Finite Mathematics
Chapter 6
Name ________________________________ Date ______________ Class ____________
Section 6-1 The Table Method: An Introduction
to the Simplex Method
Goal: To solve problems using the simplex method
Procedure: Simplex method to solve linear programming problems
Step 1: Find the feasible region by graphing.
Step 2: Find all intersection points, whether the points are in the feasible region or not.
Step 3: Using slack variables, convert the system into a system of equations.
Step 4: Fill in a table giving the basic solutions and intersection points (from step 2) and
stating whether the point is in the feasible region.
Step 5: Find the maximum value of P for point in the feasible region.
1. Consider the following linear programming problem:
Maximize: P = 12 x1 + 10 x2
subject to: 10 x1 + 6 x2 £ 30
x1 + x2 £ 4
x1 ³ 0, x2 ³ 0
a) Is the linear programming problem in standard form?
Yes, both values on the right side of the equations are positive, and both x values have
nonnegative constraints.
b) Find the feasible region by graphing.
The feasible region is shaded above.
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Chapter 6
c) Find all intersection points of lines in the graph, and state whether the points are in the
feasible region.
Based on the graph in part b, the points of intersection are (0, 0), (0, 4), (0, 5), (3, 0),
(4, 0), and ( 32 , 52 ). Only (0, 0), (0, 4), (3, 0), and ( 32 , 52 ) are in the feasible region.
d) Using slack variables, convert the system of inequalities into a system of equations.
= 30
ì 10 x1 + 6 x2 + s1
í
+ s2 = 4
î x1 + x2
e) Fill in the following table:
Basic Solutions
x1
0
0
0
3
4
x2
0
4
5
0
0
3
2
5
2
s1
30
6
0
0
–10
0
s2
4
0
–1
1
0
0
Intersection
Point
(from part c)
Is point in
Feasible
Region?
(0, 0)
(0, 4)
(0, 5)
(3, 0)
(4, 0)
( 32 , 52 )
Yes
Yes
No
Yes
No
Yes
f) Find the maximum value of P (evaluate P only at points that are in the feasible region).
Point of Intersection
(0, 0)
(0, 4)
(3, 0)
P = 12 x1 + 10 x2
P = 12(0) + 10(0) = 0
P = 12(0) + 10(4) = 40
P = 12(3) + 10(0) = 36
( 32 , 52 )
P = 12( 32 ) + 10( 52 ) = 43
Therefore, P has a maximum value of 43 at the point ( 32 , 52 ).
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Chapter 6
2. Consider the following linear programming problem:
Maximize: P = 80 x1 + 30 x2
subject to: x1 + x2 £ 6
3x1 + x2 £ 9
x1 ³ 0, x2 ³ 0
a) Is the linear programming problem in standard form?
Yes, both values on the right side of the equations are positive and both x values have
nonnegative constraints.
b) Find the feasible region by graphing.
The feasible region is shaded above.
c) Find all intersection points of lines in the graph, and state whether the points are in the
feasible region.
Based on the graph in part b, the points of intersection are (0, 0), (0, 6), (0, 9), (3, 0),
(6, 0), and ( 32 , 92 ). Only (0, 0), (0, 6), (3, 0), and ( 32 , 92 ) are in the feasible region.
d) Using slack variables, convert the system of inequalities into a system of equations.
=6
ì x1 + x2 + s1
í
+ s2 = 9
î 3 x1 + x2
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Chapter 6
e) Fill in the following table:
Basic Solutions
x1
0
0
0
3
6
x2
0
6
9
0
0
3
2
9
2
s1
6
0
–3
3
0
0
Intersection
Point
(from part c)
Is point in
Feasible
Region?
(0, 0)
(0, 6)
(0, 9)
(3, 0)
(6, 0)
( 32 , 92 )
Yes
Yes
No
Yes
No
Yes
s2
9
3
0
0
–9
0
f) Find the maximum value of P (evaluate P only at points that are in the feasible region).
Point of Intersection
(0, 0)
(0, 6)
(3, 0)
P = 80 x1 + 30 x2
P = 80(0) + 30(0) = 0
P = 80(0) + 30(6) = 180
P = 80(3) + 30(0) = 240
( 32 , 92 )
P = 80( 32 ) + 30( 92 ) = 255
Therefore, P has a maximum value of 255 at the point ( 32 , 92 ).
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Copyright © 2015 Pearson Education, Inc.
Finite Mathematics
Chapter 6
Name ________________________________ Date ______________ Class ____________
Section 6-2 The Simplex Method: Maximization
with Problem Constraints of the Form ≤
Goal: To solve maximum problems using the simplex method
Procedure:
Selecting the Pivot Element
Step 1 - Locate the most negative indicator in the bottom row of the tableau to the left
of the P column (the negative number with the largest absolute value). The column
containing this element is the pivot column.
Step 2 - Divide each positive element in the pivot column above the line into the
corresponding element in the last column. The pivot row is the row corresponding to the
smallest quotient obtained.
Step 3 - The pivot element is the element in the intersection of the pivot column and
the pivot row. (Note that the pivot element is always positive and never appears in the bottom
row.
Procedure:
Performing a Pivot Operation
A pivot operation, or pivoting, consists of performing row operations as follows:
Step 1 - Multiply the pivot row by the reciprocal of the pivot element to transform the
pivot element into a 1.
Step 2 - Add multiples of the pivot row to other rows in the tableau to transform all
other nonzero elements in the pivot column into 0’s.
1. Consider the following linear programming problem:
Maximize: P = 12 x1 + 10 x2
subject to: 10 x1 + 6 x2 £ 30
x1 + x2 £ 4
x1 ³ 0, x2 ³ 0
a. Using slack variables, convert the system of inequalities into a system of
equations.
b. Write the initial simplex tableau for the linear programming problem.
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Chapter 6
c. Choose the first Pivot element and “clear” the pivot column using this pivot
element.
d. Is there another pivot element? If so, continue until there are no more pivot
elements.
e. State the solution to the linear programming problem in sentence form.
Solution:
a.
= 30
ì 10 x1 + 6 x2 + s1
ï
í x1 + x2 + s2 = 4
ï
x1 ³ 0, x2 ³ 0
î
- 12 x1 - 10 x2 + P = 0
b - d.
é 10
6 1 0 0 30ù
ê
ú
1 0 1 0 4ú
ê 1
ê–12 –10 0 0 1 0ú
ë
û
3
1 0 0 3ù
é 1
5
10
ê
ú
1 0 1 0 4ú
ê 1
ê–12 –10 0 0 1 0ú
ú
ëê
û
é1
3
5
ê
ê0
2
5
ê
ê0 – 14
5
êë
-
1
10
1
10
6
5
é1
3
1
5
10
ê
ê0
1 - 14
ê
6
ê0 – 14
5
5
êë
é1 0 - 1
10
ê
ê0 1 - 1
4
ê
1
ê0 0
2
ëê
e.
3ù
ú
1 0 1ú
ú
0 1 36ú
ú
û
0 0
3ù
ú
5 0
5ú
2
2ú
0 1 36ú
ú
û
0 0
-
3
5
5
2
7
0
0
3ù
2ú
5ú
2ú
1 43ú
ú
û
The function P has a maximum value of 430 at the point ( 32 , 52 ).
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Chapter 6
2. Consider the following linear programming problem:
Maximize: P = 80 x1 + 30 x2
subject to: x1 + x2 £ 6
3x1 + x2 £ 9
x1 ³ 0, x2 ³ 0
a. Using slack variables, convert the system of inequalities into a system of equations.
b. Write the initial simplex tableau for the linear programming problem.
c. Choose the first Pivot element and “clear” the pivot column using this pivot
element.
d. Is there another pivot element? If so, continue until there are no more pivot
elements.
e. State the solution to the linear programming problem in sentence form.
Solution:
a.
=6
ì x1 + x2 + s1
ï
í 3 x1 + x2 + s2 = 9
ï x ³ 0, x ³ 0
î
1
2
- 80 x1 - 30 x2 + P = 0
b - d.
é 1
1 1 0 0 6ù
ê
ú
1 0 1 0 9ú
ê 3
ê–80 –30 0 0 1 0ú
ë
û
é 1
1 1 0 0 6ù
ê
ú
1 0 1 0 3
ê 1
ú
3
3
ê
ú
êë–80 –30 0 0 1 0ú
û
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Chapter 6
é0
2 1 - 1 0
3ù
3
3
ê
ú
ê1
1 0
1 0
3ú
3
3
ê
ú
ê0 – 10 0 80 1 240 ú
3
3
êë
ú
û
é0
9ù
1 32 - 12 0
2ú
ê
ê1
1 0
1 0
3ú
3
3
ê
ú
ê0 – 10 0 80 1 240ú
3
3
ú
ëê
û
3 - 1 0
9ù
é0 1
2
2
2ú
ê
3ú
1 0
ê1 0 - 1
2
2
2
ê
ú
5 25 1 255ú
ê0 0
ë
û
e.
The function P has a maximum value of 255 at the point ( 32 , 92 ).
For problems 3 and 4, construct a mathematical model in the form of a linear programming
problem. Then solve the problem using the simplex method.
3. A dietician needs to plan a diet for a patient using two different food combinations to meet
the patient’s needs. Each container of Food A contains 2 units of Additive #1 and 2 units of
Additive #2. Each container of Food B contains 2 units of Additive #1 and 4 units of
Additive #2. The patient needs at most 10 units of Additive #1 and at most 12 units of
Additive #2. If each container of Food A has 12 units of calcium and Food B has 8 units of
calcium, how many containers of each food should be used to maximize the amount of
calcium?
Let x1 represent Food A and x2 represent Food B. Based on the above information,
ì 2 x1 + 2 x2 £ 10
ï
we will try to maximize P = 12 x1 + 8x2 subject to í 2 x1 + 4 x2 £ 12.
ï x ,x ³ 0
î
1 2
Adding the slack variables, we will have the following equations:
= 10
ì 2 x1 + 2 x2 + s1
ï
í 2 x1 + 4 x2 + s2 = 12
ï
x1 ³ 0, x2 ³ 0
î
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Finite Mathematics
Chapter 6
- 12 x1 - 8x2 + P = 0
Solving the initial simplex tableau:
é 2 2 1 0 0 10ù
ê
ú
ê 2 4 0 1 0 12ú
ê–12 –8 0 0 1 0ú
ë
û
1
é1 1
2
ê
1
ê0 1 2
ê
6
êë0 4
5ù
ú
1 0
ú
1
2
ú
0 1 60 ú
û
0 0
The bottom row has all positive values, therefore the solution is a maximum of 60
units of calcium when 5 containers of Food A are used and 1 container of Food B is used.
4. Pharmer Phil has 640 acres to plant in corn and soybeans. Each acre of corn requires 45
labor-hours and each acre costs $100 in seed and fertilizer. Each acre of soybeans requires
60 labor hours and each acre costs $80 in seed and fertilizer. Phil estimates that he has
36,000 hours of labor and $60,000 capital to spend. Pharmer Phil estimates that each acre
planted in corn will yield a profit of $120 and each acre planted in soybeans will yield a
profit of $100. Under these conditions, how many acres of corn and how many acres of
soybeans should Pharmer Phil plant to maximize his profit?
Let x1 represent acres of corn and x2 represent acres of soybeans. Based on the above
x1 + x2 £ 640
ì
ï 45 x + 60 x £ 36, 000
ï
1
2
.
information, we will try to maximize P = 120 x1 + 100 x2 subject to í
ï 100 x1 + 80 x2 £ 60, 000
ïî
x1, x2 ³ 0
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Copyright © 2015 Pearson Education, Inc.
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Chapter 6
Adding the slack variables, we will have the following equations:
x1 + x2 + s1
= 640
ì
ï 45 x + 60 x + s
= 36, 000
ï
1
2
2
í
+ s3 = 60, 000
ï 100 x1 + 80 x2
ïî
x1, x2 ³ 0
Solving the initial simplex tableau:
1
1
é
ê 45
60
ê
ê 100
80
ê
ëê–120 –100
640ù
0 1 0 0 36, 000ú
ú
0 0 1 0 60, 000ú
ú
0 0 0 1
0û
ú
1 0 0 0
1
1
é1
ê0
15 - 45
ê
ê0 - 20 - 100
ê
êë0 20 120
1
é1 1
ê0 15 - 45
ê
ê0
1
5
ê
êë0 20 120
é1
ê
ê0
ê
ê0
ê
êë0
640ù
1 0 0
7200ú
ú
0 1 0 - 4000 ú
ú
0 0 1 76,800ú
û
0 0 0
640ù
1
0 0
7200ú
ú
1
0 - 20
0
200ú
ú
0
0 1 76,800ú
û
0
0 0
440ù
ú
0 - 120 1
0
4200ú
ú
1
5 0 0
200ú
ú
0
20 0
1 1 72,800 ú
û
0
-4 0
1
20
3
4
1
20
0
Therefore the solution is a maximum profit of $72,800 when 440 acres of corn is
planted and 200 acres of soybeans.
6-10
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Finite Mathematics
Chapter 6
Name ________________________________ Date ______________ Class ____________
Section 6-3 The Dual Problems: Minimization
with Problem Constraints of the Form ≥
Goal: To solve minimization problems that are in the form of a dual problem
Formation of the Dual Problem:
Given a minimization problem with > problem constraints,
Step 1.
Use the coefficients and constants in the problem constraints and the
objective function to form a matrix A with coefficients of the objective
function in the last row.
Step 2. Interchange the rows and columns of matrix A to form the matrix AT,
the transpose of A.
Step 3. Use the rows of AT to form a maximization problem with < problem
constraints.
In Problems 1–6 , find the transpose of each matrix.
é4 ù
ê7 ú
ê ú
1. ê2 ú
ê ú
ê1 ú
êë8 úû
2. [4 2 1 7 3]
AT = [4 7 2 1 8]
é4ù
ê2ú
ê ú
AT = ê1 ú
ê ú
ê7 ú
êë3 úû
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Chapter 6
é2 5 ù
3. ê7 1 ú
ê
ú
êë3 4 úû
é2 7 3 ù
AT = ê
ú
ë5 1 4û
é3 1 7 ù
4. ê
ú
ë5 8 2û
é3 5 ù
A = ê1 8 ú
ê
ú
êë7 2úû
é1 2 ù
ê3 4 ú
ê
ú
5. ê5 6 ú
ê
ú
ê7 8 ú
êë9 10úû
é1 3 5 7 9 ù
AT = ê
ú
ë2 4 6 8 10û
é 1 5 7 9 11ù
6. ê 2 4 6 8 10ú
ê
ú
ëê13 23 14 12 3 ú
û
T
é1
ê5
ê
AT = ê7
ê
ê9
êë11
13 ù
4 23ú
ú
6 14 ú
ú
8 12 ú
10 3 ú
û
2
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Finite Mathematics
Chapter 6
In problems 7–10:
a) Form the matrix A, using the coefficients and constants in
the problem constraints and objective function.
b) Find the dimensions of A, then find the dimensions of AT.
c) Find AT.
d) State the dual problem.
e) Use the simplex method to solve the dual problem.
f) Read the solution of the minimization problem from the
bottom row of the final simplex tableau.
7. Minimize: C = 5x1 + 6 x2
subject to:
8 x1 + 4 x2 ³ 16
x1 ³ 0
x2 ³ 0
a.
b.
c.
d.
é8 4 16ù
A=ê
ú
ë5 6 1 û
The matrix A is a 2 ´ 3 and the transpose is a 3 ´ 2.
é8 5 ù
T
A = ê4 6ú
ê
ú
êë16 1 ú
û
Based on the above matrix, we want to maximize P = 16 y1
subject to 8 y1 £ 5
4 y1 £ 6
y1 ³ 0
e.
f.
é 8 1
ê
ê 4 0
ê–16 0
ë
0 0 5ù
ú
1 0 6ú ®
0 1 0ú
û
1
é1
8
ê
1
ê0 2
ê
2
ê0
ë
0 0
1 0
0
é 1
ê
ê 4
ê–16
êë
1
8
0
0
0 0 85 ù
ú
1 0 6ú ®
0 1 0ú
ú
û
5ù
8ú
9ú
2
ú
1 10ú
û
The function has a minimum value of 10 at the point (2, 0).
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Chapter 6
8. Minimize: C = 20 x1 + 40 x2
subject to: 2 x1 - x2 ³ 5
x1 + x2 ³ 7
x1, x2 ³ 0
a.
é2 - 1 5 ù
A = ê1 1
7ú
ê
ú
êë20 40 1 ú
û
b.
The matrix A is a 3 ´ 3 and the transpose is a 3 ´ 3.
c.
é 2 1 20ù
A = ê- 1 1 40ú
ê
ú
êë 5 7 1 ú
û
d.
Based on the above matrix, we want to maximize P = 5 y1 + 7 y2
T
subject to: 2 y1 + y2 £ 20
- y1 + y2 £ 40
y1, y2 ³ 0
e.
é 2
1 1 0 0 20ù
ê
ú
ê - 1 1 0 1 0 40ú
ê–5 –7 0 0 1 0ú
ë
û
é 2 1 1 0 0 20ù
ê
ú
ê- 3 0 - 1 1 0 20ú
ê 9 0 7 0 1 140ú
ë
û
f.
The function has a minimum value of 140 at the point (7, 0).
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Chapter 6
9. Minimize: C = 130 x1 + 120 x2
subject to:
1
x1 - x2 ³ 2
2
x1 + x2 ³ 7
x1, x2 ³ 0
a.
é1
- 1 2ù
ê2
ú
A = ê1
1
7ú
ê130 120 1 ú
êë
ú
û
b.
The matrix A is a 3 ´ 3 and the transpose is a 3 ´ 3.
c.
é 1 1 130ù
ê2
ú
T
A = ê- 1 1 120ú
ê2 7 1 ú
ú
ëê
û
d.
Based on the above matrix, we want to maximize P = 2 y1 + 7 y2
subject to:
1 y +y
2
2 1
£ 130
- y1 + y2 £ 120
y1, y2 ³ 0
e.
f.
é 1
1 1 0 0 130ù
ê 2
ú
ê - 1 1 0 1 0 120ú ®
ê–2 –7 0 0 1
0ú
êë
ú
û
é
ê
êê
ê
êë
é1 1 1
0 0 130ù
ê2
ú
20 ú ®
ê1 0 2 - 2 0
3
3
3ú
ê
ê3 0 7
0 1 910ú
ú
ëê 2
û
1 0 380 ù
é0 1 2
3
3
3 ú
ê
20
2
2
ê1 0
ú
- 3 0
3
3
ê
ú
1 1 900ú
ê0 0 6
ë
û
1
2
3
2
3
2
1 0 0 130ù
ú
0 - 1 1 0 - 10 ú
ú
0 7 0 1 910ú
ú
û
1
The function has a minimum value of 900 at the point (6, 1).
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Chapter 6
For problem 10, construct a mathematical model in the form of a linear programming
problem. Then solve the problem by applying the simplex method to the dual problem.
10. A dietician needs to plan a diet for a patient using two different food combinations to
meet the patient’s needs. Each container of Food A contains 4 units of Additive #1 and 2
units of Additive #2. Each container of Food B contains 2 units of Additive #1 and 4 units of
Additive #2. The patient needs at least 10 units of Additive #1 and at least 12 units of
Additive #2. How many containers of each food should the dietician use to meet the patients
needs and minimize the cost if one container of Food A costs $6 and each container of Food
B costs $10.
Let x1 represent Food A and x2 represent Food B. Based on the above information,
ì 4 x1 + 2 x2 ³ 10
ï
we will minimize the function P = 6 x1 + 10 x2 subject to í 2 x1 + 4 x2 ³ 12
ï x ,x ³ 0
î
1 2
é4 2 10ù
A = ê2 4 12ú
ê
ú
ëê6 10 1 ú
û
é4 2 6ù
A = ê 2 4 10ú
ê
ú
ëê10 12 1 ú
û
T
Based on the above matrix, we want to maximize P = 10 y1 + 12 y2
subject to: 4 y1 + 2 y2 £ 6
2 y1 + 4 y2 £ 10
y1, y2 ³ 0
é 4
2 1 0 0 6ù
ê
ú
4 0 1 0 10ú ®
ê 2
ê–10 –12 0 0 1 0ú
ë
û
é 2 1 1 0 0
3ù
2
ê
ú
ê- 6 0 - 2 1 0 - 2ú
ê14 0 6 0 1 36ú
ú
ëê
û
é 2 1 1 0 0
3ù
2
ê
ú
ê- 6 0 - 2 1 0 - 2ú
ê 8 0 4 1 1 34ú
êë
ú
û
Therefore, to have a minimum cost of $34, you should use 4 containers of Food A
and 1 container of Food B.
6-16
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Chapter 6
Name ________________________________ Date ______________ Class ____________
Section 6-4 Maximization and Minimization with
Mixed Problem Constraints
Goal: To solve maximum and minimum problems with mixed problem constraints
To Form the Modified Problem:
Step 1. If any problem constraints have negative constants on the right side, multiply
both sides by –1 to obtain a constraint with a nonnegative constant. (If the
constraint is an inequality, this will reverse the direction of the inequality.)
Step 2. Introduce a slack variable in each < constraint.
Step 3. Introduce a surplus variable and an artificial variable in each > constraint.
Step 4 Introduce an artificial variable in each = constraint.
Step 5. For each artificial variable ai, add –Mai to the objective function. Use the
same constant M for all artificial variables.
Solving the problem using the Big M method:
Step 1. Form the preliminary simplex tableau for the modified problem (see above).
Step 2 Use row operations to eliminate the M’s in the bottom row of the preliminary
simplex tableau in the columns corresponding to the artificial variables. The
resulting tableau is the initial simplex tableau.
Step 3 Solve the modified problem by applying the simplex method to the initial
simplex tableau found in step 2.
Step 4. Relate the optimal solution of the modified problem to the original problem.
a) If the modified problem has no optimal solution, the original
problem has no optimal solution.
b) If all artificial variables are 0 in the optimal solution to the modified
problem, delete the artificial variables to find an optimal solution to
the original problem.
c) If any artificial variables are nonzero in the optimal solution to the
modified problem, the original problem has no optimal solution.
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Chapter 6
For the following linear programming problems:
a. Form the modified problem.
b. Write the preliminary simplex tableau for the modified problem.
c. Find the initial simplex tableau.
d. Find the optimal solution of the original problem, if it exists.
1. Maximize: P = 7 x1 + 5x2
subject to: 3x1 + 2 x2 £ 6
x1 + x2 ³ 2
x1, x2 ³ 0
a. Maximize: P = 7 x1 + 5 x2 - Ma1
subject to:
3x1 + 2 x2 + s1
x1 + x2
=6
- s2 + a1 = 2
x1, x2 , s1, s2 , a1 ³ 0
x1
x2
s1 a1 s2
P
1 0 0 0 6ù
é 3 2
b. ê
1 1
0 1 - 1 0 2ú
ê
ú
êë–7 –5 0 M
0 1 0ú
û
3
2 1 0 0 0
6ù
c. é
ê
1
1 0 1 -1 0
2ú
ê
ú
êë- M – 7 - M – 5 0 0 M 1 - 2 M ú
û
d. é 0 - 13
ê
ê3
1
ê2
ê1
0
ë2
1
3
1
2
5
2
0ù
ú
0 0 0 3ú
ú
M 0 1 15ú
û
-1 1 0
Therefore, P has a maximum value of 15 when x1 = 0 and x2 = 3.
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Chapter 6
2. Minimize: C = x1 + 4 x2
subject to: x1 + x2 £ 6
- x1 + x2 ³ 2
x1, x2 ³ 0
a. Maximize: P = - C = - x1 - 4 x2 - Ma1
subject to:
x1 + x2 + s1
- x1 + x2
=6
- s2 + a1 = 2
x1, x2 , s1, s2 , a1 ³ 0
x1
x2
s1 a1 s2
P
é 1 1 1 0 0 0 6ù
b. ê
-1 1
0 1 - 1 0 2ú
ê
ú
êë 1 4 0 M
0 1 0ú
û
c. é
1
1 1 0 0 0
6ù
ê -1
1 0 1 -1 0
2ú
ê
ú
êëM + 1 - M + 4 0 0 M 1 - 2M ú
û
d. é 2 0 1 - 1
1 0 4ù
ê- 1 1 0
1
- 1 0 2ú
ê
ú
êë 5 0 0 M - 4 4 1 - 8ú
û
Therefore, C = - P = - (- 8) = 8 has a minimum value of 8 when x1 = 0 and x2 = 2.
6-19
Copyright © 2015 Pearson Education, Inc.
Finite Mathematics
Chapter 6
6-20
Copyright © 2015 Pearson Education, Inc.