 # 3.4 Linear Programming.

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3.4 Linear Programming

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

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

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

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

(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

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

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

(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

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