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3.4 Linear Programming. Vocabulary Constraints Conditions given to variables usu. expressed as linear inequalities Feasible region The intersection of.

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Presentation on theme: "3.4 Linear Programming. Vocabulary Constraints Conditions given to variables usu. expressed as linear inequalities Feasible region The intersection of."— Presentation transcript:

1 3.4 Linear Programming

2 Vocabulary Constraints 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 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 1. Solve the SOI by graphing 2. Name the coords of the feasible region 3. Plug the coords into the function to find the max/min

5 Example 4-1a 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). 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.

6 Example 4-1a Use a table to find the maximum and minimum values of f (x, y). Substitute the coordinates of the vertices into the function. 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). (x, y)(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 Page 2 example 1:

7 Example 4-1b Answer: vertices: (1, 5), (4, 5) (4, 2) ; maximum: f(4, 2) = 10, minimum: f(1, 5) = –11 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. Example 2 :

8 Example 4-2a Graph the system of inequalities. There are only two points of intersection, (–2, 0) and (0, –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. Example 3:

9 Example 4-2a (x, y)(x, y) 2x + 3y f(x, y)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 3 page 2

10 Example 4-2b Answer: vertices: (0, –3), (6, 0) ; maximum: f(6, 0) = 6 ; no minimum 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. Example 4:


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