chapter 3: section 3-2 graphing linear inequalities 3: section 3-2 graphing linear inequalities d....

Chapter 3: Section 3-2Graphing Linear Inequalities

D. S. MalikCreighton University, Omaha, NE

D. S. Malik Creighton University, Omaha, NE ()Chapter 3: Section 3-2 Graphing Linear Inequalities 1 / 19

Geometric Approach to Solve Linear ProgrammingProblems

Graph the constraints of the problem

Find the solutions set (feasible region)

Find the corner points of the solution set

Use corner points to find a solution, if it exists

D. S. Malik Creighton University, Omaha, NE ()Chapter 3: Section 3-2 Graphing Linear Inequalities 2 / 19

Graphing Linear Inequalities

ExampleConsider the inequality: x + 3y ≥ 3.

The first step in drawing the graph of an inequality is to change theinequality into an equality and then draw the graph of the equality.

So first we draw the graph of the equality x + 3y = 3.

Two points on this line are:

x y (x , y)0 1 (0, 1)3 0 (3, 0)

D. S. Malik Creighton University, Omaha, NE ()Chapter 3: Section 3-2 Graphing Linear Inequalities 3 / 19

Example

(Two points on the line are: (0, 1) and (3, 0))

The graph of x + 3y = 3 is:

x

y

0 1 2 3­1­2­3­1­2

­3

1

2

3

x + 3y = 3

Select a point, which is not on the line. For example, (0, 2) is such apoint.

D. S. Malik Creighton University, Omaha, NE ()Chapter 3: Section 3-2 Graphing Linear Inequalities 4 / 19

Example

Determine, whether (0, 2) satisfies the given inequality, which isx + 3y ≥ 3.Now 0+ 2 · 3 = 6 ≥ 3, which is true. So the point (0, 2) is a solutionof the given inequality.

Shade the portion of the graph containing the point (0, 2) to drawthe graph of x + 3y ≥ 3.The graph of x + 3y ≥ 3 is:

x

y

0 1 2 3­1­2­3­1­2

­3

1

2

3x + 3y > 3

x + 3y = 3

D. S. Malik Creighton University, Omaha, NE ()Chapter 3: Section 3-2 Graphing Linear Inequalities 5 / 19

ExampleIn this example, we draw the graph of x − 2y < −2.

First we draw the graph of x − 2y = −2. Two points on this line are(−2, 0) and (0, 1). The graph of the line is shown in (a).(Note that the inequality is <, so we draw a dotted line.)

Next we choose a point which is not on the line. (0, 0) is one suchpoint. We can use (0, 0) as a test point.

Now 0− 2 · 0 = 0 < −2, which is false. So we shade the region thatdoes not contain the point (0, 0) and obtain the graph in (b).

x

y

0 1 2 3­1­2­3­1­2­3

123

(a)

x

y

0 1 2 3­1­2­3­1­2­3

123

x ­ 2y < ­2

(b)D. S. Malik Creighton University, Omaha, NE ()Chapter 3: Section 3-2 Graphing Linear Inequalities 6 / 19

ExampleConsider the following inequalities: 2x + 3y ≥ 6 and 1.5x − 2y ≥ −3

The graphs of these inequalities are:

x

y

0 1 2 3­1­2­3­1­2

­3

1

2

3

2x + 3y > 6

x

y

0 1 2 3­1­2­3­1­2

­3

1

2

3

1.5x ­ 2y > ­3

We determine the region in the x-y plane that simultaneously satisfiesthese inequalities.

D. S. Malik Creighton University, Omaha, NE ()Chapter 3: Section 3-2 Graphing Linear Inequalities 7 / 19

ExampleThe graphs of the inequalities: 2x + 3y ≥ 6 and 1.5x − 2y ≥ −3

We draw the graph of the these inequalities in the same x-y plane,see Figure (a).

Two find the region that simultaneously satisfies these inequalities, wechoose the commonly shaded region as shown in Figure (b).

x

y

0 1 2 3­1­2­3­1­2

­3

1

2

3

1.5x ­ 2y > ­3

x

y

0 1 2 3­1­2­3­1­2

­3

1

2

31.5x ­ 2y > ­3

2x + 3y > 6

2x + 3y > 6

D. S. Malik Creighton University, Omaha, NE ()Chapter 3: Section 3-2 Graphing Linear Inequalities 8 / 19

ExampleConsider the following inequalities: x + 3y ≥ 3, 2x − 2y ≥ −5,6x − y ≤ 3.The region that simultaneously satisfies these equations is shown inthe following figure:

x

y

0 1 2 3­1­2­3­1­2

­3

1

2

3

2x ­ 2y = ­5

6x ­ y = 3

x + 3y = 3

A

B

C

The corner points of the shaded region are A, B, and C .

D. S. Malik Creighton University, Omaha, NE ()Chapter 3: Section 3-2 Graphing Linear Inequalities 9 / 19

ExamplePoint A is the intersection of the lines x + 3y = 3 and 2x − 2y = −5.Point B is the intersection of the lines 6x − y = 3 and 2x − 2y = −5.Point C is the intersection of the lines x + 3y = 3 and 6x − y = 3.To find the coordinates of the point A, solve the equationsx + 3y = 3 and 2x − 2y = −5 for x and y to get

(− 98 ,118

).

To find the coordinates of the point B, solve the equations6x − y = 3 and 2x − 2y = −5 for x and y to get

( 1110 ,

185

).

To find the coordinates of the point C , solve the equationsx + 3y = 3 and 6x − y = 3 for x and y to get

( 1219 ,

1519

).

D. S. Malik Creighton University, Omaha, NE ()Chapter 3: Section 3-2 Graphing Linear Inequalities 10 / 19

Definition(Bounded and Unbounded Sets) A set of points in the x-y plane is saidto be bounded if it can be contained in a circle with center (0, 0).Otherwise, the set is said to be unbounded.

Definition(Solutions set) The solution set of a system of linear inequalities is theset of points that simultaneously satisfy all of the inequalities.

RemarkWhen we graph a system of inequalities, then the solution set is the set ofall points in the shaded region. In the graph, the shaded region is calledthe solution set or the feasible region.

Definition(Corner points) Let S be a solution set for a system of linear inequalities.A point C in the x-y plane is called a corner point of S if every linesegment in S that contains C has C as one of its endpoints.

D. S. Malik Creighton University, Omaha, NE ()Chapter 3: Section 3-2 Graphing Linear Inequalities 11 / 19

ExampleConsider the following system of inequalities: x + y ≤ 5, 4x − y ≥ 0,x − 3y ≤ −3The graph of these inequalities is:

x

y

0 1 2 3­1­2­3­1­2

­3

1

2

3

4x ­ y = 0

x + y = 5

x ­ 3y = ­3

AC

5

4 5

B4

D. S. Malik Creighton University, Omaha, NE ()Chapter 3: Section 3-2 Graphing Linear Inequalities 12 / 19

ExamplePoint A is the intersection of the lines 4x − y = 0 and x − 3y = −3.Point B is the intersection of the lines 4x − y = 0 and x + y = 5.Point C is the intersection of the lines x + y = 5 and x − 3y = −3.The coordinates of A are

( 311 ,

1211

).

The coordinates of B are (1, 4).

The coordinates of C are (3, 2).

D. S. Malik Creighton University, Omaha, NE ()Chapter 3: Section 3-2 Graphing Linear Inequalities 13 / 19

ExampleConsider the following system of inequalities: x ≥ 0, y ≥ 0,x + 2y ≥ 4, 5x + 2y ≥ 10.The graph of these inequalities is:

x

y

0 1 2 3­1­2­3­1­2

­3

1

2

3

5x + 2y = 10

x + 2y = 4B

C

4

4 5

A5

The solution set is unbounded with three corner points A, B, and C .

D. S. Malik Creighton University, Omaha, NE ()Chapter 3: Section 3-2 Graphing Linear Inequalities 14 / 19

ExampleA is the intersection of the lines x = 0 and 5x + 2y = 10.

B is the intersection of the lines 5x + 2y = 10 and x + 2y = 4.

C is the intersection of the lines y = 0 and x + 2y = 4.

The coordinates of A are (0, 5), and the coordinates of C are (4, 0).

The coordinates of B are( 32 ,54

)

D. S. Malik Creighton University, Omaha, NE ()Chapter 3: Section 3-2 Graphing Linear Inequalities 15 / 19

Exercise: Consider the following exercise described earlier.

A pet store specializes in cats and bunnies. Each cat costs $9 and eachbunny costs $6. The profit on each cat is $12 and on each bunny is $9.The store cannot house more than 30 animals and cannot spend morethan $216 to buy the pets. Under a special agreement the pet store musthouse at least 2 cats. How many pets of each type should be housed tomaximize the profit?

Let x be the number of cats and y be the numbers of bunnies.

The linear programming problem describing this problem is: Thelinear programming problem is:

maximize: 12x + 9ysubject to: x ≥ 2, y ≥ 0

x + y ≤ 303x + 2y ≤ 72

Let us graph the constraints and determine the feasible region.

D. S. Malik Creighton University, Omaha, NE ()Chapter 3: Section 3-2 Graphing Linear Inequalities 16 / 19

The graph of the inequalities x ≥ 2, y ≥ 0, x + y ≤ 30,3x + 2y ≤ 72 is:

x

y

0­10

10

20

30

3x + 2y = 72

x + y = 3040

10 20 30 40

C(12, 18)

B(2, 28)

A(2, 0)

x = 2

D(24, 0)

The corner points are A(2, 0), B(2, 28), C (12, 18), D(24, 0).

D. S. Malik Creighton University, Omaha, NE ()Chapter 3: Section 3-2 Graphing Linear Inequalities 17 / 19

Exercise Find the inequality that represents the following graph:

x

y

0 1 2 3­1­2­3­1­2

­3

1

2

3

Solution: The line passes through (−2, 0) and (0, 3). The slope of thisline is m = 3−0

0−(−2) =32 . The equation of the line with slope

32 and y -intercept (0, 3) is

y =32x + 3, i.e., 2y = 3x + 6, or 3x − 2y = −6.

This implies that the required inequality is 3x − 2y ≥ −6 or3x − 2y ≤ −6.

D. S. Malik Creighton University, Omaha, NE ()Chapter 3: Section 3-2 Graphing Linear Inequalities 18 / 19

x

y

0 1 2 3­1­2­3­1­2

­3

1

2

3

This implies that the required inequality is 3x − 2y ≥ −6 or3x − 2y ≤ −6. Next choose a point which is either in the shaded region orin the unshaded region. The point (0, 0) is in the unshaded region. Nextevaluate the expression 3x − 2y at (0, 0). Thus,3x − 2y = 3 · 0− 2 · 0 = 0. If the inequality is 3x − 2y ≥ −6, then (0, 0)will satisfy the inequality because at (0, 0), 3x − 2y = 0 ≥ −6 is true.However, (0, 0) is not in the shaded region, so it should not satisfy theinequality. Hence, the inequality that represents the above graph is3x − 2y ≤ −6.

D. S. Malik Creighton University, Omaha, NE ()Chapter 3: Section 3-2 Graphing Linear Inequalities 19 / 19