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Today’s Agenda Attendance / Announcements Questions from 6.1 / 6.2 Sections 7.1 / 7.2 E.C. Quiz Today

# Lecture 7.1 to 7.2 bt

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Today’s Agenda

Attendance / Announcements

Questions from 6.1 / 6.2

Sections 7.1 / 7.2

E.C. Quiz Today

Exam Schedule

Exam 5 (Ch 6.1, 7)

Wed 4/1

Exam 6 (Ch 10)

Monday 4/27

Final Exam (Cumulative)

Monday 5/4

Graphing Systems of Inequalities

1. Graph boundary lines

2. Check inequality signs

(Dashed or Solid?)

5. Test Points

4𝑥 + 𝑦 ≥ 92𝑥 + 3𝑦 ≤ 7

4. Identify Intersecting

regions (“Feasible”)

Graph the Feasible Region

5𝑦 − 2𝑥 ≤ 10𝑥 ≥ 3𝑦 ≥ 2

𝑦 ≤2

5𝑥 + 2

𝑥 ≥ 3𝑦 ≥ 2

Find “Corner Points” of the Feasible

Region3𝑥 + 2𝑦 ≤ 6−2𝑥 + 4𝑦 ≤ 8

𝑥 + 𝑦 ≥ 1𝑥 ≥ 0𝑦 ≥ 0

𝑦 ≤ −32

𝑥+3

𝑦 ≤ 12

𝑥 + 2

𝑦 ≥ −𝑥 + 1𝑥 ≥ 0𝑦 ≥ 0

Find “Corner Points” of the Feasible

Region3𝑥 + 2𝑦 ≤ 6−2𝑥 + 4𝑦 ≤ 8

𝑥 + 𝑦 ≥ 1𝑥 ≥ 0𝑦 ≥ 0

Finding Feasible RegionsFind the system

whose feasible region

is a triangle with

vertices (2,4),

(-4,0), and (2,-1)

2

46

832

x

yx

yx

Linear Programming

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 ≤5Need 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 395

•1 – 6

•7 – 45 e.o.odd

•47, 49, 53

• Page 403

•1, 3, 7, 9, 11

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