slide 8.4- 1 copyright © 2007 pearson education, inc. publishing as pearson addison-wesley

Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

OBJECTIVES

Systems of Linear Inequalities; Linear Programming

Learn to graph linear systems of inequalities in two variables.

Learn to graph a linear inequality in two variables.

Learn to apply systems of linear inequalities to linear programming.

SECTION 8.4

1

2

3


DefinitionsThe statements x + y > 4, 2x + 3y < 7, y ≥ x, and x + y ≤ 9 are examples of linear inequalities in the variables x and y.A solution of an inequality in two variables x and y is an ordered pair (a, b) that results in a true statement when x is replaced by a, and y is replaced by b in the inequality.The set of all solutions of an inequality is called the solution set of the inequality. The graph of an inequality in two variables is the graph of the solution set of the inequality.


PROCEDURE FOR GRAPHING A LINEAR INEQUALITY IN TWO VARIABLES

Step 1. Replace the inequality symbol by an

equals (=) sign.

Step 2. Sketch the graph of the corresponding

equation Step 1. Use a dashed line for

the boundary if the given inequality

sign is < or >, and a solid line if the

inequality symbol is ≤ or ≥.



Step 3. The graph in Step 2 will divide the

plane into two regions. Select a test

point in the plane. Be sure that the test

point does not lie on the graph of the

equation in Step 1.

Step 4. (i) If the coordinates of the test point

satisfy the given inequality, then so do

all the points of the region that contains



the test point. Shade the region that contains

the test point.

(ii) If the coordinates of the test point do not

satisfy the given inequality, shade the region

that does not contain the test point.

The shaded region (including the boundary if it

is solid) is the graph of the given inequality.


EXAMPLE 2 Graphing Inequalities

Sketch the graph of each of the following inequalities. a. x ≥ 2 b. y < 3 c. x + y < 4

Solution

a. Step 1 Change the ≥ to = : x = 2Step 2 Graph x = 2

with a solid line.Step 3 Test (0, 0). 0 ≥ 2 is a false statement.



Solution continued

Step 3 continuedThe region not containing (0, 0), together with the vertical line, is the solution set.

Step 4 Shade the solution set.



Solution continued

Step 2 Graph y = 3 with a dashed line.

Step 3 Test (0, 0). 0 < 3 is a true statement. The region containing (0, 0) is the solution set.


b. Step 1 Change the < to = : y < 3



Solution continued

Step 2 Graph x + y = 4 with a dashed line.

Step 3 Test (0, 0). 0 < 4 is a true statement. The region containing (0, 0) is the solution set.


c. Step 1 Change the < to = : x + y = 4


SYSTEMS OF LINEAR INEQUALITIES IN TWO VARIABLES

An ordered pair (a, b) is a solution of a system of inequalities involving two variables x and y if and only if, when x is replaced by a and y is replaced by b in each inequality of the system, all resulting statements are true.

The solution set of a system of inequalities is the intersection of the solution sets of all the inequalities in the system.


EXAMPLE 3 Graphing a System of Two Inequalities

Graph the solution set of the system of

Solution

Step 1 2x + 3y = 6Step 2 Sketch as a dashed line by joining the

points (0, 2) and (3, 0).

2x 3y 6 (1)

y x 0 (2)

inequalities:

Step 3 Test (0, 0). 2(0) + 3(0) > 6 is a false statement.




Solution continued

Now graph the second inequality.



Solution continued

Step 2 Sketch as a solid line by joining the points (0, 0) and (1, 1).

Step 3 Test (1, 0). 2(0) – 3(1) > 6 is a false statement.


Step 1 y – x = 0



Solution continued

The graph of the solution set of inequalities (1) and (2) is the region where the shading overlaps.


LINEAR PROGRAMMINGThe process of finding a maximum or minimum value of a quantity is called optimization.

A linear programming problem satisfies the following two conditions:

1. The quantity f to be maximized or minimized can be written as a linear expression in x and y. That is,

f = ax + by,where a ≠ 0, b ≠ 0 are constants.


LINEAR PROGRAMMING

The inequalities that determine the region S are called constraints, the region S is called the set of feasible solutions, and f = ax + by is called the objective function. A point in S at which f attains its maximum (or minimum) value, together with the value of f at that point, is called an optimal solution.

2. The domain of the variables x and y is restricted to a region S that is determined by (is a solution set of) a system of linear inequalities.


PROCEDURE FOR SOLVING A LINEAR PROGRAMMING PROBLEM

Step 1. Write an expression for the quantity to be maximized or minimized. This expression is the objective function.

Step 2. Write all constraints as linear inequalities.

Step 3. Graph the solution set of the constraint inequalities. This set is the set of feasible solutions.


Step 4. Find all the vertices of the solution set in Step 3 by solving all pairs of corresponding equations for the constraint inequalities.

Step 5. Find the values of the objective function at each of the vertices of Step 4.

Step 6. The largest of the values (if any) in Step 5 is the maximum value of the objective function, and the smallest of the values (if any) in Step 5 is the minimum value of the objective function.


EXAMPLE 5 Nutrition; Minimizing Calories

Fat Albert wishes to go on a crash diet and needs your help in designing a lunch menu. The menu is to include two items: soup and salad. The vitamin units (milligrams) and calorie counts in each ounce of soup and salad are given in the table.

Item Vitamin A Vitamin C Calories

SoupSalad

11

32

5040



Find the number of ounces of each item in the menu needed to provide the required vitamins with the fewest number of calories.

The menu must provide at least:10 units of Vitamin A24 units of Vitamin C

Solution

a. State the problem mathematically. Step 1 Write the objective function.

x = ounces of soupy = ounces of salad



Minimize the total number of calories. Solution continued

Step 2 Write the constraints.1 ounce soup provides 1 unit vitamin A1 ounce salad provides 1 unit vitamin ASo the total vitamin A is x + y and this must be at least 10 units so

x + y ≥ 10.

50 calories per ounce of soup: 50x40 calories per ounce of salad: 40yTotal calories = f = 50x + 40y



Solution continued

Step 2 Write the constraints. (continued)

1 ounce soup provides 3 units vitamin C

1 ounce salad provides 2 units vitamin C

So the total vitamin C is 3x + 2y and this

must be at least 24 units so

3x + 2y ≥ 24.

x and y cannot be negative so

x ≥ 0 and y ≥ 0.



Solution continued

Find x and y such that the value of f = 50x + 40y

is a minimum, with the restrictionsx y 10

3x 2y 24

x 0

y 0

b. Solve the linear programming problem.

Summarize the information.



Solution continued

Step 3 Graph the set of feasible solutions.

The set is bounded by

x y 10

3x 2y 24

x 0

y 0x y 10

3x 2y 24



Solution continued

Step 4 Find the vertices. The vertices of the feasible solutions are A(10, 0), B(4, 6) and C(0, 12).

y 0

x y 10

A (10, 0) is obtained by solving

x y 10

3x 2y 24

B (4, 6) is obtained by solving

x 0

3x 2y 24

C (0, 12) is obtained by solving



Solution continued

Step 5 Find the value of f at the vertices.

Vertex (x, y) Value of f = 50x + 40y

(10, 0)(4, 6)

(0, 12)

50(10) + 40(0) = 50050(4) + 40(0) = 44050(0) + 40(0) = 480

Step 6 Find the maximum or minimum value of f. The smallest value of f is 440, which occurs when x = 4 and y = 6.



Solution continued

c. State the conclusion

The lunch menu for Fat Albert should contain 4 ounces of soup and 6 oounces of salad. His intake of 440 calories will be as small as possible under the given constraints.

slide 8.4- 1 copyright © 2007 pearson education, inc. publishing as pearson addison-wesley

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