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Chapter 3 Section 2 Linear Programming I
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
