Graphing Linear Inequalities in Two Variables
A linear inequality in the two variables x and y looks like
ax + by <
c 
ax + by < c 
ax + by > c 
ax + by > c 
where a, b, and c are constants.
A solution to an inequality is any pair of numbers x and y that
satisfy the inequality.
The rules for finding the solution set of a linear inequality are much the
same as those for finding the solution to a linear equation.
 Add or subtract the same expression to both sides.
 Multiply or divide both sides by the same nonzero quantity; if that
quantity is negative, then the inequality must be reversed.
Example 1.
Determine the solution set of 5x + 2y < 17.
One solution to this is x = 2 and y = 3, because 5(2) + 2(3) =
16, which is indeed less than or equal to 17.
A pair of numbers that does not form a solution is x = 3 and y
= 2, because 5(3) + 2(2) = 19, which is not less than or equal to 17. The pair x
= 2 and y = 3 isn't the only solution; as a matter of fact, there are infinitely
many solutions.
Since we can't write down all possible solutions to a linear inequality, a
good way to describe the set of solutions to any linear inequality is by a
graph. If the pair of numbers x and y is a solution, then think of this pair as
a point in the plane, so the set of all solutions can be thought of as a region
in the xyplane.
To illustrate how to determine this region; first, we solve the inequality
for y in terms of x.
5x + 2y 
< 
17 
2y 
< 
~5x+ 17 
y 
< 
(5/2) x + 17/2 
Next, graph the line y = (5/2) x + 17/2.
The set of points (x,y) that lie on this line is the set of all
(x,y) such that y is exactly equal to (5/2) x +
17/2.
These points make up part of the set of solutions to the inequality,
but not all. We see that y can also be less than (5/2) x + 17/2,
so all points below the line would also be solutions. The shaded region
in Figure
1 shows the solution set.

Figure 1 
Example 2. Graph
the solution set for the inequality 3x  8y > 12.
Solve for y:
8y 
> 
3x + 12 
y 
< 
(3/8) x  3/2 
Do you notice that the inequality is reversed? That's because we divided by a
negative number, any time you multiply or divide an
inequality by a negative number, you must reverse the inequality.
Next, graph the line
Points on this line are part of the solution set; the other part consists of
all points below the line, shown if Figure 2.

Figure 2 
Example 3. Graph
the solution set for the inequality  1Ox  2y > 7.
Figure 3

Solving for y gives
2y 
> 
10x + 7 
y 
< 
5 x  7/2 (Remember the minus?) 
This example is a little different because there's no
equals sign in the inequality. But you still graph the line 7 = 5x
7/2, except draw it as a dashed lire.
This indicates that the line itself is not part of the solution set. The
actual solution set consists of all points below the dashed line. This is
because y must be strictly less than 5x  7/2.

Systems of Linear Inequalities in Two Variables
A system of linear inequalities is a set of one or more linear inequalities;
a solution is a pair of numbers (x, y) that satisfies all of the inequalities.
Example 4.
Determine the solution set to the system of linear inequalities
The pair of numbers x = 1, y = 2 is one solution because 1 + 5(2) = 11 <
20 and 3(1)+2(2) = 7 < 21.
The pair x = 0, y = 5 is not a solution because it doesn't even satisfy the
first inequality 0 + 5(5) = 25, which is not less than or equal to 20. Notice
that it does satisfy the second inequality, but in order to be a solution, it
must satisfy both.
As before, a system can have an infinite number of solutions, so we present
its solution set by a region in the plane. Solve each inequality for y:
y 
< 
(1/5)x + 4 
y 
< 
(3/2)x + 21/2 
Graph each of the lines on the same coordinate system. The solution set for
the first inequality lies in the region below and on the line y = (1/5)x + 4
and the solution set for the second inequality lies in the region on and below
the line y = (3/2)x + 21/2.
The solution set for the system lies in the region common to both, and is the
darker region shown in Figure 4.

Figure 4

The corner of this region is the intersection of the two lines, and corners
will be very important in the next chapter. In order to find this point, return
to the original system of inequalities and replace the inequality symbols with
equal signs. The resulting two equations are the equations of the two
lines:
x + 5 y 
= 20 

3x + 2y 
= 
21 
This is a system of two linear equations in two unknowns, and its solution is
precisely the point of intersection of the two lines. You know how to solve this
system by hand or using your calculator.
The solution matrix is x = 5 and y = 3. The point of
intersection, or corner of the region, is (5,3).
Example 5. Graph
the solution set to the system
7x  5y 
< 
12 
2x + 3y 
< 
18 
x 
> 
0 
y 
> 
0 
and find all corners.
The last two inequalities in this system simply restrict the solution set to
points in the first quadrant (that's the quadrant in which both coordinates are
nonnegative).
Solve the first two inequalities for y to get
y 
> 
(7/5)x 12/5 
y 
< 
(2/3)x + 6 
The solution set to the first inequality lies in the region on and above the
graph of the line y = (7/5)x 12/5, and the solution set to the second lies on
and below the line y = (2/3)x + 6.
The common region which is in the first quadrant is the solution set to the
original four inequalities.

Figure 5a

Figure 5b

Next we find corners of this region. There are four of them. One of them is
really easy; that's the origin: x = 0 and y = 0. Two others are also easy; these
are the ones on the axes. The one on the xaxis is the xintercept
of the line 7x  5y = 12. To find this, take y = 0, and solve for x, x = 12/7
(= 1.71), so this corner is (12/7,0).
On the yaxis, the corner is the yintercept of the line 2x +
3y = 18; take x = 0 to get y = 6, so the corner is (0,6).
The last point requires some work, but not much. You must find the
intersection of the lines 7x  5y = 12 and 2x + 3y = 18, which is the solution
to the linear system
7x  5 y 
= 
12 
2x + 3y 
= 
18 
The solution is x = 4.06 and y = 3.29, and the corner is (4.06, 3.29).

Example 6. Graph
the solution set to the system
x + 4y 
< 
20 
3x + 4y 
> 
28 
x  y 
< 
7 
and find all corners.
As before, graph each line and shade the appropriate region;
the solution set to the entire system is the region common to
all three, and is shown in the figure on the right, complete
with corners. 
Figure 6

Linear Programming
The primary reason for studying systems of linear inequalities at this point
is to determine solutions to mathematical problems that can be solved using
linear programming methods. Specifically, one has a function that is to
be maximized
or minimized, but the solution is subject to
certain restrictions. The function itself is called the
objective function, and the restrictions are called
constraints
and usually appear in a system of inequalities.
In order to solve a linear programming problem follow the steps below.
 Step 1. Graph the inequalities
that express the constraints and determine the solutions set as a region
in the plane; this region is called the feasible region. We will
consider only bounded regions, which insures the existence of maximum
and minimum values for the objective function.
 Step 2. Determine the corners of
the feasible region.
 Step 3. The maximum or minimum
of the objective function will occur at one of these corners. In order
to find which one, evaluate the objective function at each corner; the
maximum is the largest value and the minimum is the smallest.

Example 7.
Maximize the objective function A = 3x + 5y subject to the
constraints
2x + y 
< 
16 
16 x + 3y 
< 
18 
x 
> 
0 
y 
> 
0 
First, graph each line and shade the appropriate region; then find the
corners.
Now evaluate the objective function at each corner.
Corner

Functional Value

(0, 0)

A = 3 (0) + 5 (0) = 0

(0, 6)

A = 3(0) + 5(6) = 30

(6, 4)

A = 3(6) + 5(4) = 38

(8, 0)

A = 3(8) + 5(0) = 24

Thus, the maximum is 38 and occurs at corner (6,4).

Figure 7a

Here is a note on why you only need to look at the corners when you are
finding the maximum or minimum value of the objective function. Of course, 3x +
5y can be as large as you want, but the problem is to stay inside the feasible
region.
Figure 7b shows 3x + 5y = 50, 3x + 5y = 40, and the
one that passes through the corner at which the solution occurs. The latter line
is 3x + 5y = 38. None of the points on the line 3x + 5y = 50 are inside the
feasible region, so none of these points are valid solutions.
Notice that the line 3x + 5y = 40 is closer, and all the lines 3x + 5y =
any number
are parallel. Slide the highest line down staying parallel until you first hit a
point in the feasible region. What point did you hit? A corner! So where is the
maximum? At a corner!

Figure 7b

Example 8.
Maximize the objective function B = 8x + y subject to
2x + y 
< 
16 
16 x + 3y 
< 
18 
x 
> 
0 
y 
> 
0 
The feasible region with corners is shown in Figure 7b.
Evaluate this objective function at the corners.
Corner

Functional Value

(0, 0)

B = 8(0) + 0 = 0

(0, 6)

B = 8(0) + 6 = 6

(6,4)

B=8(6) + 4=52

(8,0)

B=8(8) + 0=64

The maximum for the objective function B is 64, and occurs at corner (8, 0).
Example 9.
Minimize the objective function C = 11x + 10y subject to the
constraints
x + y 
> 
15 
6x + 5y 
> 
80 
4x + 3y 
< 
60 
x 
> 
0 
y 
> 
0 
The graph of the feasible region with its corners is shown in
Figure 9.
Evaluate the objective function at the corners.
Corner

Functional Value

(5, 10)

C = 11(5) + 10(10) = 155

(0, 16)

C = 11(0) + 10(16) = 160

(0, 20)

C = 11(0) + 10(20) = 200

(15, 0)

C = 11(15) + 10(0) = 165

The minimum value of the objective function is 155, which occurs at (5, 10).

Figure 9 
The main deficiency of the graphical method is that it is limited to
solving LP problems with 1 or 2 decision variables only. However, the main and
useful conclusion we easily observe from the graphical methods, is as follow:
If a linear program has a nonempty,
bounded feasible region, then the optimal solution is always one of
the corner points. 
The proof of this claim, follows from the results of the the following two
facts:
Fact 1: The feasible region of any linear
program is always a convex set.
Since all of the constraints are linear, the feasible region (F.R.) is a
polygon. Moreover, this polygon is a convex set. In any higher than
twodimensional LP problem, the boundaries of F.R. are parts of the
hyperplanes, and the F.R. in this case is called polyhedra that is also convex.
A convex set is the one that if you choose two feasible points from it, then all
the points on the straight line segment joining these two points are also
feasible. The proof that the F.R. of linear programs are always convex sets
follows by contradiction. The following figures depict examples for the two
types of sets: A nonconvex and a convex set.
The set of feasible region in any linear program is called a polyhedron,
it is called a polytope if it is bounded.
Fact 2: The Isovalue of a linear program
objective function is always a linear function.
This fact follows from the nature of the objective function in any LP
problem. The following figures depict the two typical kinds of isovalued
objective functions.
Combining the above two facts, it follows that, if a linear program has a
nonempty, bounded feasible region, then the optimal solution is always one of
the corner points.
To overcome the deficiency of the graphical method, we will utilize this
useful and practical conclusion in developing an algebraic method that is
applicable to multidimensional LP problems.
The convexity of the feasible region for linear programs makes the LP
problems easy to solve. Because of this property and linearity of the objective
function, the solution is always one of the vertices. Moreover, since number of
vertices are limited, one has to find all feasible vertices, and then evaluate
the objective function at these vertices to seek the optimal point.
For nonlinear programs, the problem is much harder to solve, because the
solution could be anywhere inside the feasible region on the boundary of the
feasible region, or at a vertex.
Fortunately, most of the Business optimization problems are linear, which is
why LP is so popular. There are well over 400 computer packages in the market
today solving LP problems. Most of them are based on vertex searching, that is,
jumping from one vertex to the neighboring one in search of an optimal point.
You have already noticed that, a graph of a system of
inequalities and/or equalities is called the feasible region. These two
representations, graphical, and algebraic are
equivalent to each other, which means the coordinate of any point satisfying the
constraints is located in the feasible region, and the coordinate of any point
in the feasible region satisfies all the constraints.
A numerical Example: Find the system of
constraints representing the following feasible region.
In the above Figure, the system of coordinate is shown in gray color at the
background. By constructing the equations of the boundary lines of the feasible
region, we can verify that the following system of inequalities indeed
represents the above feasible region:
x_{1} Ò 1
x_{2} £ 1
x_{1}  x_{2} £ 1 