Chapter 3 Section 2

1
Chapter 3 Section 2

• Linear Programming I

2
Fundamental Theorem of Linear Programming

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

3
Exercise 21 (page 131)

• Minimize the objective function 3x 4y subject
  to the below constraints.
• Solution
• 2x y gt 10 y gt 2x 10
• x 2y gt 14 y gt ½ x 7
• x gt 0 x gt 0
• y gt 0 y gt 0

4
Graph of the Inequalities
x 0
I
y 2x 10
II
y ½ x 7
III
y 0
5
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 )

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 )
7
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?
• Let
• x represent the number of cans of Fruit Delight
  produced
• y represent the number of cans of Heavenly Punch
  produced

8
The Objective Function

• 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

9
Table
Fruit Delight Heavenly Punch Maximum Amount
Pineapple Juice
Orange Juice
Apricot Juice
Profit
10
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
11
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
12
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
13
Restrictions from the Table and Problem

• Restrictions that are placed on the x and y
  variables
• 10 x 10 y lt 9,000
• 3 x 2 y lt 2,400
• x 2 y lt 1,400
• x gt 0
• y gt 0

14
Change From General Form to Standard Form

• 10 x 10 y lt 9,000 y lt x 900
• 3 x 2 y lt 2,400 y lt 3/2 x 1,200
• x 2 y lt 1,400 y lt ½ x 700
• x gt 0 x gt 0
• y gt 0 y gt 0

15
Graph of the System of Equations
x 0
Not to scale
y 0
y 1/2 x 700
y 3/2 x 1200
y x 900
16
Shading for the Inequalities
x 0
Not to scale
y 0
y 1/2 x 700
y 3/2 x 1200
y x 900
17
A Modified Graph of the Feasible Set from the
Previous Slide
II
y 1/2 x 700
III
y x 900
IV
x 0
Feasible Set
y 3/2 x 1200
V
I
y 0
18
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

19
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

20
Answer

• Produce 400 cans of Fruit Delight and
• 500 cans of Heavenly Punch to maximize profits

21
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?
• Let
• x represent the number of acres of oats planted
• y represent the number of acres of corn planted
• This now defines Column Headings

22
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

23
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
24
Table
Oats Corn Available
Capital Needed ()
Labor Needed ()
Revenue ()
25
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
26
Restrictions from Table and Problem

• The inequalities that restrict the values of the
  variables
• 18 x 36 y lt 2,100
• 16 x 48 y lt 2,400
• x y lt 100
• x gt 0
• y gt 0

Why do we need this restriction?
27
Why x y lt 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.

28
Convert from General Form to Standard Form

• Restrictions
• 18 x 36 y lt 2,100 y lt ½ x 175/3
• 16 x 48 y lt 2,400 y lt 1/3 x 50
• x y lt 100 y lt x 100
• x gt 0 x gt 0
• y gt 0 y gt 0

29
Graph of the System of Equations
x 0
Not to scale
y 0
y 1/3 x 50
y x 100
y ½ x 175/3
30
Graph of the System of Inequalities
x 0
Not to scale
y 0
y 1/3 x 50
y x 100
y ½ x 175/3
31
Modified Graph of the Feasible Set form the
Previous Slide
II
y 1/3 x 50
III
y ½ x 175/3
IV
x 0
Feasible Set
y x 100
V
I
y 0
32
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

33
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

34
The Answer

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