4. Inequalities. 4.1 Solving Linear Inequalities Problem Basic fee: $20 Basic fee: $20 Per minute: 5¢ Per minute: 5¢ Budget: $40 Budget: $40 How many

4. Inequalities4. Inequalities

4.1 Solving Linear Inequalities4.1 Solving Linear Inequalities

ProblemProblem Basic fee: $20Basic fee: $20 Per minute: 5Per minute: 5¢¢ Budget: $40Budget: $40 How many minutes?: xHow many minutes?: x

20 + 0.05x ≤ 4020 + 0.05x ≤ 40

x ≤ 400x ≤ 400

NotationsNotations

Closed intervalClosed interval [a, b] = {x | a [a, b] = {x | a ≤≤ x x ≤≤ b} b}

Open intervalOpen interval (a, b) = {x | a < x < b}(a, b) = {x | a < x < b}

a b

a b

NotationsNotations

Infinite intervalInfinite interval [a, [a, ∞∞) = {x | a ) = {x | a ≤≤ x x < < ∞∞}}

Infinite intervalInfinite interval (-(-∞∞ , b] = {x | - , b] = {x | -∞∞ < x < x ≤≤ b} b}

a

b

Your TurnYour Turn

Express in set-builder notation.Express in set-builder notation. [a, b)[a, b) (- (- ∞∞ , b) , b)

Solving Inequalities in One VariableSolving Inequalities in One Variable

0.05x + 20 0.05x + 20 ≤≤ 40 400.05x 0.05x ≤≤ 20 20x x ≤≤ 20/0.05 20/0.05x x ≤≤ 400 400

[0, 400] (Interval notation)[0, 400] (Interval notation)

{x | x {x | x ≤≤ 400} (set-builder notation) 400} (set-builder notation)

Properties of InequalitiesProperties of Inequalities

AdditionAddition a < b a < b → a + c < b + c→ a + c < b + c

a < b a < b → a - c < b – c→ a - c < b – c

Positive Multiplication (c > 0)Positive Multiplication (c > 0) a < b a < b → ac < bc→ ac < bc

a < b a < b → a /c < b/c→ a /c < b/c

Negative Multiplication (c < 0)Negative Multiplication (c < 0) a < b a < b → ac ≥ bc→ ac ≥ bc

a < b a < b → a /c ≥ b/c→ a /c ≥ b/c

Example 1Example 1

-2x – 4 > x + 5-2x – 4 > x + 5

-3x > 9-3x > 9(-1/3)(-3x) < (-1/3)9(-1/3)(-3x) < (-1/3)9x < -3x < -3

(- (- ∞∞ ,-3) ,-3)

Example 2Example 2

(x + 3) (x – 2) 1(x + 3) (x – 2) 1--------- --------- ≥≥ ---------- + --- ---------- + --- 4 4 3 4 3 43(x + 3) 4(x – 2) 3 3(x + 3) 4(x – 2) 3 ------------ ------------ ≥≥ ----------- + ----- ----------- + ----- 12 12 12 12 12 12 3x + 9 3x + 9 ≥ 4x – 8 + 3≥ 4x – 8 + 3-x ≥ -14-x ≥ -14 x ≤ 14 x ≤ 14

(-∞, 14](-∞, 14]

Special CasesSpecial Cases

x > x + 1x > x + 1 {x | x > x + 1}{x | x > x + 1} What kind of set is this?What kind of set is this?

x < x + 1x < x + 1 {x } x < x + 1}{x } x < x + 1} What kind of set is this?What kind of set is this?

4.2 Compound Inequalities4.2 Compound Inequalities

Intersection of SetsIntersection of Sets Given set A and B, intersection A and B,Given set A and B, intersection A and B,

A A ∩∩ B = {x | x B = {x | x εε A A ANDAND x x εε B} B}

A

A ∩ ∩ BB

B

Compound InequalitiesCompound Inequalities

Union of SetsUnion of Sets Given set A and B, union of A and B,Given set A and B, union of A and B,

A A UU B = {x | x B = {x | x εε A A OROR x x εε B} B}

A

A UU BB

A

Intersection of SetsIntersection of Sets

Given: A = {1, 2, 3,5, 9}Given: A = {1, 2, 3,5, 9} B = {3, 5, 9, 10, 12} B = {3, 5, 9, 10, 12} A A ∩∩ B = {5, 9} B = {5, 9}

Given: A = {x | x Given: A = {x | x ≥ ≥ 3} 3} B = {x | x B = {x | x ≤ 10}≤ 10}

A A ∩∩ B = {x | x B = {x | x ≥ ≥ 3 AND x 3 AND x ≤ 10≤ 10 } } [ ] [ ]

3 10

Union and Intersection of SetsUnion and Intersection of Sets

Given:Given: A = set of all male students at CUHA = set of all male students at CUH B = set of all female students at CUHB = set of all female students at CUH C = set of all freshman students at CUHC = set of all freshman students at CUH

Draw a diagram of:Draw a diagram of: A ∩ BA ∩ B A A UU B B A ∩ CA ∩ C A A UU C C

Solving Compound InequalitySolving Compound Inequality

Given: 2x – 7 > 3 AND 5x – 4 < 6Given: 2x – 7 > 3 AND 5x – 4 < 6

What does it mean: Solve the compound What does it mean: Solve the compound inequality?inequality?

It means: Find the set of x so that both It means: Find the set of x so that both inequalities are true inequalities are true

Solution Set:Solution Set:{x | 2x – 7 > 3 AND 5x – 4 < 6}{x | 2x – 7 > 3 AND 5x – 4 < 6}

Solving an AND Compound InequalitySolving an AND Compound Inequality

2x – 7 2x – 7 ≥≥ 3 AND 5x – 4 3 AND 5x – 4 ≤≤ 6 6

2x 2x ≥≥ 3 + 7 5x 3 + 7 5x ≤≤ 6 + 4 6 + 42x 2x ≥≥ 10 5x 10 5x ≤≤ 10 10x x ≥≥ 10/2 x 10/2 x ≤≤ 10/5 10/5x x ≥≥ 5 x 5 x ≤≤ 2 2

Solution Set: { Solution Set: { ΦΦ } }

2 5

Solving an AND Compound InequalitySolving an AND Compound Inequality

-3 < 2x + 1 -3 < 2x + 1 ≤≤ 3 3

This means: This means: (-3 < 2x + 1 AND 2x + 1 (-3 < 2x + 1 AND 2x + 1 ≤≤ 3) 3)

-3 – 1 < 2x + 1 – 1 -3 – 1 < 2x + 1 – 1 ≤ ≤ 3 - 13 - 1-2 < 2x -2 < 2x ≤ ≤ 22-1 < x -1 < x ≤ 1≤ 1

Solution Set: { Solution Set: { x | x | -1 < x -1 < x ≤ 1≤ 1 } }

-1 1

Solving an OR Compound InequalitySolving an OR Compound Inequality

Given:Given:2x – 3 < 7 OR 35 – 4x 2x – 3 < 7 OR 35 – 4x ≤≤ 3 32x < 4 OR -4x 2x < 4 OR -4x ≤ -32≤ -32x < 4 OR x ≥x < 4 OR x ≥ 8 8Take the union of solution setsTake the union of solution sets

{x | {x | x < 4 U x ≥x < 4 U x ≥ 8} 8}= {x | x < 4 or x = {x | x < 4 or x ≥≥ 8} 8}

4 8

Solving an OR Compound InequalitySolving an OR Compound Inequality

Given:Given:3x – 5 3x – 5 ≤ 13 ≤ 13 OR 5x + 2 > -3 OR 5x + 2 > -3

3x 3x ≤≤ 18 OR 5x > -5 18 OR 5x > -5x ≤ 6 OR x >x ≤ 6 OR x > -1 -1Take the union of solution setsTake the union of solution sets

{x | {x | x ≤ 6 U x > -1x ≤ 6 U x > -1}}= {x | x = {x | x ≤≤ 6 or x > -1} = 6 or x > -1} = RR

-1 6

Your TurnYour Turn

Find the following setsFind the following sets1)1) {a, b, c, d, e} {a, b, c, d, e} ∩∩ {b, c, 2, 3, x, y} {b, c, 2, 3, x, y}

2)2) {a, b, c, d, e} U {b, c, 2, 3, x, y}{a, b, c, d, e} U {b, c, 2, 3, x, y}

Solve the followingSolve the following1)1) 3 3 ≤≤ 4x – 3 < 19 4x – 3 < 19

2)2) 3x < 3 or 2x > 103x < 3 or 2x > 10

4.3 Equations & Inequalities 4.3 Equations & Inequalities Involving Absolute ValuesInvolving Absolute Values

Absolute value of A -- |A| -- where A is any Absolute value of A -- |A| -- where A is any algebraic expression:algebraic expression: |A| = c |A| = c A = c or A = -c, where c > 0 A = c or A = -c, where c > 0

|2x – 3| = 11|2x – 3| = 112x – 3 = 11 or 2x – 3 = -112x – 3 = 11 or 2x – 3 = -11

0-c c

A A|A||A|

Solving Equation Involving Solving Equation Involving Absolute ValueAbsolute Value

Solve for x: |2x – 3| = 11Solve for x: |2x – 3| = 11 2x – 3 = 11 or 2x – 3 = -112x – 3 = 11 or 2x – 3 = -11 2x = 14 2x = -8 2x = 14 2x = -8 x = 7 x = - 4 {4, 7}x = 7 x = - 4 {4, 7}Solve for x: 5|1 - 4x| -15 = 0Solve for x: 5|1 - 4x| -15 = 0

|1 – 4x| = 15/5 = 3 |1 – 4x| = 15/5 = 3 1 – 4x = 3 or 1 – 4x = -3 1 – 4x = 3 or 1 – 4x = -3 -4x = 2 -4x = -4 -4x = 2 -4x = -4 x = -1/2 x = 1 {-1/2, 1}x = -1/2 x = 1 {-1/2, 1}

Equation With 2 Absolute ValuesEquation With 2 Absolute Values

Solve for x: |2x – 7| = |x + 3|Solve for x: |2x – 7| = |x + 3|

2x – 7 = (x + 3) or 2x – 7 = -(x + 3)2x – 7 = (x + 3) or 2x – 7 = -(x + 3) x = 10 2x – 7 = -x – 3 x = 10 2x – 7 = -x – 3 3x = 4 3x = 4 x = 4/3 x = 4/3 {4/3, 10} {4/3, 10}

Solving Absolute Value Inequality Solving Absolute Value Inequality (Using Boundary Points)(Using Boundary Points)

Solve and graph: |2x + 3| Solve and graph: |2x + 3| ≥≥ 5 51.1. Solve the equationSolve the equation

2x + 3 = 5 or 2x + 3 = -52x + 3 = 5 or 2x + 3 = -5 x = 1 x = -4 x = 1 x = -4

2.2. Locate the boundary pointsLocate the boundary points

3.3. Choose a test value in each interval and Choose a test value in each interval and substitute in the inequalitysubstitute in the inequality

-4 1

Solve and graph: |2x + 3| Solve and graph: |2x + 3| ≥≥ 5 5


3.3. Choose a test value in each interval and substitute in Choose a test value in each interval and substitute in inequalityinequality

- 4 1

IntervalInterval Test valueTest value CheckCheck ConclusionConclusion

(-(-∞∞ , -4) , -4) -5-5 |2 |2 ∙∙(-5) + 3| ≥ 5(-5) + 3| ≥ 5|-7| ≥ 5 true|-7| ≥ 5 true

(-(-∞∞ , -4) in , -4) in

solution setsolution set

(-4, 1)(-4, 1) 00 |2 |2 ∙ 0 + 3| ≥ 5∙ 0 + 3| ≥ 5|3| ≥ 5 false|3| ≥ 5 false

(-4, 1) not in(-4, 1) not in


(1, (1, ∞∞)) 22 |2 |2 ∙∙ 2 + 3| ≥ 5 2 + 3| ≥ 5|7| ≥ 5 true |7| ≥ 5 true

(1, (1, ∞∞) in) in


Solve and graph: |2x + 3| Solve and graph: |2x + 3| ≥≥ 5 5

- 4 1

4.4. Write the solution set. Check for Write the solution set. Check for boundaries.boundaries.

Preliminary Solution: Preliminary Solution: (-(-∞∞ , -4) , -4) UU (1, (1, ∞∞) )

Because |2x + 3| = 5, we need to include the Because |2x + 3| = 5, we need to include the solution set of this equation (i.e., boundaries): solution set of this equation (i.e., boundaries): x = -4, 1. (This was found in step 1.)x = -4, 1. (This was found in step 1.)

Final Solution: (-Final Solution: (-∞∞ , -4] , -4] UU [1, [1, ∞∞) )

- 4 1

Using Boundary PointsUsing Boundary Points

Solve and graph: |2x -5| Solve and graph: |2x -5| ≥≥ 3 31.1. Solve the equationSolve the equation

2x – 5 = 3 or 2x – 5 = -32x – 5 = 3 or 2x – 5 = -3 x = 4 x = 1 x = 4 x = 1


3.3. Choose a test value in each interval and Choose a test value in each interval and substitute in inequalitysubstitute in inequality

1 4

Solve and graph: |2x - 5| Solve and graph: |2x - 5| ≥≥ 3 32.2. Locate the boundary pointsLocate the boundary points

3.3. Choose a test value in each interval and substitute in Choose a test value in each interval and substitute in inequalityinequality

1 4

IntervalInterval Test valueTest value CheckCheck ConclusionConclusion

(-(-∞∞ , 1) , 1) 00 |2 |2 ∙∙ 0 – 5| 0 – 5| ≥≥ 3 3|-5| ≥ 3 true|-5| ≥ 3 true

(-(-∞∞ , 1) in , 1) in


(1, 4)(1, 4) 22 |2 |2 ∙ 2 - 5| ≥ 3∙ 2 - 5| ≥ 3|-1| ≥ 3 false|-1| ≥ 3 false

(1, 4) not in(1, 4) not in


(4, (4, ∞∞)) 55 |2 |2 ∙∙ 5 - 5| ≥ 3 5 - 5| ≥ 3|5| ≥ 5 true |5| ≥ 5 true

(4, (4, ∞∞) in) in


Solve and graph: |2x – 5| Solve and graph: |2x – 5| ≥≥ 3 3

1 4

4.4. Write the solution set. Check for Write the solution set. Check for boundaries.boundaries.

Preliminary Solution: Preliminary Solution: (-(-∞∞ , 1) , 1) UU (4, (4, ∞∞) )

Because |2x - 5| = 3, we need to include the Because |2x - 5| = 3, we need to include the solution set of this equation (i.e., boundaries): solution set of this equation (i.e., boundaries): x = 1, 4 (This was found in step 1.)x = 1, 4 (This was found in step 1.)

Final Solution: (-Final Solution: (-∞∞ , 1] , 1] UU [4, [4, ∞∞) ) 1 4

Solving Absolute Value Inequality Solving Absolute Value Inequality (Using Compound Inequalities)(Using Compound Inequalities)

Note:Note: Solution set of |x| < 2 is (-2, 2)Solution set of |x| < 2 is (-2, 2)

(-2, 2) (-2, 2) -2 < x < 2 -2 < x < 2

Solution set of |x| > 2 is (-Solution set of |x| > 2 is (-∞∞, -2) , -2) UU (2, (2, ∞∞))

-2 20

-2 20

(-∞, -2) U (2, ∞) (-∞, -2) U (2, ∞) x < -2 or x > 2 x < -2 or x > 2


Solve: |x – 4| < 3Solve: |x – 4| < 3

-3 < x – 4 < 3-3 < x – 4 < 31 < x < 71 < x < 7


Solve: |2x + 3| Solve: |2x + 3| ≥≥ 5 5

2x + 3 2x + 3 ≥≥ 5 or 2x + 3 5 or 2x + 3 ≤≤ -5 -5 2x 2x ≥≥ 2 or 2x 2 or 2x ≤≤ -8 -8 x x ≥≥ 1 or x 1 or x ≤≤ -4 -4

-4 1

Your TurnYour Turn

Solve inequalities using equivalent Solve inequalities using equivalent compound inequalitiescompound inequalities

1.1. |x – 2| < 5|x – 2| < 5

2.2. |2x – 5| |2x – 5| ≥ ≥ 33

4.4 Linear Inequalities in 2 Variables4.4 Linear Inequalities in 2 VariablesSolve: 2x – 3y Solve: 2x – 3y ≥≥ 6 6

1.1. Graph: 2x – 3y = 6Graph: 2x – 3y = 6

To find y-intercept To find x-intercept To find y-intercept To find x-intercept y = 0 x = 0 y = 0 x = 0 2x = 6 -3y = 6 2x = 6 -3y = 6 x = 3 y = -2 x = 3 y = -2

(0, -2)

(3, 0)

2x – 3y > 6

2x – 3y = 6

2x – 3y < 6

2.2. Choose a test point in one half-plane and Choose a test point in one half-plane and check with original inequality.check with original inequality.

2x – 3y 2x – 3y ≥≥ 6 6

Choose A (0, 0) as a test pointChoose A (0, 0) as a test point0 – 0 0 – 0 ≥≥ 6 60 0 ≥≥ 6 6false—A is outside the solution setfalse—A is outside the solution set

3.3. If A(0, 0) is not in solution set, the other half-If A(0, 0) is not in solution set, the other half-plane is the solution set of 2x – 3y plane is the solution set of 2x – 3y ≥≥ 6 6

Because of Because of ≥≥ , include the boundary line in , include the boundary line in the graph of the solution set.the graph of the solution set.

2x – 3y < 6

2x – 3y = 6

2x – 3y > 6

A(0, 0)

Graph of : {x | 2x – 3y ≥ 6}

Your TurnYour Turn

Graph the following inequality:Graph the following inequality:1.1. 4x – 2y 4x – 2y ≥≥ 8 8

2.2. x/4 + y/2 < 1x/4 + y/2 < 1

Graphing System of Linear InequalitiesGraphing System of Linear Inequalities

Graph solution set of:Graph solution set of:x – y < 1x – y < 12x + 3y 2x + 3y ≥≥ 12 12

Graph equationsGraph equations x – y = 1 2x + 3y = 12 x – y = 1 2x + 3y = 12x-intercept: x-intercept:x-intercept: x-intercept: y = 0 y = 0 y = 0 y = 0 x – 0 = 1 2x + 0 = 12 x – 0 = 1 2x + 0 = 12 x = 1 x = 6 x = 1 x = 6 (1, 0) (6, 0) (1, 0) (6, 0)

Graphing System of Linear InequalitiesGraphing System of Linear Inequalities Graph equalitiesGraph equalities

x – y = 1 2x + 3y = 12 x – y = 1 2x + 3y = 12x-intercept: x-intercept:x-intercept: x-intercept: (1, 0) (6, 0) (1, 0) (6, 0)y-intercept: y-intercept:y-intercept: y-intercept: x = 0 x = 0 x = 0 x = 0 0 – y = 1 0 + 3y = 12 0 – y = 1 0 + 3y = 12 -y = 1 3y = 12 -y = 1 3y = 12 y = -1 y = 4 y = -1 y = 4 (0, -1) ( 0, 4) (0, -1) ( 0, 4)

Points for: x – y = 1 Points for: 2x + 3y = 12Points for: x – y = 1 Points for: 2x + 3y = 12 (1, 0) (0, -1) (6, 0) (0, 4) (1, 0) (0, -1) (6, 0) (0, 4)


2x + 3y = 12 x – y = 1

(0, 0) (1, 0)

(1, 0)

(0, 4)

(6, 0)

Graphing System of Linear InequalitiesGraphing System of Linear Inequalities Choose a point and check with original Choose a point and check with original

inequalities.inequalities.

Pick (0, 0) Pick (0, 0)Pick (0, 0) Pick (0, 0)Part of x – y < 1? Part of 2x + 3y Part of x – y < 1? Part of 2x + 3y ≥≥ 12? 12? Check: Check:Check: Check: 0 – 0 < 1? 0 + 0 0 – 0 < 1? 0 + 0 ≥ 12?≥ 12? 0 < 1? 0 0 < 1? 0 ≥ 12?≥ 12? true false true false this half-plane other half-plane this half-plane other half-plane


2x + 3y = 12

x – y = 1

(0, 0)

Your TurnYour Turn

Graph the solution set of the system:Graph the solution set of the system:

x – 3y < 6x – 3y < 62x + 3y 2x + 3y ≥≥ -6 -6

4.5 Linear Programming4.5 Linear Programming

Problem:Problem: A division of a furniture company specializes inA division of a furniture company specializes in

manufacturing bookcases and computer desks.manufacturing bookcases and computer desks. The division makes $25 per bookcase and $55 The division makes $25 per bookcase and $55

per desk.per desk. To maintain quality, the division can make a To maintain quality, the division can make a

maximum of 80 bookcases and desks (total) per maximum of 80 bookcases and desks (total) per dayday


Problem (cont.)Problem (cont.) Because of customer demands, between 30 and Because of customer demands, between 30 and

80 bookcases must be made daily.80 bookcases must be made daily. Furthermore, at least 10 and not more than 30 Furthermore, at least 10 and not more than 30

desks must be made per daydesks must be made per day How many bookcases and desks must be made How many bookcases and desks must be made

each day to maximize profit?each day to maximize profit?

4.5 Linear Programming4.5 Linear ProgrammingSolutionSolution

1.1. Use variables to represent quantitiesUse variables to represent quantitiesx = number of bookcases per monthx = number of bookcases per monthy = desks per monthy = desks per monthz = profit for monthz = profit for month

2.2. Form Form objective functionobjective functionz = 25x + 55yz = 25x + 55y

3.3. Write Write constraints constraints as inequalitiesas inequalitiesx + y x + y ≤≤ 80 8030 30 ≤ ≤ x x ≤≤ 80 8010 10 ≤≤ y y ≤≤ 30 30

4.4. Graph the inequalitiesGraph the inequalities


4.4. Graph the inequalitiesGraph the inequalities1.1. x + y x + y ≤ ≤ 80 80

x + y = 80x + y = 80line passes through (80, 0) and (0, 80)line passes through (80, 0) and (0, 80)

2.2. 30 30 ≤≤ x x ≤≤ 80 80y can be any valuey can be any value

3.3. 10 10 ≤≤ y y ≤≤ 30 30x can be any valuex can be any value



(80, 0)

(0, 80)

x + y ≤ 80

30 ≤ x ≤ 80

10 ≤ y ≤ 30

A

D C

B


5.5. Determine the corners of the solution areaDetermine the corners of the solution areaTo find A: x = 30 y = 30 (30, 30)

To find B: y = 30 x + 30 = 80 x = 50 (50, 30)

To find C: y = 10 x + 10 = 80 x = 70 (70, 10)

To find D: (30, 10)

(80, 0)

(0, 80)

x + y = 80 x = 80

y = 30 A

D C

B

y = 10

x = 30

4.5 Linear Programming4.5 Linear Programming6.6. Check the objective equation with the corner pointsCheck the objective equation with the corner points

Corner (x, y)Corner (x, y) Objective FunctionObjective Functionz = 25x + 55yz = 25x + 55y

(30, 30)(30, 30) z = 25(30) + 55(30) = 2400 z = 25(30) + 55(30) = 2400

(50, 30)(50, 30) z = 25(50) + 55(30) = 2900z = 25(50) + 55(30) = 2900

(70, 10)(70, 10) z = 25(70) + 55(10) = 2300z = 25(70) + 55(10) = 2300

(30, 10)(30, 10) z = 25(30) + 55(10) = 1300z = 25(30) + 55(10) = 1300

Solution:

50 bookcases, 30 desks, resulting in $2900 profit

Linear ProgrammingLinear Programming(Another Example)(Another Example)

Problem:Problem: Food and clothing are shipped to survivors of a hurricane. Food and clothing are shipped to survivors of a hurricane.

Each carton of food will feed 12 people, while each Each carton of food will feed 12 people, while each carton of clothing will help 5 people.carton of clothing will help 5 people.

Each 20 ftEach 20 ft33 box of food weights 50 lb, and each 10 ft box of food weights 50 lb, and each 10 ft3 3 box box of food will weight 20 lbof food will weight 20 lb

Planes are bound by the following constraintsPlanes are bound by the following constraintsTotal weight per plane Total weight per plane ≤≤ 19000 lb 19000 lb

Total volume per plane Total volume per plane ≤≤ 8000 ft 8000 ft33

How many cartons of food and how many cartons How many cartons of food and how many cartons of clothing should be sent with each plane to of clothing should be sent with each plane to maximize the number of people who can be maximize the number of people who can be helped?helped?

Linear ProgrammingLinear Programming

SolutionSolution1.1. Use variables to represent quantitiesUse variables to represent quantities

x = cartons of foodx = cartons of foody = cartons of clothingy = cartons of clothingz = number of people helpedz = number of people helped

2.2. Form Form objective functionobjective functionz = 12x + 5yz = 12x + 5y

3.3. Write Write constraints constraints as inequalitiesas inequalities50x + 20y 50x + 20y ≤≤ 19,000 19,00020x + 10y 20x + 10y ≤≤ 8,000 8,000


4.4. Graph the inequalitiesGraph the inequalities1.1. 50x + 20y 50x + 20y ≤ 19000≤ 19000

50x + 20y = 1900050x + 20y = 19000

Find 2 points—e.g., y-intercept & x-interceptFind 2 points—e.g., y-intercept & x-intercept(0, 950) & (380, 0)(0, 950) & (380, 0)

2.2. 20x + 10y 20x + 10y ≤≤ 8000 800020x + 10y = 800020x + 10y = 8000

Find 2 pointsFind 2 points (0, 800) & (400, 0) (0, 800) & (400, 0)y can be any valuey can be any value



(380, 0)

(0, 950)

20x + 10y ≤ 800050x + 20y ≤ 19000

A

C

B

(400, 0)

(0, 800)


5.5. Determine the corners of the solution areaDetermine the corners of the solution areaTo find A: (0, 800)

To find B: 50x + 20y = 19000 20x + 10y = 8000

50x + 20y = 19000 -40x – 20y = -16000 10x = 3000 x = 300 y = 200 (300, 200)

To find C: (380, 0)

(380, 0)

(0, 950)

20x + 10y ≤ 800050x + 20y ≤ 19000

A

C

B

(400, 0)

(0, 800)

Linear ProgrammingLinear Programming6.6. Check the objective equation with the corner pointsCheck the objective equation with the corner points

Corner (x, y)Corner (x, y) Objective FunctionObjective Functionz = 12x + 5yz = 12x + 5y

(0, 800)(0, 800) z = 12(0) + 5(800) = 4000 z = 12(0) + 5(800) = 4000

(300, 200)(300, 200) z = 12(300) + 5(200) = 4600z = 12(300) + 5(200) = 4600

(380, 0)(380, 0) z = 12(380) + 5(0) = 4560z = 12(380) + 5(0) = 4560

Solution:

300 food cartons, 200 clothing cartons, resulting in 4600 people helped

Your turnYour turnProblem:Problem:

A theater is presenting a program on A theater is presenting a program on drinking and driving for students and their drinking and driving for students and their parents.parents.

Admission $2.00 for parents $1.00 for Admission $2.00 for parents $1.00 for students.students.

However, the situation has two constraints:However, the situation has two constraints:The theater can hold no more than 150 people,The theater can hold no more than 150 people,and every two parents must bring at least one and every two parents must bring at least one studentsstudents

How many parents and students should How many parents and students should attend to raise the maximum amount of attend to raise the maximum amount of money?money?

SolutionSolution Write the objective functionWrite the objective function Write the constraints inequalitiesWrite the constraints inequalities

SolutionsSolutions

VariablesVariables x = number of parentsx = number of parents y = number of studentsy = number of students z = total amount of moneyz = total amount of money

Objective FunctionObjective Function z = 2x + yz = 2x + y

ConstraintsConstraints x + y x + y ≤≤ 150 150 x x ≥≥ 2y 2y

SolutionSolution

x (parents)x (parents) y (students)y (students)

11 11

22 11

33 22

44 22

55 33

66 33x ≤ 2y

SolutionSolution

(0, 150)

(150, 0)

x + y = 150

x = 2y

(100, 50)

Documents

4. Inequalities. 4.1 Solving Linear Inequalities Problem Basic fee: $20 Basic fee: $20 Per minute: 5¢ Per minute: 5¢ Budget: $40 Budget: $40 How many