Let a linear inequality with two variables be given and 
(1)
If the values 
And 
considered as coordinates of points on the plane, then the set of points on the plane whose coordinates satisfy inequality (1) is called the solution region of this inequality. Consequently, the domain of solutions to inequality (1) is a half-plane with a boundary straight line 
.
Example 1.
.
Solution. Building a straight line 
by two points, for example, by the points of intersection with the coordinate axes (0; 4) and (6; 0). This line divides the plane into two parts, i.e. into two half-planes. We take any point of the plane that does not lie on the constructed line. If the coordinates of a point satisfy the given inequality, then the solution region is the half-plane in which this point is located. If we get an incorrect numerical inequality, then the solution area is the half-plane to which this point does not belong. Usually the point (0; 0) is taken for control.
Let's substitute 
And 
to the given inequality. We get 
. Consequently, the half-plane “toward zero” is the region of solutions to this inequality (shaded part of Fig. 1).
Example 2. Find the half-plane defined by the inequality
.
Solution. Building a straight line 
, for example, by points (0; 0) and (1; 3). Because the straight line passes through the origin of coordinates, the point (0; 0), then you cannot take it for control. Take, for example, the point (– 2; 0) and substitute its coordinates into the given inequality. We get 
. This is not true. This means that the region of solutions to this inequality will be the half-plane to which the control point does not belong (the shaded part of Fig. 2).
2. Solution domain of a system of linear inequalities.
Example. Find the solution area of the system of inequalities:
Solution. We find the region of solutions to the first inequality (Fig. 1) and the second inequality (Fig. 2).
All points of the part of the plane where the hatching is superimposed will satisfy both the first and second inequalities. Thus, the solution area for the given system of inequalities is obtained (Fig. 3).
If to given system inequalities add conditions 
And 
, then the solution domain of the system of inequalities 
will be located only in the I coordinate quarter (Fig. 4).
The principle of finding a solution to a system of linear inequalities does not depend on the number of inequalities included in the system.
Note : Region admissible solutions(ODR) if exists, then it is a closed or open convex polygon.
3. Algorithm for the graphical method of solving problems
If the task linear programming contains only two variables, it can be solved graphically by performing the following operations:
Example. Solve a linear programming problem graphically
max
Solution. The third and fourth restrictions of the system are double inequalities; let us transform them to a form more familiar for such problems 
, This 
And 
, That. the first of the resulting inequalities 
(or 
) refers to the condition of non-negativity, and the second 
to a system of restrictions. Likewise, 
This 
And 
.
That. the problem will take the form
max
,
Replacing the inequality signs with exact equality signs, we construct a region of admissible solutions using straight line equations:
;
;
;
.
The solution region of the inequalities is a pentagon ABCDE.
Let's build a vector 
.
Through the origin perpendicular to the vector 
draw a level line 
. And then we will move it parallel to itself in the direction of the vector 
to the point of exit from the region of feasible solutions. This will be the point WITH. Let's find the coordinates of this point by solving a system consisting of equations of the first and fourth lines:
.
Let's substitute the coordinates of the point WITH V target function and find its maximum value 
Example. Construct level lines 
And 
for a linear programming problem:
max
(min)
Solution. The region of feasible solutions is an open region (Fig. 6). Level line 
passes through a point IN. Function Z has a minimum at this point. Level line 
cannot be constructed, since there is no exit point from the region of feasible solutions, this means that 
.
Tasks for independent work.
Find the solution area of the system of inequalities:
A) 
b) 
Solve a linear programming problem graphically
min
Create an economic-mathematical model and solve graphically a linear programming problem
The company produces products of two types A and B. Products of each type are processed on two machines (I and II). The processing time of one product of each type on machines, the operating time of machines per work shift, the company’s profit from the sale of one product of type A and type B are listed in the table:
A study of the sales market showed that the daily demand for products of type B never exceeds the demand for products of type A by more than 40 units, and the demand for products of type A does not exceed 90 units per day.
Determine the product production plan that provides the greatest profit.
The system consists of inequalities in two variables:
To solve the system you need:
1. For each inequality, write down the equation corresponding to this inequality.
2. Construct straight lines, which are graphs of functions specified by equations.
3. For each line, determine the half-plane, which is given by the inequality. To do this take arbitrary point, not lying on a line, substitute its coordinates into the inequality. if the inequality is true, then the half-plane containing the chosen point is the solution to the original inequality. If the inequality is false, then the half-plane on the other side of the line is the set of solutions to this inequality.
4. To solve a system of inequalities, it is necessary to find the area of intersection of all half-planes that are the solution to each inequality of the system.
This area may turn out to be empty, then the system of inequalities has no solutions and is inconsistent. Otherwise, the system is said to be consistent. There may be solutions final number And infinite set. The area can be a closed polygon or unbounded.
Example 3. Solve the system graphically: 
Consider the equations x + y–1 = 0 and –2x – 2y + 5 = 0, corresponding to the inequalities. Let's construct straight lines given by these equations (Fig. 3).
Figure 3 – Image of straight lines
Let us define the half-planes defined by the inequalities. Let's take an arbitrary point, let (0; 0). Consider x+ y– 1 ≤ 0, substitute the point (0; 0): 0 + 0 – 1 ≤ 0. This means that in the half-plane where the point (0; 0) lies, x + y – 1 ≤ 0, i.e. . the half-plane lying below the line is a solution to the first inequality. Substituting this point (0; 0) into the second, we get: –2 ∙ 0 – 2 ∙ 0 + 5 ≤ 0, i.e. in the half-plane where the point (0; 0) lies, –2x – 2y + 5≥ 0, and we were asked where –2x – 2y + 5 ≤ 0, therefore, in the other half-plane – in the one above the straight line.
Let's find the intersection of these two half-planes. The lines are parallel, so the planes do not intersect anywhere, which means that the system of these inequalities has no solutions and is inconsistent.
Example 4. Find graphically solutions to the system of inequalities:
1. Let’s write out the equations corresponding to the inequalities and construct straight lines (Fig. 4).
x + 2y– 2 = 0 x 2 0
y – x – 1 = 0 x 0 2
y + 2 = 0; y = –2.
Figure 4 – Image of straight lines
2. Having chosen the point (0; 0), we determine the signs of inequalities in the half-planes:
0 + 2 ∙ 0 – 2 ≤ 0, i.e. x + 2y– 2 ≤ 0 in the half-plane below the straight line;
0 – 0 – 1 ≤ 0, i.e. y –x– 1 ≤ 0 in the half-plane below the straight line;
0 + 2 =2 ≥ 0, i.e. y + 2 ≥ 0 in the half-plane above the straight line.
3. The intersection of these three half-planes will be an area that is a triangle. It is not difficult to find the vertices of the region as the intersection points of the corresponding lines
Thus, A(–3; –2), B(0; 1), C(6; –2).
Let's consider another example in which the resulting solution domain of the system is unlimited.
Example 5. Solve the system graphically
Let's write out the equations corresponding to the inequalities and construct straight lines (Fig. 5).
Figure 5 – Image of straight lines
x + y – 1 = 0 x 0 1
y – x – 1 = 0 x 0 –1
Let us define signs in half-planes. Let's select the point (0; 0):
0 – 0 – 1 ≤ 0, i.e. y – x – 1 ≤ 0 below the straight line;
0 + 0 – 1 ≤ 0, i.e. x + y – 1 ≤ 0 below the straight line.
The intersection of two half-planes is an angle with its vertex at point A(0;1). This unbounded region is the solution to the original system of inequalities.
