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Questions from Yesterday

Sections 7.2

Quiz Today

Exam Schedule

Exam 4 (Ch 6,7)

Fri 11/15

Exam 5 (Ch 10)

Thur 12/5

Final Exam (All)

Thur 12/12

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

The Objective Function is

what we need to maximize or

minimize. For us, this will be a

function of 2 variables, f(x, y)

Linear Programming

The Constraints are the

inequalities that provide us with

the Feasible Region.

Linear Programming (pg 400)

The general idea… (pg 398)Find max/min values of the objective

function, subject to the constraints.

yxyxf 52),(

0,0

1

842

623

yx

yx

yx

yx

Objective Function Constraints

The general idea… (pg 398)

Graph the Feasible Region

The general idea… (pg 398)

The Feasible Region makes up the possible inputs to the Objective Function

yxyxf 52),(

Corner Point Thm (pg 400)

If a feasible region is bounded, then the objective function has both a maximum and minimum value, with each occurring at one or more corner points.

Find the minimum and maximum

value of the function f(x, y) = 3x - 2y.

We are given the constraints:

• y ≥ 2

• 1 ≤ x ≤5

• y ≤ x + 3

6

4

2

2 3 4

3

1

1

5

5

7

8

y ≤ x + 3

y ≥ 2

1 ≤ x ≤5

6

4

2

2 3 4

3

1

1

5

5

7

8

y ≤ x + 3

y ≥ 2

1 ≤ x ≤5 Need to find corner

points (vertices)

• The vertices (corners) of the

feasible region are:

(1, 2) (1, 4) (5, 2) (5, 8)

• Plug these points into the

function f(x, y) = 3x - 2y

Note: plug in BOTH x, and y values.

Evaluate the function at each vertex

to find min/max values

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

So, the optimized solution is:

• f(1, 4) = -5 minimum

• f(5, 2) = 11 maximum

Find the minimum and maximum value

of the function f(x, y) = 4x + 3y

With the constraints:

52

24

1

2

xy

xy

xy

6

4

2

53 4

5

1

1

2

3y ≥ -x + 2

y ≥ 2x -5

y ≤ 1/4x + 2

Need to find corner

points (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

Evaluate the function at each vertex

to find min/max values

• f(0, 2) = 6 minimum

• f(4, 3) = 25 maximum

So, the optimized solution is:

Classwork / Homework

• Page 403

•1, 3, 7, 9, 11