Transcript Slide 1
4. Inequalities
4.1 Solving Linear Inequalities
Problem
Basic fee:
$20
Per minute:
5¢
Budget:
$40
How many minutes?: x
20 + 0.05x ≤ 40
x ≤ 400
Notations
Closed interval
[a, b] = {x | a ≤ x ≤ b}
a
b
Open interval
(a, b) = {x | a < x < b}
a
b
Notations
Infinite interval
[a, ∞) = {x | a ≤ x < ∞}
a
Infinite interval
(-∞ , b] = {x | -∞ < x ≤ b}
b
Your Turn
Express in set-builder notation.
[a, b)
(- ∞ , b)
Your Turn
Express in set-builder notation.
[a, b) = {x | a ≤ x < b}
(- ∞ , b) = {x | ∞ < x < b}
Solving Inequalities in One Variable
0.05x + 20 ≤ 40
0.05x
≤ 20
x
≤ 20/0.05
x
≤ 400
[0, 400]
(Interval notation)
{x | x ≤ 400} (set-builder notation)
Properties of Inequalities
Addition
a<b → a+c<b+c
a<b → a- c<b–c
Positive Multiplication (c > 0)
a < b → ac < bc
a < b → a /c < b/c
Negative Multiplication (c < 0)
a < b → ac ≥ bc
a < b → a /c ≥ b/c
Example 1
-2x – 4 > x + 5
-3x > 9
(-1/3)(-3x) < (-1/3)9
x < -3
(- ∞ ,-3)
Example 2
(x + 3)
(x – 2)
1
--------- ≥ ---------- + --4
3
4
3(x + 3)
4(x – 2)
3
------------ ≥ ----------- + ----12
12
12
3x + 9 ≥ 4x – 8 + 3
-x
≥ -14
x
≤ 14
(-∞, 14]
Special Cases
x>x+1
{x | x > x + 1}
What kind of set is this?
x<x+1
{x | x < x + 1}
What kind of set is this?
Your Turn
Solve the inequality and graph the solution
set.
5(3 – x) ≤ 3x - 1
Solution
5(3 – x) ≤ 3x – 1
15 – 5x ≤ 3x – 1
-8x ≤ -16
(-8x)/(-8) ≥ (-16)/(-8)
x≥2
Solution: [2, ∞)
2
4.2 Compound Inequalities
Intersection of Sets
Given set A and B, intersection A and B,
A ∩ B = {x | x ε A AND x ε B}
E.g.,
A
B
A∩B
A = {Female students
at Chaminade}
B = {First-year students
at Chaminade}
A ∩ B = {First-year female
students at Chaminade}
Compound Inequalities
Union of Sets
Given set A and B, union of A and B,
A U B = {x | x ε A OR x ε B}
E.g.,
A
B
AUB
A = {Female students
at Chaminade}
B = {First-year students
at Chaminade}
A U B = {First-year students
OR female students
at Chaminade}
Intersection of Sets
Given: A = {1, 2, 3, 5, 9}
B = {3, 5, 9, 10, 12}
A ∩ B = {3, 5, 9}
Given: A = {x | x ≥ 3}
B = {x | x ≤ 10}
A ∩ B = {x | x ≥ 3 AND x ≤ 10 }
3
10
Union and Intersection of Sets`
Given:
A = set of all male students at CUH
B = set of all female students at CUH
C = set of all freshman students at CUH
Draw a diagram of:
1.
2.
3.
4.
A∩B
AUB
A∩C
AUC
Union and Intersection of Sets`
Given:
A = set of all male students at CUH
B = set of all female students at CUH
C = set of all freshman students at CUH
Draw a diagram of:
3. A ∩ C
1. A ∩ B
A
A
B
C
A∩C
A∩B=Φ
4. A U C
2. A U B
A
A
B
AUB
C
AUC
Solving Compound Inequality
Given: 2x – 7 > 3 AND 5x – 4 < 6
What does it mean: Solve the compound
inequality?
It means: Find the set of x so that both
inequalities are true
Solution Set:
{x | 2x – 7 > 3 AND 5x – 4 < 6}
Solving an AND Compound Inequality
2x – 7 ≥ 3 AND 5x – 4 ≤ 6
2x ≥ 3 + 7
2x ≥ 10
x ≥ 10/2
x ≥ 5
2
Solution Set: Φ
5x ≤ 6 + 4
5x ≤ 10
x ≤ 10/5
x ≤ 2
5
Solving an AND Compound Inequality
-3 < 2x + 1 ≤ 3
This means:
(-3 < 2x + 1 AND 2x + 1 ≤ 3)
-3 – 1 < 2x + 1 – 1 ≤ 3 - 1
-2 < 2x ≤ 2
-1 < x ≤ 1
-1
1
Solution Set: { x | -1 < x ≤ 1 }
Solving an OR Compound Inequality
Given:
2x – 3 < 7
OR 35 – 4x ≤ 3
2x < 4
OR -4x ≤ -32
x<4
OR x ≥ 8
Take the union of solution sets
4
8
{x | x < 4 U x ≥ 8}
= {x | x < 4 or x ≥ 8}
Solving an OR Compound Inequality
Given:
3x – 5 ≤ 13 OR 5x + 2 > -3
3x ≤ 18
OR 5x > -5
x≤6
OR x > -1
Take the union of solution sets
-1
6
{x | x ≤ 6 U x > -1}
= {x | x ≤ 6 or x > -1} = R
Your Turn
Find the following sets
1) {a, b, c, d, e} ∩ {b, c, 2, 3, x, y}
2) {a, b, c, d, e} U {b, c, 2, 3, x, y}
• Solutions
1. {b, c}
2. {a, b, c, d, e, 2, 3, x, y}
Your Turn
Solve the following inequalities and
graph the solution set.
3 ≤ 4x – 3 < 19
Solution
3 ≤ 4x – 3 AND
6 ≤ 4x
6/4 ≤ x
3/2 ≤ x
3/2
4x - 3< 19
4x < 22
x < 22/4
x < 11/2
11/2
Your Turn
Solve the following inequalities and
graph the solution set.
3x < 3 or 2x > 10
Solution
3x < 3
OR
x<1
1
2x > 10
x>5
5
4.3 Equations & Inequalities
Involving Absolute Values
Absolute value of A -- |A| -- where A is any
algebraic expression:
|A| = c A = c or A = -c, where c > 0
-c
0
|A|
c
|A|
|2x – 3| = 11
2x – 3 = 11 or 2x – 3 = -11
Solving Equation Involving
Absolute Value
Solve for x:
|2x – 3| = 11
2x – 3 = 11 or 2x – 3 = -11
2x = 14
2x = -8
x=7
x=-4
Solution: {4, 7}
Solve for x:
5|1 - 4x| -15 = 0
|1 – 4x| = 15/5 = 3 OR 1 – 4x = -3
1 – 4x = 3
-4x = -4
-4x = 2
x=1
x = -1/2
Soltion:
{-1/2, 1}
Equation With 2 Absolute Values
Solve for x:
|2x – 7| = |x + 3|
2x – 7 = (x + 3)
x = 10
or
2x – 7 = -(x + 3)
2x – 7 = -x – 3
3x = 4
x = 4/3
{4/3, 10}
Solving Absolute Value Inequality
(Using Boundary Points)
Solve and graph:
|2x + 3| ≥ 5
1. Solve the equation
2x + 3 = 5
or 2x + 3 = -5
x=1
x = -4
2. Locate the boundary points
-4
1
3. Choose a test value in each interval and
substitute in the inequality
|2x + 3| ≥ 5
Solve and graph:
2. Locate the boundary points
-4
1
3. Choose a test value in each interval and substitute in
inequality
Interval
(-∞ , -4)
Test value
-5
Check
|2 ∙(-5) + 3| ≥ 5
|-7| ≥ 5
true
(-4, 1)
0
|2 ∙ 0 + 3| ≥ 5
|3| ≥ 5
(1, ∞)
2
|2 ∙ 2 + 3| ≥ 5
|7| ≥ 5
false
Conclusion
(-∞ , -4) in
solution set
(-4, 1) not in
solution set
true
(1, ∞) in
solution set
|2x + 3| ≥ 5
Solve and graph:
-4
1
4. Write the solution set. Check for
boundaries.
Preliminary Solution: (-∞ , -4) U (1, ∞)
Because |2x + 3| = 5, we need to include the
solution set of this equation (i.e., boundaries):
x = -4, 1. (This was found in step 1.)
-4
1
Final Solution: (-∞ , -4] U [1, ∞)
Using Boundary Points
Solve and graph:
|2x -5| ≥ 3
1. Solve the equation
2x – 5 = 3
or 2x – 5 = -3
x=4
x=1
2. Locate the boundary points
1
4
3. Choose a test value in each interval and
substitute in inequality
|2x - 5| ≥ 3
Solve and graph:
2. Locate the boundary points
1
4
3. Choose a test value in each interval and substitute in
inequality
Interval
(-∞ , 1)
(1, 4)
(4, ∞)
Test value
0
2
5
Check
|2 ∙ 0 – 5| ≥ 3
|-5| ≥ 3
true
|2 ∙ 2 - 5| ≥ 3
|-1| ≥ 3
false
|2 ∙ 5 - 5| ≥ 3
|5| ≥ 5
true
Conclusion
(-∞ , 1) in
solution set
(1, 4) not in
solution set
(4, ∞) in
solution set
|2x – 5| ≥ 3
Solve and graph:
1
4
4. Write the solution set. Check for
boundaries.
Preliminary Solution: (-∞ , 1) U (4, ∞)
Because |2x - 5| = 3, we need to include the
solution set of this equation (i.e., boundaries):
x = 1, 4 (This was found in step 1.)
1
4
Final Solution: (-∞ , 1] U [4, ∞)
Solving Absolute Value Inequality
(Using Compound Inequalities)
Note:
Solution set of |x| < 2 is (-2, 2)
-2
0
2
(-2, 2) -2 < x < 2
Solution set of |x| > 2 is (-∞, -2) U (2, ∞)
-2
0
2
(-∞, -2) U (2, ∞) x < -2 or x > 2
Solving Absolute Value Inequality
(Using Compound Inequalities)
Solve: |x – 4| < 3
-3 < x – 4 < 3
1<x<7
Solving Absolute Value Inequality
(Using Compound Inequalities)
Solve: |2x + 3| ≥ 5
2x + 3 ≥ 5
2x ≥ 2
x ≥1
-4
or
or
or
1
2x + 3 ≤ -5
2x ≤ -8
x ≤ -4
Your Turn
Solve inequalities using equivalent
compound inequalities
|x – 2| < 5
Your Turn
|x – 2| < 5
1. Solve equation
x–2=5
AND x – 2 = -5
x=7
x = -3
2. Locate boundary points
-3
7
Solve: |x – 2| < 5
4. Choose test values
-3
7
a) x = 8
|8 – 2| < 5
|6| < 5
False
c) x = -4
|-4 – 2| < 5
|-6| < 5
False
5. Solution Set:
-3
Solution: [-3, 7]
7
b) x = 0
|0 – 2| < 5
|2| < 5
True
4.3 Inequalities Involving Absolute
Values
There are 2 cases
Case 1. |x| < c
Case 2. |x| > c
Case 1. |x| < c
-c
0
c
Absolute Values in Inequalities
Case 1. |x| < c
-c
|x| < c
0
means
Same as: -c < x
c
-c < x < c
AND
x<c
Absolute Values in Inequalities
Case 1 Example
Solve: |2x – 6| < 8
-8 < 2x – 6 < 8
-8 < 2x – 6
AND
2x – 6 < 8
-2 < 2x
AND
2x < 14
-1 < x
AND
x<7
Solution: (-1, ∞) ∩ (-∞ ,7) = (-1, 7)
-1
7
Absolute Values in Inequalities
Case 2. |x| > c
-c
|x| > c
0
means
c
x < -c
OR
x>c
Absolute Values in Inequalities
Case 2 Example
-2
0
2
Solve: |x – 4| ≥ 2
x-4≥ 2
OR
x≥6
OR
Solution: (-∞, 2] U [6, ∞)
x – 4 ≤ -2
x≤2
4.4 Linear Inequalities in 2 Variables
Solve: 2x – 3y ≥ 6
1. Graph: 2x – 3y = 6
To find x-intercept
y=0
2x = 6
x=3
To find y-intercept
x=0
-3y = 6
y = -2
2x – 3y > 6
(3, 0)
2x – 3y = 6
(0, -2)
2x – 3y < 6
2. Choose a test point in one half-plane and
check with original inequality.
2x – 3y ≥ 6
Choose A (0, 0) as a test point
0–0≥6
0≥6
false—A is outside the solution set
2x – 3y > 6
(3, 0)
2x – 3y = 6
(0, -2)
2x – 3y < 6
3. If A(0, 0) is not in solution set, the other halfplane is the solution set of 2x – 3y ≥ 6
Because of ≥ , include the boundary line in
the graph of the solution set.
Graph of : {x | 2x – 3y ≥ 6}
2x – 3y = 6
2x – 3y < 6
A(0, 0)
2x – 3y > 6
Why is the solution set
below the 2x – 3y = 6 line?
Line:
2x – 3y = 6
-3y = -2x + 6
y = (-2x + 6)/(-3)
y = (2/3)x – 2
Half-plane below line
2x – 3y ≥ 6
-3y ≥ -2x + 6
y ≤ (-2x + 6)/(-3)
y ≤ (2/3)x - 3
y=(2/3)x - 2
A(0, 0)
y < (2/3)x - 3
Your Turn
Graph the following inequality:
4x – 2y ≥ 8
Solution
4x – 2y ≥ 8
4x – 2y = 8
x-intercept:
y=0
4x - 2(0) = 8
4x = 8
x=2
(2, 0)
y-intercept:
x=0
4(0) – 2y = 8
-2y = 8
y = -4
(0, -4)
Solution
4x – 2y ≥ 8
4x – 2y = 8
x-intercept:
(2, 0)
y-intercept:
(0, -4)
4x – 2y = 8)
(0, 0)
(2, 0)
(0,-4)
Solution
4x – 2y ≥ 8
4x – 2y = 8
Is (0, 0) on the line? No.
Thus, solution set is below the line
4x – 2y = 8)
(0, 0)
(2, 0)
(0,-4)
Your Turn
Graph the solution set of the following
inequality.
x/4 + y/2 < 1
Consider: x/4 + y/2 = 1
x-intercept
y-intercept
y=0
x=0
x/4 = 1
y/2 = 1
x=4
y=2
(4, 0)
(0, 2)
Your Turn
x/4 + y/2 < 1
2 points: (4, 0) and (0, 2)
(0, 2)
(4, 0)
Test point, e.g., (0, 0)
0/4 + 0/2 < 1
True
Solution set: half plane below the line.
Graphing System of Linear Inequalities
Graph solution set of:
x–y<1
2x + 3y ≥ 12
Graph equations
x–y=1
x-intercept:
y=0
x–0=1
x=1
(1, 0)
2x + 3y = 12
x-intercept:
y=0
2x + 0 = 12
x=6
(6, 0)
Graphing System of Linear Inequalities
Graph equalities
x–y=1
x-intercept:
(1, 0)
y-intercept:
x=0
0–y=1
-y = 1
y = -1
(0, -1)
Points for: x – y = 1
(1, 0) and (0, -1)
2x + 3y = 12
x-intercept:
(6, 0)
y-intercept:
x=0
0 + 3y = 12
3y = 12
y=4
( 0, 4)
Points for: 2x + 3y = 12
(6, 0) and (0, 4)
Graphing System of Linear Inequalities
x–y=1
2x + 3y = 12
(0, 4)
(0, 0)
(1, 0)
(0, -1)
(6, 0)
Graphing System of Linear Inequalities
Choose a point and check with original
inequalities.
Pick (0, 0)
Part of x – y < 1?
Check:
0 – 0 < 1?
0 < 1?
true
this half-plane
Pick (0, 0)
Part of 2x + 3y ≥ 12?
Check:
0 + 0 ≥ 12?
0 ≥ 12?
false
other half-plane
Graphing System of Linear Inequalities
x–y=1
2x + 3y = 12
(0, 0)
Your Turn
Graph the solution set of the system:
x – 3y < 6
2x + 3y ≥ -6
Your Turn
x – 3y < 6
2x + 3y ≥ -6
x – 3y < 6
AND
Graph the line
x – 3y = 6
Let x = 0
Then -3y = 6
y = -2
(0, -2)
2x + 3y ≥ -6
2x + 3y = -6
Let x = 0
Then 3y = -6
y = -2
(0, -2)
Your Turn
x – 3y < 6
Graph line
x – 3y = 6
Let y = 0
Then x = 6
(6, 0)
2 points are
(0, -2) & (6, 0)
AND
2x + 3y ≥ -6
2x + 3y = -6
Let y = 0
Then 2x = -6
x = -3
(-3, 0)
(0, -2) & (-3, 0)
Your Turn
x – 3y < 6
AND
2 points for line are
(0, -2) & (6, 0)
(-3, 0)
2x + 3y ≥ -6
(0, -2) & (-3, 0)
(6, 0)
(0,-2)
Review
Given:
a)
b)
c)
d)
3 < 7 - 4x < 19
Express this as a compound inequality.
Solve the two inequalities.
Express the solution set in interval notation.
Graph the solution set.
Given: 3 < 7 - 4x < 19
a) Express this as a compound inequality.
3 < 7 – 4x
AND
7 – 4x < 19
b) Solve the two inequalities.
- 4 < - 4x
-4x < 12
1>x
x > -3
c) Express the solution set in interval notation.
(-∞ , 1) AND (-3, ∞)
d) Graph the solution set.
-3
1
4.5 Linear Programming
Problem:
A division of a furniture company specializes in
manufacturing bookcases and computer desks.
The division makes $25 per bookcase and $55
per desk.
To maintain quality, the division can make a
maximum of 80 bookcases and desks (total) per
day
4.5 Linear Programming
Problem (cont.)
Because of customer demands, between 30 and
80 bookcases must be made daily.
Furthermore, at least 10 and not more than 30
desks must be made per day
How many bookcases and desks must be made
each day to maximize profit?
4.5 Linear Programming
Solution
1.
Use variables to represent quantities
x = number of bookcases per month
y = desks per month
z = profit for month
2.
Form objective function
z = 25x + 55y
3.
Write constraints as inequalities
x + y ≤ 80
30 ≤ x ≤ 80
10 ≤ y ≤ 30
4.
Graph the inequalities
4.5 Linear Programming
4. Graph the inequalities
1. x + y ≤ 80
x + y = 80
line passes through (80, 0) and (0, 80)
2. 30 ≤ x ≤ 80
y can be any value
3. 10 ≤ y ≤ 30
x can be any value
4.5 Linear Programming
4. Graph the inequalities
30 ≤ x ≤ 80
(0, 80)
A
B
10 ≤ y ≤ 30
D
C
(80, 0)
(0, 0)
x + y ≤ 80
4.5 Linear Programming
5. Determine the corners of the solution area
x + y = 80
x = 80
x = 30
To find A:
x = 30
y = 30
(30, 30)
(0, 80)
To find B:
y = 30
x + 30 = 80
x = 50
(50, 30)
y = 30
A
B
y = 10
D
C
(80, 0)
To find C:
y = 10
x + 10 = 80
x = 70
(70, 10)
To find D:
(30, 10)
4.5 Linear Programming
6. Check the objective equation with the corner
points
Corner (x, y)
(30, 30)
(50, 30)
(70, 10)
(30, 10)
Objective Function
z = 25x + 55y
z = 25(30) + 55(30) = 2400
z = 25(50) + 55(30) = 2900
z = 25(70) + 55(10) = 2300
z = 25(30) + 55(10) = 1300
Solution:
50 bookcases, 30 desks, resulting in $2900 profit
Linear Programming
(Another Example)
Problem:
Food and clothing are shipped to survivors of a hurricane.
Each carton of food will feed 12 people, while each carton
of clothing will help 5 people.
Each 20 ft3 box of food weights 50 lb, and each 10 ft3 box
of food will weight 20 lb
Planes are bound by the following constraints
Total weight per plane ≤ 19000 lb
Total volume per plane ≤ 8000 ft3
How many cartons of food and how many cartons
of clothing should be sent with each plane to
maximize the number of people who can be
helped?
Linear Programming
Solution
1. Use variables to represent quantities
x = cartons of food
y = cartons of clothing
z = number of people helped
2. Form objective function
z = 12x + 5y
3. Write constraints as inequalities
50x + 20y ≤ 19,000
20x + 10y ≤ 8,000
Linear Programming
4. Graph the inequalities
1. 50x + 20y ≤ 19000
50x + 20y = 19000
Find 2 points—e.g., y-intercept & x-intercept
(0, 950) & (380, 0)
2. 20x + 10y ≤ 8000
20x + 10y = 8000
Find 2 points
(0, 800) & (400, 0)
Linear Programming
4. Graph the inequalities
(0, 950)
A
(0, 800)
B
(400, 0)
(380, 0) C
50x + 20y ≤ 19000
20x + 10y ≤ 8000
Linear Programming
5. Determine the corners of the solution area
To find A:
(0, 800)
(0, 950)
To find B:
50x + 20y = 19000
20x + 10y = 8000
A
(0, 800)
50x + 20y = 19000
-40x – 20y = -16000
10x = 3000
x = 300
y = 200
(300, 200)
B
(400, 0)
(380, 0) C
50x + 20y ≤ 19000
20x + 10y ≤ 8000
To find C:
(380, 0)
Linear Programming
6. Check the objective equation with the corner
points
Corner (x, y)
(0, 800)
(300, 200)
(380, 0)
Objective Function
z = 12x + 5y
z = 12(0) + 5(800) = 4000
z = 12(300) + 5(200) = 4600
z = 12(380) + 5(0) = 4560
Solution:
300 food cartons, 200 clothing cartons, resulting in 4600
people helped
Your turn
Problem:
A theater is presenting a program on drinking
and driving for students and their parents.
Admission $2.00 for parents $1.00 for
students.
However, the situation has two constraints:
The theater can hold no more than 150 people.
Every two parents must bring at least one student.
How many parents and students should
attend to raise the maximum amount of
money?
Solution
Write the objective function
Write the constraints inequalities
Solutions
Variables
x = number of parents
y = number of students
z = total amount of money
Objective Function
z = 2x + y
Constraints
x + y ≤ 150
x ≥ 2y
Solution
x (parents)
y (students)
1
1
2
1
3
2
4
2
5
3
6
3
x ≤ 2y
Solution
(0, 150)
(100, 50)
x = 2y
(150, 0)
x + y = 150
More Example of L.P.
Suppose that you have 240 acres of land,
on which you plan to grow corn and oats.
You estimate that corn yields $40/acre
profit; and oats, $30/acre profit.
At harvest time, you have only 320 hours
to work. Corn takes 2 hr/acre to harvest;
and oats, 1 hr/acre.
How may acres of corn and oats should
you plant in order to maximize your profit?
Solution
1. Use variables to represent quantities
being sought
x: acres of corn
y: acres of oats
2. Find objective function
z = 40x + 30y
3. Set up constraints
x ≥ 0; y ≥ 0
x + y ≤ 240
2x + y ≤ 320
Solution
4. Solve inequalities
x + y ≤ 240 AND
x + y = 240
If x = 0,
then y = 0
(0, 240)
If y = 0,
then x = 240
(240, 0)
2x + y ≤ 320
2x + y = 320
If x = 0,
then y = 320
(0, 320)
If y = 0,
then x = 160
(160, 0)
Solution
5. Graph inequality solutions
x + y ≤ 240
(0, 240) & (240, 0)
2x + y ≤ 320
320
(0, 320) & (160, 0)
240 A
B
D
C
160
240
Solution
5. Find the coordinates of corners.
A. (0, 240)
C. (160, 0)
D. (0, 0)
B. Where 2 lines intersect
2x + y = 320
x + y = 240
320
240 A
B
D
2x + y = 320
- (x + y) = -240
C
160
240
x = 80
y = 240 – x = 160
(80, 160)
Solution
6. Check the objective function with the corner
values
320
240 A
B
D
Corner
coordinates
Objective function
z = 40x + 30y
A (0, 240)
z = 40(0) + 30(240) = 7200
B (80, 160)
z = 40(80) + 30(160) = 8000
C (160, 0)
z = 40(160) + 30(0) = 6400
D (0, 0)
z=0
C
160
240
Thus,
x. = 80 acres (corn) and
y = 160 acres (oats)
will yield the maximum profit
Another Exercise on L.P.
Given
Objective function: z = 2x + 3y
Constraints:
a) x ≥ 0, y ≥ 0
b) x + y ≤ 8
c) 3x + 2y ≥ 6
a) Graph the region determined by the
constraints
b) Find the maximum value of the objective
function, subject o the constraints.
Objective function: z = 2x + 3y
Constraints:
a) x ≥ 0, y ≥ 0
b) x + y ≤ 8
c) 3x + 2y ≥ 6
a)
Objective function: z = 2x + 3y
Constraints:
a) x ≥ 0, y ≥ 0
b) x + y ≤ 8
c) 3x + 2y ≥ 6
b)
x+y=8
y = -x + 8
Objective function: z = 2x + 3y
Constraints:
a) x ≥ 0, y ≥ 0
b) x + y ≤ 8
c) 3x + 2y ≥ 6
c)
3x + 2y = 6
2y = -3x + 6
y = (-3/2)x + 3
Intersection of all constraints
A (0, 8)
D (0, 3)
C (2, 0)
B (8, 0)
Check the objective equation with the corner
points
Corner (x, y)
A (0, 8)
B (8, 0)
C (2, 0)
D (0, 3)
Objective Function
z = 2x + 3y
z = 2(0) + 3(8) = 24
z = 2(8) + 3(0) = 16
Z = 2(2) + 2(0) = 4
z = 2(0) + 3(3) = 9
Solution:
X = 0, y = 8, resulting in max z of 24
Last Example, L.P.
A manufacturer makes 2 types of jet skis,
regular and deluxe. The profit on each
regular model is $200; deluxe model, $250.
At least 50 regular model can be made per
week; at least 75 deluxe model can be made
per week
To maintain quality, the total number of jet
skis cannot exceed 150 per week.
How many of each model should be made to
maximize the profit?
Solution
1. Set variables to represent quantities sought
x: number of regular model
y: number of deluxe model
2. Write objective function
z = 200x + 250y
3. Write constraints
x ≥ 50
y ≥ 75
x + y ≤ 150
Solution
4. Solve inequalities and graph solution sets
x ≥ 50
(0,150)
y ≥ 75
x + y ≤ 150
A
x + y = 150
If x = 0
Then y = 150.
(0, 150)
If y = 0
Then x = 150
(150, 0)
(0, 75)
(0,0)
C
(50,0)
B
(150,0)
Solution
5. Find coordinates of corners.
C: (50, 75)
A: (50, ?)
x + y = 150
50 + y = 150
y = 100
(50, 100)
B: (?, 75)
x + 75 = 150
x = 75
(75, 75)
(0,150)
A
(0, 75)
C
(50,0)
B
(150,0)
Solution
6. Check the objective equation with the corner
points
Corner (x, y)
Objective Function
z = 200x + 250y
A (50, 100)
B (75, 75)
z = 200(50) + 250(100) = 26,000
z = 200(75) + 250(75) = 38,750
C (50, 75)
z = 200(50) + 250(75) = 28,800
Solution:
75 regular models, 75 deluxe models, resulting in $38,750
profit.