Math 210richard/Math210/Lp/lp2.pdf · provides certain minimum levels of protein, calories, and vitamin B12 (riboa vin). One cup of uncooked rice costs 21 cents and contains 15 grams

Math 210Chapter 3

Linear Programming Problems

Professor Richard [email protected]

Dept. of Mathematical SciencesNorthern Illinois University

Math 210 Website:http://math.niu.edu/courses/math210

– p. 1/29

Nutrition ProblemSarah decides to make rice and soybeans part of herstaple diet.

The object is to design a lowest–cost diet thatprovides certain minimum levels of protein, calories,and vitamin B12 (riboflavin).One cup of uncooked rice costs 21 cents and contains15 grams of protein, 810 calories, and 1

9mg of

riboflavin.One cup of uncooked soybeans costs 14 cents andcontains 22.5 grams of protein, 270 calories, and 1

3mg

of riboflavin.The minimum daily requirements are: 90 grams ofprotein, 1620 calories, and 1 mg of B12.Design the lowest cost diet that meets theserequirements.

– p. 2/29

Nutrition ProblemSarah decides to make rice and soybeans part of herstaple diet.The object is to design a lowest–cost diet thatprovides certain minimum levels of protein, calories,and vitamin B12 (riboflavin).

One cup of uncooked rice costs 21 cents and contains15 grams of protein, 810 calories, and 1

9mg of


3mg


– p. 2/29

Nutrition ProblemSarah decides to make rice and soybeans part of herstaple diet.The object is to design a lowest–cost diet thatprovides certain minimum levels of protein, calories,and vitamin B12 (riboflavin).One cup of uncooked rice costs 21 cents and contains15 grams of protein, 810 calories, and 1

9mg of

riboflavin.

One cup of uncooked soybeans costs 14 cents andcontains 22.5 grams of protein, 270 calories, and 1

3mg


– p. 2/29


9mg of


3mg

of riboflavin.

The minimum daily requirements are: 90 grams ofprotein, 1620 calories, and 1 mg of B12.Design the lowest cost diet that meets theserequirements.

– p. 2/29


9mg of


3mg

of riboflavin.The minimum daily requirements are: 90 grams ofprotein, 1620 calories, and 1 mg of B12.

Design the lowest cost diet that meets theserequirements.

– p. 2/29


9mg of


3mg

of riboflavin.The minimum daily requirements are: 90 grams ofprotein, 1620 calories, and 1 mg of B12.Design the lowest cost diet that meets theserequirements. – p. 2/29

Set the VariablesLet

• x = the number of cups of uncooked rice in herdiet

• y = the number of cups of uncooked soybeans inher diet

The problem is to find the values of x and y which will

• minimize the cost and• provide the daily requirements of protein,

calories, and riboflavin.

– p. 3/29







– p. 3/29







– p. 3/29







– p. 3/29





• minimize the cost and

• provide the daily requirements of protein,calories, and riboflavin.

– p. 3/29







– p. 3/29

Organizing the DataCategory Rice Soybeans Requirement

Protein 15 22.5 90Calories 810 270 1620

Riboflavin 1

9

1

31

Cost 21 14 C

Nutrition Inequalities:• Protein 15x + 22.5y ≥ 90

• Calories 810x + 270y ≥ 1620

• Riboflavin 1

9x + 1

3y ≥ 1

Cost Equation:• Cost C = 21x + 14y

– p. 4/29

Organizing the DataCategory Rice Soybeans RequirementProtein

15 22.5 90Calories 810 270 1620

Riboflavin 1

9

1

31

Cost 21 14 C


• Calories 810x + 270y ≥ 1620

• Riboflavin 1

9x + 1

3y ≥ 1


– p. 4/29


15 22.5 90

Calories

810 270 1620Riboflavin 1

9

1

31

Cost 21 14 C


• Calories 810x + 270y ≥ 1620

• Riboflavin 1

9x + 1

3y ≥ 1


– p. 4/29


15 22.5 90

Calories

810 270 1620

Riboflavin

1

9

1

31

Cost 21 14 C


• Calories 810x + 270y ≥ 1620

• Riboflavin 1

9x + 1

3y ≥ 1


– p. 4/29


15 22.5 90

Calories

810 270 1620

Riboflavin

1

9

1

31

Cost

21 14 C


• Calories 810x + 270y ≥ 1620

• Riboflavin 1

9x + 1

3y ≥ 1


– p. 4/29

Organizing the DataCategory Rice Soybeans RequirementProtein 15 22.5

90

Calories

810 270 1620

Riboflavin

1

9

1

31

Cost

21 14 C


• Calories 810x + 270y ≥ 1620

• Riboflavin 1

9x + 1

3y ≥ 1


– p. 4/29


90

Calories 810 270

1620

Riboflavin

1

9

1

31

Cost

21 14 C


• Calories 810x + 270y ≥ 1620

• Riboflavin 1

9x + 1

3y ≥ 1


– p. 4/29


90

Calories 810 270

1620

Riboflavin 1

9

1

3

1

Cost

21 14 C


• Calories 810x + 270y ≥ 1620

• Riboflavin 1

9x + 1

3y ≥ 1


– p. 4/29


90

Calories 810 270

1620

Riboflavin 1

9

1

3

1

Cost 21 14

C


• Calories 810x + 270y ≥ 1620

• Riboflavin 1

9x + 1

3y ≥ 1


– p. 4/29

Organizing the DataCategory Rice Soybeans RequirementProtein 15 22.5 90Calories 810 270

1620

Riboflavin 1

9

1

3

1

Cost 21 14

C


• Calories 810x + 270y ≥ 1620

• Riboflavin 1

9x + 1

3y ≥ 1


– p. 4/29

Organizing the DataCategory Rice Soybeans RequirementProtein 15 22.5 90Calories 810 270 1620

Riboflavin 1

9

1

3

1

Cost 21 14

C


• Calories 810x + 270y ≥ 1620

• Riboflavin 1

9x + 1

3y ≥ 1


– p. 4/29


Riboflavin 1

9

1

31

Cost 21 14

C


• Calories 810x + 270y ≥ 1620

• Riboflavin 1

9x + 1

3y ≥ 1


– p. 4/29


Riboflavin 1

9

1

31

Cost 21 14 C


• Calories 810x + 270y ≥ 1620

• Riboflavin 1

9x + 1

3y ≥ 1


– p. 4/29


Riboflavin 1

9

1

31

Cost 21 14 C

Nutrition Inequalities:

• Protein 15x + 22.5y ≥ 90

• Calories 810x + 270y ≥ 1620

• Riboflavin 1

9x + 1

3y ≥ 1


– p. 4/29


Riboflavin 1

9

1

31

Cost 21 14 C


• Calories 810x + 270y ≥ 1620

• Riboflavin 1

9x + 1

3y ≥ 1


– p. 4/29


Riboflavin 1

9

1

31

Cost 21 14 C


• Calories 810x + 270y ≥ 1620

• Riboflavin 1

9x + 1

3y ≥ 1


– p. 4/29


Riboflavin 1

9

1

31

Cost 21 14 C


• Calories 810x + 270y ≥ 1620

• Riboflavin 1

9x + 1

3y ≥ 1


– p. 4/29


Riboflavin 1

9

1

31

Cost 21 14 C


• Calories 810x + 270y ≥ 1620

• Riboflavin 1

9x + 1

3y ≥ 1


– p. 4/29


Riboflavin 1

9

1

31

Cost 21 14 C


• Calories 810x + 270y ≥ 1620

• Riboflavin 1

9x + 1

3y ≥ 1


– p. 4/29

Simplify the inequalitiesNutrition Inequalities:

(1) Protein: 15x + 22.5y ≥ 90(2) Calories: 810x + 270y ≥ 1620(3) Riboflavin: 1

9x + 1

3y ≥ 1

To remove the decimal 22.5 in (1), multiply (1) by 2:(1′) Protein: 30x + 45y ≥ 180Now notice that 30, 45, and 180 are all divisible by15. So divide (1′) by 15:(1′′) Protein: 2x + 3y ≥ 12In (2) notice that 810, 270, and 1620 are all divisibleby 270. So divide (2) by 270:(2′) Calories: 3x + y ≥ 6To remove the denominators in (3), multiply (3) by 9:(3′) Riboflavin: x + 3y ≥ 9

– p. 5/29

Simplify the inequalitiesNutrition Inequalities:(1) Protein: 15x + 22.5y ≥ 90(2) Calories: 810x + 270y ≥ 1620(3) Riboflavin: 1

9x + 1

3y ≥ 1


– p. 5/29


9x + 1

3y ≥ 1

To remove the decimal 22.5 in (1), multiply (1) by 2:

(1′) Protein: 30x + 45y ≥ 180Now notice that 30, 45, and 180 are all divisible by15. So divide (1′) by 15:(1′′) Protein: 2x + 3y ≥ 12In (2) notice that 810, 270, and 1620 are all divisibleby 270. So divide (2) by 270:(2′) Calories: 3x + y ≥ 6To remove the denominators in (3), multiply (3) by 9:(3′) Riboflavin: x + 3y ≥ 9

– p. 5/29


9x + 1

3y ≥ 1

To remove the decimal 22.5 in (1), multiply (1) by 2:(1′) Protein: 30x + 45y ≥ 180

Now notice that 30, 45, and 180 are all divisible by15. So divide (1′) by 15:(1′′) Protein: 2x + 3y ≥ 12In (2) notice that 810, 270, and 1620 are all divisibleby 270. So divide (2) by 270:(2′) Calories: 3x + y ≥ 6To remove the denominators in (3), multiply (3) by 9:(3′) Riboflavin: x + 3y ≥ 9

– p. 5/29


9x + 1

3y ≥ 1

To remove the decimal 22.5 in (1), multiply (1) by 2:(1′) Protein: 30x + 45y ≥ 180Now notice that 30, 45, and 180 are all divisible by15. So divide (1′) by 15:

(1′′) Protein: 2x + 3y ≥ 12In (2) notice that 810, 270, and 1620 are all divisibleby 270. So divide (2) by 270:(2′) Calories: 3x + y ≥ 6To remove the denominators in (3), multiply (3) by 9:(3′) Riboflavin: x + 3y ≥ 9

– p. 5/29


9x + 1

3y ≥ 1

To remove the decimal 22.5 in (1), multiply (1) by 2:(1′) Protein: 30x + 45y ≥ 180Now notice that 30, 45, and 180 are all divisible by15. So divide (1′) by 15:(1′′) Protein: 2x + 3y ≥ 12

In (2) notice that 810, 270, and 1620 are all divisibleby 270. So divide (2) by 270:(2′) Calories: 3x + y ≥ 6To remove the denominators in (3), multiply (3) by 9:(3′) Riboflavin: x + 3y ≥ 9

– p. 5/29


9x + 1

3y ≥ 1

To remove the decimal 22.5 in (1), multiply (1) by 2:(1′) Protein: 30x + 45y ≥ 180Now notice that 30, 45, and 180 are all divisible by15. So divide (1′) by 15:(1′′) Protein: 2x + 3y ≥ 12In (2) notice that 810, 270, and 1620 are all divisibleby 270. So divide (2) by 270:

(2′) Calories: 3x + y ≥ 6To remove the denominators in (3), multiply (3) by 9:(3′) Riboflavin: x + 3y ≥ 9

– p. 5/29


9x + 1

3y ≥ 1

To remove the decimal 22.5 in (1), multiply (1) by 2:(1′) Protein: 30x + 45y ≥ 180Now notice that 30, 45, and 180 are all divisible by15. So divide (1′) by 15:(1′′) Protein: 2x + 3y ≥ 12In (2) notice that 810, 270, and 1620 are all divisibleby 270. So divide (2) by 270:(2′) Calories: 3x + y ≥ 6

To remove the denominators in (3), multiply (3) by 9:(3′) Riboflavin: x + 3y ≥ 9

– p. 5/29


9x + 1

3y ≥ 1

To remove the decimal 22.5 in (1), multiply (1) by 2:(1′) Protein: 30x + 45y ≥ 180Now notice that 30, 45, and 180 are all divisible by15. So divide (1′) by 15:(1′′) Protein: 2x + 3y ≥ 12In (2) notice that 810, 270, and 1620 are all divisibleby 270. So divide (2) by 270:(2′) Calories: 3x + y ≥ 6To remove the denominators in (3), multiply (3) by 9:

(3′) Riboflavin: x + 3y ≥ 9

– p. 5/29


9x + 1

3y ≥ 1


– p. 5/29

Nutrition Words → MathMinimize the cost

C = 21x + 14ysubject to the constraints

• 2x + 3y ≥ 12

• 3x + y ≥ 6

• x + 3y ≥ 9

• x ≥ 0, y ≥ 0

– p. 6/29

Nutrition Words → MathMinimize the cost C = 21x + 14ysubject to the constraints

• 2x + 3y ≥ 12

• 3x + y ≥ 6

• x + 3y ≥ 9

• x ≥ 0, y ≥ 0

– p. 6/29


• 2x + 3y ≥ 12

• 3x + y ≥ 6

• x + 3y ≥ 9

• x ≥ 0, y ≥ 0

– p. 6/29


• 2x + 3y ≥ 12

• 3x + y ≥ 6

• x + 3y ≥ 9

• x ≥ 0, y ≥ 0

– p. 6/29


• 2x + 3y ≥ 12

• 3x + y ≥ 6

• x + 3y ≥ 9

• x ≥ 0, y ≥ 0

– p. 6/29


• 2x + 3y ≥ 12

• 3x + y ≥ 6

• x + 3y ≥ 9

• x ≥ 0, y ≥ 0

– p. 6/29

Protein HalfplaneHalfplane: 2x + 3y ≥ 12

Line: 2x + 3y = 12x–intercept: (6, 0)y–intercept: (0, 4)Test point equation: 0 ≥ 12Position: above line

(0,0)s

Line 1:2x + 3y = 12

6

4

s

s

Halfplane 2x + 3y ≥ 12

– p. 7/29

Protein HalfplaneHalfplane: 2x + 3y ≥ 12Line: 2x + 3y = 12

x–intercept: (6, 0)y–intercept: (0, 4)Test point equation: 0 ≥ 12Position: above line

(0,0)s

Line 1:2x + 3y = 12

6

4

s

s


– p. 7/29

Protein HalfplaneHalfplane: 2x + 3y ≥ 12Line: 2x + 3y = 12x–intercept: (6, 0)

y–intercept: (0, 4)Test point equation: 0 ≥ 12Position: above line

(0,0)s

Line 1:2x + 3y = 12

6

4

s

s


– p. 7/29

Protein HalfplaneHalfplane: 2x + 3y ≥ 12Line: 2x + 3y = 12x–intercept: (6, 0)y–intercept: (0, 4)

Test point equation: 0 ≥ 12Position: above line

(0,0)s

Line 1:2x + 3y = 12

6

4

s

s


– p. 7/29

Protein HalfplaneHalfplane: 2x + 3y ≥ 12Line: 2x + 3y = 12x–intercept: (6, 0)y–intercept: (0, 4)Test point equation: 0 ≥ 12

Position: above line

(0,0)s

Line 1:2x + 3y = 12

6

4

s

s


– p. 7/29

Protein HalfplaneHalfplane: 2x + 3y ≥ 12Line: 2x + 3y = 12x–intercept: (6, 0)y–intercept: (0, 4)Test point equation: 0 ≥ 12Position: above line

(0,0)s

Line 1:2x + 3y = 12

6

4

s

s


– p. 7/29


(0,0)s

Line 1:2x + 3y = 12

6

4

s

s


– p. 7/29


(0,0)s

Line 1:2x + 3y = 12

6

4

s

s


– p. 7/29


(0,0)s

Line 1:2x + 3y = 12

6

4

s

s


– p. 7/29

Calorie HalfplaneHalfplane: 3x + y ≥ 6

Line: 3x + y = 6x–intercept: (2, 0)y–intercept: (0, 6)Test point equation: 0 ≥ 6Position: above line

(0,0)s

Line 2:3x + y = 6

2

6

s

s

Halfplane 3x + y ≥ 6

– p. 8/29

Calorie HalfplaneHalfplane: 3x + y ≥ 6Line: 3x + y = 6


(0,0)s

Line 2:3x + y = 6

2

6

s

s


– p. 8/29

Calorie HalfplaneHalfplane: 3x + y ≥ 6Line: 3x + y = 6x–intercept: (2, 0)


(0,0)s

Line 2:3x + y = 6

2

6

s

s


– p. 8/29

Calorie HalfplaneHalfplane: 3x + y ≥ 6Line: 3x + y = 6x–intercept: (2, 0)y–intercept: (0, 6)


(0,0)s

Line 2:3x + y = 6

2

6

s

s


– p. 8/29

Calorie HalfplaneHalfplane: 3x + y ≥ 6Line: 3x + y = 6x–intercept: (2, 0)y–intercept: (0, 6)Test point equation: 0 ≥ 6


(0,0)s

Line 2:3x + y = 6

2

6

s

s


– p. 8/29

Calorie HalfplaneHalfplane: 3x + y ≥ 6Line: 3x + y = 6x–intercept: (2, 0)y–intercept: (0, 6)Test point equation: 0 ≥ 6Position: above line

(0,0)s

Line 2:3x + y = 6

2

6

s

s


– p. 8/29


(0,0)s

Line 2:3x + y = 6

2

6

s

s


– p. 8/29


(0,0)s

Line 2:3x + y = 6

2

6

s

s


– p. 8/29


(0,0)s

Line 2:3x + y = 6

2

6

s

s


– p. 8/29

Riboflavin HalfplaneHalfplane: x + 3y ≥ 9

Line: x + 3y = 9x–intercept: (9, 0)y–intercept: (0, 3)Test point equation: 0 ≥ 9Position: above line

(0,0)s

Line 3:x + 3y = 9

9

3

s

sHalfplane x + 3y ≥ 9

– p. 9/29

Riboflavin HalfplaneHalfplane: x + 3y ≥ 9Line: x + 3y = 9


(0,0)s

Line 3:x + 3y = 9

9

3

s


– p. 9/29

Riboflavin HalfplaneHalfplane: x + 3y ≥ 9Line: x + 3y = 9x–intercept: (9, 0)


(0,0)s

Line 3:x + 3y = 9

9

3

s


– p. 9/29

Riboflavin HalfplaneHalfplane: x + 3y ≥ 9Line: x + 3y = 9x–intercept: (9, 0)y–intercept: (0, 3)


(0,0)s

Line 3:x + 3y = 9

9

3

s


– p. 9/29

Riboflavin HalfplaneHalfplane: x + 3y ≥ 9Line: x + 3y = 9x–intercept: (9, 0)y–intercept: (0, 3)Test point equation: 0 ≥ 9


(0,0)s

Line 3:x + 3y = 9

9

3

s


– p. 9/29

Riboflavin HalfplaneHalfplane: x + 3y ≥ 9Line: x + 3y = 9x–intercept: (9, 0)y–intercept: (0, 3)Test point equation: 0 ≥ 9Position: above line

(0,0)s

Line 3:x + 3y = 9

9

3

s


– p. 9/29


(0,0)s

Line 3:x + 3y = 9

9

3

s


– p. 9/29


(0,0)s

Line 3:x + 3y = 9

9

3

s

s

Halfplane x + 3y ≥ 9

– p. 9/29


(0,0)s

Line 3:x + 3y = 9

9

3

s


– p. 9/29

Where do Protein and CalorieLines Intersect?

• Write the equations:

Protein 2x + 3y = 12

Calorie 3x + y = 6

• Multiply Calorie equation by 3Calorie 9x + 3y = 18


Subtract 7x = 6

• So x = 6

7

• Plug x = 6

7back into the Calorie equation

• y = 6 − 3x = 6 − 3 ·6

7= 42

7−

18

7= 24

7

• The intersection point is(

6

7, 24

7

)

– p. 10/29


• Write the equations:Protein 2x + 3y = 12

Calorie 3x + y = 6



Subtract 7x = 6

• So x = 6

7

• Plug x = 6


• y = 6 − 3x = 6 − 3 ·6

7= 42

7−

18

7= 24

7


6

7, 24

7

)

– p. 10/29



Calorie 3x + y = 6

• Multiply Calorie equation by 3

Calorie 9x + 3y = 18


Subtract 7x = 6

• So x = 6

7

• Plug x = 6


• y = 6 − 3x = 6 − 3 ·6

7= 42

7−

18

7= 24

7


6

7, 24

7

)

– p. 10/29



Calorie 3x + y = 6



Subtract 7x = 6

• So x = 6

7

• Plug x = 6


• y = 6 − 3x = 6 − 3 ·6

7= 42

7−

18

7= 24

7


6

7, 24

7

)

– p. 10/29



Calorie 3x + y = 6



Subtract 7x = 6

• So x = 6

7

• Plug x = 6


• y = 6 − 3x = 6 − 3 ·6

7= 42

7−

18

7= 24

7


6

7, 24

7

)

– p. 10/29



Calorie 3x + y = 6



Subtract 7x = 6

• So x = 6

7

• Plug x = 6


• y = 6 − 3x = 6 − 3 ·6

7= 42

7−

18

7= 24

7


6

7, 24

7

)

– p. 10/29



Calorie 3x + y = 6



Subtract 7x = 6

• So x = 6

7

• Plug x = 6


• y = 6 − 3x

= 6 − 3 ·6

7= 42

7−

18

7= 24

7


6

7, 24

7

)

– p. 10/29



Calorie 3x + y = 6



Subtract 7x = 6

• So x = 6

7

• Plug x = 6


• y = 6 − 3x = 6 − 3 ·6

7

= 42

7−

18

7= 24

7


6

7, 24

7

)

– p. 10/29



Calorie 3x + y = 6



Subtract 7x = 6

• So x = 6

7

• Plug x = 6


• y = 6 − 3x = 6 − 3 ·6

7= 42

7−

18

7

= 24

7


6

7, 24

7

)

– p. 10/29



Calorie 3x + y = 6



Subtract 7x = 6

• So x = 6

7

• Plug x = 6


• y = 6 − 3x = 6 − 3 ·6

7= 42

7−

18

7= 24

7


6

7, 24

7

)

– p. 10/29



Calorie 3x + y = 6



Subtract 7x = 6

• So x = 6

7

• Plug x = 6


• y = 6 − 3x = 6 − 3 ·6

7= 42

7−

18

7= 24

7


6

7, 24

7

)

– p. 10/29

Where do Protein and Ri-boflavin Lines Intersect?



Ribofl x + 3y = 9

Subtract x = 3

• Plug x = 3 back into the Riboflavin equation

• y =9 − x

3=

9 − 3

3=2

• The intersection point is (3, 2)

– p. 11/29



Ribofl x + 3y = 9

Subtract x = 3


• y =9 − x

3=

9 − 3

3=2


– p. 11/29



Ribofl x + 3y = 9

Subtract x = 3


• y =9 − x

3=

9 − 3

3=2


– p. 11/29



Ribofl x + 3y = 9

Subtract x = 3


• y =9 − x

3

=9 − 3

3=2


– p. 11/29



Ribofl x + 3y = 9

Subtract x = 3


• y =9 − x

3=

9 − 3

3

=2


– p. 11/29



Ribofl x + 3y = 9

Subtract x = 3


• y =9 − x

3=

9 − 3

3=2


– p. 11/29



Ribofl x + 3y = 9

Subtract x = 3


• y =9 − x

3=

9 − 3

3=2


– p. 11/29

Where do Calorie and Ri-boflavin Lines Intersect?


Calorie 3x + y = 6

Ribofl x + 3y = 9


Ribofl x + 3y = 9

Subtract 8x = 9

• So x = 9

8

• Plug x = 6


• y = 6 − 3x = 6 − 3 ·9

8= 48

8−

27

8= 21

8


9

8, 21

8

)

– p. 12/29


• Write the equations:Calorie 3x + y = 6

Ribofl x + 3y = 9


Ribofl x + 3y = 9

Subtract 8x = 9

• So x = 9

8

• Plug x = 6


• y = 6 − 3x = 6 − 3 ·9

8= 48

8−

27

8= 21

8


9

8, 21

8

)

– p. 12/29



Ribofl x + 3y = 9

• Multiply Calorie equation by 3

Calorie 9x + 3y = 18

Ribofl x + 3y = 9

Subtract 8x = 9

• So x = 9

8

• Plug x = 6


• y = 6 − 3x = 6 − 3 ·9

8= 48

8−

27

8= 21

8


9

8, 21

8

)

– p. 12/29



Ribofl x + 3y = 9


Ribofl x + 3y = 9

Subtract 8x = 9

• So x = 9

8

• Plug x = 6


• y = 6 − 3x = 6 − 3 ·9

8= 48

8−

27

8= 21

8


9

8, 21

8

)

– p. 12/29



Ribofl x + 3y = 9


Ribofl x + 3y = 9

Subtract 8x = 9

• So x = 9

8

• Plug x = 6


• y = 6 − 3x = 6 − 3 ·9

8= 48

8−

27

8= 21

8


9

8, 21

8

)

– p. 12/29



Ribofl x + 3y = 9


Ribofl x + 3y = 9

Subtract 8x = 9

• So x = 9

8

• Plug x = 6


• y = 6 − 3x = 6 − 3 ·9

8= 48

8−

27

8= 21

8


9

8, 21

8

)

– p. 12/29



Ribofl x + 3y = 9


Ribofl x + 3y = 9

Subtract 8x = 9

• So x = 9

8

• Plug x = 6


• y = 6 − 3x

= 6 − 3 ·9

8= 48

8−

27

8= 21

8


9

8, 21

8

)

– p. 12/29



Ribofl x + 3y = 9


Ribofl x + 3y = 9

Subtract 8x = 9

• So x = 9

8

• Plug x = 6


• y = 6 − 3x = 6 − 3 ·9

8

= 48

8−

27

8= 21

8


9

8, 21

8

)

– p. 12/29



Ribofl x + 3y = 9


Ribofl x + 3y = 9

Subtract 8x = 9

• So x = 9

8

• Plug x = 6


• y = 6 − 3x = 6 − 3 ·9

8= 48

8−

27

8

= 21

8


9

8, 21

8

)

– p. 12/29



Ribofl x + 3y = 9


Ribofl x + 3y = 9

Subtract 8x = 9

• So x = 9

8

• Plug x = 6


• y = 6 − 3x = 6 − 3 ·9

8= 48

8−

27

8= 21

8


9

8, 21

8

)

– p. 12/29



Ribofl x + 3y = 9


Ribofl x + 3y = 9

Subtract 8x = 9

• So x = 9

8

• Plug x = 6


• y = 6 − 3x = 6 − 3 ·9

8= 48

8−

27

8= 21

8


9

8, 21

8

)

– p. 12/29

Feasible Region

(0,0)v

Line 1:2x + 3y = 12

6

4

v

v

Line 2:3x + y = 6

2

6

v

v

Line 3:x + 3y = 9

9

3

v

v

(3, 2)v

(

9

8, 21

8

)v

(

6

7, 24

7

)

v

Feasible Region

– p. 13/29

Feasible Region

(0,0)v

Line 1:2x + 3y = 12

6

4

v

v

Line 2:3x + y = 6

2

6

v

v

Line 3:x + 3y = 9

9

3

v

v

(3, 2)v

(

9

8, 21

8

)v

(

6

7, 24

7

)

v

Feasible Region

– p. 13/29

Feasible Region

(0,0)v

Line 1:2x + 3y = 12

6

4

v

v

Line 2:3x + y = 6

2

6

v

v

Line 3:x + 3y = 9

9

3

v

v

(3, 2)v

(

9

8, 21

8

)v

(

6

7, 24

7

)

v

Feasible Region

– p. 13/29

Feasible Region

(0,0)v

Line 1:2x + 3y = 12

6

4

v

v

Line 2:3x + y = 6

2

6

v

v

Line 3:x + 3y = 9

9

3

v

v

(3, 2)v

(

9

8, 21

8

)v

(

6

7, 24

7

)

v

Feasible Region

– p. 13/29

Feasible Region

(0,0)v

Line 1:2x + 3y = 12

6

4

v

v

Line 2:3x + y = 6

2

6

v

v

Line 3:x + 3y = 9

9

3

v

v

(3, 2)v

(

9

8, 21

8

)v

(

6

7, 24

7

)

v

Feasible Region

– p. 13/29

Feasible Region

(0,0)v

Line 1:2x + 3y = 12

6

4

v

v

Line 2:3x + y = 6

2

6

v

v

Line 3:x + 3y = 9

9

3

v

v

(3, 2)v

(

9

8, 21

8

)v

(

6

7, 24

7

)

v

Feasible Region

– p. 13/29

Solving the Nutrition ProblemThe list of corner points is:

(0, 6),(

6

7, 24

7

)

, (3, 2), (9, 0)

The intersection(

9

8, 21

8

)

of Lines 2 and 3 is not acorner point because it is not on halfplane2x + 3y ≥ 12:2 ·

9

8+ 3 ·

21

8= 18

8+ 63

8= 81

8= 10.125 < 12

x y C = 21x + 14y

0 6 846

7

24

718 + 48 = 66

3 2 919 0 189

Minimum value of C is 66 at the point(

6

7, 24

7

)

– p. 14/29

Solving the Nutrition ProblemThe list of corner points is:(0, 6),

(

6

7, 24

7

)

, (3, 2), (9, 0)

The intersection(

9

8, 21

8

)


9

8+ 3 ·

21

8= 18

8+ 63

8= 81

8= 10.125 < 12

x y C = 21x + 14y

0 6 846

7

24

718 + 48 = 66

3 2 919 0 189


6

7, 24

7

)

– p. 14/29


(

6

7, 24

7

)

, (3, 2), (9, 0)

The intersection(

9

8, 21

8

)

of Lines 2 and 3 is not acorner point because it is not on halfplane2x + 3y ≥ 12:

2 ·9

8+ 3 ·

21

8= 18

8+ 63

8= 81

8= 10.125 < 12

x y C = 21x + 14y

0 6 846

7

24

718 + 48 = 66

3 2 919 0 189


6

7, 24

7

)

– p. 14/29


(

6

7, 24

7

)

, (3, 2), (9, 0)

The intersection(

9

8, 21

8

)


9

8+ 3 ·

21

8= 18

8+ 63

8= 81

8= 10.125 < 12

x y C = 21x + 14y

0 6 846

7

24

718 + 48 = 66

3 2 919 0 189


6

7, 24

7

)

– p. 14/29


(

6

7, 24

7

)

, (3, 2), (9, 0)

The intersection(

9

8, 21

8

)


9

8+ 3 ·

21

8= 18

8+ 63

8= 81

8= 10.125 < 12

x y C = 21x + 14y

0 6

846

7

24

718 + 48 = 66

3 2 919 0 189


6

7, 24

7

)

– p. 14/29


(

6

7, 24

7

)

, (3, 2), (9, 0)

The intersection(

9

8, 21

8

)


9

8+ 3 ·

21

8= 18

8+ 63

8= 81

8= 10.125 < 12

x y C = 21x + 14y

0 6

84

6

7

24

7

18 + 48 = 66

3 2 919 0 189


6

7, 24

7

)

– p. 14/29


(

6

7, 24

7

)

, (3, 2), (9, 0)

The intersection(

9

8, 21

8

)


9

8+ 3 ·

21

8= 18

8+ 63

8= 81

8= 10.125 < 12

x y C = 21x + 14y

0 6

84

6

7

24

7

18 + 48 = 66

3 2

919 0 189


6

7, 24

7

)

– p. 14/29


(

6

7, 24

7

)

, (3, 2), (9, 0)

The intersection(

9

8, 21

8

)


9

8+ 3 ·

21

8= 18

8+ 63

8= 81

8= 10.125 < 12

x y C = 21x + 14y

0 6

84

6

7

24

7

18 + 48 = 66

3 2

91

9 0

189


6

7, 24

7

)

– p. 14/29


(

6

7, 24

7

)

, (3, 2), (9, 0)

The intersection(

9

8, 21

8

)


9

8+ 3 ·

21

8= 18

8+ 63

8= 81

8= 10.125 < 12

x y C = 21x + 14y

0 6 846

7

24

7

18 + 48 = 66

3 2

91

9 0

189


6

7, 24

7

)

– p. 14/29


(

6

7, 24

7

)

, (3, 2), (9, 0)

The intersection(

9

8, 21

8

)


9

8+ 3 ·

21

8= 18

8+ 63

8= 81

8= 10.125 < 12

x y C = 21x + 14y

0 6 846

7

24

718 + 48 = 66

3 2

91

9 0

189


6

7, 24

7

)

– p. 14/29


(

6

7, 24

7

)

, (3, 2), (9, 0)

The intersection(

9

8, 21

8

)


9

8+ 3 ·

21

8= 18

8+ 63

8= 81

8= 10.125 < 12

x y C = 21x + 14y

0 6 846

7

24

718 + 48 = 66

3 2 919 0

189


6

7, 24

7

)

– p. 14/29


(

6

7, 24

7

)

, (3, 2), (9, 0)

The intersection(

9

8, 21

8

)


9

8+ 3 ·

21

8= 18

8+ 63

8= 81

8= 10.125 < 12

x y C = 21x + 14y

0 6 846

7

24

718 + 48 = 66

3 2 919 0 189


6

7, 24

7

)

– p. 14/29


(

6

7, 24

7

)

, (3, 2), (9, 0)

The intersection(

9

8, 21

8

)


9

8+ 3 ·

21

8= 18

8+ 63

8= 81

8= 10.125 < 12

x y C = 21x + 14y

0 6 846

7

24

718 + 48 = 66

3 2 919 0 189

Minimum value of C is

66 at the point(

6

7, 24

7

)

– p. 14/29


(

6

7, 24

7

)

, (3, 2), (9, 0)

The intersection(

9

8, 21

8

)


9

8+ 3 ·

21

8= 18

8+ 63

8= 81

8= 10.125 < 12

x y C = 21x + 14y

0 6 846

7

24

718 + 48 = 66

3 2 919 0 189

Minimum value of C is 66 at the point

(

6

7, 24

7

)

– p. 14/29


(

6

7, 24

7

)

, (3, 2), (9, 0)

The intersection(

9

8, 21

8

)


9

8+ 3 ·

21

8= 18

8+ 63

8= 81

8= 10.125 < 12

x y C = 21x + 14y

0 6 846

7

24

718 + 48 = 66

3 2 919 0 189


6

7, 24

7

)

– p. 14/29


(

6

7, 24

7

)

, (3, 2), (9, 0)

The intersection(

9

8, 21

8

)


9

8+ 3 ·

21

8= 18

8+ 63

8= 81

8= 10.125 < 12

x y C = 21x + 14y

0 6 846

7

24

718 + 48 = 66

3 2 919 0 189


6

7, 24

7

)

– p. 14/29


(

6

7, 24

7

)

, (3, 2), (9, 0)

The intersection(

9

8, 21

8

)


9

8+ 3 ·

21

8= 18

8+ 63

8= 81

8= 10.125 < 12

x y C = 21x + 14y

0 6 846

7

24

718 + 48 = 66

3 2 919 0 189


6

7, 24

7

)

– p. 14/29

Transportation ProblemA Chicago TV dealer has stores in Schaumburg andAurora and warehouses in DeKalb and Elkhart,Indiana.

The cost of shipping a 48 inch flat screen TVset from DeKalb to Schaumburg is $6; from DeKalbto Aurora is $3; from Elkhart to Schaumburg, $9; andfrom Elkhart to Aurora is $5. Suppose the Shuamburgstore orders 25 TV sets and the Aurora store orders30. The DeKalb warehouse has a stock of 45 TV setsand the Elkhart warehouse has 40.What is the most economical way to supply therequired sets to the two stores?

– p. 15/29

Transportation ProblemA Chicago TV dealer has stores in Schaumburg andAurora and warehouses in DeKalb and Elkhart,Indiana. The cost of shipping a 48 inch flat screen TVset from DeKalb to Schaumburg is $6; from DeKalbto Aurora is $3; from Elkhart to Schaumburg, $9; andfrom Elkhart to Aurora is $5.

Suppose the Shuamburgstore orders 25 TV sets and the Aurora store orders30. The DeKalb warehouse has a stock of 45 TV setsand the Elkhart warehouse has 40.What is the most economical way to supply therequired sets to the two stores?

– p. 15/29

Transportation ProblemA Chicago TV dealer has stores in Schaumburg andAurora and warehouses in DeKalb and Elkhart,Indiana. The cost of shipping a 48 inch flat screen TVset from DeKalb to Schaumburg is $6; from DeKalbto Aurora is $3; from Elkhart to Schaumburg, $9; andfrom Elkhart to Aurora is $5. Suppose the Shuamburgstore orders 25 TV sets and the Aurora store orders30.

The DeKalb warehouse has a stock of 45 TV setsand the Elkhart warehouse has 40.What is the most economical way to supply therequired sets to the two stores?

– p. 15/29

Transportation ProblemA Chicago TV dealer has stores in Schaumburg andAurora and warehouses in DeKalb and Elkhart,Indiana. The cost of shipping a 48 inch flat screen TVset from DeKalb to Schaumburg is $6; from DeKalbto Aurora is $3; from Elkhart to Schaumburg, $9; andfrom Elkhart to Aurora is $5. Suppose the Shuamburgstore orders 25 TV sets and the Aurora store orders30. The DeKalb warehouse has a stock of 45 TV setsand the Elkhart warehouse has 40.

What is the most economical way to supply therequired sets to the two stores?

– p. 15/29

Transportation ProblemA Chicago TV dealer has stores in Schaumburg andAurora and warehouses in DeKalb and Elkhart,Indiana. The cost of shipping a 48 inch flat screen TVset from DeKalb to Schaumburg is $6; from DeKalbto Aurora is $3; from Elkhart to Schaumburg, $9; andfrom Elkhart to Aurora is $5. Suppose the Shuamburgstore orders 25 TV sets and the Aurora store orders30. The DeKalb warehouse has a stock of 45 TV setsand the Elkhart warehouse has 40.What is the most economical way to supply therequired sets to the two stores?

– p. 15/29

Solution: Determine VariablesAt first it appears that 4 variables will be needed,since there are two stores and two warehouses, hence4 possible shipping combinations.

A closer look shows that only two variables x and yare needed.For if x represents the number of TV sets to beshipped from DeKalb to Schaumburg, then sinceSchaumburg needs 25 sets, the number of TV sets tobe shipped from Elkhart to Schaumburg is 25 − x.Similarly, if y represents the number of TV sets to beshipped from DeKalb to Aurora, then since Auroraneeds 30 sets, the number of TV sets to be shippedfrom Elkhart to Aurora is 30 − y.

– p. 16/29

Solution: Determine VariablesAt first it appears that 4 variables will be needed,since there are two stores and two warehouses, hence4 possible shipping combinations.A closer look shows that only two variables x and yare needed.

For if x represents the number of TV sets to beshipped from DeKalb to Schaumburg, then sinceSchaumburg needs 25 sets, the number of TV sets tobe shipped from Elkhart to Schaumburg is 25 − x.Similarly, if y represents the number of TV sets to beshipped from DeKalb to Aurora, then since Auroraneeds 30 sets, the number of TV sets to be shippedfrom Elkhart to Aurora is 30 − y.

– p. 16/29

Solution: Determine VariablesAt first it appears that 4 variables will be needed,since there are two stores and two warehouses, hence4 possible shipping combinations.A closer look shows that only two variables x and yare needed.For if x represents the number of TV sets to beshipped from DeKalb to Schaumburg,

then sinceSchaumburg needs 25 sets, the number of TV sets tobe shipped from Elkhart to Schaumburg is 25 − x.Similarly, if y represents the number of TV sets to beshipped from DeKalb to Aurora, then since Auroraneeds 30 sets, the number of TV sets to be shippedfrom Elkhart to Aurora is 30 − y.

– p. 16/29

Solution: Determine VariablesAt first it appears that 4 variables will be needed,since there are two stores and two warehouses, hence4 possible shipping combinations.A closer look shows that only two variables x and yare needed.For if x represents the number of TV sets to beshipped from DeKalb to Schaumburg, then sinceSchaumburg needs 25 sets,

the number of TV sets tobe shipped from Elkhart to Schaumburg is 25 − x.Similarly, if y represents the number of TV sets to beshipped from DeKalb to Aurora, then since Auroraneeds 30 sets, the number of TV sets to be shippedfrom Elkhart to Aurora is 30 − y.

– p. 16/29

Solution: Determine VariablesAt first it appears that 4 variables will be needed,since there are two stores and two warehouses, hence4 possible shipping combinations.A closer look shows that only two variables x and yare needed.For if x represents the number of TV sets to beshipped from DeKalb to Schaumburg, then sinceSchaumburg needs 25 sets, the number of TV sets tobe shipped from Elkhart to Schaumburg is 25 − x.

Similarly, if y represents the number of TV sets to beshipped from DeKalb to Aurora, then since Auroraneeds 30 sets, the number of TV sets to be shippedfrom Elkhart to Aurora is 30 − y.

– p. 16/29

Solution: Determine VariablesAt first it appears that 4 variables will be needed,since there are two stores and two warehouses, hence4 possible shipping combinations.A closer look shows that only two variables x and yare needed.For if x represents the number of TV sets to beshipped from DeKalb to Schaumburg, then sinceSchaumburg needs 25 sets, the number of TV sets tobe shipped from Elkhart to Schaumburg is 25 − x.Similarly, if y represents the number of TV sets to beshipped from DeKalb to Aurora,

then since Auroraneeds 30 sets, the number of TV sets to be shippedfrom Elkhart to Aurora is 30 − y.

– p. 16/29

Solution: Determine VariablesAt first it appears that 4 variables will be needed,since there are two stores and two warehouses, hence4 possible shipping combinations.A closer look shows that only two variables x and yare needed.For if x represents the number of TV sets to beshipped from DeKalb to Schaumburg, then sinceSchaumburg needs 25 sets, the number of TV sets tobe shipped from Elkhart to Schaumburg is 25 − x.Similarly, if y represents the number of TV sets to beshipped from DeKalb to Aurora, then since Auroraneeds 30 sets,

the number of TV sets to be shippedfrom Elkhart to Aurora is 30 − y.

– p. 16/29

Solution: Determine VariablesAt first it appears that 4 variables will be needed,since there are two stores and two warehouses, hence4 possible shipping combinations.A closer look shows that only two variables x and yare needed.For if x represents the number of TV sets to beshipped from DeKalb to Schaumburg, then sinceSchaumburg needs 25 sets, the number of TV sets tobe shipped from Elkhart to Schaumburg is 25 − x.Similarly, if y represents the number of TV sets to beshipped from DeKalb to Aurora, then since Auroraneeds 30 sets, the number of TV sets to be shippedfrom Elkhart to Aurora is 30 − y.

– p. 16/29

Cost TableWarehouse–Store Number Cost per set

DeKalb–Schaumburg x $6DeKalb–Aurora y $3

Elkhart–Schaumburg 25 − x $9Elkhart–Aurora 30 − y $5

Total Cost:C = 6x + 3y + 9(25 − x) + 5(30 − y)

= 6x + 3y + 225 − 9x + 150 − 5y= −3x − 2y + 375

– p. 17/29


DeKalb–Schaumburg

x $6DeKalb–Aurora y $3


Total Cost:C = 6x + 3y + 9(25 − x) + 5(30 − y)

= 6x + 3y + 225 − 9x + 150 − 5y= −3x − 2y + 375

– p. 17/29


DeKalb–Schaumburg

x $6

DeKalb–Aurora

y $3Elkhart–Schaumburg 25 − x $9

Elkhart–Aurora 30 − y $5

Total Cost:C = 6x + 3y + 9(25 − x) + 5(30 − y)

= 6x + 3y + 225 − 9x + 150 − 5y= −3x − 2y + 375

– p. 17/29


DeKalb–Schaumburg

x $6

DeKalb–Aurora

y $3

Elkhart–Schaumburg

25 − x $9Elkhart–Aurora 30 − y $5

Total Cost:C = 6x + 3y + 9(25 − x) + 5(30 − y)

= 6x + 3y + 225 − 9x + 150 − 5y= −3x − 2y + 375

– p. 17/29


DeKalb–Schaumburg

x $6

DeKalb–Aurora

y $3


25 − x $9

Elkhart–Aurora

30 − y $5

Total Cost:C = 6x + 3y + 9(25 − x) + 5(30 − y)

= 6x + 3y + 225 − 9x + 150 − 5y= −3x − 2y + 375

– p. 17/29


DeKalb–Schaumburg x

$6

DeKalb–Aurora

y $3


25 − x $9

Elkhart–Aurora

30 − y $5

Total Cost:C = 6x + 3y + 9(25 − x) + 5(30 − y)

= 6x + 3y + 225 − 9x + 150 − 5y= −3x − 2y + 375

– p. 17/29


DeKalb–Schaumburg x $6DeKalb–Aurora

y $3


25 − x $9

Elkhart–Aurora

30 − y $5

Total Cost:C = 6x + 3y + 9(25 − x) + 5(30 − y)

= 6x + 3y + 225 − 9x + 150 − 5y= −3x − 2y + 375

– p. 17/29


DeKalb–Schaumburg x $6DeKalb–Aurora y

$3


25 − x $9

Elkhart–Aurora

30 − y $5

Total Cost:C = 6x + 3y + 9(25 − x) + 5(30 − y)

= 6x + 3y + 225 − 9x + 150 − 5y= −3x − 2y + 375

– p. 17/29




25 − x $9

Elkhart–Aurora

30 − y $5

Total Cost:C = 6x + 3y + 9(25 − x) + 5(30 − y)

= 6x + 3y + 225 − 9x + 150 − 5y= −3x − 2y + 375

– p. 17/29



Elkhart–Schaumburg 25 − x

$9

Elkhart–Aurora

30 − y $5

Total Cost:C = 6x + 3y + 9(25 − x) + 5(30 − y)

= 6x + 3y + 225 − 9x + 150 − 5y= −3x − 2y + 375

– p. 17/29



Elkhart–Schaumburg 25 − x $9Elkhart–Aurora

30 − y $5

Total Cost:C = 6x + 3y + 9(25 − x) + 5(30 − y)

= 6x + 3y + 225 − 9x + 150 − 5y= −3x − 2y + 375

– p. 17/29



Elkhart–Schaumburg 25 − x $9Elkhart–Aurora 30 − y

$5

Total Cost:C = 6x + 3y + 9(25 − x) + 5(30 − y)

= 6x + 3y + 225 − 9x + 150 − 5y= −3x − 2y + 375

– p. 17/29




Total Cost:C = 6x + 3y + 9(25 − x) + 5(30 − y)

= 6x + 3y + 225 − 9x + 150 − 5y= −3x − 2y + 375

– p. 17/29




Total Cost:C = 6x + 3y + 9(25 − x) + 5(30 − y)

= 6x + 3y + 225 − 9x + 150 − 5y= −3x − 2y + 375

– p. 17/29




Total Cost:C = 6x + 3y + 9(25 − x) + 5(30 − y)

= 6x + 3y + 225 − 9x + 150 − 5y

= −3x − 2y + 375

– p. 17/29




Total Cost:C = 6x + 3y + 9(25 − x) + 5(30 − y)

= 6x + 3y + 225 − 9x + 150 − 5y= −3x − 2y + 375

– p. 17/29

ConstraintsThere are two kinds of constraints:

• none of x, y, 25 − x, 30 − y can be negativeEquivalently, x ≥ 0, y ≥ 0, x ≤ 25, y ≤ 30

• a warehouse cannot ship more TVs than it has instock.Since DeKalb ships x + y sets and has 45 instock, we get the constraint: x + y ≤ 45Since Elkhart ships (25 − x) + (30 − y) sets andhas 40 in stock, we get the constraint:(25 − x) + (30 − y) ≤ 40or 55 − x − y ≤ 40or 15 ≤ x + y

– p. 18/29


• none of x, y, 25 − x, 30 − y can be negative

Equivalently, x ≥ 0, y ≥ 0, x ≤ 25, y ≤ 30


– p. 18/29




– p. 18/29



• a warehouse cannot ship more TVs than it has instock.

Since DeKalb ships x + y sets and has 45 instock, we get the constraint: x + y ≤ 45Since Elkhart ships (25 − x) + (30 − y) sets andhas 40 in stock, we get the constraint:(25 − x) + (30 − y) ≤ 40or 55 − x − y ≤ 40or 15 ≤ x + y

– p. 18/29



• a warehouse cannot ship more TVs than it has instock.Since DeKalb ships x + y sets and has 45 instock, we get the constraint:

x + y ≤ 45Since Elkhart ships (25 − x) + (30 − y) sets andhas 40 in stock, we get the constraint:(25 − x) + (30 − y) ≤ 40or 55 − x − y ≤ 40or 15 ≤ x + y

– p. 18/29



• a warehouse cannot ship more TVs than it has instock.Since DeKalb ships x + y sets and has 45 instock, we get the constraint: x + y ≤ 45

Since Elkhart ships (25 − x) + (30 − y) sets andhas 40 in stock, we get the constraint:(25 − x) + (30 − y) ≤ 40or 55 − x − y ≤ 40or 15 ≤ x + y

– p. 18/29



• a warehouse cannot ship more TVs than it has instock.Since DeKalb ships x + y sets and has 45 instock, we get the constraint: x + y ≤ 45Since Elkhart ships (25 − x) + (30 − y) sets andhas 40 in stock, we get the constraint:

(25 − x) + (30 − y) ≤ 40or 55 − x − y ≤ 40or 15 ≤ x + y

– p. 18/29



• a warehouse cannot ship more TVs than it has instock.Since DeKalb ships x + y sets and has 45 instock, we get the constraint: x + y ≤ 45Since Elkhart ships (25 − x) + (30 − y) sets andhas 40 in stock, we get the constraint:(25 − x) + (30 − y) ≤ 40or

55 − x − y ≤ 40or 15 ≤ x + y

– p. 18/29



• a warehouse cannot ship more TVs than it has instock.Since DeKalb ships x + y sets and has 45 instock, we get the constraint: x + y ≤ 45Since Elkhart ships (25 − x) + (30 − y) sets andhas 40 in stock, we get the constraint:(25 − x) + (30 − y) ≤ 40or 55 − x − y ≤ 40or

15 ≤ x + y

– p. 18/29




– p. 18/29

Transportation Words → MathMinimize the cost

C = −3x − 2y + 375subject to the constraints

• x ≤ 25, y ≤ 30

• 15 ≤ x + y

• x + y ≤ 45

• x ≥ 0, y ≥ 0

– p. 19/29

Transportation Words → MathMinimize the cost C = −3x − 2y + 375subject to the constraints

• x ≤ 25, y ≤ 30

• 15 ≤ x + y

• x + y ≤ 45

• x ≥ 0, y ≥ 0

– p. 19/29


• x ≤ 25, y ≤ 30

• 15 ≤ x + y

• x + y ≤ 45

• x ≥ 0, y ≥ 0

– p. 19/29


• x ≤ 25, y ≤ 30

• 15 ≤ x + y

• x + y ≤ 45

• x ≥ 0, y ≥ 0

– p. 19/29


• x ≤ 25, y ≤ 30

• 15 ≤ x + y

• x + y ≤ 45

• x ≥ 0, y ≥ 0

– p. 19/29


• x ≤ 25, y ≤ 30

• 15 ≤ x + y

• x + y ≤ 45

• x ≥ 0, y ≥ 0

– p. 19/29

The HalfplanesHalfplane Line Intercept(s) Position15 ≤ x + y

15 = x + y (15, 0) (0, 15) abovex + y ≤ 45 x + y = 45 (45, 0) (0, 45) below

y ≤ 30 y = 30 (0, 30) belowx ≤ 25 x = 25 (25, 0) left

The constraints x, y ≥ 0 imply that the region lies inthe First Quadrant.

– p. 20/29

The HalfplanesHalfplane Line Intercept(s) Position15 ≤ x + y 15 = x + y

(15, 0) (0, 15) abovex + y ≤ 45 x + y = 45 (45, 0) (0, 45) below

y ≤ 30 y = 30 (0, 30) belowx ≤ 25 x = 25 (25, 0) left


– p. 20/29

The HalfplanesHalfplane Line Intercept(s) Position15 ≤ x + y 15 = x + y (15, 0) (0, 15)

abovex + y ≤ 45 x + y = 45 (45, 0) (0, 45) below

y ≤ 30 y = 30 (0, 30) belowx ≤ 25 x = 25 (25, 0) left


– p. 20/29

The HalfplanesHalfplane Line Intercept(s) Position15 ≤ x + y 15 = x + y (15, 0) (0, 15) above

x + y ≤ 45 x + y = 45 (45, 0) (0, 45) belowy ≤ 30 y = 30 (0, 30) belowx ≤ 25 x = 25 (25, 0) left


– p. 20/29

The HalfplanesHalfplane Line Intercept(s) Position15 ≤ x + y 15 = x + y (15, 0) (0, 15) abovex + y ≤ 45

x + y = 45 (45, 0) (0, 45) belowy ≤ 30 y = 30 (0, 30) belowx ≤ 25 x = 25 (25, 0) left


– p. 20/29

The HalfplanesHalfplane Line Intercept(s) Position15 ≤ x + y 15 = x + y (15, 0) (0, 15) abovex + y ≤ 45 x + y = 45

(45, 0) (0, 45) belowy ≤ 30 y = 30 (0, 30) belowx ≤ 25 x = 25 (25, 0) left


– p. 20/29

The HalfplanesHalfplane Line Intercept(s) Position15 ≤ x + y 15 = x + y (15, 0) (0, 15) abovex + y ≤ 45 x + y = 45 (45, 0) (0, 45)

belowy ≤ 30 y = 30 (0, 30) belowx ≤ 25 x = 25 (25, 0) left


– p. 20/29

The HalfplanesHalfplane Line Intercept(s) Position15 ≤ x + y 15 = x + y (15, 0) (0, 15) abovex + y ≤ 45 x + y = 45 (45, 0) (0, 45) below

y ≤ 30 y = 30 (0, 30) belowx ≤ 25 x = 25 (25, 0) left


– p. 20/29


y ≤ 30

y = 30 (0, 30) belowx ≤ 25 x = 25 (25, 0) left


– p. 20/29


y ≤ 30 y = 30

(0, 30) belowx ≤ 25 x = 25 (25, 0) left


– p. 20/29


y ≤ 30 y = 30 (0, 30)

belowx ≤ 25 x = 25 (25, 0) left


– p. 20/29


y ≤ 30 y = 30 (0, 30) below

x ≤ 25 x = 25 (25, 0) leftThe constraints x, y ≥ 0 imply that the region lies inthe First Quadrant.

– p. 20/29


y ≤ 30 y = 30 (0, 30) belowx ≤ 25

x = 25 (25, 0) leftThe constraints x, y ≥ 0 imply that the region lies inthe First Quadrant.

– p. 20/29


y ≤ 30 y = 30 (0, 30) belowx ≤ 25 x = 25

(25, 0) leftThe constraints x, y ≥ 0 imply that the region lies inthe First Quadrant.

– p. 20/29


y ≤ 30 y = 30 (0, 30) belowx ≤ 25 x = 25 (25, 0)

leftThe constraints x, y ≥ 0 imply that the region lies inthe First Quadrant.

– p. 20/29


y ≤ 30 y = 30 (0, 30) belowx ≤ 25 x = 25 (25, 0) left


– p. 20/29


y ≤ 30 y = 30 (0, 30) belowx ≤ 25 x = 25 (25, 0) left


– p. 20/29

Feasible Region

(0,0)p

Line 1:x + y = 15

15

15

p

p

Line 1:x + y = 45

45

45

p

p

Line 3:y = 30

(15, 30)30 pp

Line 4:x = 25

(25, 20)

25p

p

– p. 21/29

Feasible Region

(0,0)p

Line 1:x + y = 15

15

15

p

p

Line 1:x + y = 45

45

45

p

p

Line 3:y = 30

(15, 30)30 pp

Line 4:x = 25

(25, 20)

25p

p

– p. 21/29

Feasible Region

(0,0)p

Line 1:x + y = 15

15

15

p

p

Line 1:x + y = 45

45

45

p

p

Line 3:y = 30

(15, 30)30 pp

Line 4:x = 25

(25, 20)

25p

p

– p. 21/29

Feasible Region

(0,0)p

Line 1:x + y = 15

15

15

p

p

Line 1:x + y = 45

45

45

p

p

Line 3:y = 30

(15, 30)30 pp

Line 4:x = 25

(25, 20)

25p

p

– p. 21/29

Feasible Region

(0,0)p

Line 1:x + y = 15

15

15

p

p

Line 1:x + y = 45

45

45

p

p

Line 3:y = 30

(15, 30)30 pp

Line 4:x = 25

(25, 20)

25p

p– p. 21/29

Feasible Region

(0,0)p

Line 1:x + y = 15

15

15

p

p

Line 1:x + y = 45

45

45

p

p

Line 3:y = 30

(15, 30)30 pp

Line 4:x = 25

(25, 20)

25p

p– p. 21/29

Solving the TransportationProblemThe list of corner points is:

(0, 15), (0, 30), (15, 30), (25, 20), (25, 0) and (15, 0)

x y C = 375 − 3x − 2y

0 15 3450 30 315

15 30 27025 20 26025 0 30015 0 330

Minimum value of C is• 260 and occurs at the point (25, 20)

– p. 22/29

Solving the TransportationProblemThe list of corner points is:(0, 15), (0, 30), (15, 30), (25, 20), (25, 0) and (15, 0)

x y C = 375 − 3x − 2y

0 15 3450 30 315

15 30 27025 20 26025 0 30015 0 330


– p. 22/29


x y C = 375 − 3x − 2y

0 15

3450 30 315

15 30 27025 20 26025 0 30015 0 330


– p. 22/29


x y C = 375 − 3x − 2y

0 15

345

0 30

31515 30 27025 20 26025 0 30015 0 330


– p. 22/29


x y C = 375 − 3x − 2y

0 15

345

0 30

315

15 30

27025 20 26025 0 30015 0 330


– p. 22/29


x y C = 375 − 3x − 2y

0 15

345

0 30

315

15 30

270

25 20

26025 0 30015 0 330


– p. 22/29


x y C = 375 − 3x − 2y

0 15

345

0 30

315

15 30

270

25 20

260

25 0

30015 0 330


– p. 22/29


x y C = 375 − 3x − 2y

0 15

345

0 30

315

15 30

270

25 20

260

25 0

300

15 0

330


– p. 22/29


x y C = 375 − 3x − 2y

0 15 3450 30

315

15 30

270

25 20

260

25 0

300

15 0

330


– p. 22/29


x y C = 375 − 3x − 2y

0 15 3450 30 315

15 30

270

25 20

260

25 0

300

15 0

330


– p. 22/29


x y C = 375 − 3x − 2y

0 15 3450 30 315

15 30 27025 20

260

25 0

300

15 0

330


– p. 22/29


x y C = 375 − 3x − 2y

0 15 3450 30 315

15 30 27025 20 26025 0

300

15 0

330


– p. 22/29


x y C = 375 − 3x − 2y

0 15 3450 30 315

15 30 27025 20 26025 0 30015 0

330


– p. 22/29


x y C = 375 − 3x − 2y

0 15 3450 30 315

15 30 27025 20 26025 0 30015 0 330


– p. 22/29


x y C = 375 − 3x − 2y

0 15 3450 30 315

15 30 27025 20 26025 0 30015 0 330

Minimum value of C is

• 260 and occurs at the point (25, 20)

– p. 22/29


x y C = 375 − 3x − 2y

0 15 3450 30 315

15 30 27025 20 26025 0 30015 0 330

Minimum value of C is• 260 and occurs at the point

(25, 20)

– p. 22/29


x y C = 375 − 3x − 2y

0 15 3450 30 315

15 30 27025 20 26025 0 30015 0 330


– p. 22/29

Simplex Method:• For each inequality introduce a slack variable to

convert the inequality into an equation.

• Write each slack equation as a row in the SimplexTableau (French for “matrix”)

• Write the function to be maximized as the bottomrow of the Simplex Tableau

• Find the Pivot Point (row and column)• Pivot and check the tableau to see if you are done.

– p. 23/29


convert the inequality into an equation.• Write each slack equation as a row in the Simplex

Tableau (French for “matrix”)

• Write the function to be maximized as the bottomrow of the Simplex Tableau


– p. 23/29



Tableau (French for “matrix”)• Write the function to be maximized as the bottom

row of the Simplex Tableau


– p. 23/29




row of the Simplex Tableau• Find the Pivot Point (row and column)

• Pivot and check the tableau to see if you are done.

– p. 23/29




row of the Simplex Tableau• Find the Pivot Point (row and column)• Pivot and check the tableau to see if you are done.

– p. 23/29

Step 1: Slack VariablesThis is easier to illustrate that to explain.

Consider the inequality:

x + y ≤ 16

Rewrite this inequality as the equation:

x + y + u = 16

Note that

• u = 16 − x − y

• x + y ≤ 16 is equivalent to 0 ≤ 16 − x − y oru ≥ 0

– p. 24/29

Step 1: Slack VariablesThis is easier to illustrate that to explain.Consider the inequality:

x + y ≤ 16


x + y + u = 16

Note that

• u = 16 − x − y


– p. 24/29


x + y ≤ 16


x + y + u = 16

Note that

• u = 16 − x − y


– p. 24/29


x + y ≤ 16


x + y + u = 16

Note that

• u = 16 − x − y


– p. 24/29


x + y ≤ 16


x + y + u = 16

Note that

• u = 16 − x − y


– p. 24/29


x + y ≤ 16


x + y + u = 16

Note that

• u = 16 − x − y


– p. 24/29


x + y ≤ 16


x + y + u = 16

Note that

• u = 16 − x − y


– p. 24/29


x + y ≤ 16


x + y + u = 16

Note that

• u = 16 − x − y


– p. 24/29

Step 1: Slack VariablesEach inequality in your system requires its own slackvariable.

The system of inequalities

1. x + y ≤ 16

2. 5x + 2y ≤ 50

3. y ≤ 12

can be rewritten as

1. x + y + u = 16

2. 5x + 2y + v = 50

3. y + w = 12

– p. 25/29

Step 1: Slack VariablesEach inequality in your system requires its own slackvariable.The system of inequalities

1. x + y ≤ 16

2. 5x + 2y ≤ 50

3. y ≤ 12

can be rewritten as

1. x + y + u = 16

2. 5x + 2y + v = 50

3. y + w = 12

– p. 25/29


1. x + y ≤ 16

2. 5x + 2y ≤ 50

3. y ≤ 12

can be rewritten as

1. x + y + u = 16

2. 5x + 2y + v = 50

3. y + w = 12

– p. 25/29


1. x + y ≤ 16

2. 5x + 2y ≤ 50

3. y ≤ 12

can be rewritten as

1. x + y + u = 16

2. 5x + 2y + v = 50

3. y + w = 12

– p. 25/29


1. x + y ≤ 16

2. 5x + 2y ≤ 50

3. y ≤ 12

can be rewritten as

1. x + y + u = 16

2. 5x + 2y + v = 50

3. y + w = 12

– p. 25/29


1. x + y ≤ 16

2. 5x + 2y ≤ 50

3. y ≤ 12

can be rewritten as

1. x + y + u = 16

2. 5x + 2y + v = 50

3. y + w = 12

– p. 25/29


1. x + y ≤ 16

2. 5x + 2y ≤ 50

3. y ≤ 12

can be rewritten as

1. x + y + u = 16

2. 5x + 2y + v = 50

3. y + w = 12

– p. 25/29


1. x + y ≤ 16

2. 5x + 2y ≤ 50

3. y ≤ 12

can be rewritten as

1. x + y + u = 16

2. 5x + 2y + v = 50

3. y + w = 12

– p. 25/29


1. x + y ≤ 16

2. 5x + 2y ≤ 50

3. y ≤ 12

can be rewritten as

1. x + y + u = 16

2. 5x + 2y + v = 50

3. y + w = 12

– p. 25/29

Step 2: The Simplex TableauThe Simplex Tableau has a row for each inequality inthe system,

plus an additonal row at the bottom for theobjective function.There is a column for each regular variable, for eachslack variable, and for the objective function variable(usually Profit or Cost).The last column holds the constants to the right of theequal sign for each equation.

– p. 26/29

Step 2: The Simplex TableauThe Simplex Tableau has a row for each inequality inthe system, plus an additonal row at the bottom for theobjective function.

There is a column for each regular variable, for eachslack variable, and for the objective function variable(usually Profit or Cost).The last column holds the constants to the right of theequal sign for each equation.

– p. 26/29

Step 2: The Simplex TableauThe Simplex Tableau has a row for each inequality inthe system, plus an additonal row at the bottom for theobjective function.There is a column for each regular variable,

for eachslack variable, and for the objective function variable(usually Profit or Cost).The last column holds the constants to the right of theequal sign for each equation.

– p. 26/29

Step 2: The Simplex TableauThe Simplex Tableau has a row for each inequality inthe system, plus an additonal row at the bottom for theobjective function.There is a column for each regular variable, for eachslack variable, and

for the objective function variable(usually Profit or Cost).The last column holds the constants to the right of theequal sign for each equation.

– p. 26/29

Step 2: The Simplex TableauThe Simplex Tableau has a row for each inequality inthe system, plus an additonal row at the bottom for theobjective function.There is a column for each regular variable, for eachslack variable, and for the objective function variable(usually Profit or Cost).

The last column holds the constants to the right of theequal sign for each equation.

– p. 26/29

Step 2: The Simplex TableauThe Simplex Tableau has a row for each inequality inthe system, plus an additonal row at the bottom for theobjective function.There is a column for each regular variable, for eachslack variable, and for the objective function variable(usually Profit or Cost).The last column holds the constants to the right of theequal sign for each equation.

– p. 26/29

Simplex Tableau IllustrationThe first three rows of the Simplex Tableau for

x + y + u = 165x + 2y + v = 50y + w = 12arex y u v w P const1 1 1 0 0 0 165 2 0 1 0 0 500 1 0 0 1 0 12

– p. 27/29

Simplex Tableau IllustrationThe first three rows of the Simplex Tableau forx + y + u = 165x + 2y + v = 50y + w = 12

arex y u v w P const1 1 1 0 0 0 165 2 0 1 0 0 500 1 0 0 1 0 12

– p. 27/29

Simplex Tableau IllustrationThe first three rows of the Simplex Tableau forx + y + u = 165x + 2y + v = 50y + w = 12arex y u v w P const

1 1 1 0 0 0 165 2 0 1 0 0 500 1 0 0 1 0 12

– p. 27/29

Simplex Tableau IllustrationThe first three rows of the Simplex Tableau forx + y + u = 165x + 2y + v = 50y + w = 12arex y u v w P const1

1 1 0 0 0 165 2 0 1 0 0 500 1 0 0 1 0 12

– p. 27/29

Simplex Tableau IllustrationThe first three rows of the Simplex Tableau forx + y + u = 165x + 2y + v = 50y + w = 12arex y u v w P const1 1

1 0 0 0 165 2 0 1 0 0 500 1 0 0 1 0 12

– p. 27/29

Simplex Tableau IllustrationThe first three rows of the Simplex Tableau forx + y + u = 165x + 2y + v = 50y + w = 12arex y u v w P const1 1 1

0 0 0 165 2 0 1 0 0 500 1 0 0 1 0 12

– p. 27/29

Simplex Tableau IllustrationThe first three rows of the Simplex Tableau forx + y + u = 165x + 2y + v = 50y + w = 12arex y u v w P const1 1 1 0

0 0 165 2 0 1 0 0 500 1 0 0 1 0 12

– p. 27/29

Simplex Tableau IllustrationThe first three rows of the Simplex Tableau forx + y + u = 165x + 2y + v = 50y + w = 12arex y u v w P const1 1 1 0 0

0 165 2 0 1 0 0 500 1 0 0 1 0 12

– p. 27/29

Simplex Tableau IllustrationThe first three rows of the Simplex Tableau forx + y + u = 165x + 2y + v = 50y + w = 12arex y u v w P const1 1 1 0 0 0

165 2 0 1 0 0 500 1 0 0 1 0 12

– p. 27/29

Simplex Tableau IllustrationThe first three rows of the Simplex Tableau forx + y + u = 165x + 2y + v = 50y + w = 12arex y u v w P const1 1 1 0 0 0 16

5 2 0 1 0 0 500 1 0 0 1 0 12

– p. 27/29

Simplex Tableau IllustrationThe first three rows of the Simplex Tableau forx + y + u = 165x + 2y + v = 50y + w = 12arex y u v w P const1 1 1 0 0 0 165

2 0 1 0 0 500 1 0 0 1 0 12

– p. 27/29

Simplex Tableau IllustrationThe first three rows of the Simplex Tableau forx + y + u = 165x + 2y + v = 50y + w = 12arex y u v w P const1 1 1 0 0 0 165 2

0 1 0 0 500 1 0 0 1 0 12

– p. 27/29

Simplex Tableau IllustrationThe first three rows of the Simplex Tableau forx + y + u = 165x + 2y + v = 50y + w = 12arex y u v w P const1 1 1 0 0 0 165 2 0

1 0 0 500 1 0 0 1 0 12

– p. 27/29

Simplex Tableau IllustrationThe first three rows of the Simplex Tableau forx + y + u = 165x + 2y + v = 50y + w = 12arex y u v w P const1 1 1 0 0 0 165 2 0 1

0 0 500 1 0 0 1 0 12

– p. 27/29

Simplex Tableau IllustrationThe first three rows of the Simplex Tableau forx + y + u = 165x + 2y + v = 50y + w = 12arex y u v w P const1 1 1 0 0 0 165 2 0 1 0

0 500 1 0 0 1 0 12

– p. 27/29

Simplex Tableau IllustrationThe first three rows of the Simplex Tableau forx + y + u = 165x + 2y + v = 50y + w = 12arex y u v w P const1 1 1 0 0 0 165 2 0 1 0 0

500 1 0 0 1 0 12

– p. 27/29

Simplex Tableau IllustrationThe first three rows of the Simplex Tableau forx + y + u = 165x + 2y + v = 50y + w = 12arex y u v w P const1 1 1 0 0 0 165 2 0 1 0 0 50

0 1 0 0 1 0 12

– p. 27/29

Simplex Tableau IllustrationThe first three rows of the Simplex Tableau forx + y + u = 165x + 2y + v = 50y + w = 12arex y u v w P const1 1 1 0 0 0 165 2 0 1 0 0 500

1 0 0 1 0 12

– p. 27/29

Simplex Tableau IllustrationThe first three rows of the Simplex Tableau forx + y + u = 165x + 2y + v = 50y + w = 12arex y u v w P const1 1 1 0 0 0 165 2 0 1 0 0 500 1

0 0 1 0 12

– p. 27/29

Simplex Tableau IllustrationThe first three rows of the Simplex Tableau forx + y + u = 165x + 2y + v = 50y + w = 12arex y u v w P const1 1 1 0 0 0 165 2 0 1 0 0 500 1

0 0

1

0 12

– p. 27/29

Simplex Tableau IllustrationThe first three rows of the Simplex Tableau forx + y + u = 165x + 2y + v = 50y + w = 12arex y u v w P const1 1 1 0 0 0 165 2 0 1 0 0 500 1 0 0 1 0

12

– p. 27/29

Simplex Tableau IllustrationThe first three rows of the Simplex Tableau forx + y + u = 165x + 2y + v = 50y + w = 12arex y u v w P const1 1 1 0 0 0 165 2 0 1 0 0 500 1 0 0 1 0 12

– p. 27/29

Rewriting the Objective Func-tionSuppose the objective function is

P = 7x + 5y

We rewrite this as

−7x + −5y + P = 0

and add this equation as the bottom row of the

Simplex Tableau

x y u v w P const1 1 1 0 0 0 165 2 0 1 0 0 500 1 0 0 1 0 12−7 −5 0 0 0 1 0

– p. 28/29


P = 7x + 5y

We rewrite this as

−7x + −5y + P = 0


Simplex Tableau

x y u v w P const1 1 1 0 0 0 165 2 0 1 0 0 500 1 0 0 1 0 12−7 −5 0 0 0 1 0

– p. 28/29


P = 7x + 5y

We rewrite this as

−7x + −5y + P = 0


Simplex Tableau

x y u v w P const1 1 1 0 0 0 165 2 0 1 0 0 500 1 0 0 1 0 12−7 −5 0 0 0 1 0

– p. 28/29


P = 7x + 5y

We rewrite this as

−7x + −5y + P = 0


Simplex Tableau

x y u v w P const1 1 1 0 0 0 165 2 0 1 0 0 500 1 0 0 1 0 12−7 −5 0 0 0 1 0

– p. 28/29


P = 7x + 5y

We rewrite this as

−7x + −5y + P = 0


Simplex Tableau

x y u v w P const

1 1 1 0 0 0 165 2 0 1 0 0 500 1 0 0 1 0 12−7 −5 0 0 0 1 0

– p. 28/29


P = 7x + 5y

We rewrite this as

−7x + −5y + P = 0


Simplex Tableau

x y u v w P const1 1 1 0 0 0 16

5 2 0 1 0 0 500 1 0 0 1 0 12−7 −5 0 0 0 1 0

– p. 28/29


P = 7x + 5y

We rewrite this as

−7x + −5y + P = 0


Simplex Tableau

x y u v w P const1 1 1 0 0 0 165 2 0 1 0 0 50

0 1 0 0 1 0 12−7 −5 0 0 0 1 0

– p. 28/29


P = 7x + 5y

We rewrite this as

−7x + −5y + P = 0


Simplex Tableau

x y u v w P const1 1 1 0 0 0 165 2 0 1 0 0 500 1 0 0 1 0 12

−7 −5 0 0 0 1 0

– p. 28/29


P = 7x + 5y

We rewrite this as

−7x + −5y + P = 0


Simplex Tableau

x y u v w P const1 1 1 0 0 0 165 2 0 1 0 0 500 1 0 0 1 0 12−7

−5 0 0 0 1 0

– p. 28/29


P = 7x + 5y

We rewrite this as

−7x + −5y + P = 0


Simplex Tableau

x y u v w P const1 1 1 0 0 0 165 2 0 1 0 0 500 1 0 0 1 0 12−7 −5

0 0 0 1 0

– p. 28/29


P = 7x + 5y

We rewrite this as

−7x + −5y + P = 0


Simplex Tableau

x y u v w P const1 1 1 0 0 0 165 2 0 1 0 0 500 1 0 0 1 0 12−7 −5 0

0 0 1 0

– p. 28/29


P = 7x + 5y

We rewrite this as

−7x + −5y + P = 0


Simplex Tableau

x y u v w P const1 1 1 0 0 0 165 2 0 1 0 0 500 1 0 0 1 0 12−7 −5 0 0

0 1 0

– p. 28/29


P = 7x + 5y

We rewrite this as

−7x + −5y + P = 0


Simplex Tableau

x y u v w P const1 1 1 0 0 0 165 2 0 1 0 0 500 1 0 0 1 0 12−7 −5 0 0 0

1 0

– p. 28/29


P = 7x + 5y

We rewrite this as

−7x + −5y + P = 0


Simplex Tableau

x y u v w P const1 1 1 0 0 0 165 2 0 1 0 0 500 1 0 0 1 0 12−7 −5 0 0 0 1

0

– p. 28/29


P = 7x + 5y

We rewrite this as

−7x + −5y + P = 0


Simplex Tableau

x y u v w P const1 1 1 0 0 0 165 2 0 1 0 0 500 1 0 0 1 0 12−7 −5 0 0 0 1 0

– p. 28/29

Basic and Non-basic VariablesNotice that in the Simplex Tableau

x y u v w P const1 1 1 0 0 0 165 2 0 1 0 0 500 1 0 0 1 0 12−7 −5 0 0 0 1 0

the columns under

the variables u, v, w, and P are unit columns meaningthe column consists of a single 1 and the other entriesare zero. The variables corresponding to unitcolumns—u, v, w, and P—are called basic variables;the other variables—x and y are called non-basicvariables.If we set the non-basic variables to zero, then it iseasy to read the values of the basic variables:u = 16, v = 50, w = 12, P = 0

– p. 29/29


x y u v w P const1 1 1 0 0 0 165 2 0 1 0 0 500 1 0 0 1 0 12−7 −5 0 0 0 1 0

the columns under


– p. 29/29


x y u v w P const1 1 1 0 0 0 165 2 0 1 0 0 500 1 0 0 1 0 12−7 −5 0 0 0 1 0

the columns under

the variables u, v, w, and P are unit columns

meaningthe column consists of a single 1 and the other entriesare zero. The variables corresponding to unitcolumns—u, v, w, and P—are called basic variables;the other variables—x and y are called non-basicvariables.If we set the non-basic variables to zero, then it iseasy to read the values of the basic variables:u = 16, v = 50, w = 12, P = 0

– p. 29/29


x y u v w P const1 1 1 0 0 0 165 2 0 1 0 0 500 1 0 0 1 0 12−7 −5 0 0 0 1 0

the columns under

the variables u, v, w, and P are unit columns meaningthe column consists of a single 1 and the other entriesare zero.

The variables corresponding to unitcolumns—u, v, w, and P—are called basic variables;the other variables—x and y are called non-basicvariables.If we set the non-basic variables to zero, then it iseasy to read the values of the basic variables:u = 16, v = 50, w = 12, P = 0

– p. 29/29


x y u v w P const1 1 1 0 0 0 165 2 0 1 0 0 500 1 0 0 1 0 12−7 −5 0 0 0 1 0

the columns under

the variables u, v, w, and P are unit columns meaningthe column consists of a single 1 and the other entriesare zero. The variables corresponding to unitcolumns—u, v, w, and P—are called basic variables;

the other variables—x and y are called non-basicvariables.If we set the non-basic variables to zero, then it iseasy to read the values of the basic variables:u = 16, v = 50, w = 12, P = 0

– p. 29/29


x y u v w P const1 1 1 0 0 0 165 2 0 1 0 0 500 1 0 0 1 0 12−7 −5 0 0 0 1 0

the columns under

the variables u, v, w, and P are unit columns meaningthe column consists of a single 1 and the other entriesare zero. The variables corresponding to unitcolumns—u, v, w, and P—are called basic variables;the other variables—x and y are called non-basicvariables.

If we set the non-basic variables to zero, then it iseasy to read the values of the basic variables:u = 16, v = 50, w = 12, P = 0

– p. 29/29


x y u v w P const1 1 1 0 0 0 165 2 0 1 0 0 500 1 0 0 1 0 12−7 −5 0 0 0 1 0

the columns under

the variables u, v, w, and P are unit columns meaningthe column consists of a single 1 and the other entriesare zero. The variables corresponding to unitcolumns—u, v, w, and P—are called basic variables;the other variables—x and y are called non-basicvariables.If we set the non-basic variables to zero, then it iseasy to read the values of the basic variables:

u = 16, v = 50, w = 12, P = 0

– p. 29/29


x y u v w P const1 1 1 0 0 0 165 2 0 1 0 0 500 1 0 0 1 0 12−7 −5 0 0 0 1 0

the columns under

the variables u, v, w, and P are unit columns meaningthe column consists of a single 1 and the other entriesare zero. The variables corresponding to unitcolumns—u, v, w, and P—are called basic variables;the other variables—x and y are called non-basicvariables.If we set the non-basic variables to zero, then it iseasy to read the values of the basic variables:u = 16,

v = 50, w = 12, P = 0

– p. 29/29


x y u v w P const1 1 1 0 0 0 165 2 0 1 0 0 500 1 0 0 1 0 12−7 −5 0 0 0 1 0

the columns under

the variables u, v, w, and P are unit columns meaningthe column consists of a single 1 and the other entriesare zero. The variables corresponding to unitcolumns—u, v, w, and P—are called basic variables;the other variables—x and y are called non-basicvariables.If we set the non-basic variables to zero, then it iseasy to read the values of the basic variables:u = 16, v = 50,

w = 12, P = 0

– p. 29/29


x y u v w P const1 1 1 0 0 0 165 2 0 1 0 0 500 1 0 0 1 0 12−7 −5 0 0 0 1 0

the columns under

the variables u, v, w, and P are unit columns meaningthe column consists of a single 1 and the other entriesare zero. The variables corresponding to unitcolumns—u, v, w, and P—are called basic variables;the other variables—x and y are called non-basicvariables.If we set the non-basic variables to zero, then it iseasy to read the values of the basic variables:u = 16, v = 50, w = 12,

P = 0

– p. 29/29


x y u v w P const1 1 1 0 0 0 165 2 0 1 0 0 500 1 0 0 1 0 12−7 −5 0 0 0 1 0

the columns under


– p. 29/29

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Math 210richard/Math210/Lp/lp2.pdf · provides certain minimum levels of protein, calories, and vitamin B12 (riboa vin). One cup of uncooked rice costs 21 cents and contains 15 grams