Chapter 3 Section 2

Linear Programming I

Fundamental Theorem of Linear Programming

• The maximum (or minimum) value of the objective function is achieved at one of the vertices

Exercise 21 (page 131)

• Minimize the objective function 3x + 4y subject to the below constraints.

• Solution:2x + y > 10 y > – 2x + 10

x + 2y > 14 y > – ½ x + 7

x > 0 x > 0

y > 0 y > 0

Graph of the Inequalities

y = – 2x + 10

y = – ½ x + 7

y = 0

x = 0

I

II

III

Finding the Vertices (by hand)

Vertex I Vertex II Vertex III

x = 0 y = – 2x + 10 y = – ½ x + 7

y = – 2x + 10 y = – ½ x + 7 y = 0

y = – 2(0) + 10 – 2x + 10 = – ½ x + 7 ( 0 ) = – ½ x + 7

y = 10 3/2 x = 3 ½ x = 7

x = 2 x = 14

( 0 , 10 )

y = – 2( 2 ) + 10 ( 14 , 0 )

y = 6

( 2 , 6 )

Find the Optimal Point

Vertex Objective Function: 3 x + 4 y

( 0 , 10 ) 3 ( 0 ) + 4 ( 10 ) = 40

( 2 , 6 ) 3 ( 2 ) + 4 ( 6 ) = 30

( 14 , 0 ) 3 ( 14 ) + 4 ( 0 ) = 42

The minimum value of the objective function occurs at the vertex ( 2 , 6 )

Exercise 33 (page 132)

• Define the variables being used. Look at the question being asked!

• Last sentence: “How many cans of Fruit Delight and Heavenly Punch should be produced…?”

• Letx represent the number of cans of Fruit Delight producedy represent the number of cans of Heavenly Punch produced

• Last sentence: “How many cans of Fruit Delight and Heavenly Punch should be produced each week to maximize profits?”

• The objective function is:

Profit = 0.20 · x + 0.30 · y

The Objective Function

Table

Fruit

Delight

Heavenly

Punch

Maximum

Amount

Pineapple

Juice

Orange

Juice

Apricot

Juice

Profit

Table

Fruit

Delight

Heavenly

Punch

Maximum

Amount

Pineapple

Juice

10

oz/can

Orange

Juice

3

oz/can

Apricot

Juice

1

oz/can

Profit 0.20

$

Table

Fruit

Delight

Heavenly

Punch

Maximum

Amount

Pineapple

Juice

10

oz/can

10

oz/can

Orange

Juice

3

oz/can

2

oz/can

Apricot

Juice

1

oz/can

2

oz/can

Profit 0.20

$

0.30

$

Table

Fruit

Delight

Heavenly

Punch

Maximum

Amount

Pineapple

Juice

10

oz/can

10

oz/can

9,000

oz

Orange

Juice

3

oz/can

2

oz/can

2,400

oz

Apricot

Juice

1

oz/can

2

oz/can

1,400

oz

Profit 0.20

$

0.30

$

Restrictions from the Table and Problem

• Restrictions that are placed on the x and y variables:

10 x + 10 y < 9,000

3 x + 2 y < 2,400

x + 2 y < 1,400

x > 0

y > 0

Change From General Form to Standard Form

10 x + 10 y < 9,000 y < – x + 900

3 x + 2 y < 2,400 y < – 3/2 x + 1,200

x + 2 y < 1,400 y < – ½ x + 700

x > 0 x > 0

y > 0 y > 0

Graph of the System of Equations

Not to scale

y = – 3/2 x + 1200 y = – x + 900

y = – 1/2 x + 700

y = 0

x = 0

Shading for the Inequalities

Not to scale

y = – 3/2 x + 1200 y = – x + 900

y = – 1/2 x + 700

y = 0

x = 0

A Modified Graph of the Feasible Set from the Previous Slide

y = 0

y = – 1/2 x + 700

y = – x + 900

y = – 3/2 x + 1200

x = 0

I

II

III

IV

V

Feasible Set

Finding the Vertices (using a calculator when possible)

• Vertex I x = 0 ( 0 , 0 )y = 0

• Vertex II x = 0 ( 0 , 700 )y = – ½ x + 700

• Vertex III y = – ½ x + 700 ( 400 , 500 )y = – x + 900

• Vertex IV y = – x + 900 ( 600 , 300 )y = – 3/2 x + 1200

• Vertex V y = – 3/2 x + 1200 ( 800 , 0 )y = 0

Find the Optimal Point

Vertex Objective Function 0.2 x + 0.3 y

( 0 , 0 ) 0.2 ( 0 ) + 0.3 ( 0 ) = 0

( 0 , 700 ) 0.2 ( 0 ) + 0.3 ( 700 ) = 210

( 400 , 500 ) 0.2 ( 400 ) + 0.3 ( 500 ) = 230

( 600 , 300 ) 0.2 ( 600 ) + 0.3 ( 300 ) = 210

( 800 , 0 ) 0.2 ( 800 ) + 0.3 ( 0 ) = 160

• ( 400 , 500 ) maximizes the objective function

Answer

Produce 400 cans of Fruit Delight and

500 cans of Heavenly Punch to maximize profits

Exercise 35 (page 133)

• Define the variables being used. Look at the question being asked!

• Key Question: “…, what planting combination will produce the greatest total profit?”

• Letx represent the number of acres of oats plantedy represent the number of acres of corn planted

• This now defines Column Headings

The Objective Function

First we need to recognize these relationships

• Profit = Revenue + Left over capital cash + Left over labor costs

• Where – Revenue = 55 x + 125 y– Left over capital cash = 2100 – 18 x – 36 y– Left over labor cash = 2400 – 16 x – 48 y

The Objective Function

• Now add revenue, left over capital cash, and left over labor costs together [i.e. ( 55 x + 125 y ) + (2100 – 18 x – 36 y ) + ( 2400 – 16 x – 48 y )] to get the profit and the objective function becomes:

Profit = 4500 + 21 x + 41 y

Table

Oats Corn Available

Capital

Needed ($)

Labor

Needed ($)

Revenue ($)

Table

Oats Corn Available

Capital

Needed ($)

18

$/acre

36

$/acre

2,100

$

Labor

Needed ($)

16

$/acre

48

$/acre

2,400

$

Revenue ($) 55

$/acre

125

$/acre

Restrictions from Table and Problem

• The inequalities that restrict the values of the variables:

18 x + 36 y < 2,100

16 x + 48 y < 2,400

x + y < 100

x > 0

y > 0

Why do we need this restriction?

Why x + y < 100?

• The farmer has only 100 acres available for the two crops. The amount of oats plus the amount of corn that the farmer plants has to be 100 acres or less.

Convert from General Form to Standard Form

• Restrictions:

18 x + 36 y < 2,100 y < – ½ x + 175/3

16 x + 48 y < 2,400 y < – 1/3 x + 50

x + y < 100 y < – x + 100

x > 0 x > 0

y > 0 y > 0

Graph of the System of Equations

Not to scale

y = – x + 100y = – ½ x + 175/3

y = – 1/3 x + 50

y = 0

x = 0

Graph of the System of Inequalities

Not to scale

y = – x + 100y = – ½ x + 175/3

y = – 1/3 x + 50

y = 0

x = 0

Modified Graph of the Feasible Set form the Previous Slide

y = 0

y = – 1/3 x + 50

y = – ½ x + 175/3

y = – x + 100

x = 0

I

II

III

IV

V

Feasible Set

Finding the Vertices (using a calculator when possible)

• Vertex I x = 0 ( 0 , 0 )

y = 0

• Vertex II x = 0 ( 0 , 50 )

y = – 1/3 x + 50

• Vertex III y = – 1/3 x + 50 ( 50 , 100/3 )

y = – ½ x + 175/3

• Vertex IV y = – ½ x + 175/3 ( 250/3 , 50/3)

y = – x + 100

• Vertex V y = – x + 100 ( 100 , 0 )

y = 0

Finding the Optimal Point

Vertex Objective Function 4500 + 21x + 41y( 0 , 0 ) 4500 + 21 ( 0 ) + 41 ( 0 ) = 4,500.00

( 0 , 50 ) 4500 + 21 ( 0 ) + 41 ( 50 ) = 6,550.00

( 50 , 100/3 ) 4500 + 21 ( 50 ) + 41 ( 100/3 ) ~ 6,916.67

( 250/3 , 50/3 ) 4500 + 21 ( 250/3 ) + 41 ( 50/3 ) ~ 6,933.33

( 100 , 0 ) 4500 + 21 ( 100 ) + 41 ( 0 ) = 6,600.00

• ( 250/3 , 50/3 ) = ( 83 1/3 , 16 2/3 ) maximizes the objective function

The Answer

Plant 83-and-a-third acres of oats and 16-and-two-thirds acres of corn to maximize the profit