3-4 Linear ProgrammingAlgebra II Unit 3 Linear Systems© Tentinger

Essential Understanding and Objectives

●Essential Understanding: Some real-world problems involve multiple linear relationships. Linear programming accounts for all of these linear relationships and gives the solution to the problem.

●Objectives:●Students will be able to:

●Solving problems using linear programming●Define constraint, linear programming, feasible region, and objective function

Iowa Core Curriculum• Algebra• A.CED.3 Represent constraints by equations or inequalities,

and by systems of equations and/or inequalities, and interpret solutions as viable or nonviable options in a modeling context. For example, represent inequalities describing nutritional and cost constraints on combinations of different foods.

Linear Programming

Businesses use linear programming to find out how to maximize profit or minimize costs. Most have constraints on what they can use or buy.

Linear Programming

Linear programming is a process of finding a maximum or minimum of a function by using coordinates of the polygon formed by the graph of the constraints.

What is a constraint?

A restriction...

A boundary…

A limitation…

What is the feasible region?

The feasible region is the area of the graph in which all the constraints are met.

Objective Function• The quantity you are trying to maximize or minimize is

modeled by this.• Usually this quantity is the cost or profit• Looks something like this C = ax + by, where a and b are real

numbers

Vertex Principle• If there is a maximum or minimum value of the linear objective

function, it occurs at one or more vertices of the feasible region.

Online Example

Find the minimum and maximumvalue of the function f(x, y) = 3x - 2y.

We are given the constraints:• y ≥ 2• 1 ≤ x ≤5 • y ≤ x + 3

Linear Programming

•Find the minimum and maximum values by graphing the inequalities and finding the vertices of the polygon formed. •Substitute the vertices into the function and find the largest and smallest values.

6

4

2

2 3 4

3

1

1

5

5

7

8

y ≤ x + 3

y ≥ 2

1 ≤ x ≤5

Linear Programming

•The vertices of the quadrilateral formed are:

(1, 2) (1, 4) (5, 2) (5, 8) •Plug these points into the objective function f(x, y) = 3x - 2y

Linear Programming

f(x, y) = 3x - 2y• f(1, 2) = 3(1) - 2(2) = 3 - 4 = -1 • f(1, 4) = 3(1) - 2(4) = 3 - 8 = -5• f(5, 2) = 3(5) - 2(2) = 15 - 4 = 11• f(5, 8) = 3(5) - 2(8) = 15 - 16 = -1

Linear Programming

•f(1, 4) = -5 minimum•f(5, 2) = 11 maximum

Find the minimum and maximum value of the function f(x, y) = 4x + 3y

We are given the constraints:• y ≥ -x + 2

• y ≤ x + 2

• y ≥ 2x -5

1

4

6

4

2

53 4

5

1

1

2

3y ≥ -x + 2

y ≥ 2x -5

y x 1

24

Vertices

f(x, y) = 4x + 3y• f(0, 2) = 4(0) + 3(2) = 6• f(4, 3) = 4(4) + 3(3) = 25

• f( , - ) = 4( ) + 3(- ) = -1 = 7

3

1

3

1

3

7

3

28

3

25

3

Linear Programming

•f(0, 2) = 6 minimum•f(4, 3) = 25 maximum

A farmer has 25 days to plant cotton and soybeans. The cotton can be planted at a rate of 9 acres per day, and the soybeans can be planted at a rate of 12 acres a day. The farmer has 275 acres available. If the profit for soybeans is $18 per acre and the profit for cotton is $25 per acre, how many acres of each crop should be planted to maximize profits?

Example 1

Step 1: Define the variables

What are the unknown values?

Let c = number of acres of cotton to plant

Let s = number of acres of soybeans to plant

Write the constraints. What are the limitations given in the problem?

0c

The total number of acres planted must be less than or equal to 275.

The time available for planting must be less than or equal to 25 days.

The number of acres planted in cotton must be greater than or equal to 0.

The number of acres planted in soybeans must be greater than or equal to 0.

s0

275c s

259 12

c s

Step 2: Write a System of Inequalities

Step 3: Graph the Inequalities

c

s

The purple area is the feasible region.

Step 4: Name the Vertices of the Feasible Region

(0,275)

(225,0)

(0,0)

(75,200)

Find the coordinates of the vertices of the feasible region, the area inside the polygon.

Step 5: Write an Equation to be Maximized or Minimizedp(c,s) = 25c + 18s

Maximum profit = $25 times the number of acres of cotton planted + $18 times the number of acres of soybeans planted.

Step 6: Substitute the Coordinates into the Equation

(c,s) 25c + 18s f(c,s) (0,275) 25(0) + 18(275) 4950 (225,0) 25(225) + 18(0) 5625 (0,0) 25(0) + 18(0) 0 (75,200) 25(75) + 18(200) 5475

Substitute the coordinates of the vertices into the maximum profit equation.

Step 7: Find the Maximum

225 acres of cotton and 0 acres of soybeans should be planted for a maximum profit of $5,625.

(c,s) 25c + 18s f(c,s)

(0,275) 25(0) + 18(275) 4950

(225,0) 25(225) + 18(0) 5625

(0,0) 25(0) + 18(0) 0

(75,200) 25(75) + 18(200) 5475

Example 2:

The Bethlehem Steel Mill can convert steel into girders and rods. The mill can produce at most 100 units of steel a day. At least 20 girders and at least 60 rods are required daily by regular customers. If the profit on a girder is $8 and the profit on a rod is $6, how many units of each type of steel should the mill produce each day to maximize the profits?

Step 1: Define the Variables

Let x = number of girders

Let y = number of rods

Step 2: Write a System of Inequalities

60y

100x y

20x At least 20 girders are required daily.

At least 60 rods are required daily.

The mill can produce at most 100 units of steel a day.

100

80

60

40

20

-50 50 100

Step 3: Graph the Inequalities20, 60, 100and x y x y

The purple region represents the feasible region.

Step 4: Name the Vertices of the Feasible Region Find the coordinates of the vertices of the feasible region, the

area inside the polygon.

(20, 60)

(20, 80)

(40, 60)

Step 5: Write an Equation to be Maximized or Minimized

p(x,y) = 8x + 6y

Maximum profit = $8 times the number of girders produced + $6 times the number of rods produced

Step 6: Substitute the Coordinates into the Equation

Substitute the coordinates of the vertices

into the maximum profit equation.

(x,y) 8x + 6y p(x,y) (20, 60) 8(20) + 6(60) 520 (20, 80) 8(20) + 6(80) 640 (40, 60) 8(40) + 6(60) 680

Step 7: Find the Maximum

40 girders and 60 rods of steel should be produced for a maximum profit of $680.

(x,y) 8x + 6y p(x,y) (20, 60) 8(20) + 6(60) 520 (20, 80) 8(20) + 6(80) 640 (40, 60) 8(40) + 6(60) 680

Homework• Pg 160-161

#10-13, 16, 17-19• 8 problems