5.6 - 1 Copyright © 2013, 2009, 2005 Pearson Education, Inc. 1 5 Systems and Matrices Copyright ©...

5.6 - 1Copyright © 2013, 2009, 2005 Pearson Education, Inc. 1

5Systems and Matrices

Copyright © 2013, 2009, 2005 Pearson Education, Inc.

Copyright © 2013, 2009, 2005 Pearson Education, Inc. 2

5.6Systems of Inequalities and Linear Programming

•Solving Linear Inequalities•Solving Systems of Inequalities•Linear Programming


Solving Linear Inequalities

A linear inequality in two variables is an inequality of the form

,Ax By C

where A, B, and C are real numbers, with A and B not both equal to 0. (The symbol could be replaced with , <, or >.) The graph of a linear inequality is a half-plane, perhaps with its boundary.


Graph 3 2 6.x y

First graph the boundary, 3x – 2y = 6, as shown in the figure.

Since the points of the line 3x – 2y = 6 satisfy 3x – 2y < 6, this line is part of the solution set.

Example 1GRAPHING A LINEAR INEQUALITY

Solution


To decide which half-plane (the one above theline 3x – 2y = 6 or the one below the line) is part of the solution set, solve the original inequality for y.

3 2 6x y

2 3 6y x Subtract 3x.

33

2y x Divide by − 2.

Reverse the inequality symbol when dividing by a negative number.



For a particular value of x, the inequality will be satisfied by all values of y that are greater than or equal to Thus, the solution set contains the half-plane above the line.

33.

2x


Coordinates for x and y from the solution set (the shaded region) satisfy the original inequality, while coordinates outside the solution set do not.


Caution A linear inequality must be in slope-intercept form (solved for y) to determine, from the presence of a < symbol or a > symbol, whether to shade the lower or upper half-plane. In the previous slide, the upper half-plane is shaded, even though the inequality is 3x – 2y 6 is (with a < symbol) in standardform. Only when we write the inequality as(slope-intercept form) does the > symbol indicate to shade the upper half-plane.

33

2y x



Graph x + 4y > 4.Solution

The boundary of the graph is the straight line x + 4y = 4. Since points on this line do not satisfy x + 4y > 4, it is customary to make the line dashed. To decide which half-plane represents the solution set, solve for y.



4 4x y

4 4y x Subtract x.

11

4y x Divide by 4.

Since y is greater than the graph of the solution set is the half-plane above the boundary.

11,

4x



4 4x y ?

0 04( ) 4

Original inequality.

0 4

Use (0, 0) as a test point.

Alternatively, or as a check, choose a test point not on the boundary line and substitute into the inequality. The point (0, 0) is a good choice if it does not lieon the boundary, since the substitution is easily done.

False



Since the point (0, 0) is below the boundary, the points that satisfy the inequality must be above the boundary, which agrees with the result.


Graphing an Inequality in Two Variables

Method 1 If the inequality is or can be solved for y, then the following hold.

• The graph of y < (x) consists of all the points that are below the graph of y = (x).

• The graph of y > (x) consists of all the points that are above the graph of y = (x).


Graphing an Inequality in Two Variables

Method 2 If the inequality is not or cannot be solved for y, choose a test point not on the boundary.

• If the test point satisfies the inequality, the graph includes all points on the same side of the boundary as the test point.

• If the test point does not satisfy the inequality, the graph includes all points on the other side of the boundary.


Solving Systems of Inequalities

The solution set of a system of inequalities, such as

6 2x y 2 2 ,x y

is the intersection of the solution sets of its members. We find this intersection by graphing the solution sets of all inequalities on the same coordinate axes and identifying, by shading, the region common to all graphs.


Example 3GRAPHING SYSTEMS OF INEQUALITIES

Solution

(a)

Graph the solution set of each system.

6 2x y 2 2x y

The figures show the graphs of x > 6 – 2y and x2 < 2y. The methods presented earlier in this chapter can be used to show that the boundaries intersect at the points (2, 2) and (–3, 9/2).



Solution

(a)

Graph the solution set of each system.

6 2x y 2 2x y

The solution set of the system is shown. Since the points on the boundaries of x > 6 – 2y and x2 < 2y do not belong to the graph of the solution set, the boundaries are dashed.



Solution

(b)Graph the solution set of each system.

3x 0y

1y x

Writing x 3 as – 3 x 3 shows that this inequality is satisfied by points in the region between and including x = – 3 and x = 3.



The set of points that satisfies y 0 includes the points below or on the x-axis.



Graph y = x + 1 and use a test point to verify that the solutions of y x + 1 are on or above the boundary.



Since the solution sets of y 0 and y x +1 have no points in common, the solution set of the system is ø.

The solution set of the system is ø, because there are no points common to all three regions.


Note While we gave three graphs in the solutions of Example 3, in practicewe usually give only a final graph showing the solution set of the system.


Example 4 FINDING A MAXIMUM PROFIT

The Charlson Company makes two products: MP3 players and DVD players. Each MP3 gives a profit of $30, while each DVD player produces $70 profit. The company must manufacture at least 10 MP3 players per day to satisfy one of its customers, but it can manufacture no more than 50 per day because of production restrictions. The number of DVD players produced cannot exceed 60 per day, and the number of MP3 players cannot exceed the number of DVD players. How many of each should the company manufacture to obtain maximum profit?


Example 4

Solution First we translate the statement of the problem into symbols.Let x = number of MP3 players to be produced daily,and y = number of DVD players to be produced daily.

The company must produce at least 10 MP3 players (10 or more), so

10.x

Since no more than 50 MP3 players may be produced,

50.x

FINDING A MAXIMUM PROFIT


Example 4

Solution

60.y

No more than 60 DVD players may be made in one day, so

The number of MP3 players may not exceed the number of DVD players, so

.x y

The numbers of MP3 players and of DVD players cannot be negative, so

0 and 0.x y



Example 4

Solution

10, 50, 60, , 0, 0x x y x y x y

These restrictions, or constraints, form the following system of inequalities.

Each MP3 player gives a profit of $30, so the daily profit from production of x MP3 players is 30x dollars. Also, the profit from production of y DVD players will be 70y dollars per day. Therefore, the total daily profit is

profit 30 70 .x y This equation defines the objective function to be maximized.



Example 4

Solution

To find the maximum possible profit, subject to these constraints, we sketch the graph of each constraint.



Example 4

Solution

The only feasible values of x and y are those thatsatisfy all constraints—that is, the values that lie in the intersection of thegraphs.



Example 4

Solution

Any point lying inside the shaded region or on the boundary satisfies therestrictions as to the number of MP3 and DVD players that can be produced.



Example 4

Solution

(For practical purposes, however, only points with integer coefficients are useful.) This region is called the region of feasible solutions. The vertices (singular vertex) or corner points of the region of feasible solutions have coordinates(10,10), (10,60), (50,50),

and (50,60).



Example 4

( , ) : 30( ) 70(10 1 )01 10 00 10 0

We must find the value of the objective function 30x + 70y for each vertex. We want the vertex that produces the maximum possible value of 30x + 70y.

( , ) : 30( ) 70(60 6 )01 10 00 45 0

( , ) : 30( ) 70(50 5 )05 50 00 50 0

( , ) : 30( ) 70(60 6 )05 50 00 57 0 Maximum

The maximum profit, obtained when 50 MP3 players and 60 DVD players are produced each day, will be 30(50) + 70(60) = $5700 dollars per day.



Example 4

The Charlson Company needed to find values of x and y in the shaded region that produce the maximum profit—that is, the maximum value of 30x + 70y. To locate the point (x, y) that gives the maximum profit, add to the graph lines corresponding to arbitrarily chosen profits of $0, $1000, $3000, and $7000:

30 70 0, 30 70 1000,x y x y

30 70 3000, and 30 70 7000.x y x y

To justify the procedure used in Example 4, consider the following.



Example 4

For instance, each point on the line 30x + 70y = 3000 corresponds to production values that yield a profit of $3000.

The figure shows the region of feasible solutions, together with these lines. The lines are parallel, and the higher the line, the greater the profit. The line 30x + 70y = 7000 yields the greatest profit but does not contain any points of the region of feasible solutions.



Example 4

To find the feasible solution of greatest profit, lower the line 30x + 70y = 7000 until it contains a feasible solution—that is, until it just touches the region of feasible solutions. This occurs at point A, a vertex of the region.

This discussion can be generalized.



Fundamental Theorem of Linear ProgrammingIf an optimal value for a linear programming problem exists, it occurs at a vertex of the region of feasible solutions.


Solving a Linear Programming ProblemStep 1 Write the objective function and all

necessary constraints.Step 2 Graph the region of feasible solutions.Step 3 Identify all vertices (corner points).Step 4 Find the value of the objective function at

each vertex.Step 5 The solution is given by the vertex

producing the optimal value of the objective function.


Example 5 FINDING A MINIMUM COST

Robin takes multivitamins each day. She wants at least 16 units of vitamin A, at least 5 units of vitamin B1 and at least 20 units of vitamin C. Capsules, costing $0.10 each, contain 8 units of A, 1 of B1 and 2 of C. Chewable tablets, costing $0.20 each, contain 2 units of A, 1 of B1 and 7 of C. How many of each should she take each day to minimize her cost and yet fulfill her daily requirements?


Example 5

Step 1 Let x represent the number of capsules to take each day, and let y represent the number of chewable tablets to take. Then the cost in pennies per day is

cost 10 20 .x y

FINDING A MINIMUM COST

Objective function

Solution


Example 5

Robin takes x of the $0.10 capsules and y of the $0.20 chewable tablets, and she gets 8 units of vitamin A from each capsule and 2 units of vitamin A from each tablet. Altogether she gets 8x + 2y units of A per day. Since she wants at least 16 units,

8 2 16.x y



Example 5

Each capsule and each tablet supplies 1 unit of vitamin B1. Robin wants at least 5 units per day, so 5.x y

For vitamin C, the inequality is

2 7 20.x y

Since Robin cannot take negative numbers of multivitamins,


x 0 and y 0.


Example 5

Step 2 The figure shows the intersection of the graphs of8 2 16, 5, 2 7 20,x y x y x y

0, and 0.x y



Example 5

Step 3 The vertices are (0, 8), (1, 4), (3, 2), and (10, 0).



Example 5

Steps 4 and 5 The minimum cost occurs at (3, 2).

Point Cost = 10x + 20y

(0, 8) 10(0) + 20(8) = 160

(1, 4) 10(1) + 204) = 90

(3, 2) 10(3) + 20(2) = 70

(10, 0) 10(10) + 20(0) = 100

Minimum

Robin’s best choice is to take 3 capsules and 2 chewable tablets, for a total cost of $0.70 per day. She receives just the minimum amounts of vitamins B1 and C, and an excess of vitamin A.


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