contains all zeros, and it is not belowany row having at least one nonzero element. Perform the row operation R2 ↔R3 to obtain the reduced form:

(D) Condition 2 is violated: The leftmost nonzero element in row 2 is not a 1. Per-form the row operation R2 → R2 to obtain the reduced form:

Matched Problem 1 The matrices below are not in reduced form. Indicate which condition in the defini-tion is violated for each matrix. State the row operation(s) required to transform thematrix to reduced form and find the reduced form.

(A) (B)

(C) (D) �1

0

0

2

0

0

0

0

1

� 304��0

1

0

1

0

0

0

0

1

� �3

0

2�

�1

0

0

5

1

0

4

2

0

� 3

�1

0��1

0

0

3 � 2

�6�

�1

0

0

0

1

0

0

0

1

� �132

�5�

12

�1

0

0

0

1

0

� �3

�2

0�

�1

0

2

0

0

1 � 1

�1�

�1

0

0

1 � 3

�2�

�1

0

0

0

2

0

0

0

1

� �1

3

�5��

1

0

0

0

0

1

� �3

0

�2�

�1

0

2

0

�2

1 � 3

�1��0

1

1

0 � �2

3�


We are now ready to outline the Gauss–Jordan elimination method for solving sys-tems of linear equations. The method systematically transforms an augmented matrixinto a reduced form. The system corresponding to a reduced augmented coefficientmatrix is called a reduced system. As we will see, reduced systems are easy to solve.

The Gauss–Jordan elimination method is named after the German mathematicianCarl Friedrich Gauss (1777–1885) and the German geodesist Wilhelm Jordan(1842–1899). Gauss, one of the greatest mathematicians of all time, used a methodof solving systems of equations that was later generalized by Jordan to solve prob-lems in large-scale surveying.

EXAMPLE 2 Solving a System Using Gauss–Jordan Elimination

Solve by Gauss–Jordan elimination: 2x1 � 2x2 � x3 � 33x1 � x2 � x3 � 7x1 � 3x2 � 2x3 � 0

Solution Write the augmented matrix and follow the steps indicated at the right to produce areduced form.

2

3

1

�2

1

�3

1

�1

2

3

7

0

R1 ↔ R3

� ��Need a 1 here.

1

3

2

�3

1

�2

2

�1

1

0

7

3

(�3)R1 � R2 → R2

(�2)R1 � R3 → R3

� ��Need 0’s here.

1

0

0

�3

10

4

2

�7

�3

0

7

3

0.1R2 → R2� ��Need a 1 here.

1

0

0

�3

1

4

2

�0.7

�3

0

0.7

3 (�4)R2 � R3 → R3

3R2 � R1 → R1� ��Need 0’s here.

1

0

0

0

1

0

�0.1

�0.7

�0.2

2.1

0.7

0.2 (�5)R3 → R3

� ��Need a 1 here.

1

0

0

0

1

0

�0.1

�0.7

1

2.1

0.7

�1

0.7R3 � R2 → R2

0.1R3 � R1 → R1� ��Need 0’s here.

Step 1: Choose the leftmost nonzero column and get a 1 at the top.

Step 2: Use multiples of the row containing the 1 from step 1 to get zeros in all remaining places in the column containing this 1.

Step 3: Repeat step 1 with the submatrix formed by (mentally) deleting the top row.

Step 3: Repeat step 1 with the submatrix formed by (mentally) deleting the top two rows.

Step 4: Repeat step 2 with the entire matrix.

Step 4: Repeat step 2 with the entire matrix.

• Solving Systems byGauss––Jordan

Elimination


x1 x2 x3 � 2

x1 x2 x3 � 0

x1 x2 x3 � �1

The solution to this system is x1 � 2, x2 � 0, x3 � �1. You should check this solu-tion in the original system.

Remarks1. Even though each matrix has a unique reduced form, the sequence of steps (algo-

rithm) presented here for transforming a matrix into a reduced form is not unique.That is, other sequences of steps (using row operations) can produce a reducedmatrix. (For example, it is possible to use row operations in such a way that com-putations involving fractions are minimized.) But we emphasize again that we arenot interested in the most efficient hand methods for transforming small matricesinto reduced forms. Our main interest is in giving you a little experience with amethod that is suitable for solving large-scale systems on a computer or graphingutility.

2. Most graphing utilities have the ability to find reduced forms, either directly orwith some programming. Figure 1 illustrates the solution of Example 2 on a graph-ing calculator that has a built-in routine for finding reduced forms. Notice that inrow 2 and column 4 of the reduced form the graphing calculator has displayed thevery small number �3.5E-13 instead of the exact value 0. This is a common occur-rence on a graphing calculator and causes no problems. Just replace any very smallnumbers displayed in scientific notation with 0.

Gauss–Jordan Elimination

Step 1. Choose the leftmost nonzero column and use appropriate row opera-tions to get a 1 at the top.

Step 2. Use multiples of the row containing the 1 from step 1 to get zeros inall remaining places in the column containing this 1.

Step 3. Repeat step 1 with the submatrix formed by (mentally) deleting therow used in step 2 and all rows above this row.

Step 4. Repeat step 2 with the entire matrix, including the mentally deletedrows. Continue this process until it is impossible to go further.

[Note: If at any point in this process we obtain a row with all zeros to the leftof the vertical line and a nonzero number to the right, we can stop, since wewill have a contradiction: 0 � n, n � 0. We can then conclude that the systemhas no solution.]

1

0

0

0

1

0

0

0

1

2

0

�1� �� The matrix is now in

reduced form, and we can proceed to solve the corresponding reduced system.


Matched Problem 2 Solve by Gauss–Jordan elimination: 3x1 � x2 � 2x3 � 2x1 � 2x2 � x3 � 3

2x1 � x2 � 3x3 � 3


Solve by Gauss–Jordan elimination:

Solution

The system is inconsistent and has no solution.

Matched Problem 3 Solve by Gauss–Jordan elimination:

�2x1 � 4x2 � x3 � 11 4x1 � 8x2 � 3x3 � 4 2x1 � 4x2 � x3 � �8

0.5R1 → R1

1

0

0

�2

0

0

0.5

5

�2

0.2R2 → R2�

2

4

�2

�4

�8

4

1

7

�3

�4

2

5� ��

(�4)R1 � R2 → R2

2R1 � R3 → R3

1

4

�2

�2

�8

4

0.5

7

�3

�2

2

5� ��

�2

10

1��

1

0

0

�2

0

0

0.5

1

�2

(�0.5)R2 � R1 → R1

2R2 � R3 → R3

��2

2

1��

1

0

0

�2

0

0

0

1

0�

�3

2

5��

Note that column 3 is the leftmost nonzero column in this submatrix.

We stop the Gauss–Jordan elimination, even though the matrix is not in reduced form, since the last row produces a contradiction.

�2x1 � 4x2 � 3x3 � �5 4x1 � 8x2 � 7x3 � �2 2x1 � 4x2 � x3 � �4

FIGURE 1 Gauss–Jordan elimina-tion on a graphing calculator.


CAUTION



Solution

Note that the leftmost variable in each equation appears in one and only one equa-

1

0

0

2

0

1

�3

0

�2

R2 ↔ R3�

3

2

�2

6

4

�3

�9

�6

4

15

10

�6� ��

(�2)R1 � R2 → R2

2R1 � R3 → R3

1

2

�2

2

4

�3

�3

�6

4

5

10

�6� ��

5

0

4��

1

0

0

2

1

0

�3

�2

0

(�2)R2 � R1 → R1�5

4

0��

1

0

0

x1 � x3 � �3

x2 � 2x3 � 4

0

1

0

1

�2

0�

�3

4

0��

Note that we must interchange rows 2 and 3 to obtain a nonzero entry at the top of the second column of this submatrix.

This matrix is now in reduced form. Write the corresponding reduced system and solve.

We discard the equation corresponding to the third (all 0) row in the reduced form, since it is satisfied by all values of x1, x2, and x3.

R1 → R113

�2x1 � 3x2 � 4x3 � �6 2x1 � 4x2 � 6x3 � 10 3x1 � 6x2 � 9x3 � 15

FIGURE 2 Recognizing contradic-tions on a graphing calculator.

Figure 2 shows the solution to Example 3 on a graphing calculator with abuilt-in reduced form routine. Notice that the graphing calculator does notstop when a contradiction first occurs, as we did in the solution to Example3, but continues on to find the reduced form. Nevertheless, the last row in thereduced form still produces a contradiction, indicating that the system has nosolution.


tion. We solve for the leftmost variables x1 and x2 in terms of the remaining vari-able x3:

This dependent system has an infinite number of solutions. We will use a parameterto represent all the solutions. If we let x3 � t, then for any real number t,

x1 � �t � 3

x2 � 2t � 4

x3 � t

You should check that (�t � 3, 2t � 4, t) is a solution of the original system for anyreal number t. Some particular solutions are

t � 0 t � �2 t � 3.5

(�3, 4, 0) (�1, 0, �2) (�6.5, 11, 3.5)

Matched Problem 4 Solve by Gauss–Jordan elimination:

In general,

If the number of leftmost 1’s in a reduced augmented coefficient matrixis less than the number of variables in the system and there are nocontradictions, then the system is dependent and has infinitely manysolutions.

There are many different ways to use the reduced augmented coefficient matrixto describe the infinite number of solutions of a dependent system. We will alwaysproceed as follows: Solve each equation in a reduced system for its leftmost variableand then introduce a different parameter for each remaining variable. As the solutionto Example 4 illustrates, this method produces a concise and useful representation ofthe solutions to a dependent system. Example 5 illustrates a dependent system wheretwo parameters are required to describe the solution.

EXPLORE-DISCUSS 2 Explain why the definition of reduced form ensures that each leftmost variable ina reduced system appears in one and only one equation and no equation containsmore than one leftmost variable. Discuss methods for determining if a consistentsystem is independent or dependent by examining the reduced form.

�2x1 � 3x2 � x3 � 7 3x1 � 3x2 � 6x3 � �3 2x1 � 2x2 � 4x3 � �2

x2 � 2x3 � 4

x1 � �x3 � 3




Solution

(�2)R1 � R2 → R2

(�1)R1 � R3 → R3

R2 ↔ R3

(�2)R2 � R1 → R1

(�3)R3 � R1 → R1

R3 � R2 → R2

Matrix is in reduced form.

x1 x0 � 2x3 x0 � 3x5 � �7

x0 x2 � 3x3 x0 � 2x5 � �3

x0 x0 � 0x0 x4 � 2x5 � �0

Solve for the leftmost variables x1, x2, and x4 in terms of the remaining variables x3

and x5:

x1 � �2x3 � 3x5 � 7

x2 � �3x3 � 2x5 � 3

x4 � �2x5

If we let x3 � s and x5 � t, then for any real numbers s and t,

x1 � 2s � 3t � 7

x2 � �3s � 2t � 3

x3 � s

x4 � 2t

x5 � t

is a solution. The check is left for you to perform.

�1

0

0

0

1

0

�2

3

0

0

0

1

�3

2

�2

� 7

�3

0�

�1

0

0

0

1

0

�2

3

0

3

�1

1

�9

4

�2

� 7

�3

0�

�1

0

0

2

1

0

4

3

0

1

�1

1

�1

4

�2

� 1

�3

0�

�1

0

0

2

0

1

4

0

3

1

1

�1

�1

�2

4

� 1

0

�3�

�1

2

1

2

4

3

4

8

7

1

3

0

�1

�4

3

� 1

2

�2�

x1 � 3x2 � 7x3 � 3x5 � �2 2x1 � 4x2 � 8x3 � 3x4 � 4x5 � 2

x1 � 2x2 � 4x3 � x4 � x5 � 1


Matched Problem 5 Solve by Gauss–Jordan elimination: �0x1 � 0x2 � 2x3 � x0 � 2x5 � �3�2x1 � 2x2 � 4x3 � x4 � 0x5 � �5�3x1 � 3x2 � 7x3 � x4 � 4x5 � �6

We now consider an application that involves a dependent system of equations.

EXAMPLE 6 Purchasing

A chemical manufacturer wants to purchase a fleet of 24 railroad tank cars with acombined carrying capacity of 250,000 gallons. Tank cars with three different carry-ing capacities are available: 6,000 gallons, 8,000 gallons, and 18,000 gallons. Howmany of each type of tank car should be purchased?

Solution Let

x1 � Number of 6,000-gallon tank cars



Then

0,000x1 � 0,000x2 � 00,000x3 � 000,024 Total number of tank cars

6,000x1 � 8,000x2 � 18,000x3 � 250,000 Total carrying capacity

Now we can form the augmented matrix of the system and solve by using Gauss–Jordan elimination:

R2 → R2 (simplify R2)

(�6)R1 � R2 → R2

R2 → R2

(�1) R2 � R1 → R1

Matrix is in reduced form

x1 x2 � 5x3 � �29 or x1 � �5x3 � 29

x1 x2 � 6x3 � �53 or x2 � �6x3 � 53

�1

0

0

1

�5

6 � �29

53�

�1

0

1

1

1

6 � 24

53�

12

�1

0

1

2

1

12 � 24

106�

�1

6

1

8

1

18 � 24

250�

11,000

� 1

6,000

1

8,000

1

18,000 � 24

250,000�

• Application


Let x3 � t. Then for t any real number,

x1 � 5t � 29

x2 � �6t � 53

x3 � t

is a solution—or is it? Since the variables in this system represent the number of tankcars purchased, the values of x1, x2, and x3 must be nonnegative integers. Thus, thethird equation requires that t must be a nonnegative integer. The first equation requiresthat 5t � 29 � 0, so t must be at least 6. The middle equation requires that �6t �53 � 0, so t can be no larger than 8. Thus, 6, 7, and 8 are the only possible valuesfor t. There are only three possible combinations that meet the company’s specifica-tions of 24 tank cars with a total carrying capacity of 250,000 gallons, as shown inTable 1:

6,000-gallon 8,000-gallon 18,000-gallontank cars tank cars tank cars

t x1 x2 x3

6 01 17 6

7 06 11 7

8 11 05 8

The final choice would probably be influenced by other factors. For example, thecompany might want to minimize the cost of the 24 tank cars.

Matched Problem 6 A commuter airline wants to purchase a fleet of 30 airplanes with a combined carry-ing capacity of 960 passengers. The three available types of planes carry 18, 24, and42 passengers, respectively. How many of each type of plane should be purchased?

Answers to Matched Problems

1. (A) Condition 2 is violated: The 3 in row 2 and column 2 should be a 1. Perform the operation R2 → R2 to obtain:

(B) Condition 3 is violated: The 5 in row 1 and column 2 should be a 0. Perform the operation(�5)R2 � R1 → R1 to obtain:

�1

0

0

0

1

0

�6

2

0

� 8

�1

0�

�1

0

0

1 � 2

�2�

13

TABLE 1


(C) Condition 4 is violated: The leftmost 1 in the second row is not to the right of the leftmost 1 inthe first row. Perform the operation R1 ↔ R2 to obtain:

(D) Condition 1 is violated: The all-zero second row should be at the bottom. Perform the operationR2 ↔ R3 to obtain:

2. x1 � 1, x2 � �1, x3 � 03. Inconsistent; no solution4. x1 � 5t � 4, x2 � 3t � 5, x3 � t, t any real number5. x1 � s � 7, x2 � s, x3 � t � 2, x4 � �3t � 1, x5 � t, s and t any real numbers6.

18-passenger 24-passenger 42-passengerplanes planes planes

t x1 x2 x3

14 02 14 14

15 05 10 15

16 08 06 16

17 11 02 17

�1

0

0

2

0

0

0

1

0

� 340��1

0

0

0

1

0

0

0

1

� 0

�3

2�

EXERCISE 6-2AIn Problems 1–8, indicate whether each matrix is in reducedform.

1. 2.

3. 4.

5. 6.

7. 8.

In Problems 9–16, write the linear system corresponding toeach reduced augmented matrix and solve.

9. 10. �1

0

0

0

0

1

0

0

0

0

1

0

0

0

0

1

� �2

0

1

3��

1

0

0

0

1

0

0

0

1

� �2

3

0�

�0

0

0

0

1

0 � 0

0��0

0

1

0

6

0

0

1 � �8

1�

�1

0

0

�2

0

0

4

1

0

� 1

�3

0��

0

0

1

0

1

0

1

0

0

� 2

�5

4�

�1

0

0

�1

0

0

4

0

0

� 001��0

0

0

1

0

0

�2

0

0

� 0

1

0�

�1

0

0

1 � 5

�3��1

0

0

2 � �1

6�

11. 12.

13. 14.

15. 16.

BUse row operations to change each matrix in Problems17–22 to reduced form.

17. 18.

19. 20. �1

0

0

0

1

0

4

�3

�2

� 0

�1

2��

1

0

0

0

1

0

�3

2

3

� 1

0

�6�

�1

0

3

2 � 1

�4��1

0

2

1 � �1

3�

�1

0

0

1

�2

�1

3

2 � 4

�1��1

0

�2

0

0

1

�3

3 � �5

2�

�1

0

0

0

1

0

� 5

�3

0��

1

0

0

0

1

0

� 0

0

1�

�1

0

0

�2

0

0

0

1

0

� �3

5

0��

1

0

0

0

1

0

�2

1

0

� 3

�5

0�


21. 22.

Solve Problems 23–42 using Gauss–Jordan elimination.

23. 2x1 � 4x2 � 10x3 � �2 24. 3x1 � 5x2 � x3 � �73x1 � 9x2 � 21x3 � 0 x1 � x2 � x3 � �1x1 � 5x2 � 12x3 � 1 2x1 � x0 � 11x3 � 7

25. 3x1 � 8x2 � x3 � �18 26. 2x1 � 7x2 � 15x3 � �122x1 � x2 � 5x3 � 8 4x1 � 7x2 � 13x3 � �102x1 � 4x2 � 2x3 � �4 3x1 � 6x2 � 12x3 � �9

27. 2x1 � x2 � 3x3 � 8 28. 2x1 � 4x2 � 6x3 � 10x1 � 2x2 � 0x0 � 7 3x1 � 3x2 � 3x3 � 6

29. 2x1 � x2 � 0 30. 2x1 � x2 � 03x1 � 2x2 � 7 3x1 � 2x2 � 7x1 � x2 � �1 x1 � x2 � �2

31. 3x1 � 4x2 � x3 � 1 32. 3x1 � 7x2 � x3 � 112x1 � 3x2 � x3 � 1 x1 � 2x2 � x3 � 3x1 � 2x2 � 3x3 � 2 2x1 � 4x2 � 2x3 � 10

33. �2x1 � x2 � 3x3 � �7 34. �2x1 � 5x2 � 4x3 � �7� x1 � 4x2 � 2x3 � 0 �4x1 � 5x2 � 2x3 � 9� x1 � 3x2 � x3 � 1 �2x1 � x2 � 4x3 � 3

35. �2x1 � 2x2 � 4x3 � �2 36. �2x1 � 8x2 � 6x3 � 4�3x1 � 3x2 � 6x3 � 3 �3x1 � 12x2 � 9x3 � �6

37. �4x1 � x2 � 2x3 � 3�4x1 � x2 � 3x3 � �10�8x1 � 2x2 � 9x3 � �1

38. �4x1 � 2x2 � 2x3 � 5�6x1 � 3x2 � 3x3 � �210x1 � 5x2 � 9x3 � 4

39. �2x1 � 5x2 � 3x3 � 7�4x1 � 10x2 � 2x3 � 6�6x1 � 15x2 � x3 � �19

40. �4x1 � 8x2 � 10x3 � �6�6x1 � 12x2 � 15x3 � 9�8x1 � 14x2 � 19x3 � �8

41. 5x1 � 3x2 � 2x3 � 13 42. 4x1 � 2x2 � 3x3 � 32x1 � x2 � 3x3 � 1 3x1 � x2 � 2x3 � �104x1 � 2x2 � 4x3 � 12 2x1 � 4x2 � x3 � �1

43. Consider a consistent system of three linear equations inthree variables. Discuss the nature of the solution set for thesystem if the reduced form of the augmented coefficientmatrix has(A) One leftmost 1(B) Two leftmost 1’s(C) Three leftmost 1’s(D) Four leftmost 1’s

�0

2

0

�2

�2

�1

8

6

4

� 1

�412��

1

0

0

2

3

�1

�2

�6

2

� �1

1

�13�

44. Consider a system of three linear equations in three vari-ables. Give examples of two reduced forms that are not rowequivalent if the system is(A) Consistent and dependent(B) Inconsistent

CSolve Problems 45–50 using Gauss–Jordan elimination.

45. x1 � 2x2 � 4x3 � x4 � 72x1 � 5x2 � 9x3 � 4x4 � 16x1 � 5x2 � 7x3 � 7x4 � 13

46. 2x1 � 4x2 � 5x3 � 4x4 � 8x1 � 2x2 � 2x3 � x4 � 3

47. � x1 � x2 � 3x3 � 2x4 � 1�2x1 � 4x2 � 3x3 � x4 � 0.5�3x1 � x2 � 10x3 � 4x4 � 2.9�4x1 � 3x2 � 8x3 � 2x4 � 0.6

48. x1 � x2 � 4x3 � x4 � 1.3�x1 � x2 � x3 � 0x0 � 1.12x1 � x0 � 0x3 � 3x4 � �4.42x1 � 5x2 � 11x3 � 3x4 � 5.6

49. �0x1 � 2x2 � x3 � x4 � 2x5 � 2�2x1 � 4x2 � 2x3 � 2x4 � 2x5 � 0�3x1 � 6x2 � x3 � x4 � 5x5 � 40�x1 � 2x2 � 3x3 � x4 � x5 � 3

50. x1 � 3x2 � x3 � x4 � 2x5 � 2�x1 � 5x2 � 2x3 � 2x4 � 2x5 � 02x1 � 6x2 � 2x3 � 2x4 � 4x5 � 4

�x1 � 3x2 � x3 � 0x0 � 0x5 � �3

APPLICATIONS

Solve Problems 51–72 using Gauss–Jordan elimination.

51. Puzzle. A friend of yours came out of the post office afterspending $14.00 on 15¢, 20¢, and 35¢ stamps. If shebought 45 stamps in all, how many of each type did shebuy?

52. Puzzle. A parking meter accepts only nickels, dimes, andquarters. If the meter contains 32 coins with a total value of$6.80, how many of each type are there?

53. Chemistry. A chemist can purchase a 10% saline solutionin 500 cubic centimeter containers, a 20% saline solution in500 cubic centimeter containers, and a 50% saline solutionin 1,000 cubic centimeter containers. He needs 12,000 cu-bic centimeters of 30% saline solution. How many contain-ers of each type of solution should he purchase in order toform this solution?

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54. Chemistry. Repeat Problem 53 if the 50% saline solution isavailable only in 1,500 cubic centimeter containers.

55. Geometry. Find a, b, and c so that the graph of the parabolawith equation y � a � bx � cx2 passes through the points(�2, 3), (�1, 2), and (1, 6).

56. Geometry. Find a, b, and c so that the graph of the parabolawith equation y � a � bx � cx2 passes through the points(1, 3), (2, 2), and (3, 5).

57. Geometry. Find a, b, and c so that the graph of the circlewith equation x2 � y2 � ax � by � c � 0 passes throughthe points (6, 2), (4, 6), and (�3, �1).

58. Geometry. Find a, b, and c so that the graph of the circlewith equation x2 � y2 � ax � by � c � 0 passes throughthe points (�4, 1), (�1, 2), and (3, �6).

59. Production Scheduling. A small manufacturing plantmakes three types of inflatable boats: one-person, two-person, and four-person models. Each boat requires the ser-vices of three departments, as listed in the table. The cut-ting, assembly, and packaging departments have available amaximum of 380, 330, and 120 labor-hours per week, re-spectively. How many boats of each type must be producedeach week for the plant to operate at full capacity?

One- Two- Four-person person personboat boat boat

Cuttingdepartment 0.5 h 1.0 h 1.5 h

Assemblydepartment 0.6 h 0.9 h 1.2 h

Packagingdepartment 0.2 h 0.3 h 0.5 h

60. Production Scheduling. Repeat Problem 59 assuming thecutting, assembly, and packaging departments have avail-able a maximum of 350, 330, and 115 labor-hours perweek, respectively.

61. Production Scheduling. Rework Problem 59 assuming thepackaging department is no longer used.

62. Production Scheduling. Rework Problem 60 assuming thepackaging department is no longer used.

63. Production Scheduling. Rework Problem 59 assuming thefour-person boat is no longer produced.

64. Production Scheduling. Rework Problem 60 assuming thefour-person boat is no longer produced.

65. Nutrition. A dietitian in a hospital is to arrange a specialdiet using three basic foods. The diet is to include exactly340 units of calcium, 180 units of iron, and 220 units of vi-tamin A. The number of units per ounce of each special in-gredient for each of the foods is indicated in the table. Howmany ounces of each food must be used to meet the dietrequirements?

Units per ounce

Food A Food B Food C

Calcium 30 10 20

Iron 10 10 20

Vitamin A 10 30 20

66. Nutrition. Repeat Problem 65 if the diet is to include ex-actly 400 units of calcium, 160 units of iron, and 240 unitsof vitamin A.

67. Nutrition. Solve Problem 65 with the assumption that foodC is no longer available.

68. Nutrition. Solve Problem 66 with the assumption that foodC is no longer available.

69. Nutrition. Solve Problem 65 assuming the vitamin A re-quirement is deleted.

70. Nutrition. Solve Problem 66 assuming the vitamin A re-quirement is deleted.

71. Sociology. Two sociologists have grant money to studyschool busing in a particular city. They wish to conduct anopinion survey using 600 telephone contacts and 400 housecontacts. Survey company A has personnel to do 30 tele-phone and 10 house contacts per hour; survey company Bcan handle 20 telephone and 20 house contacts per hour.How many hours should be scheduled for each firm to pro-duce exactly the number of contacts needed?

72. Sociology. Repeat Problem 71 if 650 telephone contactsand 350 house contacts are needed.

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a coffee manufacturer uses colombian and brazilian coffee ... · resource allocation.a coffee...

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