FM Ch 3 Notes O’Brien F09
FM 9th ed Lial
3.1
Graphing Linear Inequalities
I.
Inequalities
A.
Introduction
Many mathematical descriptions of real situations are best expressed as inequalities rather
than equations. For example, a firm might be able to use a machine no more than 12 hours
a
day, while production of at least 500 cases of the product might be required to meet a
contract.
B.
Linear Inequalities
A linear inequality in two variables is an inequality of the form Ax + By < C where A, B, & C
are real numbers and the inequality symbol could be <, ≤, >, ≥.
Examples: 2x – 3y < 5
7x + 6y ≥ 9
y≤
1
2
x–4
The graph of a linear inequality is a half-plane, including the boundary line only if the
inequality is ≤ or ≥. Inequalities such as x < a or x ≥ a have vertical borders and y > b or
y ≤ b have horizontal borders. The simplest way to see the solution of an inequality in two
variables is to draw its graph.
II.
Graphing an Inequality
A.
The Solution Set of an Inequality
The solution set of an inequality is the set of all ordered pairs that make the inequality true.
The graph of an inequality is a pictorial representation of its solution set.
B.
How to Graph a Linear Inequality
1.
Write the “related equation” by replacing the inequality symbol with an equal sign.
2.
Make a t-chart with 3 points (the x & y intercepts and one other point) for a “related
equation” representing an oblique border line. You do not have to make a t-chart for a
vertical or horizontal border line.
3.
Plot your 3 points. Connect them with a solid line if the inequality is ≤ or ≥ or a dotted
line if the inequality is < or >.
4.
Test a point not on the border and shade the true region.
Example 1
Graph the linear inequality 2x + 3y ≤ 6.
related equation:
[10]
2x + 3y = 6
t-chart:
test (0, 0):
2(0) + 3(0) ≤ 6
0 ≤ 6 true
shade to include the origin
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FM Ch 3 Notes O’Brien F09
FM 9th ed Lial
III.
Systems of Inequalities
A.
The Feasible Region of a System of Inequalities
The solution of a system of inequalities is the set of points which make all of the inequalities
true simultaneously. This is called the feasible region. It is the area where the solutions of
the individual inequalities overlap.
B.
How to Graph a System of Inequalities
1.
Graph each individual inequality.
2.
Find and label the feasible region, the region where all of the inequalities are
simultaneously true. [Hint: Sometimes it is easier to find the feasible region of a
system of inequalities by shading the false side of each inequality rather than the true
side. This will leave the feasible region blank, making it easy to identify. If you use
this method, be sure to indicate that you are shading false.]
Example 2
Graph the feasible region for the following system of inequalities.
x – 2y ≥ 4
A.
related eqns:
3x + 5y ≤ 30
x – 2y ≥ 4
3x + 5y ≤ 30
B.
x – 2y = 4
0≤x≤9
C.
3x + 5y = 30
x≥0
x=0
y-axis
T 
shade right
test (0, 0):
0≥4
[32]
y≥0
D.
x≤9
E.
x=9
y=0
vertical line
T 
shade left
y≥0
x-axis
T 
shade up
0 ≤ 30
F
T
shade to exclude origin
shade to include origin
The feasible region is shown shaded in the graph on the left (shading true) and unshaded in
the graph on the right (shading false).
shading true
shading false
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FM Ch 3 Notes O’Brien F09
FM 9th ed Lial
IV.
Applications
Many realistic problems lead to systems of linear inequalities.
I.
How to Setup and Solve an Application Problem
1.
Read the problem carefully and identify the unknowns. Assign each unknown a variable.
2.
If possible, organize the given information into a table. Each column of the table will be
labeled with one variable and each row will be labeled with a constrained element.
3.
From your table, write the constraints in the form of linear inequalities.
[Hint: Don’t forget the non-negativity constraints, i.e. x ≥ 0, y ≥ 0.]
4.
Graph the system of constraints. Find the feasible region.
Example 3 Production Scheduling
[44]
A cement manufacturer produces at least 3.2 million barrels of cement annually. He is told by the
Environmental Protection Agency (EPA) that his operation emits 2.5 lb of dust for each barrel
produced. The EPA has ruled that annual emissions must be reduced to no more than 1.8 million
pounds. To do this, the manufacturer plans to replace the present dust collectors with two types of
electronic precipitators. One type would reduce emissions to .5 lb per barrel and operating costs
would be 16¢ per barrel. The other would reduce dust to .3 lb per barrel and operating costs would
be 20¢ per barrel. The manufacturer does not want to spend more than .8 million dollars in
operating costs on the precipitators. He needs to know how many barrels he could produce with
each type.
x = the number of barrels (in millions) produced with the first type of precipitator
y = the number of barrels (in millions) produced with the second type of precipitator
Production
Emissions
Cost
x (type 1)
1
.5
.16
y (type 2)
1
.3
.20
limit (in millions)
3.2
1.8
.8
x + y ≥ 3.2
Constraint on volume of production:
Constraint on emissions:
.5x + .3y ≤ 1.8
Constraint on cost:
.16x + .20y ≤ .8
x≥0
Non-negativity constraints:
A.
x + y ≥ 3.2
related eqns:
x + y = 3.2
B.
.5x + .3y ≤ 1.8
y≥0
C.
.5x + .3y = 1.8
.16x + .20y ≤ .8
D. x ≥ 0
.16x + .20y = .8
x=0
y-axis
T 
shade right
test (0, 0):
0 ≥ 3.2
F
shade to exclude origin
0 ≤ 1.8
0 ≤ .8
T
T
shade to include origin
E.
y≥0
y=0
x-axis
T 
shade up
shade to include origin
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FM 9th ed Lial
FM Ch 3 Notes O’Brien F09
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3.2
Solving Linear Programming Problems Graphically
I.
Introduction to Linear Programming Problems
A.
Terminology
Linear programming problems: problems where we must find the optimum (minimum or
maximum) value of a function, subject to certain restrictions
Objective function: the function we are trying to optimize
Constraints: the set of restrictions, given by linear inequalities, which the solution must
satisfy
Non-negativity constraints: x  0 , y  0 These constraints limit the feasible region to the
first quadrant.
Feasible Region: the area where all of the inequalities are simultaneously true.
A feasible region which extends indefinitely in one direction is called an unbounded region.
A bounded region is hemmed in on all sides by borders.
Corner points: the points where the borders of the feasible region cross.
To find the corner points of the feasible region you will need to solve a system of linear
equations. To do this, you may use the substitution method, the elimination method, or the
SOLVSYS program.
B.
Corner Point Theorem
If an optimum value (min or max) of the objective function exists, it will occur at one or more
of the corner points of the feasible region or on the line segment between two corner points.
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FM Ch 3 Notes O’Brien F09
FM 9th ed Lial
Example 1
The following graph shows a region of feasible solutions. Use this region to find maximum and
minimum values of the given objective functions.
[3]
a.
z = .40x + .75y
b.
z = 1.50x + .25y
The minimum value of z1 = 0. It occurs when x- 0 and y = 0. (Note: This is often a “trivial” solution.
The maximum value of z1 = 9. It occurs when x = 0 and y = 12.
The minimum value of z2 = 0. It occurs when x- 0 and y = 0. (Note: This is often a “trivial” solution.
The maximum value of z2 = 12. It occurs when x = 8 and y = 0.
II.
How to Solve a Linear Programming Problem Graphically
1.
If necessary, identify the unknowns, label the variables, write the objective function, and write
the constraints.
2.
Graph the system of constraints. Find the feasible region and the corner points.
3.
Find the value of the objective function for each corner point. Identify the minimum and / or
maximum value of the objective function and the point (or points) where this optimum value
occurs.
Example 2
Use the graphical method to solve the following linear programming problem.
Maximize z = 3x + y subject to x ≥ 1, y ≥ 0, 2x + y ≤ 6.
constraints:
A
related equations:
x≥1
B
x=1
y≥0
C
[Similar to 7 – 14]
2x + y ≤ 6
y=0
2x + y = 6
vertical line
x-axis
test (0, 0):
true 
true 
2(0) + 0 ≤ 6
0 ≤ 6 true
shade toward origin
The feasible region is a bounded region with 3 corner points.
corner points
objective function values
AB (1, 0)
f(1, 0) = 3(1) + 0 =3 (min)
AC (1, 4)
f(1, 4) = 3(1) + 4 = 7
BC (3, 0)
f(3, 0) = 3(3) + 0 = 9 (max)
Solution:
A maximum value of f = 9 occurs at (3, 0) and
a minimum value of f = 3 occurs at (1, 0).
***********************************************************************************************************************
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FM Ch 3 Notes O’Brien F09
FM 9th ed Lial
3.3
Applications of Linear Programming
I.
How to Solve a Linear Programming Application Problem
1.
Read the problem carefully and identify the unknowns. Assign each unknown a variable.
2.
If possible, organize the given information into a table. Each column of the table will be
labeled with one variable and each row will be labeled with a constrained element.
3.
From your table, write the objective function and all necessary constraints.
[Hint: Don’t forget the non-negativity constraints, i.e. x ≥ 0, y ≥ 0.]
4.
Graph the system of constraints. Find the feasible region and the corner points.
5.
Find the value of the objective function for each corner point. Identify the minimum
and / or maximum value of the objective function and the point where this optimum value
occurs.
Example 1
An international relief organization wants to ship food and clothing to hurricane victims in Mexico.
Commercial carriers have volunteered to transport the packages, provided they fit in the available
cargo space. Each 20-ft3 box of food weighs 40 lb and each 30-ft3 box of clothing weighs 10 lb.
The total weight cannot exceed 16,000 lb and the total volume must be less than 18,000 ft 3. Each
carton of food will feed 10 people, while each carton of clothing will help 8 people. How many
cartons of food and clothing should be sent to maximize the number of people helped? What is the
maximum number of people helped?
x = number of cartons of food
y = number of cartons of clothing
N = number of people helped
x
y
limit
weight
40
10
16000
volume
20
30
18000
objective function:
constraints:
related eqns:
t-charts:
A.
N = 10x + 8y (maximize)
40x + 10y ≤ 16000
40x + 10y = 16000
40(0) + 10(0) ≤ 16000
0 ≤ 16000 True
shade towards origin
tests:
B.
20x + 30y ≤ 18000
20x + 30y = 18000
x≥0
x=0
y-axis
true 
D.
y≤0
y=0
x-axis
true 
20(0) + 30(0) ≤ 18000
0 ≤ 18000 True
shade towards origin
corner points
N value
CD
BC
AB
AD
N = 10(0) + 8(0) = 0
N = 10(600) + 8(0) = 6000
N = 10(300) + 8(400) = 6200 (max)
N = 10(400) + 8(0) = 4000
(0, 0)
(0, 600)
(300, 400)
(400, 0)
C.
Solution:
300 cartons of food and 400 cartons of clothing should be sent to help a maximum of 6200 people
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FM Ch 3 Notes O’Brien F09
FM 9th ed Lial
Example 2
A rancher raises goats and llamas on his 400-acre ranch. Each goat needs 2 acres of land and
requires $100 of veterinary care per year, while each llama needs 5 acres of land and requires
$80 of veterinary care per year. The rancher can afford no more than $13,200 for veterinary care
this year. If the expected profit is $60 for each goat and $90 for each llama, how many of each
animal should he raise to obtain the greatest possible profit?
x = number of goats
y = number of llamas
x
2
100
60
acres of land
cost of vet care
profit
y
5
80
90
limit
400
13,200
objective function: P = 60x + 90y (maximize)
constraints: 2x + 5y  400
x  0
y  0
corner points: (0, 80)
100x + 80 y  13,200
(100, 40)
(132, 0)
P (0, 80) = 60(0) + 90 (80) = $7,200
P (100, 40) = 60(100) + 90 (40) = $9,600 (max)
P (132, 0) = 60(132) + 90 (0) = $7,920
The rancher should raise 100 goats and 40 llamas
to earn a maximum profit of $9,600.
Example 3
The Movie Mania store is about to place an order for VHS tapes and DVD’s. The distributor it
orders from requires that an order contain at least 250 items. The prices Movie Mania must pay
are $12 for each tape and $16 for each DVD. The distributor also requires that at least 30% of
the order be DVD’s. How many tapes and how many DVD’s should Movie Mania order so that its
costs will be kept to a minimum?
x = number of VHS tapes
y = number of DVD’s
minimum total order
x
y
limit
1
1
250
1
.3(x + y)
minimum # of DVD’s
Cost
C = total cost of order
12
16
objective function: C = 12x + 16y (minimize)
constraints: x + y  250 -.3x + .7y  0 x  0
y  0
y  .3(x  y)  y  .3x  .3y  -.3x  .7y  0
corner points: (0, 250) (175, 75)
C (0, 250) = 12(0) + 16 (250) = $4,000
C (175, 75) = 12(175) + 16(75) = $3,300 (min)
175 VHS tapes and 75 DVD’s should be ordered
to minimize cost to $3,300
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