ICAI Notes- Linear Inequalities | Quantitative Aptitude for CA Foundation PDF Download

 Page 1


LINEAR INEQUALITIES
3
CHAPTER
Development of Inequalities
from the Descriptive Problem
Forms of Linear inequalities
in one variable and solution
space with various condition
like >0, >0, <0 and <0
Graphing of linear inequation
Determination of
Common region
Optimal solution
One of the widely used decision making problems, nowadays, is to decide on the optimal mix
of scarce resources in meeting the desired goal.  In simplest form, it uses several linear
inequations in two variables derived from the description of the problem.
The objective in this section is to make a foundation of the working methodology for the
above by way of introduction of the idea of :
? development of inequations from the descriptive problem;
? graphing of linear inequations; and
? determination of common region satisfying the inequations.
Feasible region or
Feasible points
CHAPTER OVERVIEW
© The Institute of Chartered Accountants of India
Page 2


LINEAR INEQUALITIES
3
CHAPTER
Development of Inequalities
from the Descriptive Problem
Forms of Linear inequalities
in one variable and solution
space with various condition
like >0, >0, <0 and <0
Graphing of linear inequation
Determination of
Common region
Optimal solution
One of the widely used decision making problems, nowadays, is to decide on the optimal mix
of scarce resources in meeting the desired goal.  In simplest form, it uses several linear
inequations in two variables derived from the description of the problem.
The objective in this section is to make a foundation of the working methodology for the
above by way of introduction of the idea of :
? development of inequations from the descriptive problem;
? graphing of linear inequations; and
? determination of common region satisfying the inequations.
Feasible region or
Feasible points
CHAPTER OVERVIEW
© The Institute of Chartered Accountants of India
BUSINESS MATHEMATICS
3. 2
Inequalities are statements where two quantities are unequal but a relationship exists between
them. These type of inequalities occur in business whenever there is a limit on supply, demand,
sales etc. For example, if a producer requires a certain type of raw material for his factory and
there is an upper limit in the availability of that raw material, then any decision which he takes
about production should involve this constraint also. We will see in this chapter more about such
situations.
Any linear function that involves an inequality sign is a linear inequality. It may be of one variable,
or, of more than one variable. Simple example of linear inequalities are those of one variable
only; viz., x > 0, x < 0 etc.
x ? 0
– 3 – 2 – 1 0 1 2 3
x > 0
– 3 – 2 – 1 0 1 2 3
x >O x > O
x >O
y >O
x > O
y > O
y y y y
x x x x
The inequality mentioned above is true for certain pairs of
numbers (x, y) that satisfy 3x + y < 6. By trial, we may arbitrarily
find such a pair to be (1,1) because 3 ? 1 + 1 = 4, and 4 < 6.
Linear inequalities in two variables may be solved easily by
extending our knowledge of straight lines.
For this purpose, we replace the inequality by an equality and
seek the pairs of number that satisfy 3x + y = 6. We may write
3x + y = 6 as y = 6 – 3x, and draw the graph of this linear
The values of the variables that satisfy an inequality are called the solution space, and is abbreviated
as S.S. The solution spaces for (i) x > 0,  (ii) x ? 0 are shaded in the above diagrams, by using deep
lines.
Linear inequalities in two variables: Now we turn to linear inequalities in two variables x and y
and shade a few S.S.
Let us now consider a linear inequality in two variables given by 3x + y < 6
Y
A
O
B
X
(2, 0)
(x,  y)
(0, 6)
6 – 3x
{
© The Institute of Chartered Accountants of India
Page 3


LINEAR INEQUALITIES
3
CHAPTER
Development of Inequalities
from the Descriptive Problem
Forms of Linear inequalities
in one variable and solution
space with various condition
like >0, >0, <0 and <0
Graphing of linear inequation
Determination of
Common region
Optimal solution
One of the widely used decision making problems, nowadays, is to decide on the optimal mix
of scarce resources in meeting the desired goal.  In simplest form, it uses several linear
inequations in two variables derived from the description of the problem.
The objective in this section is to make a foundation of the working methodology for the
above by way of introduction of the idea of :
? development of inequations from the descriptive problem;
? graphing of linear inequations; and
? determination of common region satisfying the inequations.
Feasible region or
Feasible points
CHAPTER OVERVIEW
© The Institute of Chartered Accountants of India
BUSINESS MATHEMATICS
3. 2
Inequalities are statements where two quantities are unequal but a relationship exists between
them. These type of inequalities occur in business whenever there is a limit on supply, demand,
sales etc. For example, if a producer requires a certain type of raw material for his factory and
there is an upper limit in the availability of that raw material, then any decision which he takes
about production should involve this constraint also. We will see in this chapter more about such
situations.
Any linear function that involves an inequality sign is a linear inequality. It may be of one variable,
or, of more than one variable. Simple example of linear inequalities are those of one variable
only; viz., x > 0, x < 0 etc.
x ? 0
– 3 – 2 – 1 0 1 2 3
x > 0
– 3 – 2 – 1 0 1 2 3
x >O x > O
x >O
y >O
x > O
y > O
y y y y
x x x x
The inequality mentioned above is true for certain pairs of
numbers (x, y) that satisfy 3x + y < 6. By trial, we may arbitrarily
find such a pair to be (1,1) because 3 ? 1 + 1 = 4, and 4 < 6.
Linear inequalities in two variables may be solved easily by
extending our knowledge of straight lines.
For this purpose, we replace the inequality by an equality and
seek the pairs of number that satisfy 3x + y = 6. We may write
3x + y = 6 as y = 6 – 3x, and draw the graph of this linear
The values of the variables that satisfy an inequality are called the solution space, and is abbreviated
as S.S. The solution spaces for (i) x > 0,  (ii) x ? 0 are shaded in the above diagrams, by using deep
lines.
Linear inequalities in two variables: Now we turn to linear inequalities in two variables x and y
and shade a few S.S.
Let us now consider a linear inequality in two variables given by 3x + y < 6
Y
A
O
B
X
(2, 0)
(x,  y)
(0, 6)
6 – 3x
{
© The Institute of Chartered Accountants of India
3 . 3 LINEAR INEQUALITIES
function.
Let x = 0 so that y = 6. Let y = 0, so that x = 2.
Any pair of numbers (x, y) that satisfies the equation y = 6 – 3x falls on the line AB.
Note: The pair of inequalities x ? 0, y ? 0 play an important role in linear programming problems.
Therefore, if y is to be less than 6 – 3x for the same value of x, it must assume a value that is less
than the ordinate of length 6 – 3x.
All such points (x, y) for which the ordinate is less than 6 – 3x lie below the line AB.
The region where these points fall is indicated by an
arrow and is shaded too in the adjoining diagram. Now
we consider two inequalities 3x + y ? 6  and  x – y ? –  2
being satisfied simultaneously by x and y. The pairs of
numbers (x, y) that satisfy both the inequalities may be
found by drawing the graphs of the two lines y = 6 – 3x
and y = 2 + x, and determining the region where both
the inequalities hold. It is convenient to express each
equality with y on the left-side and the remaining terms
in the right side. The first inequality 3x + y ? 6 is
equivalent to y ? 6 – 3x and it requires the value of y for
each x to be less than or equal to that of and on 6 – 3x.
The inequality is therefore satisfied by all points lying
below the line y = 6 – 3x. The region where these points
fall has been shaded in the adjoining diagram.
We consider the second inequality x – y ?  –2, and note that this is equivalent to y ? 2 + x. It
requires the value of  y  for each x to be larger than or equal to that of  2 + x. The inequality is,
therefore, satisfied by all points lying on and above the line y = 2 + x.
The region of interest is indicated by an arrow on the line y = 2 + x in the diagram below.
For x = 0, y = 2 + 0 = 2;
For y = 0, 0 = 2 + x   i.e,   x = –2.
y = 2 + x
x
0
(-2, 0)
(0, 2)
y
Y
X
B
A
y = 6 – 3 x
O
© The Institute of Chartered Accountants of India
Page 4


LINEAR INEQUALITIES
3
CHAPTER
Development of Inequalities
from the Descriptive Problem
Forms of Linear inequalities
in one variable and solution
space with various condition
like >0, >0, <0 and <0
Graphing of linear inequation
Determination of
Common region
Optimal solution
One of the widely used decision making problems, nowadays, is to decide on the optimal mix
of scarce resources in meeting the desired goal.  In simplest form, it uses several linear
inequations in two variables derived from the description of the problem.
The objective in this section is to make a foundation of the working methodology for the
above by way of introduction of the idea of :
? development of inequations from the descriptive problem;
? graphing of linear inequations; and
? determination of common region satisfying the inequations.
Feasible region or
Feasible points
CHAPTER OVERVIEW
© The Institute of Chartered Accountants of India
BUSINESS MATHEMATICS
3. 2
Inequalities are statements where two quantities are unequal but a relationship exists between
them. These type of inequalities occur in business whenever there is a limit on supply, demand,
sales etc. For example, if a producer requires a certain type of raw material for his factory and
there is an upper limit in the availability of that raw material, then any decision which he takes
about production should involve this constraint also. We will see in this chapter more about such
situations.
Any linear function that involves an inequality sign is a linear inequality. It may be of one variable,
or, of more than one variable. Simple example of linear inequalities are those of one variable
only; viz., x > 0, x < 0 etc.
x ? 0
– 3 – 2 – 1 0 1 2 3
x > 0
– 3 – 2 – 1 0 1 2 3
x >O x > O
x >O
y >O
x > O
y > O
y y y y
x x x x
The inequality mentioned above is true for certain pairs of
numbers (x, y) that satisfy 3x + y < 6. By trial, we may arbitrarily
find such a pair to be (1,1) because 3 ? 1 + 1 = 4, and 4 < 6.
Linear inequalities in two variables may be solved easily by
extending our knowledge of straight lines.
For this purpose, we replace the inequality by an equality and
seek the pairs of number that satisfy 3x + y = 6. We may write
3x + y = 6 as y = 6 – 3x, and draw the graph of this linear
The values of the variables that satisfy an inequality are called the solution space, and is abbreviated
as S.S. The solution spaces for (i) x > 0,  (ii) x ? 0 are shaded in the above diagrams, by using deep
lines.
Linear inequalities in two variables: Now we turn to linear inequalities in two variables x and y
and shade a few S.S.
Let us now consider a linear inequality in two variables given by 3x + y < 6
Y
A
O
B
X
(2, 0)
(x,  y)
(0, 6)
6 – 3x
{
© The Institute of Chartered Accountants of India
3 . 3 LINEAR INEQUALITIES
function.
Let x = 0 so that y = 6. Let y = 0, so that x = 2.
Any pair of numbers (x, y) that satisfies the equation y = 6 – 3x falls on the line AB.
Note: The pair of inequalities x ? 0, y ? 0 play an important role in linear programming problems.
Therefore, if y is to be less than 6 – 3x for the same value of x, it must assume a value that is less
than the ordinate of length 6 – 3x.
All such points (x, y) for which the ordinate is less than 6 – 3x lie below the line AB.
The region where these points fall is indicated by an
arrow and is shaded too in the adjoining diagram. Now
we consider two inequalities 3x + y ? 6  and  x – y ? –  2
being satisfied simultaneously by x and y. The pairs of
numbers (x, y) that satisfy both the inequalities may be
found by drawing the graphs of the two lines y = 6 – 3x
and y = 2 + x, and determining the region where both
the inequalities hold. It is convenient to express each
equality with y on the left-side and the remaining terms
in the right side. The first inequality 3x + y ? 6 is
equivalent to y ? 6 – 3x and it requires the value of y for
each x to be less than or equal to that of and on 6 – 3x.
The inequality is therefore satisfied by all points lying
below the line y = 6 – 3x. The region where these points
fall has been shaded in the adjoining diagram.
We consider the second inequality x – y ?  –2, and note that this is equivalent to y ? 2 + x. It
requires the value of  y  for each x to be larger than or equal to that of  2 + x. The inequality is,
therefore, satisfied by all points lying on and above the line y = 2 + x.
The region of interest is indicated by an arrow on the line y = 2 + x in the diagram below.
For x = 0, y = 2 + 0 = 2;
For y = 0, 0 = 2 + x   i.e,   x = –2.
y = 2 + x
x
0
(-2, 0)
(0, 2)
y
Y
X
B
A
y = 6 – 3 x
O
© The Institute of Chartered Accountants of India
BUSINESS MATHEMATICS
3. 4
By superimposing the above two graphs we determine the common region ACD in which the
pairs (x, y) satisfy both inequalities.
Y
X
O
A
y = 2 + x
C
D
y = 6 – 3x
Example 1: We now consider the problem of drawing graphs of the following inequalities
x ? 0,  y ? 0,  x ? 6,  y ? 7,   x + y ? 12
and shading the common region.
Note: [1] The inequalities 3x + y ? 6 and x – y ? –2 differ from the preceding ones in that these
also include equality signs. It means that the points lying on the corresponding lines
are also included in the region.
[2] The procedure may be extended to any number of inequalities.
We note that the given inequalities may be grouped as follows :
x ? 0, y ? 0
x ? 6, y ? 7, x + y ? 12
Y
X
O
y > 0, y < 7
y =7
x > 0, x < 6
X
Y
9 = x
O
x + y < 12
Y
X
© The Institute of Chartered Accountants of India
Page 5


LINEAR INEQUALITIES
3
CHAPTER
Development of Inequalities
from the Descriptive Problem
Forms of Linear inequalities
in one variable and solution
space with various condition
like >0, >0, <0 and <0
Graphing of linear inequation
Determination of
Common region
Optimal solution
One of the widely used decision making problems, nowadays, is to decide on the optimal mix
of scarce resources in meeting the desired goal.  In simplest form, it uses several linear
inequations in two variables derived from the description of the problem.
The objective in this section is to make a foundation of the working methodology for the
above by way of introduction of the idea of :
? development of inequations from the descriptive problem;
? graphing of linear inequations; and
? determination of common region satisfying the inequations.
Feasible region or
Feasible points
CHAPTER OVERVIEW
© The Institute of Chartered Accountants of India
BUSINESS MATHEMATICS
3. 2
Inequalities are statements where two quantities are unequal but a relationship exists between
them. These type of inequalities occur in business whenever there is a limit on supply, demand,
sales etc. For example, if a producer requires a certain type of raw material for his factory and
there is an upper limit in the availability of that raw material, then any decision which he takes
about production should involve this constraint also. We will see in this chapter more about such
situations.
Any linear function that involves an inequality sign is a linear inequality. It may be of one variable,
or, of more than one variable. Simple example of linear inequalities are those of one variable
only; viz., x > 0, x < 0 etc.
x ? 0
– 3 – 2 – 1 0 1 2 3
x > 0
– 3 – 2 – 1 0 1 2 3
x >O x > O
x >O
y >O
x > O
y > O
y y y y
x x x x
The inequality mentioned above is true for certain pairs of
numbers (x, y) that satisfy 3x + y < 6. By trial, we may arbitrarily
find such a pair to be (1,1) because 3 ? 1 + 1 = 4, and 4 < 6.
Linear inequalities in two variables may be solved easily by
extending our knowledge of straight lines.
For this purpose, we replace the inequality by an equality and
seek the pairs of number that satisfy 3x + y = 6. We may write
3x + y = 6 as y = 6 – 3x, and draw the graph of this linear
The values of the variables that satisfy an inequality are called the solution space, and is abbreviated
as S.S. The solution spaces for (i) x > 0,  (ii) x ? 0 are shaded in the above diagrams, by using deep
lines.
Linear inequalities in two variables: Now we turn to linear inequalities in two variables x and y
and shade a few S.S.
Let us now consider a linear inequality in two variables given by 3x + y < 6
Y
A
O
B
X
(2, 0)
(x,  y)
(0, 6)
6 – 3x
{
© The Institute of Chartered Accountants of India
3 . 3 LINEAR INEQUALITIES
function.
Let x = 0 so that y = 6. Let y = 0, so that x = 2.
Any pair of numbers (x, y) that satisfies the equation y = 6 – 3x falls on the line AB.
Note: The pair of inequalities x ? 0, y ? 0 play an important role in linear programming problems.
Therefore, if y is to be less than 6 – 3x for the same value of x, it must assume a value that is less
than the ordinate of length 6 – 3x.
All such points (x, y) for which the ordinate is less than 6 – 3x lie below the line AB.
The region where these points fall is indicated by an
arrow and is shaded too in the adjoining diagram. Now
we consider two inequalities 3x + y ? 6  and  x – y ? –  2
being satisfied simultaneously by x and y. The pairs of
numbers (x, y) that satisfy both the inequalities may be
found by drawing the graphs of the two lines y = 6 – 3x
and y = 2 + x, and determining the region where both
the inequalities hold. It is convenient to express each
equality with y on the left-side and the remaining terms
in the right side. The first inequality 3x + y ? 6 is
equivalent to y ? 6 – 3x and it requires the value of y for
each x to be less than or equal to that of and on 6 – 3x.
The inequality is therefore satisfied by all points lying
below the line y = 6 – 3x. The region where these points
fall has been shaded in the adjoining diagram.
We consider the second inequality x – y ?  –2, and note that this is equivalent to y ? 2 + x. It
requires the value of  y  for each x to be larger than or equal to that of  2 + x. The inequality is,
therefore, satisfied by all points lying on and above the line y = 2 + x.
The region of interest is indicated by an arrow on the line y = 2 + x in the diagram below.
For x = 0, y = 2 + 0 = 2;
For y = 0, 0 = 2 + x   i.e,   x = –2.
y = 2 + x
x
0
(-2, 0)
(0, 2)
y
Y
X
B
A
y = 6 – 3 x
O
© The Institute of Chartered Accountants of India
BUSINESS MATHEMATICS
3. 4
By superimposing the above two graphs we determine the common region ACD in which the
pairs (x, y) satisfy both inequalities.
Y
X
O
A
y = 2 + x
C
D
y = 6 – 3x
Example 1: We now consider the problem of drawing graphs of the following inequalities
x ? 0,  y ? 0,  x ? 6,  y ? 7,   x + y ? 12
and shading the common region.
Note: [1] The inequalities 3x + y ? 6 and x – y ? –2 differ from the preceding ones in that these
also include equality signs. It means that the points lying on the corresponding lines
are also included in the region.
[2] The procedure may be extended to any number of inequalities.
We note that the given inequalities may be grouped as follows :
x ? 0, y ? 0
x ? 6, y ? 7, x + y ? 12
Y
X
O
y > 0, y < 7
y =7
x > 0, x < 6
X
Y
9 = x
O
x + y < 12
Y
X
© The Institute of Chartered Accountants of India
3 . 5 LINEAR INEQUALITIES
By superimposing the above three graphs, we determine the common region in the  xy plane
where all the five inequalities are simultaneously satisfied.
O
X
(0, 7)
Y
(5,7)
(6,0)
0,0
This common region is known as feasible region or the solution set (or the polygonal convex
sets).
A region is said to be bounded if it can be totally included within a (very large) circle. The
shaded region enclosed by deep lines in the previous diagram is bounded, since it can be
included within a circle.
The objective function attains a maximum or a minimum value at one of the corner points
of the feasible solution known as extreme points of the solution set. Once these extreme
points (the points of intersection of lines bounding the region) are known, a compact matrix
representation of these points is possible. We shall denote the matrix of the extreme points
by E.
The coefficients of the objective function may also be represented by a column vector. We
shall represent this column vector by C.
The elements in the product matrix EC shows different values, which the objective function
attains at the various extreme points. The largest and the smallest elements in matrix EC are
respectively the maximum and the minimum values of the objective function. The row in
matrix EC in which this happens is noted and the elements in that row indicate the appropriate
pairing and is known as the optimal solution.
In the context of the problem under consideration.
X Y
0 0
0 7
1 x
E , C 5 7
2 y
6 0
6 6
? ?
? ?
? ?
? ?
? ? ? ?
? ?
? ?
? ?
? ?
? ?
? ?
(6, 6)
© The Institute of Chartered Accountants of India