Chapter 4 Section 1 Graphing Linear Inequalities in Two Variables

Transcription

1 Chapter 4 Section 1 Graphing Linear Inequalities in Two Variables

2 Epressions of the tpe and 3 > 6 are called linear inequalities in two variables. A solution of a linear inequalit in two variables is an ordered pair (, ) which makes the inequalit true. Eample: (1, 3) is a solution to since (1) + 2(3) = 7 8. Mathematical Applications b Harshbarger (8th ed) Copright b Houghton Mifflin Compan 2

3 The solution set, or feasible set, of a linear inequalit in two variables is the set of all solutions. Eample: The solution set for is the shaded region. 2 2 The solution set is a half-plane. It consists of the line and all the points below and to its left. The line is called the boundar line of the half-plane. Mathematical Applications b Harshbarger (8th ed) Copright b Houghton Mifflin Compan 3

4 If the inequalit is or, the boundar line is solid; its points are solutions. Eample: The boundar line of the solution set of 3 2 is solid. 3 < 2 3 = 2 3 > 2 If the inequalit is < or >, the boundar line is dotted; its points are not solutions. Eample: The boundar line of the solution set of + < 2 is dotted. Mathematical Applications b Harshbarger (8th ed) Copright b Houghton Mifflin Compan 4

5 A test point can be selected to determine which side of the half-plane to shade. Eample: For graph the boundar line. The solution set is a half-plane. Use (0, 0) as a test point. (0, 0) is a solution. So all points on the (0, 0) side of the boundar line are also solutions. Shade above and to the left of the line. -2 (0, 0) 2 Mathematical Applications b Harshbarger (8th ed) Copright b Houghton Mifflin Compan 5

6 To graph the solution set for a linear inequalit: 1. Graph the boundar line. 2. Select a test point, not on the boundar line, and determine if it is a solution. 3. Shade a half-plane. Mathematical Applications b Harshbarger (8th ed) Copright b Houghton Mifflin Compan 6

7 Eample: Graph the solution set for > Graph the boundar line = 2 as a dotted line. 2. Select a test point not on the line, sa (0, 0). (0) 0 = 0 > 2 is false. (0, 0) (0, -2) (2, 0) 3. Since this is a not a solution, shade in the half-plane not containing (0, 0). Mathematical Applications b Harshbarger (8th ed) Copright b Houghton Mifflin Compan 7

8 Solution sets for inequalities with onl one variable can be graphed in the same wa. Eample: Graph the solution set for < Eample: Graph the solution set for Mathematical Applications b Harshbarger (8th ed) Copright b Houghton Mifflin Compan 8

9 Chapter 4 Section 2 Linear Programming

10 Linear programming is a strateg for finding the optimum value either maimum or minimum - of a linear function that is subject to certain constraints. These constraints, or restrictions, are stated as a sstem of linear inequalities. Eample: Find the maimum value of z, given: objective function z = and constraints Eample continued Mathematical Applications b Harshbarger (8th ed) Copright b Houghton Mifflin Compan 10

11 The sstem of linear inequalities determines a set of feasible solutions. The graph of this set is the feasible region. Eample continued: Graph the feasible region determined b the sstem of constraints = 21 0 = 0 feasible region = = 0 25 Eample continued Mathematical Applications b Harshbarger (8th ed) Copright b Houghton Mifflin Compan 11

12 If a linear programming problem has a solution, then the solution is at a verte of the feasible region. Eample continued: Maimize the value of z = 2 +3 over the feasible region. (0, 5) feasible region (0, 0) z = 2(0) + 3(0) = 0 z = 2(0) + 3(5) = 15 (5, 3) z = 2(5) + 3(3) = 19 (7, 0) z = 2(7) + 3(0) = 14 Maimum value of z Test the value of z at each of the vertices. The maimum value of z is 19. This occurs at the point (5, 3) or when =5 and = 3. Mathematical Applications b Harshbarger (8th ed) Copright b Houghton Mifflin Compan 12

13 Solving Linear Programming Problems Graphicall 1. Graph the feasible region. 2. Find the vertices of the region. 3. Evaluate the objective function at each verte. 4. Select the vertices that optimize the objective function. a) If the feasible region is bounded the objective function will have both a maimum and a minimum. b) If the feasible region is unbounded and the objective function has an optimal value, the optimal value will occur at a verte of the feasible region. Note: If the optimum value occurs at two vertices, its value is the same at both vertices and along the line segment joining them. Mathematical Applications b Harshbarger (8th ed) Copright b Houghton Mifflin Compan 13

14 Eample: Find the maimum and minimum value of z = + 3 subject to the constraints + 3 6, 3 6, and Graph the feasible region. (0, 2) z = 0 + 3(2) = 6 2. Find the vertices. z = 3 + 3(3) = 12 (3, 3) Maimum value of z + 3 = 6 (6, 0) z = 6 + 3(0) = 6 3 = 6 + = 6 3. Evaluate the objective function at each verte. 4. The maimum value of z is 12 and occurs at (3, 3). The minimum value of z is 6 and occurs at both (0, 2) and (6, 0) and at ever point along the line joining them. Mathematical Applications b Harshbarger (8th ed) Copright b Houghton Mifflin Compan 14

15 z z z Rewriting the objective function z = + 3 in slope-intercept 1 z form gives = +. This equation represents a famil of 3 3 parallel lines, one for each value of z. 1 = 6 = = 3 = (0, 2) 1 = 0 = - 3 (3, 3) (6, 0) z = 15 1 = z = 12 1 = z = 9 1 = As z increases through values 0, 3, 6, 9, 12, and 15, the corresponding line passes through the feasible region. The point at which the famil of lines first meets the feasible region gives the minimum value of z, and the point at which the famil of lines leaves the feasible region gives the maimum. Mathematical Applications b Harshbarger (8th ed) Copright b Houghton Mifflin Compan 15

16 Eample: Minimize z = 3 subject to 1, + 5, 0, and 0. verte value of z at verte (0, 5) =1 (0, 0) z = 3(0) (0) = 0 (1, 0) z = 3(1) (0) = 3 (3, 2) z = 3(3) (2) = 7 (0, 5) z = 3(0) (5) = 5 (0, 0) = 0 (1, 0) (3, 2) + = 0 = 5 The minimum value of z is 5 and this occurs at (0, 5). Mathematical Applications b Harshbarger (8th ed) Copright b Houghton Mifflin Compan 16

17 Eample: Maimize z = 2 + subject to 3 + 6, + 4, 0, and 0. (0, 6) (1, 3) (4, 0) At (, 0), z = = 6 + = 4 Since the feasible region is unbounded there ma be no maimum value of z. For 4, (, 0) is a feasible solution. Therefore as increases without bound, z increases without bound and there is no maimum value of z. Mathematical Applications b Harshbarger (8th ed) Copright b Houghton Mifflin Compan 17

18 Eample: Sarah makes bracelets and necklaces to sell at a craft store. Each bracelet makes a profit of $7, takes 1 hour to assemble, and costs $2 for materials. Each necklace makes a profit of $12, takes 2 hour to assemble, and costs $3 for materials. Sarah has 48 hours available to assemble bracelets and necklaces. If she has $78 available to pa for materials, how man bracelets and necklaces should she make to maimize her profit? To formulate this as a linear programming problem: 1. Identif the variables. 2. Write the objective function. 3. Write the constraints. Eample continued Mathematical Applications b Harshbarger (8th ed) Copright b Houghton Mifflin Compan 18

19 Eample continued: 1. Let = the number of bracelets Sarah makes Let = the number of necklaces Sarah makes 2. Epress the profit as a function of and. p = Epress the constraints as inequalities. Cost of materials: Time limitation: Function to be maimized Since Sarah cannot make a negative number of bracelets or necklaces, 0 and 0 must also hold. Eample continued Mathematical Applications b Harshbarger (8th ed) Copright b Houghton Mifflin Compan 19

20 Eample continued: Maimize p = subject to the constraints , , 0, and = 78 (0, 24) z = 288 (12, 18) z = 300 (0, 0) z = 0 (39, 0) z = = 48 Sarah should make 12 bracelets and 18 necklaces for a maimum profit of $300. Mathematical Applications b Harshbarger (8th ed) Copright b Houghton Mifflin Compan 20