3-5: linear programming. learning target i can solve linear programing problem

3-5: Linear Programming

Learning Target I can solve linear programing

problem.

Terms Optimization – is finding the minimum or

maximum value of some quantity. Linear programming is a form of

optimization where you optimize an objective function with a system of linear inequalities called constraints.

The overlapped shaded region is called the feasible region.

Solving a linear programming problem

1. Graph the constraints.2. Locate the ordered pairs of the vertices of the feasible region.3. If the feasible region is bounded (or closed), it will have a minimum & a maximum.

If the region is unbounded (or open), it will have only one (a minimum OR a maximum).4. Plug the vertices into the linear equation (C=) to find the min. and/or max.

• If the region is unbounded, but has a top on it, there will be a maximum only.

• If the region is unbounded, but has a bottom, there will be a minimum only.

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.

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

78

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

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

14

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

313

13

73

283

253

Linear Programming

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

Pair-sharePair-shareWork on page 166 to 167 on #9 to 17

HomeworkHomeworkHomework on page 167 #18 to #20