Download presentation

Presentation is loading. Please wait.

1
3.4 Linear Programming

2
**Vocabulary Constraints Feasible region Conditions given to variables**

usu. expressed as linear inequalities Feasible region The intersection of the graphs in a system of inequalities (SOI) Feasible Region

3
**Bounded Unbounded Vertices**

Describes a closed feasible region formed by a SOI Unbounded Describes an open feasible region formed by a SOI Vertices Points located at the angles of the feasible region Vertices

4
**Steps to Linear Programming**

Solve the SOI by graphing Name the coords of the feasible region Plug the coords into the function to find the max/min

5
**Find the vertices of the region. Graph the inequalities.**

Graph the following system of inequalities. Name the coordinates of the vertices of the feasible region. Find the maximum and minimum values of the function for this region. Find the vertices of the region. Graph the inequalities. The polygon formed is a triangle with vertices at (–2, 4), (5, –3), and (5,4). Example 4-1a

6
**(x, y) 3x – 2y f(x, y) (–2, 4) 3(– 2) – 2(4) (5, –3) 3(5) – 2(–3) 21**

Page 2 example 1: Use a table to find the maximum and minimum values of f (x, y). Substitute the coordinates of the vertices into the function. (x, y) 3x – 2y f(x, y) (–2, 4) 3(– 2) – 2(4) – 14 (5, –3) 3(5) – 2(–3) 21 (5, 4) 3(5) – 2(4) 7 Answer: The vertices of the feasible region are (–2, 4), (5, –3), and (5, 4). The maximum value is 21 at (5, –3). The minimum value is –14 at (–2, 4). Example 4-1a

7
Example 2: Graph the following system of inequalities. Name the coordinates of the vertices of the feasible region. Find the maximum and minimum values of the function for this region. Answer: vertices: (1, 5), (4, 5) (4, 2); maximum: f(4, 2) = 10, minimum: f(1, 5) = –11 Example 4-1b

8
Example 3: Graph the following system of inequalities. Name the coordinates of the vertices of the feasible region. Find the maximum and minimum values of the function for this region. Graph the system of inequalities. There are only two points of intersection, (–2, 0) and (0, –2). Example 4-2a

9
**(x, y) 2x + 3y f(x, y) (–2, 0) 2(–2) + 3(0) –4 (0, –2) 2(0) + 3(–2) –6**

Example 3 page 2 (x, y) 2x + 3y f(x, y) (–2, 0) 2(–2) + 3(0) –4 (0, –2) 2(0) + 3(–2) –6 The minimum value is –6 at (0, –2). Although f(–2, 0) is –4, it is not the maximum value since there are other points that produce greater values. For example, f(2,1) is 7 and f(3, 1) is 10. It appears that because the region is unbounded, f (x, y) has no maximum value. Answer: The vertices are at (–2, 0) and (0, –2). There is no maximum value. The minimum value is –6 at (0, –2). Example 4-2a

10
Example 4: Graph the following system of inequalities. Name the coordinates of the vertices of the feasible region. Find the maximum and minimum values of the function for this region. Answer: vertices: (0, –3), (6, 0); maximum: f(6, 0) = 6; no minimum Example 4-2b

Similar presentations

Presentation is loading. Please wait....

OK

Get out your Vertices Worksheet!

Get out your Vertices Worksheet!

© 2017 SlidePlayer.com Inc.

All rights reserved.

Ads by Google

Ppt on as 14 amalgamation process Ppt on dhaka stock exchange Ppt on inventory turnover ratio Download ppt on android Ppt on poverty and hunger in india Ppt on rural and urban livelihood Ppt on amplitude shift keying applications Ppt on diode circuits tutorial Download ppt on areas of parallelograms and triangles Ppt on production planning and control