Chapter 3
Linear Inequalities and Linear
Programming
3.1
Formulating Linear Programming Problems
In the business world, the objective of a manufacturer is to produce goods subject to certain constraints, minimize the costs, and maximize the prots. For example, consider the following problem:
A toy manufacturing company makes two types of toys—airplanes and trains. Each airplane
requires 6 ounces of plastic and 3 ounces of steel. Each train requires 4 ounces of plastic and 5
ounces of steel. Each week the company has 20 pounds of plastic and 15 pounds of steel available
to make toys. The prot from selling each airplane is $2=50 and each train is $2=00= How many toys
of each type should be produced so that the prot is maximum?
Let us express this problem using variables and expressions. Suppose that the company makes
{ number of planes and | number of trains. We can express the requirements of the problem using
a table as follows:
Each airplanes
Each train
Total
Variable
{
|
Plastic
6
4
20 lb = 320 oz
Steel
3
5
15 lb = 240 oz
Prot
2=50
2=00
Each plane requires 6 ounces of plastic, so { planes require 6{ ounces of plastic. Each train requires
4 ounces of plastic, so | trains require 4| ounces of plastic. It follows that { planes and | trains
together require 6{ + 4| ounces of plastic. Now the total plastic available each week is 20 pounds =
320 ounces. Hence, we must have
6{ + 4| 320=
In a similar manner, { planes and | trains together require 3{ + 5| ounces of steel. The total steel
available each week is 15 pounds = 240 ounces. So we must have
3{ + 5| 240=
Furthermore, because the number of planes and trains cannot be negative, we must have { 0 and
| 0=
The four inequalities, namely { 0> | 0> 6{ + 4| 320> and 3{ + 5| 240> that we have
formulated so far are called the constraints of the problem.
The prot from each plane is $2=50> so the total prot from { planes is 2=50{= Similarly, the total
prot from | trains is 2=00|= The total prot from planes and trains together is 2=5{ + 2|= This is
known as the objective function.
113
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3. Linear Inequalities and Linear Programming
We must nd { and | that satisfy the following conditions
{0
|0
6{ + 4| 320
3{ + 5| 240
and maximize the prot given by the expression
2=5{ + 2|=
These types of problems fall in the area of mathematics commonly known as linear programming.
The objective of this chapter is to use some of the simple but well known techniques to solve such
problems. In particular, we will use the geometric approach to solve these problems. We will create
a graph depicting the constraints of the problem and then use that graph to maximize or minimize
the objective.
Next we give various examples illustrating how to formulate a linear programming problem.
Example 3.1.1 A hiker wants to take a snack mix of peanuts and raisins. The hiker wants 1000
calories and 120 grams of fat from the mix. Each gram of peanuts contains 8 calories and 0=4 grams
of fat, and costs 7 cents. Each gram of raisins contains 2 calories and 0=6 grams of fat, and costs
4 cents. How many grams of each food should the hiker take so that the cost of the snack is the
minimum?
Suppose that the hiker takes { grams of peanuts and | grams of raisins. The given information
can be displayed in the following table:
Each gram of peanuts
Each gram of raisins
Total (required)
Calories
8
2
1000
Fat
0=4
0=6
120
Cost
7
4
Each gram of peanuts contains 8 calories, so { grams of peanuts contain 8{ calories. Each gram
of raisins contains 2 calories, so | grams of raisins contain 2| calories. It follows that { grams of
peanuts and | grams of raisins contain 8{ + 2| calories. Now the total calories must be 1000= Hence,
we must have
8{ + 2| 1000=
In a similar manner, { grams of peanuts and | grams of raisins contain 0=4{ + 0=6| grams of fat.
Because the total fat must be 120 grams, we must have
0=4{ + 0=6| 120=
Because the calories and fat cannot be negative, we must have { 0 and | 0=
The cost of each gram of peanuts is 7 cents, so the total cost of { grams of peanuts is 7{= Similarly,
the total cost of | grams of raisins is 4|= Thus, the total cost of peanuts and raisins is 7{ + 4|=
Hence, the linear programming problem is:
minimize 7{ + 4|
subject to
{0
|0
8{ + 2| 1000
0=4{ + 0=6| 120=
3.1. Formulating Linear Programming Problems
115
Remark 3.1.2 Consider the following table of Example 3=1=1 :
Each gram of peanuts
Each gram of raisins
Total (required)
Calories
8
2
1000
Fat
0=4
0=6
120
Cost
7
4
Once you have constructed this table, you can write the constraints of the problem directly from this
table after inserting the variables and writing the total in the last row as follows:
Peanuts
Raisins
Total
Variable
{
|
{ 0> | 0
Calories in each gram
8
2
8{ + 2| 1000
Fat in each gram
0=4
0=6
0=4{ + 0=6| 120
Cost
7
4
7{ + 4|
The last column gives the objective function, which in this case is to minimize 7{ + 4|= Columns 2>
3> and 4 give the constraints of the problem.
Note that if the table becomes too wide, then it will not t on a page of this text. In that case, in
the table, we will show the variables, the limiting conditions, and the objective function. Below the
table we will write the constraints. For example, the previous table can also be shown as:
Peanuts
Raisins
Total
Variable
{
|
Calories in each gram
8
2
1000
Fat in each gram
0=4
0=6
120
Cost
7
4
7{ + 4|
and the constraints are: { 0> | 0> 8{ + 2| 1000> 0=4{ + 0=6| 120=
In Example 3=1=3> we use these techniques to formulate the linear programming problem.
Example 3.1.3 Linda and Gary’s ice-cream shop specializes in making two types of ice-cream,
tornado splash and chocolate blizzard. The main ingredients are milk, sugar, and cream. Each
gallon of tornado splash requires 0=5 gallons of milk, 1 pound of sugar, and 0=3 gallons of cream.
Each gallon of chocolate blizzard requires 0=6 gallons of milk, 0=4 pound of sugar, and 0=2 gallons
of cream. For each batch of ice cream they have 750 gallons of milk, 500 pounds of sugar, and 450
gallons of cream. The prot on each gallon of tornado splash is $1=10 and on each gallon of chocolate
blizzard is $1=40= Linda and Gary would like to know how many gallons of each type of ice-cream
should be produced so that the prot is maximized.
Let us formulate a linear programming problem that describes this situation.
Suppose that Linda and Gary make { gallons of tornado splash and | gallons of chocolate blizzard.
The given information can be displayed in the following table: (In this table D stands for the tornado
splash ice cream and E stands for the chocolate blizzard ice cream=)
Each gal. of D
Each gal. of E
Total
Var
{
|
{ 0>
|0
Milk
0=5
0=6
Sugar
1
0=4
Cream
0=3
0=2
Prot
1=10
1=40
0=5{ + 0=6| 750
{ + 0=4| 500
0=3{ + 0=2| 450
1=10{ + 1=40|
Because a gallon of ice cream cannot be negative, we must have { 0 and | 0=
From this table, we get the following constraints:
{ 0>
| 0>
0=5{ + 0=6| 750>
{ + 0=4| 500>
0=3{ + 0=2| 450=
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3. Linear Inequalities and Linear Programming
Also, the objective function is 1=10{ + 1=40|=
Hence, the linear programming problem is:
maximize 1=10{ + 1=40|
subject to
{0
|0
0=5{ + 0=6| 750
{ + 0=4| 500
0=3{ + 0=2| 450
Remark 3.1.4 (A general technique to formulate a linear programming problem) In general, you can follow the following steps to formulate a linear programming problem.
1. Read the problem carefully.
2. Identify the items that you want to determine. (For example, in Example 3=1=3> we want to
nd the number of gallons of tornado splash and chocolate blizzard ice cream).
3. Assign a variable to each item that you identied in step 2. (For example, in Example 3=1=3>
we assigned the variable { to tornado splash ice cream and the variable | to chocolate blizzard
ice cream.)
4. Draw a row for each item that you want to nd. Label each row by the item you want to nd.
In addition to these rows, draw two more rows, one at the top and one at the bottom. Then
draw a column and label it Var (Variable). In this column write the variables, that you assigned
in step 3, in front of the items that you identied. Also label the last row as Total.
5. Start building the columns of the table as follows:
(a) Read the problem again and identify the things that you need to build or create or form
the items that you determined in step (ii). (For example, in Example 3=1=3> you need
milk, sugar, and cream to make ice cream.)
(b) Next create a column for each of the things that you identied in (a) and label these
columns by these things.
(c) Create a column for the objective function and appropriately label it.
(d) Using the quantities of the things that you need to create for each unit of the items, ll-in
the numbers in the row of each item you identied in step 2.
(e) In the last row, starting with the third column, write the limiting conditions of each thing
that you identied in (a).
(f ) Starting with the third column, until the last column, multiply each number in the column
by the variable listed in the second column and in the same row, then add the resulting expressions, and in the last row write the nal expression together with the limiting
conditions.
(g) In the last row and the last column, write the objective function by multiplying the numbers
in that column with the variables and then adding the resulting expressions.
Using the table, write as many inequalities as possible and the objective function. Next, read the
problem again and see if there is some information that you could not put in the table. Using the
unused information, create additional inequalities. For example, typically, the variables of a linear
programming problem are nonnegative. So add inequalities such as { 0 and | 0=
3.1. Formulating Linear Programming Problems
117
Example 3.1.5 A bicycle company manufactures two types of bicycles—regular bikes and mountain
bikes. Each type of bike requires three types of operations—frame assembly, wheel installation, and
special paint. The time for each type of operation for each type of bike is given by the following table.
Regular bike
Mountain bike
Frame Assembly
15 minutes
30 minutes
Wheel Installation
12 minutes
18 minutes
Special Paint
8 minutes
16 minutes
Each day, the company has 100 hours available for frame assembly, 75 hours available for wheel
installation, and 60 hours available for painting. The prot on each regular bike is $14 and on each
mountain bike is $25= Every day, how many bikes of each type should be made so that the prot is
maximized?
Let us formulate a linear programming problem that describes this situation.
In this linear programming problem, we want to determine the number of regular bikes and mountain bikes. So we assign the variable { to regular bikes and the variable | to mountain bikes. We
can now proceed as follows:
Suppose that the company makes { number of regular bikes and | number of mountain bikes.
Note that 100 hours = 100 · 60 = 6000 minutes, 75 hours = 75 · 60 = 4500 minutes, and 60 hours
= 60 · 60 = 3600 minutes. Now the given information can be displayed in the following table:
Regular bike
Mountain bike
Total
Var
{
|
{ 0>
|0
Frame Assembly
15 minutes
30 minutes
Wheel Installation
12 minutes
18 minutes
Special Paint
8 minutes
16 minutes
Prot
14
25
15{ + 30| 6000
12{ + 18| 4500
8{ + 16| 3600
14{ + 25|
Because the number of bikes cannot be negative, we must have { 0 and | 0 as shown in the
table. Moreover, the objective function is 14{ + 25|=
Hence, the linear programming problem is:
maximize 14{ + 25|
subject to
{0
|0
15{ + 30| 6000
12{ + 18| 4500
8{ + 16| 3600
Example 3.1.6 To control a u outbreak, school ocials decide to vaccinate the entire student
population free of charge. To help expedite the vaccination, 250 doctors and 600 nurses volunteered
their time. Moreover to make the vaccination available to every student, teams of doctors and nurses
would set up tables in each school. Each table has one doctor and 2 or 4 nurses. A table with 1
doctor and 2 nurses can vaccinate 100 students per hour; a table with 1 doctor and 4 nurses can
vaccinate 220 students per hour. How many tables of each type should be set up to maximize the
number of vaccination per hour?
Let us formulate a linear programming problem that describes this situation.
Let us call the table with 1 doctor and 2 nurses table type I; and the table with 1 doctor and
4 nurses table type II. Suppose that the school ocials set up { number of tables of type I and |
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3. Linear Inequalities and Linear Programming
number of tables of type II. Then the given information can be displayed in the following table:
Table type I
Table type II
Total
Variable
{
|
{ 0>
|0
Doctors
1
1
Nurses
2
4
Number of Inoculations
100
220
{ + | 250
2{ + 4| 600
100{ + 220|
Because the number of tables cannot be negative, we must have { 0 and | 0> as shown in the
table. Also, the objective function is
100{ + 220|=
Hence, the linear programming problem is:
maximize 100{ + 220|
subject to
{0
|0
{ + | 250
2{ + 4| 600
Example 3.1.7 To raise money to sponsor a varsity dance team for the national competition, the
student council board decides to wash cars on the weekend. The student council board asks volunteers
to sell tickets in their neighborhood. They can oer three types of wash—regular, premium, and deluxe.
Each type of wash requires soap, wax, and dry. The soap, wax, and the time required for each type
of wash is given in the following table:
Regular
Premium
Deluxe
Soap
2 ounces
5 ounces
7 ounces
Wax
1=5 ounces
4 ounces
6 ounces
Dry time
3=5 minutes
5 minutes
6 minutes
Cost
$4=00
$7=50
$9=00
A local hardware store donated 20 pounds of soap and 5 pounds of wax. Also, the students have a
total of 10 hours of dry time. Furthermore, the cost of a regular wash is $4=00> premium wash is
$7=50> and deluxe wash is $9=00= How many tickets of each type should be sold so that the income is
maximized?
Let us formulate a linear programming problem that describes this situation.
Suppose that the students sell { number of regular wash tickets, | number of premium wash
tickets, and } number of deluxe wash tickets. Then the given information can be displayed in the
following table:
Regular
Premium
Deluxe
Total
Var
{
|
}
Soap
2 ounces
5 ounces
7 ounces
2{ + 5| + 7} 320
Wax
1=5 ounces
4 ounces
6 ounces
1=5{ + 4| + 6} 80
Dry time
3=5 minutes
5 minutes
6 minutes
3=5{ + 5| + 6} 600
Cost
$4=00
$7=50
$9=00
4{ + 7=5| + 9}
Note that 20 pounds = 20 · 16 = 320 ounces, and so on.
Because the number of washes cannot be negative, we must have { 0> | 0> and } 0=
Also, the objective function is 4{ + 7=5| + 9}=
Hence, the linear programming problem is:
maximize 4{ + 7=5| + 9}
3.1. Formulating Linear Programming Problems
subject to
119
{0
|0
}0
2{ + 5| + 7} 320
1=5{ + 4| + 6} 80
3=5{ + 5| + 6} 600
As remarked, after formulating the constraints of the problem, we create a graph describing the
constraints. So the rst step is to learn how to graph linear inequalities. The next section explains
how to draw the graphs of linear inequalities, more specically, linear inequalities in two variables.
When we formulate a linear programming problem, we usually deal with a set of linear inequalities
and a function that is to be maximized or minimized. We called the system of linear inequalities
as the constraints of the problem and the function to be maximized or minimized as the objective
functions. Moreover, in general, there will be an innite number of points that will satisfy the
constraints of the problem. However, our objective will be to nd a point (or points) that satises
the constraints of the problem and also gives a maximum or minimum value of the objective function.
Let us formally dene the terms that are typically used in a linear programming problem.
Denition 3.1.8 Let O be a linear programming problem.
(i) The inequalities that give the valid values of the variables are called the constraints of the
problem.
(ii) The function that is to be maximized or minimized is called the objective function.
(iii) The set of points that satisfy the constraints of the linear programming problem is called the
feasible set for the problem.
Worked-Out Exercises
Exercise 1 Formulate, but do not solve, a linear programming problem that describes the following situation. Identify the variables, the constraints, and the objective function.
A nutritionist prepares a special meal consisting of chicken and rice. The meal must contain
at least 45 units of protein, at least 80 units of carbohydrates, and at least 40 units of iron.
One serving of chicken contains 5 units of protein, 1 unit of carbohydrate, 1 unit of iron, and 5
units of fat. One serving of rice contains 1 unit of protein, 2 units of carbohydrate, 2 units of
iron, and 6 units of fat. Find the number of servings of chicken and rice that the menu should
contain so that the amount of fat is minimum.
Solution: Let { be the number of servings of chicken and | be the number of servings of rice. The
following table shows the given information
Chicken
Rice
Total
Variable
{
|
{ 0>
|0
Protein
5
1
Carbohydrate
1
2
Iron
1
2
Fat
5
6
5{ + | 45
{ + 2| 80
{ + 2| 40
5{ + 6|
Because the number of servings cannot be negative, we must have { 0 and | 0> as shown
in the table=
The objective function is 4{ + 5|=
The linear programming problem is:
minimize 5{ + 6|
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3. Linear Inequalities and Linear Programming
subject to the following constraints
{ 0>
| 0>
5{ + | 45>
{ + 2| 80>
{ + 2| 40=
Exercise 2 Formulate a linear programming problem for the following situation.
A bicycle manufacturer supplies bicycles to two stores, one in Omaha and another in Lincoln.
One truck makes all of the deliveries from the manufacturing warehouse to the stores in Omaha
and Lincoln. It takes 2 hours to make a round trip to the store in Omaha and 4 hours to make
a round trip to the store in Lincoln. The cost of a round trip to the store in Omaha is $50
and the store in Lincoln is $200= The prot of each truckload sold through the store in Omaha
is $250 and the prot is $600 for each truckload sold to the store in Lincoln. How many
shipments per week should be scheduled to each store in order to maximize the prot if the
manufacturer is limited to no more than 40 hours of travel time and a total delivery cost of
$1> 600=
Solution: Let { be the number of shipments to Omaha and | be the number of shipments to Lincoln.
Then the given information can be displayed in the following table.
Omaha
Lincoln
Total
Variable
{
|
Travel time
2
4
2{ + 4| 40
Cost
50
200
50{ + 200| 1600
Prot
250
600
250{ + 600|
Because the number of shipments cannot be negative, we must have { 0 and | 0= Also
the objective function is 250{ + 600|=
Hence, the linear programming problem is:
maximize 250{ + 600|
subject to the following constraints
{ 0>
| 0>
2{ + 4| 40>
50{ + 200| 1600=
Note that the constraints can be simplied as follows:
{ 0>
| 0>
{ + 2| 20>
{ + 4| 32=
Exercise 3 New Haven Cruise line oers 3-day and 7-day cruises. They have 3 ships—Katrina, King Ocean,
and Santa Maria. Each ship has regular, premium, and deluxe cabins. The number of rooms
in each ship and the operating cost per trip is given by the following table.
Katrina
King Ocean
Santa Maria
Regular Rooms
350
500
600
Premium Rooms
250
450
500
Deluxe Rooms
150
200
250
Cost
$200> 000
$350> 000
$475> 000
3.1. Formulating Linear Programming Problems
121
For a particular month, they have requests for 4000 regular rooms, 3000 premium rooms, and
2000 deluxe rooms. How many weeks must each ship be operated so that the total operating
cost is minimized? (Formulate a linear programming problem that describes this situation.)
Solution: Suppose that Katrina operates { weeks, King Ocean operates | weeks, and Santa Maria operates } weeks. Then the given information can be displayed in the following table:
Katrina
King Ocean
Santa Maria
Total
Variable
{
|
}
Regular Rooms
350
500
600
4000
Premium Rooms
250
450
500
3000
Deluxe Rooms
150
200
250
2000
Cost
$200> 000
$350> 000
$475> 000
From this table we get the following constraints:
350{ + 500| + 600} 4000
250{ + 450| + 500} 3000
150{ + 200| + 250} 2000
Because the number of weeks cannot be negative, we must have { 0> | 0> and } 0=
Also, the objective function is 200000{ + 350000| + 475000}=
Hence, the linear programming problem is:
minimize 200000{ + 350000| + 475000}
subject to
{ 0>
| 0>
} 0>
350{ + 500| + 600} 4000>
250{ + 450| + 500} 3000>
150{ + 200| + 250} 2000=
Exercises
In the following exercises, formulate (do not solve) a linear programming problem describing the
given situation.
1. Billy’s video game company manufactures two types of video games—car racing and boxing.
Each car racing game requires 25 minutes to assemble the hardware and 20 minutes to load the
software. Each boxing game requires 30 minutes to assemble the hardware and 15 minutes to
load the software. Each day the company has 45 hours of manpower to assemble the hardware
and 30 hours of manpower to load the software. The prot on each car racing game is $12 and
on each boxing game is $9=50= How many games of each type should be manufactured each
day to maximize the prot?
2. Fast Deli makes sandwiches using bread and meat. Each regular sandwich uses 6 inches of
bread and 3 ounces of meat, while each large sandwich uses 12 inches of bread and 5 ounces
of meat. The prot on a small sandwich is $1=00 and the prot on a large sandwich is $1=70=
Each day the deli has 600 inches of bread and 270 ounces of meat. How many sandwiches of
each size should be made to maximize the prot?
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3. Linear Inequalities and Linear Programming
3. The power fertilizer company produces 50 pounds of fertilizer for garden and lawn use. The
mixture in each lawn bag contains 20% nitrate, 10% phosphate, and 15% potassium, labeled
as 20-10-15= The mixture in each garden bag contains 10% nitrate, 16% phosphate, and 14%
potassium, labeled as 10-16-14= The company has 12> 000 pounds of nitrate, 10> 000 pounds
of phosphate, and 10> 500 pounds of potassium. The prot on each bag of lawn fertilizer is
$6=00 and on each bag of garden fertilizer is $9=50= How many bags of each fertilizer should be
produced to maximize the prot?
4. To determine the popularity of weekend programs, a cable company hires men and women
to watch certain programs and then record their reactions. For each hour, a man spends 40
minutes to watch sports and 10 minutes to watch other programs. For each hour, a woman
spends 10 minutes to watch sports and 35 minutes to watch other programs. Each man is paid
$8=00 per hour and each woman is paid $8=25 per hour. Each weekend, the company wants to
hire people to watch at most 6 hours of sports and 4 hours of other programs. How many men
and women should be hired to minimize the cost?
5. A pet store specializes in cats and bunnies. Each cat costs $9 and each bunny costs $6= The
prot on each cat is $12 and on each bunny is $9= The store cannot house more than 30 animals
and cannot spend more than $210 to buy the pets. How many pets of each type should be
housed to maximize the prot?
6. A farmer has 100 acres of farmland. The farmer is planning to plant tomatoes, beans, and
potatoes. The prot per acre of tomatoes is $1100> per acre of beans is $800> and per acre of
potatoes is $500= The cost of fertilizing each acre of tomatoes is $8> each acre of beans is $6>
and each acre of potatoes is $7= The maintenance time required per week per acre for each
crop is 2 hours for tomatoes, 1 hours for beans, and 45 minutes for potatoes. The farmer
has budgeted a total of $650 for fertilizer and has 150 hours per week of labor available for
maintenance. How many acres of each crop should be planted to maximize the prot?
7. A consulting company maintains two types of teams—software and hardware. Each software
team consists of 4 programmers, 1 architect, and 1 networking person. A hardware team
consists of 2 programmers, 2 architects, and 3 networking persons. The company has 15
programmers, 10 architects, and 12 networking persons. The revenue per week from a software
team is $4> 000 and from a hardware team is $5> 000= A new business requires at least 4 software
and hardware teams to congure its computers. How many teams of each type should be made
to maximize the revenue?
8. A travel agency sold 1500 weekend packages for the Super Bowl. A package includes accommodations, a game ticket, and airfare. The agency can rent three types of charter airplanes—small,
medium, and jumbo jet. Each small airplane can carry 75 passengers, each medium-sized airplane can carry 125 passengers, and each jumbo jet can carry 150 passengers. The cost of
leasing a small airplane is $7> 000> a medium-sized airplane is $14> 000> and a jumbo jet is
$16> 000= The travel agency can lease at most 12 airplanes. How many airplanes of each type
should be leased to minimize the cost?
9. A local business manufactures desks and ling cabinets. Each ling cabinet requires 40 pounds
of steel and 4 hours of labor. Each desk requires 80 pounds of steel, one desk top, and 3 hours
of labor. The business gets a special order to supply ling cabinets and desks. The stockroom
has 10 tons of sheet steel, 200 desk tops, and they have a total of 1200 hours of labor available.
The prot on each ling cabinet is $30 and on each desk is $35. Find the number of ling
cabinets and desk that should be produced so that the prot is maximized? Identify the
variables, the constraints, and the objective function.
3.1. Formulating Linear Programming Problems
123
10. A local business manufactures three types of tents—small, medium, and large. Each small tent
requires 75 minutes to cut, 3 hours to sew, and 15 minutes to box. Each medium tent requires
90 minutes to cut, 4 hours to sew, and 20 minutes to box. Each large tent requires 120 minutes
to cut, 3 hours to sew, and 35 minutes to box. The company has 350 hours to cut, 720 hours
to sew, and 100 hours to box. The prot on each small tent is $25> on each medium tent is
$45> and on each large tent is $50= How many tents of each type should be manufactured to
maximize the prot?
11. A fruit juice company makes 3 types of juices—fruit punch, soft delight, and super juice. Each
bottle of fruit punch contains 6 ounces of mango juice, 7 ounces of orange juice, and 5 ounces
of strawberry juice. Each bottle of soft delight contains 3 ounces of mango juice, 8 ounces of
orange juice, and 4 ounces of strawberry juice. Each bottle of super juice contains 7 ounces
of mango juice, 2 ounces of orange juice, and 6 ounces of strawberry juice. Each week, the
company has 5> 000 ounces of mango juice, 7> 000 ounces of orange juice, and 2> 000 ounces of
strawberry juice. The prot on each bottle of fruit punch is 20 cents, on each bottle of soft
delight is 35 cents, and on each bottle of super juice is 30 cents. How many bottles of each
type of juice should be produced each week so that the prot is maximized?
12. A car leasing company leases four types of cars—compact, midsize, large, and luxury. Leasing
a car requires paperwork, washing, vacuuming, and inspection. The following table gives the
time in minutes required to lease a car.
Compact
Midsize
Large
Luxury
Paperwork
15
12
10
20
Washing
10
15
12
18
Vacuuming
10
5
8
12
Inspection
3
4
2
5
Each day the leasing company has 170 minutes for paperwork, 70 minutes for washing, 50
minutes for vacuuming, and 40 minutes for inspection. The prot on each compact car is $20>
on each midsize car is $35> on each large car is $30> and on each luxury car is $50= How many
cars of each type should be leased each day to maximize the prot?
13. A travel agency sold 2000 weekend packages for the Super Bowl. A package includes accommodation, game ticket, and airfare. The agency can rent three types of charter airplanes—small,
medium, and jumbo jet. Each small airplane can carry 100 passengers, each medium size
airplane can carry 155 passengers, and each jumbo jet can carry 200 passengers. The cost
of leasing a small airplane is $10> 000> a medium size airplane is $17> 000> and a jumbo jet
is $22> 000= Each small plane requires 2 persons to serve the meals, each medium size plane
requires 4 persons to serve the meals, and each jumbo jet requires 5 persons to serve the meals.
The leasing company has at most 50 people available to serve the meals and the travel agency
can lease at most 15 airplanes. How many airplanes of each type should be leased to minimize
the cost?
14. A farmer has 250 acres of farmland. The farmer plants three types of vegetables—tomatoes,
beans, and potatoes. The prot on each acre of tomatoes is $1200> on each acre of beans is
$800> and on each acre of potatoes is $750= The cost of fertilizing each acre of tomatoes is
$8> each acre of beans is $7> and each acre of potatoes is $6= The cost per acre to maintain
tomatoes is $5=00> beans is $6=00> and potatoes is $3=00= It requires 3 hours to maintain one
acre of tomatoes, 2 hours to maintain one acre of beans, and 30 minutes to maintain one acre
of potatoes. The farmer has $2> 000 for the fertilizer, $1> 500 to maintain the vegetables and
10 hours per week for maintenance. How many acres of each vegetable should be planted to
maximize the prot?
124
3. Linear Inequalities and Linear Programming
15. An elementary school provides free breakfast, for all of the students, that consists of cereal and
milk. They mix two types of cereal—type I and type II. Each package of type I cereal contains
2 units of protein, 1 unit of iron, and 2 units of thiamine. Each package of type II cereal
contains 1 unit of protein, 1 unit of iron, and 3 units of thiamine. The school must create a
mixture that contains at least 10 units of protein, 7 units of iron, and 15 units of thiamine.
The cost of each pack of type I cereal is $0=45 and type II cereal is $0=55= How many packages
of each type should be mixed so that the cost is minimum?
16. A small manufacturing plant makes steel pipes in dierent shapes. They use two machines — D
to cut and shape the pipe, and E to polish. A large pipe requires 5 minutes to cut and shape
and 3 minutes to polish. A small pipe requires 2 minutes to cut and shape and 2 minutes to
polish. Each day machine A can operate no more than 4 hours and machine E operates no
more than 3 hours. The prot on each large pipe is $25 and the prot on each small pipe is
$18= How many pipes of each type should be made each day so that the prot is maximum?
17. The mathematics department arranges math help labs on two locations in the university—in
the department and in the student center. The help lab within the department consists of 2
adjunct instructors and 4 senior students, and the help lab in the student center consists of 1
adjunct instructor and 6 senior students. The department has at least 42 individuals available
to work in the labs. Each department lab runs for 3 hours and costs $62 per hour. Each
student center lab runs for 2 hours and costs $58= The mathematics department can provide a
maximum of 30 hours of help labs per week, but must provide at least 6 lab hours within the
department. How many labs of each type should be provided to minimize the cost?
3.2
Graphing Linear Inequalities
In this section, we describe how to draw the graph of linear inequalities in two variables.
Let us rst consider the following inequality: { + 3| 3=
The rst step in drawing the graph of an inequality is to change the inequality into an equality
and then draw the graph of the equality. So rst draw the graph of the equality { + 3| = 3= Two
points on this line are:
{ | ({> |)
0 1 (0> 1)
3 0 (3> 0)
The graph of { + 3| = 3 is given in Figure 3.1:
y
3
2
x + 3y = 3
1
-3 -2 -1 0
-1
1
2
3
x
-2
-3
Figure 3.1
Note that the graph of the equation { + 3| = 3 divides the {-| plane into two regions—lower and
upper (also called left and right).
3.2. Graphing Linear Inequalities
125
Consider the point (0> 0)= Let us determine, if (0> 0) is a solution of the given inequality. So
substitute { = 0 and | = 0 to get
{ + 3| 3
0 + 3 · 0 3> substitute { = 0 and | = 0
0 3>
which is false. Thus, (0> 0) is not a solution. Similarly, we can show that every point below the
graph of the line { + 3| = 3 is not a solution of the inequality.
Now consider the point (0> 2)= Let us determine if this point is a solution. So substitute { = 0
and | = 2= Thus,
{ + 3| 3
0 + 3 · 2 3> substitute { = 0 and | = 2
6 3>
which is true. Hence, the point (0> 2) is a solution of { + 3| 3= Similarly, we can show that the
every point above the graph of { + 3| 3 is a solution of { + 3| 3=
Let us also consider a point on the graph of the line, say (0> 1), and determine if this is a solution
of the inequality { + 3| 3= Now
{ + 3| 3
0 + 3 · 1 3> substitute { = 0 and | = 1
3 3>
which is true. Thus, (0> 1) is a solution of the inequality { + 3| 3=
We thus see that every point on the graph of the line and every point above the graph of the line
is a solution. Finally, to draw the graph of the given inequality, we shade the region that contains
the points that satises the given inequality. We thus, obtain the graph given in Figure 3.2
y
3
2
x + 3y = 3
x + 3y > 3
1
-3 -2 -1 0
-1
1
2
3
x
-2
-3
Figure 3.2
Remark 3.2.1 (Drawing the graph of an inequality) To draw the graph of an inequality in
two variables we do the following:
1. Change the inequality to an equality and then draw the graph of the line.
2. If the inequality is or > then draw the line as a solid line.
3. If the inequality is strict, (A or ?), then draw the line as a dotted-line.
4. Next choose a point on one side of the line. If that point satises the given inequality, then
shade the region containing the point, otherwise shade the other region.
Example 3.2.2 Let us draw the graph of 2{ 3| 4=
126
3. Linear Inequalities and Linear Programming
First we draw the graph of 2{ 3| = 4= Two points on this line are (2> 0) and (1> 2)= The
graph of the line is shown in Figure 3.3(a). (Note that the inequality is > so we draw a solid line.)
y
y
3
2x - 3y = 4
2
3
2x - 3y < 4 2
1
1
-3 -2 -1 0
-1
1
2 3
x
-3 -2 -1 0
-1
-2
-2
-3
-3
(a)
(b)
2x - 3y = 4
1
x
2 3
Figure 3.3
Next we choose a point on one side of the inequality. Notice that the line does not pass through
(0> 0)> so we can use (0> 0) as a test point. Now,
2{ 3| 4
2 · 0 3 · 0 4> substitute { = 0 and | = 0
0 4>
which is true. Thus, (0> 0) is solution. So we shade the region that contains the point (0> 0) and
obtain the graph in Figure 3.3(b).
Example 3.2.3 Let us draw the graph of { 2| ? 2=
First we draw the graph of { 2| = 2= Two points on this line are (2> 0) and (0> 1)= The graph
of the line is shown in Figure 3.4(a). (Note that the inequality is ?> so we draw a dotted line.)
y
3
y
3
2
2
1
x - 2y < -2 1
-3 -2 -1 0
-1
1
2
3
x
-3 -2 -1 0
-1
-2
-2
-3
-3
(a)
(b)
1
2
3
x
Figure 3.4
Next we choose a point on one side of the inequality. Notice that the line does not pass through
(0> 0)> so we can use (0> 0) as a test point. Now,
{ 2| ? 2
0 2 · 0 ? 2> substitute { = 0 and | = 0
0 ? 2>
3.2. Graphing Linear Inequalities
127
which is false. Thus, (0> 0) is not a solution. So we shade the region that does not contain the point
(0> 0) and obtain the graph in Figure 3.4(b).
Example 3.2.4 Let us draw the graph of { ? 1=
First we draw the graph of { = 1= The graph of the line is shown in Figure 3.5(a). (Note that
the inequality is ?> so we draw a dotted line.)
y
y
3
3
2
2
x < -1
1
-3 -2 -1 0
-1
1
2
3
x
1
-3 -2 -1 0
-1
-2
-2
-3
-3
(a)
(b)
1
2
x
3
Figure 3.5
Next we choose a point on one side of the inequality. Notice that the line does not pass through
(0> 0)> so we can use (0> 0) as a test point. Now,
{ ? 1
0 ? 1> substitute { = 0 and | = 0
which is false. Thus, (0> 0) is not a solution. So we shade the region that does not contain the point
(0> 0) and obtain the graph in Figure 3.5(b).
Example 3.2.5 Consider the following inequalities:
2{ + 3|
1=5{ 2|
6
3
We determine the region in the {-| plane that simultaneously satises these inequalities.
First we draw the graph of the these inequalities in the same {-| plane and obtain the graph in
Figure 3.6(a).
y
3
2x + 3y > 6
y
3
2
2
1
1
-3 -2 -1 0
-1
-2
1
2
3
1.5x - 2y > -3
x
-3 -2 -1 0
-1
-2
-3
-3
(a)
(b)
Figure 3.6
2x + 3y > 6
1.5x - 2y > -3
1
2
3
x
128
3. Linear Inequalities and Linear Programming
Next we choose the region that is shaded by both of the inequalities and obtain the graph in Figure
3.6(b). Hence, the region in the {-| plane that simultaneously satises the given inequalities is shown
in Figure 3.6(b).
Remark 3.2.6 To determine the region in the {-| plane that simultaneously satises a given set of
inequalities, rst draw the graph of each inequality in the same {-| plane. Next choose the region
that is shaded by all the inequalities.
Example 3.2.7 Consider the following inequalities: { 0 and | 0= We determine the region in
the {-| plane that simultaneously satises these inequalities. First we draw the graph of the these
inequalities in the same {-| plane and obtain the graph in Figure 3.7(a).
y
3
y
3
y>0 2
2
1
-3 -2 -1 0
-1
-2
x>0
y>0
1
1
2
3
x
-3 -2 -1 0
-1
x>0
1
2
3
x
-2
-3
-3
(b)
(a)
Figure 3.7
Next we choose the region that is shaded by both of the inequalities and obtain the graph in Figure
3.7(b). Hence, the region in the {-| plane that simultaneously satises the given inequalities is shown
in Figure 3.7(b).
Example 3.2.8 Consider the following inequalities:
{ + 3|
2{ 2|
6{ |
3
5
3
We determine the region in the {-| plane that simultaneously satises these inequalities.
First we draw the graph of the these inequalities in the same {-| plane. The graph of { + 3| 3
is shown in Figure 3.8(a); the graph of 2{ 2| 5 is shown in Figure 3.8(b); and the graph of
3.2. Graphing Linear Inequalities
129
6{ | 3 is shown in Figure 3.8(c).
x + 3y = 3
y
y
3
3
2
2
1
1
-3 -2 -1 0
-1
1
2
3
x
-3 -2 -1 0
2x - 2y = -5
-1
-2
-2
-3
-3
(a)
(b)
y
y
3
x + 3y = 3
2
-2
-3
2
3
1
2
3
x
3
2
1
1
-3 -2 -1 0
-1
1
1
2
3
x
-3 -2 -1 0
-1
2x - 2y = -5
-2
6x - y = 3
-3
x
6x - y = 3
(d)
(c)
Figure 3.8
Finally, Figure 3.8(d) shows the region that is simultaneously shaded by the three inequalities.
Solutions Sets and the Corner Points
In Example 3.2.8, we obtained the region that is simultaneously satised by the inequalities
{ + 3|
2{ 2|
6{ |
3
5
3
(3.1)
130
3. Linear Inequalities and Linear Programming
Suppose that o1 denotes the line { + 3| = 3> o2 denotes the line 2{ 2| = 5> and o3 denotes the
line 6{ | = 3= Then each pair of these lines intersect at a point, see Figure 3.9.
y
B
3
x + 3y = 3
2
A 1
C
-3 -2 -1 0
-1
2x - 2y = -5
-2
1
-3
2
x
3
6x - y = 3
Figure 3.9
Note that
• Point D is the intersection of the lines { + 3| = 3 and 2{ 2| = 5=
• Point E is the intersection of the lines 6{ | = 3 and 2{ 2| = 5=
• Point F is the intersection of the lines { + 3| = 3 and 6{ | = 3=
The shaded region is bounded by the line segments DE> EF> and DF= These line segments are
called the boundaries of the shaded region. Moreover, the shaded region is the set of all points
that simultaneously satisfy the linear inequalities given in (3.1).
Points D> E> and F are the corner points of the shaded region. Next we show how to nd the
coordinates of these points.
To nd the coordinates of the D> E> and F> we solve certain systems of equations. For example,
to nd the coordinates of D> we solve the system of equations { + 3| = 3 and 2{ 2| = 5= Next
we determine the coordinates of points D> E> and F=
Point D :
Point D is the intersection of the lines {+ 3| = 3 and 2{ 2| = 5= Multiply equation { +3| = 3
by 2 and add it to the second equation, i.e.,
{
2{
+
3|
2|
=
=
3
5
This implies that 8| = 11 or | =
2{
2{
(× 2)
11
8 =
Substitute | =
11
8
{ + 3| = 3
{ + 3 · 11
8 =3
{+
33
8
=3
{=3
{=
33
8
2433
8
{ = 98 =
6|
2|
8|
=
=
=
6
5
11
add
into { + 3| = 3 and solve for {= i.e.,
3.2. Graphing Linear Inequalities
131
¡
¢
Hence, the coordinates of D are 98 > 11
8 =
Similarly, solving
the¢ system of equations 6{ | = 3 and 2{ 2| = 5> we get the coordinates
¡
18
>
of E> which are 33
30 5 = Solving
¡ 27 219the
¢ system of equations { + 3| = 3 and 6{ | = 3> we get the
coordinates of F> which are 19 > 19 =
Denition 3.2.9 (Bounded and Unbounded Sets) A set of points in the {-| plane is said to
be bounded if it can be contained in a circle with center (0> 0)= Otherwise, the set is said to be
unbounded.
Denition 3.2.10 (Solutions set) The solution set of a system of linear inequalities is the set
of points that simultaneously satisfy all of the inequalities.
Remark 3.2.11 When we graph a system of inequalities, then the solution set is the set of all points
in the shaded region. In the graph, the shaded region is called the solution set or the feasible
region.
Denition 3.2.12 (Corner points) Let V be a solution set for a system of linear inequalities. A
point F in the {-| plane is called a corner point of V if every line segment in V that contains F
has F as one of its endpoints.
Example 3.2.13 Consider the following system of inequalities:
{+|
4{ |
{ 3|
5
0
3
We determine the solution set, in the {-| plane, for these inequalities, the corner points of the
solution set, and the coordinates of the corner points.
First we draw the graph that satises these inequalities, see Figure 3.10.
y
4x - y = 0
5
B
4
3
x - 3y = -3
2
1
-3 -2 -1 0
-1
C
A
1
2
3
4
5
x
x+y=5
-2
-3
Figure 3.10
In Figure 3.10, the shaded region is the solution set. Note that the solution set is bounded and it has
three corner points D> E> and F as shown in Figure 3.10.
• Point D is the intersection of the lines 4{ | = 0 and { 3| = 3=
132
3. Linear Inequalities and Linear Programming
• Point E is the intersection of the lines 4{ | = 0 and { + | = 5=
• Point F is the intersection of the lines { + | = 5 and { 3| = 3=
To determine the coordinates of D> we solve the system of equations 4{ | = 0 and { 3| = 3
for { and |= Now,
4{
{
|
3|
Thus 11| = 12> so | =
=
=
12
11 =
0
3
4{
4{
(× 4)
Substitute | =
12
11
+
|
12|
11|
=
=
=
0
12
12
add
into 4{ | = 0 and solve for {> so
4{ | = 0
4{ 12
11 = 0
4{ = 12
11
{ = 14 · 12
11
3
{ = 11
=
¡ 3 12 ¢
> 11 =
Hence, the coordinates of D are 11
Similarly, we can show that the coordinates of E are (1> 4) and the coordinates of F are (3> 2)=
Example 3.2.14 Consider the following system of inequalities:
{
|
{ + 2|
5{ + 2|
0
0
4
10
We determine the solution set, in the {-| plane, for these inequalities, the corner points of the
solution set, and the coordinates of the corner points.
First we draw the graph that satises these inequalities, see Figure 3.11.
y
5
A
4
3
x + 2y = 4 2
B
1
-3 -2 -1 0
-1
-2
C
1
2
3
4
5
x
5x + 2y = 10
-3
Figure 3.11
In Figure 3.11, the shaded region is the solution set. Note that the solution set is unbounded and it
has three corner points D> E> and F as shown in the gure.
Note that D is the intersection of the lines { = 0 and 5{ + 2| = 10; E is the intersection of the
lines 5{ + 2| = 10 and { + 2| = 4; and F is the intersection of the lines | = 0 and { + 2| = 4=
3.2. Graphing Linear Inequalities
133
It follows easily that the coordinates of D are (0> 5)> and the coordinates of F are (4> 0)=
Point E is the intersection of the lines 5{ + 2| = 10 and { + 2| = 4= So to nd the coordinates
of E> we solve the system of equations 5{ + 2| = 10 and { + 2| = 4 for { and |= Now,
5{
{
+
+
2|
2|
=
=
Thus 8| = 10> so | =
10
8
10
4
5{
5{
(× 5)
= 54 = Substitute | =
Hence, the coordinates of E are
¡3
5
2> 4
¢
5
4
+
2|
10|
8|
=
=
=
10
20
10
add
into { + | = 2 and solve for {> so
{ + 2| = 4
{ + 2 · 54 = 4
{ + 52 = 4
{ = 4 52
{ = 32 =
=
Worked-Out Exercises
Exercise 1 Graph the following inequality:
(a) { 2| 3=
(b) 2{ 5|=
Solution: (a) First we draw the graph of the line { 2| = 3= Two points on this line are given in the
following table:
{
|
¡({> |)3 ¢
0 32
0> 2
3
0
(3> 0)
The graph of the line { 2| = 3 is shown in Figure 3.12(a).
y
y
3
x - 2y < 3
2
1
-3 -2 -1 0
-1
-3
2
1
1
2
3
x
-3 -2 -1 0
-1
-2
x - 2y = 3
3
1
2
3
x
-2
x - 2y = 3
(a)
-3
(b)
Figure 3.12
As shown in Figure 3.12(a), the graph of the line { 2| = 3> divides the {-| plane into
two regions. We need to shade one of these regions. Let us choose a test point. Because
134
3. Linear Inequalities and Linear Programming
the line does not pass through (0> 0), we can choose (0> 0) as the test point. So substitute
{ = 0 and | = 0 in the inequality { 2| 3= Now
{ 2| 3
02·03
0 3>
which is true. Thus, (0> 0) is a solution of {2| 3= So we shade the region that contains
the point (0> 0)= The graph of { 2| 3> is shown in Figure 3.12(b).
(b) First we draw the graph of the line 2{ = 5|= Two points on this line are given in the
following table:
{ | ({> |)
0 0 ¡(0> 0)¢
5
5
1
2
2> 1
The graph of the line 2{ = 5| is shown in Figure 3.13(a)
y
y
3
3
2
2
1
1
-3 -2 -1 0
-1
2x = 5y
1
2
3
x
-3 -2 -1 0
-1
-2
-2
-3
-3
(a)
(b)
2x = 5y
1
2
3
x
2x > 5y
Figure 3.13
As shown in Figure 3.13(a), the graph of the line 2{ = 5|> divides the {-| plane into two
regions. We need to shade one of these regions. Let us choose a test point. Because the
line passes through (0> 0), we cannot choose (0> 0) as the test point. The point (1> 0) is
not on the line, so let us choose (1> 0) as a test point. So substitute { = 1 and | = 0 in
the inequality 2{ 5|= Now
2{ 5|
2·15·0
2 0>
which is true. Thus, (1> 0) is a solution of 2{ 5|= So shade the region that contains the
point (1> 0)= The graph of 2{ 5|> is shown in Figure 3.13(b).
Exercise 2 Consider the following inequalities:
3{ 2|
{+|
{ + 3|
Determine the solution set for these inequalities.
6
2
9
3.2. Graphing Linear Inequalities
135
Solution: First we draw the graph of the these inequalities in the {-| plane. The graph of 3{ 2| 6
is shown in Figure 3.14(a); the graph of { + | 2 is shown in Figure 3.14(b); and the graph
of { + 3| 9 is shown in Figure 3.14(c).
3x - 2y = 6
y
y
3
3
2
2
1
1
-3 -2 -1 0
-1
1
2
3
x
-3 -2 -1 0
-1
-2
-2
-3
-3
(a)
(b)
2
3
x
x+y=2
y
y
3
3
2
2
1
2
3
x + 3y = 9
1
x + 3y = 9
1
-3 -2 -1 0
-1
1
x
-3 -2 -1 0
-1
-2
-2
1
2
3
x
x+y=2
-3
-3
3x - 2y = 6
(c)
(d)
Figure 3.14
Finally, Figure 3.14(d) shows the region that is simultaneously shaded by the three inequalities.
Exercise 3 Consider the following system of inequalities:
{
|
{ + 3|
{+|
0
0
3
2
Determine the solution set for these inequalities, the corner points of the solutions set, and
the coordinates of the corner points.
Solution: First we draw the graph of these inequalities and then shade the region that simultaneously
136
3. Linear Inequalities and Linear Programming
satises these inequalities, see Figure 3.15.
y x=0
3
x + 3y = 3
2
1
B
A
-3 -2 -1 0 O 1 C2
-1
-2
3
y=0
x
x+y=2
-3
Figure 3.15
In Figure 3.15, the shaded region is the solution set. Note that the solution set is bounded
and it has four corner points R> D> E> and F as shown in the gure.
• Point R is the intersection of the lines { = 0 and | = 0=
• Point D is the intersection of the lines { = 0 and { + 3| = 3=
• Point E is the intersection of the lines { + 3| = 3 and { + | = 2=
• Point F is the intersection of the lines | = 0 and { + | = 2=
It follows easily that the coordinates of R are (0> 0)> the coordinates of D are (0> 1)> and the
coordinates of F are (2> 0)=
Point E is the intersection of the lines { + 3| = 3 and { + | = 2= So to nd the coordinates of
E> we solve the system of equations { + 3| = 3 and { + | = 2 for { and |= Now,
{
{
+
+
3|
|
=
=
3
2
{
{
(× 1)
Thus 2| = 1> so | = 12 = Substitute | =
1
2
+
3|
|
2|
=
=
=
3
2
1
add
into { + | = 2 and solve for {> so
{+| =2
{ + 12 = 2
{ = 2 12
{ = 32 =
Hence, the coordinates of E are
¡3
1
2> 2
¢
=
Exercise 4 Consider the following system of inequalities:
{|
{ 5|
{+|
1
5
3
Determine the solution set for these inequalities, the corner points of the solutions set, and
the coordinates of the corner points.
3.2. Graphing Linear Inequalities
137
Solution: First we draw the graph of these inequalities and then shade the region that simultaneously
satises these inequalities, see Figure 3.16.
y
4
x - y = -1
3
2
B
1
x - 5y = 5
-3 -2 -1 0
-1
A
-2
1
2
3 4
C
x
5
x+y=3
-3
Figure 3.16
In Figure 3.16, the shaded region is the solution set. Note that the solution set is bounded
and it has three corner points D> E> and F> as shown in the gure.
• Point D is the intersection of the lines { 5| = 5 and { | = 1=
• Point E is the intersection of the lines { | = 1 and { + | = 3=
• Point F is the intersection of the lines { 5| = 5 and { + | = 3=
To determine the coordinates of D> we solve the system of equations {5| = 5 and {| = 1
for { and |= Now,
{
{
5|
|
=
=
5
1
(× 1)
{
{
+
5|
|
4|
=
=
=
5
1
6
add
Thus 4| = 6> so | = 64 = 32 = Substitute | = 32 into { | = 1 and solve for {> so
{ |¡ = 1
¢
{ 32 = 1
{ + 32 = 1
{ = 1 32
{ = 52 =
¡
¢
Hence, the coordinates of D are 52 > 32 =
¢ we can show that the coordinates of E are (1> 2) and the coordinates of F are
¡Similarly,
10
1
3 > 3 =
Exercises
In Exercises 1 to 13, graph the inequalities.
1. { + | 3
2. { 4
3. | ? 3
138
3. Linear Inequalities and Linear Programming
4. { ? 2
5. | 3=5
6. { | + 2
7. 2{ 3| 6
8. 3{ + | A 5
9. { 2| ? 2
10. { + 2| 0
11. { A |
12. | ? 3{
13. 4{ + 3| 12
In Exercises 14 to 24, graph the system of inequalities.
14. { 0> { + | 2=
15. { | 1
{+| 2
16. { + | 1
2{ | 4
17. { + | 3
{| 4
3{ + 2| 6
18. { 3
{ 2
2{ + | 4
|2
19. { 0
{ + 3| 3
2{ | 4
20. | 0
4{ + 2| 5
21. { 0
|0
3{ 4| 12
|2
2{ + 3| 6
3.2. Graphing Linear Inequalities
139
22. { | 0
{+| 0
{3
23. 2{ 3| 6
3{ + | 3
{ + 2| 2
24. 2{ 5| 10 0> 4{ + 3| + 12 0> { + 3| 3 0
In Exercises 25 to 29, graph the system of inequalities and nd the coordinates of the corner
points of the solution set (feasible region).
25. { + 3| 6
2{ + | 4
{ + | 1
26. { 3
{ + 3| 6
{ + | 3
27. 3{ + 4| 12
{+| 2
2{ + | 3
28. { + | 3
{| 3
{ + | 3
{ | 3
29. { 0
|0
{ + 2| 4
{+| 5
30. A solution set is dened by the following inequalities.
| 1>
2{ + 5| 10>
9{ 4| 6=
Graph these inequalities and nd the coordinates of the solution set.
31. A solution set is dened by the following inequalities.
| 1>
2{ + 5| 10>
9{ 4| 6=
Graph these inequalities and nd the coordinates of the solution set.
140
3. Linear Inequalities and Linear Programming
32. A solution set is dened by the following inequalities.
{0
|1
{ + 2| 4
4{ + 2| 9 0
Graph these inequalities and nd the coordinates of the solution set.
33. The solution set of a system of inequalities is given in Figure 3.17. Find the corner points of
the solution set.
y
x+y=4
-3x + 2y = 6
5
4
3
x - 3y = -3
2
1
-3 -2 -1 0
-1
1
2
3
4
x
5
-2
-3
Figure 3.17
34. The solution set of a system of inequalities is given in Figure 3.18. Find the corner points of
the solution set.
y
4x + 5y = 20
5
4
3
y=1
2
1
-3 -2 -1 0
-1
-2
1
2
3
4
5
x
5x + 3y = 15
-3
Figure 3.18
35. The solution set of a system of inequalities is given in Figure 3.19. Find the corner points of
3.2. Graphing Linear Inequalities
141
the solution set.
y
3
y=2
2
1
-3 -2 -1 0
-1
1
2
3
-2
y = -2
-3
x = -1
x
4
x=1
Figure 3.19
36. The solution set of a system of inequalities is given in Figure 3.20. Find the corner points of
the solution set.
y
5
4
3
x-y=3
2
1
-3 -2 -1 0
-1
1
2
3
4
x
x+y=4
-2
-3
5
x=2
Figure 3.20
37. The solution set of a system of inequalities is given in Figure 3.21. Find the corner points of
the solution set.
y
5
4
3
2
1
-1 0
-1
1
2 3 4 5
5x + 3y = 15
Figure 3.21
x
4x + 5y = 20
142
3. Linear Inequalities and Linear Programming
38. The corner points of a solution set are given in Figure 3.22. Determine the system of linear
inequalities for this solution set.
y
4
(3, 4)
3
(-2, 2)
2
(4, 1)
1
-2 -1 0
-1
1
2
3
4
x
5
Figure 3.22
39. The corner points of a solution set are given in Figure 3.23. Determine the system of linear
inequalities for this solution set.
y
4
(0, 4)
3
2
(4/3, 4/3)
1
-1 0
-1
(4, 0)
1
2
3
4
5
x
Figure 3.23
40. The corner points of a solution set are given in Figure 3.24. Determine the system of linear
inequalities for this solution set.
y
5
4
3
(3, 5)
(4/3, 3)
(4, 7/2)
2
(2, 5/3)
1
-1 0
-1
1
2
3
4
5
x
Figure 3.24
41. A stock broker’s client wants to invest at most $100> 000= Considering the current market
situation, the broker advised the client to invest no more than 75% of the total amount in
3.3. Solutions of Linear Programming Problems
143
bonds. The client was also advised to invest the amount in bonds at least as large as the
amount in stock. Find the linear inequalities to describe this situation, graph the inequalities,
and nd the solution set and the corner points of the solution set.
42. A hiker wants to take a snack mixture of chocolate and nuts with at least 750 calories. She
nds that each ounce of chocolate can supply 150 calories and each ounce of nuts can supply
100 calories. Determine a system of inequalities describing this situation, graph the system of
inequalities, and nd the solution set and the corner points of the solution set.
3.3
Solutions of Linear Programming Problems
In the previous sections, you learned how to formulate a linear programming problem, graph a set of
inequalities, and nd the corner points of the solution set. All of these things are essential to solve a
linear programming problem via the geometric approach. This section discusses how to accomplish
this.
Recall that the objective of a linear programming problem is to maximize or minimize an objective
function, subject to certain constraints.
In the rst section of this chapter, we gave several examples illustrating how to formulate the
constraints and the objective function of a linear programming problem. Note that in the constraints
the inequalities use either the symbol or the symbol > i.e., none of the inequalities are strict.
This is important because the rst step in solving a linear programming problem using the geometric
approach is to draw the graphs of the inequalities and determine the region that satises the linear
inequalities, i.e., determine the solution set. Next we determine the corner points of the solution set.
If an inequality is strict, then a corner point given by that inequality may not be in the solution set.
Thus, in all of the linear programming problems that we consider, the linear inequalities will not be
strict.
Bounded Feasible Region
The feasible region (solution set) of a problem is either bounded or unbounded. If the region is
bounded, then the following theorem shows how to nd a solution of the problem.
Theorem 3.3.1 (Bounded Region) Let I be the feasible region of a linear programming problem
and I 6= = Let i be the objective function of the linear programming problem. Suppose that I is
bounded. Then
(i) i attains its maximum value at a corner point of I>
(ii) i attains its minimum value at a corner point of I=
Remark 3.3.2 Procedure to nd a solution of a linear programming problem when the
feasible region is bounded.
1. Graph the linear inequalities and determine the feasible region (solution set).
2. Find the corner points of the feasible region.
3. Evaluate the objective function at each of the corner points.
4. If the problem is to maximize the objective function, then choose the largest value of the
objective function. The coordinates of the corner point, that gives the largest value, species the
values of the variables that give the largest value of the objective function. (If there is more than one
corner point that gives the largest value of the objective function, then choose one of those points.)
5. If the problem is to minimize the objective function, then choose the smallest value of the
objective function. The coordinates of the corner point, that gives the smallest value, species the
values of the variables that give the smallest value of the objective function. (If there is more than
144
3. Linear Inequalities and Linear Programming
one corner point that gives the smallest value of the objective function, then choose one of those
points.)
Example 3.3.3 Consider the following constraints of a linear programming problem:
{ 3>
{ + | 0>
{ | 0>
{ + 2| 5=
The region in the {-| plane that satises these inequalities is shown in Figure 3.25:
y
x=3
5
B
4
3
A
-x + 2y = 5
C
2
1
-5 -4 -3 -2 -1 0
-1
x-y=0
O
1 2 3 4 5
-2
x
x+y=0
-3
Figure 3.25
Note that the feasible region is bounded and the corner points are D> E> F> and R=
• Point D is the intersection of the lines { + 2| = 5 and { + | = 0=
• Point E is the intersection of the lines { + 2| = 5 and { = 3=
• Point F is the intersection of the lines { | = 0 and { = 3=
• Point R is the intersection of the lines { + | = 0 and { | = 0=
It is easy to see that the coordinates of R are (0> 0)> the coordinates of E are (3> 4)> and the
coordinates of F are (3> 3)= Next we determine the coordinates of D=
To nd the coordinates of D> we solve the system of equations { + 2| = 5 and { + | = 0= Now
{
{
+
+
2|
|
3|
5
Thus, 3| = 5> so | =
¡ 35= Next
¢ substitute | =
coordinates of D are 3 > 53 =
=
=
=
5
3
5
0
5
add
into { + | = 0> to get { = | = 53 = Hence, the
3.3. Solutions of Linear Programming Problems
145
(a) Let us minimize 6{ 3| subject to the given constraints. Because the feasible region is
bounded, the minimum occurs at a corner point. Let us evaluate 6{ 3| at the corner points. Now
Value of the Objective function: 6{ 3|
6·03·0=0
6 · ( 53 ) 3 · 53 = 10 5 = 15
6 · 3 3 · 4 = 18 12 = 6
6 · 3 3 · 3 = 18 9 = 9
¡
¢
The minimum value of 6{ 3| is 15 and it occurs at D 53 > 53 =
(b) Let us maximize 5{ + 2| subject to the given constraints. Because the feasible region is
bounded, the maximum occurs at a corner point. Let us evaluate 5{ + 2| at the corner points. Now
Corner Point
R(0>
¡ 0) ¢
D 53 > 53
E(3> 4)
F(3> 3)
Corner Point
R(0>
¡ 0) ¢
D 53 > 53
E(3> 4)
F(3> 3)
Value of the Objective function: 5{ + 2|
5·0+2·0=0
10
15
5 · ( 53 ) + 2 · 53 = 25
3 + 3 = 3
5 · 3 + 2 · 4 = 15 + 8 = 23
5 · 3 + 2 · 3 = 15 + 6 = 21
The maximum value of 5{ + 2| is 23 and it occurs at E(3> 4)=
Example 3.3.4 A bakery makes two types of cake—soft-chocolate and super-chocolate. The main
ingredients are our, butter, and chocolate. Each soft-chocolate cake requires 400 grams of our,
200 grams of butter, and 100 grams of chocolate. Each super-chocolate cake requires 300 grams of
our, 300 grams of butter, and 300 grams of chocolate. Each day, the bakery has 9=6 kilograms of
our, 6 kilograms of butter, and 5=4 kilograms of chocolate. The prot on each soft-chocolate cake is
$3=00 and on each super-chocolate cake is $4=00= How many cakes of each type should be baked each
day so that the prot is maximized?
Suppose that the bakery makes { soft-chocolate cakes and | super-chocolate cakes. Then the given
information can be shown in the following table: (In this table, D stands for the soft-chocolate cake
and E stands for the super-chocolate cake.)
Each D
Each E
Total
Var
{
|
Flour
400
300
400{ + 300| 9600
Butter
200
300
200{ + 300| 6000
Chocolate
100
300
100{ + 300| 5400
Prot
$3
$4
3{ + 4|
Also, because the number of cakes cannot be negative, { 0 and | 0= Moreover, the objective
function is 3{ + 4|=
Hence,
maximize 3{ + 4|
subject to
{ 0>
| 0>
400{ + 300| 9600>
200{ + 300| 6000>
100{ + 300| 5400=
Note that the last three inequalities can be simplied to obtain the following equivalent system of
inequalities.
{ 0>
| 0>
4{ + 3| 96>
2{ + 3| 60>
{ + 3| 54=
146
3. Linear Inequalities and Linear Programming
We graph these inequalities and obtain the graph as shown in Figure 3.26:
y
40
4x + 3y = 96
2x + 3y = 60 30
20
A(0, 18)
B(6, 16)
10
C(18, 8)
x + 3y = 54
O(0, 0)
0
10
20
30
D(24, 0)
40
50
60
x
Figure 3.26
Note that the feasible region is bounded and the corner points are D> E> F> G and R=
• Point D is the intersection of the lines { = 0 and { + 3| = 54=
• Point E is the intersection of the lines 2{ + 3| = 60 and { + 3| = 54=
• Point F is the intersection of the lines 2{ + 3| = 60 and 4{ + 3| = 96=
• Point G is the intersection of the lines | = 0 and 4{ + 3| = 96=
• Point R is the intersection of the lines { = 0 and | = 0=
It is easy to see that the coordinates of R are (0> 0)> the coordinates of D are (0> 18)> and the
coordinates of G are (24> 0)= Next we determine the coordinates of E=
To nd the coordinates of E> we solve the system of equations 2{ + 3| = 60 and { + 3| = 54=
Now,
2{
{
+
+
3|
3|
=
=
60
54
(× 2)
2{
2{
+
3|
6|
3|
=
=
=
60
108
48
add
Thus, 3| = 48> so | = 16= Substitute | = 16 into { + 3| = 54 and solve for {= Thus,
{ + 3| = 54
{ = 54 3|
{ = 54 3 · 16
{ = 54 48
{ = 6=
Thus, the coordinates of E are (6> 16)=
Similarly, the coordinates of F are (18> 8)=
Because the feasible region is bounded, the maximum occurs at a corner point. Let us evaluate
3{ + 4| at the corner points. Now
Corner point
R(0> 0)
D(0> 18)
E(6> 16)
F(18> 8)
G(24> 0)
Value of the objective function: 3{ + 4|
3·0+4·0=0
3 · 0 + 4 · 18 = 72
3 · 6 + 4 · 16 = 18 + 64 = 82
3 · 18 + 4 · 8 = 54 + 32 = 86
3 · 24 + 4 · 0 = 72
3.3. Solutions of Linear Programming Problems
147
The maximum value of 3{ + 4| is 86 and it occurs at F(18> 8)= Hence, to maximize the prot, the
bakery should make 18 soft-chocolate cakes and 8 super-chocolate cakes.
Unbounded Feasible Set
If the region is bounded, then the following theorem shows how to nd a solution of the problem.
Theorem 3.3.5 (Unbounded Feasible Region) Let I be the feasible region of a linear programming problem and I 6= = Let i be the objective function of the linear programming problem. Suppose
that I is unbounded.
(i) If i attains its maximum value, then the maximum value occurs at a corner point of I>
(ii) If i attains its minimum value, then the minimum value occurs at a corner point of I=
Remark 3.3.6 If the feasible region is bounded, then the preceding theorem does not guarantee that
an objective function will have a maximum or minimum. It only guarantees that, if the maximum
or minimum occurs, then it would occur at a corner point. So how do we know that a maximum or
minimum will occur? To give a general answer to this problem, we will state a theorem at the end
of this section. However, there are some special cases. Let us consider some special cases that you
will encounter frequently when solving a real-world problem.
Suppose that the feasible region is restricted to the rst quadrant and it extends indenitely in
the rst quadrant. Consider the object function 2{ + 3|= Because the objective function extends
indenitely in the rst quadrant, the value of 2{ + 3| increases without bounds when the values of
{ and | increase in the feasible region. That is, 2{ + 3| has no xed largest value. So 2{ + 3| has
no maximum in the feasible region. In fact, if the objective function is d{ + e|> where d 0> e 0>
and both d and e are not zero, then the objective function d{ + e| has no maximum.
On the other hand, suppose that the feasible region is restricted to the rst quadrant and it extends
indenitely in the rst quadrant. However, the objective function, say 3{ + 5|> is to be minimized.
In this case, the minimum will occur at a corner point, because the value of 3{ + 5| cannot be made
arbitrarily small. Thus, in general, in this case, if the objective function is d{ + e|> where d 0>
e 0> and both d and e are not zero, then the objective function d{ + e| has the minimum, and the
minimum will occur at a corner point.
In a similar manner, suppose that the feasible region is restricted to the third quadrant and it
extends indenitely in the third quadrant. Suppose that the objective function is d{+e|> where d 0>
e 0> and both d and e are not zero. Then d{ + e| has the maximum, and the maximum will occur
at a corner point. Furthermore, d{ + e| has no minimum.
Example 3.3.7 In this example, we solve the linear programming problem formulated in Example
3=1=1= We rewrite the constraints and objective function for easy reference. That is,
minimize 7{ + 4|
subject to
{ 0>
| 0>
8{ + 2| 1000>
0=4{ + 0=6| 120=
Note that the inequality 8{ + 2| 1000 is equivalent to 4{ + | 500= Let us simplify the second
equality. Now
0=4{ + 0=6| 120
5 · (0=4{ + 0=6|) 5 · 120> multiply both sides by 5
2{ + 3| 600=
148
3. Linear Inequalities and Linear Programming
Hence, the constraints of the problem are
{0
|0
4{ + | 500
2{ + 3| 600=
We graph these inequalities and obtain the graph shown in Figure 3=27 :
y
600
500
A(0, 500)
400
300
2x + 3y = 600
200
B(90, 140)
100
C(300, 0)
0
100
200
300
400
600
500
700
x
4x + y = 500
Figure 3.27
Note that the feasible region is unbounded and the corner points are D> E> and F= Point D is the
intersection of the lines { = 0 and 4{ + | = 500= It follows easily that the coordinates of D are
(0> 500)= Point F is the intersection of the lines | = 0 and 2{ + 3| = 600= It follows easily that the
coordinates of F are (300> 0)=
Point E is the intersection of the lines 2{+3| = 600 and 4{+| = 500= Let us nd the coordinates
of E= Now
2{
4{
+
+
3|
|
=
=
600
500
Thus, 10{ = 900> so { =
900
10
(× 3)
2{
12{
10{
+
3|
3|
=
=
=
600
1500
900
add
= 90= Substitute { = 90 into 4{ + | = 500 and solve for |= So
4{ + | = 500
| = 500 4{
| = 500 4 · 90
| = 500 360
| = 140=
Hence, the coordinates of E are (90> 140)=
The feasible region is unbounded and extends indenitely in the upper right corner. However, the
objective function is to be minimized. Hence, the minimum would occur at a corner point. Let us
3.3. Solutions of Linear Programming Problems
149
evaluate the objective function at the corner points. Thus,
Corner point
D(0> 500)
E(90> 140)
F(300> 0)
Value of the objective function: 7{ + 4|
7 · 0 + 4 · 500 = 2000
7 · 90 + 4 · 140 = 630 + 560 = 1190
7 · 300 + 4 · 0 = 2100
From this table, the minimum value is 1190 and this value occurs at E(90> 140)= Thus, the minimum
occurs when { = 90 and | = 140= Thus, the hiker must take 90 grams of peanuts and 140 grams of
raisins to minimize the cost.
Now that we know how to maximize or minimize an objective function in special cases of unbounded feasible regions, next we only state the theorem that gives a general solution when the
solution set is unbounded.
Theorem 3.3.8 (Solutions of Linear Programming Problems with Unbounded Feasible
Sets). Let O be a linear programming problem and V be the feasible region of O= Let i be the objective
function of O= Suppose that V is unbounded.
(i) Suppose that i is to be maximized. Let W be the set of all corner points of V= Let D be a
corner point in W that gives the maximum value of i= Let o1 and o2 be the boundary lines of V such
that o1 and o2 intersect at D= On each line o1 and o2 > choose a point in the feasible region that is not
a corner point and evaluate i at that point. If the value of i at that point is less than or equal to
the value of i at D> then the linear programming problem has a solution, and it is the largest value
that is attained at a corner point. Otherwise the problem has no solution.
(ii) Suppose that i is to be minimized. Let W be the set of all corner points of V= Let D be a
corner point in W that gives the minimized value of i= Let o1 and o2 be the boundary lines of V such
that o1 and o2 intersect at D= On each line o1 and o2 > choose a point in the feasible region that is not
a corner point and evaluate i at that point. If the value of i at that point is greater than or equal
to the value of i at D> then the linear programming problem has a solution, and it is the smallest
value that is attained at a corner point. Otherwise the problem has no solution.
Worked-Out Exercises
Exercise 1 Let V be the set of points that satisfy the following inequalities:
{ | 1>
2{ + | 13>
{ + 2| 8=
(a) Minimize 4{ + 5| subject to the given constraints.
(b) Minimize 5{ + 11| subject to the given constraints.
(c) Maximize 8{ + 6| subject to the given constraints.
(d) Maximize 13{ + 2| subject to the given constraints.
Solution: First we graph the given inequalities and determine the set V= The graph of these inequalities
150
3. Linear Inequalities and Linear Programming
is given in Figure 3.28.
y
6
5
x + 2y = 8
B(4,5)
4
3
A(2,3)
2
C(6,1)
1
-2 -1 0
-1
x - y = -1
-2
1
2
3
4
5
6
7
x
8
2x + y = 13
Figure 3.28
The set V is the set of all the points in the shaded region. Note the region is bounded and the
corner points are D> E> and F=
• Point D is the point of intersection of the lines { | = 1 and { + 2| = 8=
• Point E is the point of intersection of the lines { | = 1 and 2{ + | = 13=
• Point F is the point of intersection of the lines { + 2| = 8 and 2{ + | = 13=
Let us determine the coordinates of D= For this we solve the system of equations { | = 1
and { + 2| = 8=
{
{
+
|
2|
=
=
1
8
(× 2)
2{
{
3{
+
2|
2|
=
=
=
2
8
6
add
Thus, 3{ = 6> so { = 2= Substitute { = 2 in { | = 1 and solve for |= Now
{ | = 1
2 | = 1
| = 1 2
| = 3
| = 3=
Hence, the coordinates of D are (2> 3)= Similarly, the coordinates of E are (4> 5) and F are
(6> 1)=
(a) Because the feasible region is bounded, the minimum occurs at a corner point. Let us
evaluate 4{ + 5| at the corner points. Now
Corner point
D(2> 3)
E(4> 5)
F(6> 1)
Value of the objective function: 4{ + 5|
4 · 2 + 5 · 3 = 8 + 15 = 23
4 · 4 + 5 · 5 = 16 + 25 = 41
4 · 6 + 5 · 1 = 24 + 5 = 29
The minimum value of 4{ + 5| is 23 and it occurs at D(2> 3)=
3.3. Solutions of Linear Programming Problems
151
(b) Because the feasible region is bounded, the minimum occurs at a corner point. Let us
evaluate 5{ + 11| at the corner points. Now
Corner point
D(2> 3)
E(4> 5)
F(6> 1)
Value of the objective function: 5{ + 11|
5 · 2 + 11 · 3 = 43
5 · 4 + 11 · 5 = 75
5 · 6 + 11 · 1 = 41
The minimum value of 5{ + 11| is 41 and it occurs at F(6> 1)=
(c) Because the feasible region is bounded, the maximum occurs at a corner point. Let us
evaluate 8{ + 6| at the corner points. Now
Corner point
D(2> 3)
E(4> 5)
F(6> 1)
Value of the objective function: 8{ + 6|
8 · 2 + 6 · 3 = 16 + 18 = 34
8 · 4 + 6 · 5 = 32 + 30 = 62
8 · 6 + 6 · 1 = 48 + 6 = 54=
The maximum value of 8{ + 6| is 62 and it occurs at E(4> 5)=
(d) Because the feasible region is bounded, the maximum occurs at a corner point. Let us
evaluate 13{ + 2| at the corner points. Now
Corner point
D(2> 3)
E(4> 5)
F(6> 1)
Value of the objective function: 13{ + 2|
13 · 2 + 2 · 3 = 26 + 6 = 32
13 · 4 + 2 · 5 = 52 + 10 = 62
13 · 6 + 2 · 1 = 78 + 2 = 80=
The maximum value of 13{ + 2| is 80 and it occurs at F(6> 1)=
Exercise 2 Scott and Mary are co-treasurers of the student council. They need to raise money to sponsor
the annual homecoming event. A student survey shows that a T-shirt and a sweatshirt with
the school logo are becoming very popular. They can order a maximum of 60 T-shirts and
sweatshirts and no more than 20 sweatshirts. Each T-shirt costs $10 and each sweatshirt costs
$25= They have $750 available to buy T-shirts and sweatshirts. The prot on each T-shirt is
$12 and the prot on each sweatshirt is 25= How many shirts of each type should be ordered
so that the prot is maximized?
Solution: Suppose that Scott and Mary order { number of T-shirts and | number of sweatshirts. Then
the given information can be shown in the following table:
Each T-shirt
Each sweatshirt
Total
Variable
{
|
Cost
$10
$25
10{ + 25| 750
Prot
$12
$25
12{ + 25|
The total number of shirts that can be ordered is 60> so { + | 60= Because, the number
of sweatshirts cannot be more than 20> | 20= Also because the number of shirts cannot be
negative, { 0 and | 0=
The objective function is 12{ + 25|= Hence,
maximize 12{ + 25|
152
3. Linear Inequalities and Linear Programming
subject to
{ 0>
| 0>
| 20>
{ + | 60>
10{ + 25| 750=
We graph these inequalities and obtain the graph shown in Figure 3.29:
y
70
x + y = 60
60
50
10x + 25y = 750 40
30
y = 20
20
B(25, 20)
A(0, 20)
C(50, 10)
10
0
O(0, 0)
10
20
30
40
50
60
70
D(60, 0)
x
Figure 3.29
Note that the feasible region is bounded and the corner points are D> E> F> G and R=
• Point D is the intersection of the lines { = 0 and | = 20=
• Point E is the intersection of the lines 10{ + 25| = 750 and | = 20=
• Point F is the intersection of the lines 10{ + 25| = 750 and { + | = 60=
• Point G is the intersection of the lines | = 0 and { + | = 60=
• Point R is the intersection of the lines { = 0 and | = 0=
It is easy to see that the coordinates of R are (0> 0)> the coordinates of D are (0> 20)> and the
coordinates of G are (60> 0)= Next we determine the coordinates of E=
To nd the coordinates of E> we solve the system of equations 10{ + 25| = 750 and | = 20=
Substitute | = 20 into 10{ + 25| = 750 to get 10{ + 25 · 20 = 750> i.e., 10{ + 500 = 750 or
10{ = 250= This implies that { = 25= Thus, the coordinates of E are (25> 20)=
To nd the coordinates of F> we solve the system of equations 10{ + 25| = 750 and { + | = 60=
Now
10{
{
+
+
25|
|
=
=
750
60
(× 10)
10{
10{
+
25|
10|
15|
=
=
=
750
600
150
(× 10)
add
3.3. Solutions of Linear Programming Problems
153
Thus, 15| = 150> so | = 10= Substitute | = 10 into { + | = 60 to get { + 10 = 60> i.e., { = 50=
Thus, the coordinates of F are (50> 10)=
Because the feasible region is bounded, the maximum occurs at a corner point. Let us evaluate
12{ + 25| at the corner points. Now
Corner Point
R(0> 0)
D(0> 20)
E(25> 20)
F(50> 10)
G(60> 0)
Value of the Objective function: 12{ + 25|
12 · 0 + 25 · 0 = 0
12 · 0 + 25 · 20 = 500
12 · 25 + 25 · 20 = 300 + 500 = 800
12 · 50 + 25 · 10 = 600 + 250 = 850
12 · 60 + 25 · 20 = 720
The maximum value of 12{ + 25| is 850 and it occurs at F(50> 10)= Hence, to maximize the
prot, Scott and Mary should order 50 T-shirts and 10 sweatshirts.
Exercises
In Exercises 1 to 4, graph the system of inequalities, nd the corner points, and determine whether
the feasible region is bounded or unbounded.
{ 0>
1. 3{ 2| 4>
{ + | 3=
{ + | 3>
2. { |>
| { 2=
{ 0>
| 0>
3.
{ + | 4>
{ 2| 2=
{ + | 2>
| 4>
4.
{ | 2>
{ + | 3=
5. A feasible region is dened by the following system of inequalities: { 0> { 4> | 3>
{ + 3| 5= Determine the feasible region and nd the maximum value of 2{ + 4| and the
minimum value of 7{ + | on the feasible region.
6. Find the maximum and minimum values of 4{ 6| subject to the constraints | 0> { |>
{ 5=
7. Find the maximum and minimum values of 7{ 4| subject to the constraints { 0> | 0>
2{ + 3| 6> 2{ + | 4=
8. Maximize 9{ + 7| and minimize 2{ 8| subject to the constraints { 0> | 0> 2{ 3| 6>
4{ + 5| 20=
9. Minimize 5{ + 7| subject to the constraints { 0> | 0> 2{ + | 4> { + 2| 6=
10. Solve the linear programming problem 1 of Section 3.1.
154
3. Linear Inequalities and Linear Programming
11. Solve the linear programming problem 2 of Section 3.1.
12. Solve the linear programming problem 3 of Section 3.1.
13. Solve the linear programming problem 9 of Section 3.1.
14. Solve the linear programming problem 15 of Section 3.1.
15. A toy manufacturer makes and sells two types of toys—trucks and airplanes. Each truck requires
4 ounces of steel, 2 ounces of plastic, and 9 inches of wood. Each airplane requires 3 ounces
of steel, 4 ounces of plastic, and 12 inches of wood. The manufacturer has 192 ounces of steel,
144 ounces of plastic, and 504 inches of wood. The prot on each truck is $6 and the prot
on each airplane is $8=5= How many toys of each type should be made so that the prot is
maximized?
16. In Exercise 15, assume that the prot on each truck is $6 and the prot on each airplane is
$12= Also assume that the other conditions remain the same. How many toys of each type
should be made so that the prot is maximized?
17. An oil producing company has two reneries—one in Texas and another in Alabama. The
reneries produce three types of gasoline—regular grade, premium grade, and ultra grade. Each
day the renery in Texas can produce 100 barrels of regular grade, 180 barrels of premium
grade, and 150 barrels of ultra grade gasoline. Each day the renery in Alabama can produce
150 barrels of regular grade, 150 barrels of premium grade, and 75 barrels of ultra grade
gasoline. The cost of operating the renery in Texas is $6> 000 per day and the cost of operating
the renery in Alabama is $7> 500= The company received an order for 1800 barrels of regular
grade, 2> 700 barrels of premium grade, and 1> 500 barrels of ultra grade gasoline. How many
days each renery should be operated so that the current demand can be met at the minimum
cost?
18. In Exercise 17, assume that the cost of operating the renery in Texas is $9> 000 per day and
the cost of operating the renery in Alabama is $7> 500= Also assume that the other conditions
remain the same. How many days should each renery be operated so that the current demand
can be met at the minimum cost?
19. In Exercise 17, assume that the cost of operating the renery in Texas is $5> 000 per day and
the cost of operating the renery in Alabama is $7> 000= Also assume that the other conditions
remain the same. How many days should each renery be operated so that the current demand
can be met at the minimum cost?
20. A local computer manufacturer assembles, loads the software, and sells two types of computers—
desktop and notebook. The prot on each desktop computer is $125 on each notebook is $160=
The hardware assembly time, software loading time, and testing time for each type of computer
is given in the following table:
Desktop
Notebook
Hardware assembly
5 hours
3 hours
Software loading
3 hours
5 hours
Testing
1 hour
1 hour
The manufacturer has 7> 500 hours of assembly time, 7> 500 hours of software loading time,
and 1> 700 hours of testing time. How many computers of each type should be made so that
the prot is maximized?
3.3. Solutions of Linear Programming Problems
155
21. Angela and Brad won a lottery worth $1> 000> 000= They want to invest no more than $750> 000
into their retirement fund. Their nancial consultant advised to invest in bonds and mutual
funds. The annual return on bonds is 8% and on mutual funds is 7%= Angela and Brad wants
to invest at least $300> 000 in bonds and at least $200> 000 in mutual funds. The annual fee
on bonds is $50 per $100> 000> and in mutual funds is $30 per $100> 000= They do not want to
spend more than $300 in annual fees. How much money should be invested in each fund so
that the annual return is maximized? What is the maximum annual return?
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3. Linear Inequalities and Linear Programming