Examples of solving inequalities graphically. Graphical solution of inequalities, systems of sets of inequalities with two variables

see also Solving a linear programming problem graphically, Canonical form of linear programming problems

The system of constraints for such a problem consists of inequalities in two variables:
and the objective function has the form F = C 1 x + C 2 y, which is to be maximized.

Let's answer the question: what pairs of numbers ( x; y) are solutions to the system of inequalities, i.e., do they satisfy each of the inequalities simultaneously? In other words, what does it mean to solve a system graphically?
First you need to understand what is the solution of one linear inequality with two unknowns.
To solve a linear inequality with two unknowns means to determine all pairs of values of the unknowns for which the inequality is satisfied.
For example, inequality 3 x – 5y≥ 42 satisfy the pairs ( x , y) : (100, 2); (3, –10), etc. The problem is to find all such pairs.
Consider two inequalities: ax + by≤ c, ax + by≥ c. Straight ax + by = c divides the plane into two half-planes so that the coordinates of the points of one of them satisfy the inequality ax + by >c, and the other inequality ax + +by <c.
Indeed, take a point with coordinate x = x 0; then a point lying on a straight line and having an abscissa x 0 , has an ordinate

Let for definiteness a<0, b>0, c>0. All points with abscissa x 0 above P(e.g. dot M), have y M>y 0 , and all points below the point P, with abscissa x 0 , have yN<y 0 . Insofar as x 0 is an arbitrary point, then there will always be points on one side of the line for which ax+ by > c, forming a half-plane, and on the other hand, points for which ax + by< c.

Picture 1

The inequality sign in the half-plane depends on the numbers a, b , c.
This implies the following method for graphical solution of systems of linear inequalities in two variables. To solve the system, you need:

For each inequality, write down the equation corresponding to the given inequality.
Construct lines that are graphs of functions given by equations.
For each straight line, determine the half-plane, which is given by the inequality. To do this, take an arbitrary point that does not lie on a straight 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.
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 in the system.

This area may turn out to be empty, then the system of inequalities has no solutions, it is inconsistent. Otherwise, the system is said to be consistent.
Solutions can be a finite number and an infinite set. The area can be a closed polygon or it can be unlimited.

Let's look at three relevant examples.

Example 1. Graphically solve the system:
x + y- 1 ≤ 0;
–2x- 2y + 5 ≤ 0.

consider the equations x+y–1=0 and –2x–2y+5=0 corresponding to the inequalities;
let us construct the straight lines given by these equations.

Figure 2

Let us define the half-planes given by the inequalities. Take an arbitrary point, let (0; 0). Consider x+ y– 1 0, we substitute the point (0; 0): 0 + 0 – 1 ≤ 0. hence, in the half-plane where the point (0; 0) lies, x + y – 1 ≤ 0, i.e. the half-plane lying below the straight line is the solution to the first inequality. Substituting this point (0; 0) into the second one, we get: –2 ∙ 0 – 2 ∙ 0 + 5 ≤ 0, i.e. in the half-plane where the point (0; 0) lies, -2 x – 2y+ 5≥ 0, and we were asked where -2 x – 2y+ 5 ≤ 0, therefore, in another half-plane - in the one above the straight line.
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, it is inconsistent.

Example 2. Find graphically solutions to the system of inequalities:

Figure 3
1. Write down the equations corresponding to the inequalities and construct straight lines.
x + 2y– 2 = 0

x	2	0
y	0	1

y – x – 1 = 0

x	0	2
y	1	3

y + 2 = 0;
y = –2.
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 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 points of intersection of the corresponding lines

Thus, BUT(–3; –2), AT(0; 1), With(6; –2).

Let us consider one more example, in which the resulting domain of the solution of the system is not limited.

The graphical method consists in constructing a set of feasible LLP solutions, and finding in this set a point corresponding to the max/min objective function.

Due to the limited possibilities of a visual graphical representation, this method is used only for systems of linear inequalities with two unknowns and systems that can be reduced to this form.

In order to visually demonstrate the graphical method, we will solve the following problem:

1. At the first stage, it is necessary to construct the area of feasible solutions. For this example, it is most convenient to choose X2 for the abscissa, and X1 for the ordinate, and write the inequalities in the following form:

Since both the graphs and the area of admissible solutions are in the first quarter. In order to find the boundary points, we solve equations (1)=(2), (1)=(3) and (2)=(3).

As can be seen from the illustration, the polyhedron ABCDE forms an area of feasible solutions.

If the domain of admissible solutions is not closed, then either max(f)=+ ? or min(f)= -?.

2. Now we can proceed to directly finding the maximum of the function f.

Alternately substituting the coordinates of the vertices of the polyhedron into the function f and comparing the values, we find that f(C)=f (4; 1)=19 - the maximum of the function.

This approach is quite beneficial for a small number of vertices. But this procedure can be delayed if there are quite a lot of vertices.

In this case, it is more convenient to consider a level line of the form f=a. With a monotonous increase in the number a from -? to +? straight lines f=a are displaced along the normal vector. If, with such a displacement of the level line, there exists some point X - the first common point of the area of feasible solutions (polyhedron ABCDE) and the level line, then f(X) is the minimum of f on the set ABCDE. If X is the last point of intersection of the level line and the set ABCDE, then f(X) is the maximum on the set of feasible solutions. If for a>-? the line f=a intersects the set of admissible solutions, then min(f)= -?. If this happens when a>+?, then max(f)=+?.

Goals:

1. Repeat knowledge about the quadratic function.

2. Get acquainted with the method of solving a quadratic inequality based on the properties of a quadratic function.

Equipment: multimedia, presentation “Solving square inequalities”, cards for independent work, table “Algorithm for solving square inequalities”, control sheets with carbon paper.

DURING THE CLASSES

I. Organizational moment (1 min).

II. Updating of basic knowledge(10 minutes).

1. Plotting a quadratic function y \u003d x 2 -6x + 8<Рисунок 1. Приложение >

determination of the direction of the branches of the parabola;
determining the coordinates of the parabola vertex;
determination of the axis of symmetry;
determination of intersection points with coordinate axes;
finding additional points.

2. Determine from the drawing the sign of the coefficient a and the number of roots of the equation ax 2 +in+c=0.<Рисунок 2. Приложение >

3. According to the graph of the function y \u003d x 2 -4x + 3, determine:

What are the zeros of the function;
Find the intervals on which the function takes positive values;
Find the intervals on which the function takes negative values;
At what values of x does the function increase, and at what values does it decrease?<Рисунок 3>

4. Learning new knowledge (12 min.)

Task 1: Solve the inequality: x 2 +4x-5 > 0.

The inequality is satisfied by the x values at which the values of the function y=x 2 +4x-5 are equal to zero or positive, that is, those x values at which the points of the parabola lie on the x-axis or above this axis.

Let's build a graph of the function y \u003d x 2 + 4x-5.

With the x-axis: X 2 + 4x-5 \u003d 0. According to the Vieta theorem: x 1 \u003d 1, x 2 \u003d -5. Points(1;0),(-5;0).

With the y-axis: y(0)=-5. Point (0;-5).

Additional points: y(-1)=-8, y(2)=7.<Рисунок 4>

Bottom line: The values of the function are positive and equal to zero (non-negative) when

Is it necessary to plot a quadratic function in detail every time to solve an inequality?
Do I need to find the coordinates of the vertex of the parabola?
What is important? (a, x 1, x 2)

Conclusion: To solve a quadratic inequality, it is enough to determine the zeros of the function, the direction of the branches of the parabola and build a sketch of the graph.

Task 2: Solve the inequality: x 2 -6x + 8 < 0.

Solution: Let's determine the roots of the equation x 2 -6x+8=0.

According to the Vieta theorem: x 1 \u003d 2, x 2 \u003d 4.

a>0 - the branches of the parabola are directed upwards.

Let's build a sketch of the graph.<Рисунок 5>

We mark with signs “+” and “–” the intervals on which the function takes positive and negative values. Let's choose the interval we need.

Answer: X€.

5. Consolidation of new material (7 min).

No. 660 (3). The student decides on the board.

Solve inequality-x 2 -3x-2<0.

X 2 -3x-2=0; x 2 +3x+2=0;

the roots of the equation: x 1 \u003d -1, x 2 \u003d -2.

a<0 – ветви вниз. <Рисунок 6>

No. 660 (1) - Working with a hidden board.

Solve the inequality x 2 -3x + 2 < 0.

Solution: x 2 -3x+2=0.

Let's find the roots: ; x 1 =1, x 2 =2.

a>0 - branches up. We build a sketch of the graph of the function.<Рисунок 7>

Algorithm:

Find the roots of the equation ax 2 + in + c \u003d 0.
Mark them on the coordinate plane.
Determine the direction of the branches of the parabola.
Sketch a chart.
Mark with signs “+” and “-”, the intervals on which the function takes positive and negative values.
Select the desired interval.

6. Independent work (10 min.).

(Reception - carbon paper).

The control sheet is signed and handed over to the teacher for verification and correction determination.

Board self-check.

Additional task:

№ 670. Find the values of x at which the function takes values not greater than zero: y=x 2 +6x-9.

7. Homework (2 min).

№ 660 (2, 4), № 661 (2, 4).

Fill in the table:

D	Inequality	a	Drawing	Decision
D>0	ax 2 + in + s > 0	a>0
D>0	ax 2 + in + s > 0	a<0
D>0	ax 2 + in + s < 0	a>0
D>0	ax 2 + in + s < 0	a<0

8. Summary of the lesson (3 min).

Reproduce the algorithm for solving inequalities.
Who did a great job?
What seemed difficult?

One of the most convenient methods for solving quadratic inequalities is the graphical method. In this article, we will analyze how quadratic inequalities are solved graphically. First, let's discuss what the essence of this method is. And then we give the algorithm and consider examples of solving quadratic inequalities graphically.

Page navigation.

The essence of the graphic method

Generally graphical way to solve inequalities with one variable is used not only to solve square inequalities, but also inequalities of other types. The essence of the graphical method for solving inequalities next: consider the functions y=f(x) and y=g(x) that correspond to the left and right parts of the inequality, build their graphs in the same rectangular coordinate system and find out at what intervals the graph of one of them is located below or above the other. Those intervals where

the graph of the function f above the graph of the function g are solutions to the inequality f(x)>g(x) ;
the graph of the function f not lower than the graph of the function g are solutions to the inequality f(x)≥g(x) ;
the graph of the function f below the graph of the function g are solutions to the inequality f(x)
the graph of the function f not above the graph of the function g are solutions to the inequality f(x)≤g(x) .

Let's also say that the abscissas of the intersection points of the graphs of functions f and g are solutions to the equation f(x)=g(x) .

Let us transfer these results to our case – to solve the quadratic inequality a x 2 +b x+c<0 (≤, >, ≥).

We introduce two functions: the first y=a x 2 +b x+c (in this case f(x)=a x 2 +b x+c) corresponds to the left side of the quadratic inequality, the second y=0 (in this case g (x)=0 ) corresponds to the right side of the inequality. schedule quadratic function f is a parabola and the graph permanent function g is a straight line coinciding with the abscissa axis Ox .

Further, according to the graphical method for solving inequalities, it is necessary to analyze at what intervals the graph of one function is located above or below the other, which will allow us to write the desired solution of the quadratic inequality. In our case, we need to analyze the position of the parabola relative to the axis Ox.

Depending on the values of the coefficients a, b and c, the following six options are possible (for our needs, a schematic representation is sufficient, and it is possible not to depict the Oy axis, since its position does not affect the solution of the inequality):

In this drawing, we see a parabola whose branches are directed upwards and which intersects the axis Ox at two points, the abscissas of which are x 1 and x 2 . This drawing corresponds to the variant when the coefficient a is positive (it is responsible for the upward direction of the branches of the parabola), and when the value is positive discriminant of a square trinomial a x 2 +b x + c (in this case, the trinomial has two roots, which we denoted as x 1 and x 2, and we assumed that x 1 0 , D=b 2 −4 a c=(−1) 2 −4 1 (−6)=25>0, x 1 =−2 , x 2 =3 .

For clarity, let's draw in red the parts of the parabola located above the abscissa axis, and in blue - located below the abscissa axis.

Now let's find out what gaps correspond to these parts. The following drawing will help determine them (in the future, we will mentally make such selections in the form of rectangles):

So on the abscissa axis, two intervals (−∞, x 1) and (x 2, +∞) were highlighted in red, on them the parabola is higher than the axis Ox, they make up the solution of the quadratic inequality a x 2 +b x+c>0 , and the interval (x 1 , x 2) is highlighted in blue, on it the parabola is below the axis Ox , it is a solution to the inequality a x 2 + b x + c<0 . Решениями нестрогих квадратных неравенств a·x 2 +b·x+c≥0 и a·x 2 +b·x+c≤0 будут те же промежутки, но в них следует включить числа x 1 и x 2 , отвечающие равенству a·x 2 +b·x+c=0 .

And now briefly: for a>0 and D=b 2 −4 a c>0 (or D"=D/4>0 for an even coefficient b)

the solution to the quadratic inequality a x 2 +b x+c>0 is (−∞, x 1)∪(x 2 , +∞) or, in another way, x x2;
the solution to the quadratic inequality a x 2 +b x+c≥0 is (−∞, x 1 ]∪ or in other notation x 1 ≤x≤x 2 ,

where x 1 and x 2 are the roots of the square trinomial a x 2 + b x + c, and x 1

Here we see a parabola, the branches of which are directed upwards, and which touches the abscissa axis, that is, it has one common point with it, let's denote the abscissa of this point as x 0. The presented case corresponds to a>0 (the branches are directed upwards) and D=0 (the square trinomial has one root x 0 ). For example, we can take the quadratic function y=x 2 −4 x+4 , here a=1>0 , D=(−4) 2 −4 1 4=0 and x 0 =2 .

The drawing clearly shows that the parabola is located above the Ox axis everywhere, except for the point of contact, that is, at the intervals (−∞, x 0) , (x 0 , ∞) . For clarity, we select areas in the drawing by analogy with the previous paragraph.

We draw conclusions: for a>0 and D=0

the solution to the quadratic inequality a x 2 +b x+c>0 is (−∞, x 0)∪(x 0 , +∞) or in other notation x≠x 0 ;
the solution to the quadratic inequality a x 2 +b x+c≥0 is (−∞, +∞) or, in another notation, x∈R ;
quadratic inequality a x 2 +b x+c<0 не имеет решений (нет интервалов, на которых парабола расположена ниже оси Ox );
the quadratic inequality a x 2 +b x+c≤0 has a unique solution x=x 0 (it is given by the tangent point),

where x 0 is the root of the square trinomial a x 2 + b x + c.

In this case, the branches of the parabola are directed upwards, and it has no common points with the abscissa axis. Here we have the conditions a>0 (the branches are directed upwards) and D<0 (квадратный трехчлен не имеет действительных корней). Для примера можно построить график функции y=2·x 2 +1 , здесь a=2>0 , D=0 2 −4 2 1=−8<0 .

Obviously, the parabola is located above the Ox axis throughout its entire length (there are no intervals where it is below the Ox axis, there is no point of contact).

Thus, for a>0 and D<0 решением квадратных неравенств a·x 2 +b·x+c>0 and a x 2 +b x+c≥0 is the set of all real numbers, and the inequalities a x 2 +b x+c<0 и a·x 2 +b·x+c≤0 не имеют решений.

And there are three options for the location of the parabola with branches directed downwards, and not upwards, relative to the axis Ox. In principle, they may not be considered, since multiplying both parts of the inequality by −1 allows us to pass to an equivalent inequality with a positive coefficient at x 2 . However, it does not hurt to get an idea about these cases. The reasoning here is similar, so we write down only the main results.

Solution algorithm

The result of all previous calculations is algorithm for solving square inequalities graphically:

A schematic drawing is performed on the coordinate plane, which depicts the Ox axis (it is not necessary to depict the Oy axis) and a sketch of a parabola corresponding to a quadratic function y=a x 2 + b x + c. To construct a sketch of a parabola, it is enough to find out two points:

First, by the value of the coefficient a, it is found out where its branches are directed (for a>0 - upwards, for a<0 – вниз).
And secondly, by the value of the discriminant of the square trinomial a x 2 + b x + c, it turns out whether the parabola intersects the x-axis at two points (for D> 0), touches it at one point (for D=0), or has no common points with the Ox axis (for D<0 ). Для удобства на чертеже указываются координаты точек пересечения или координата точки касания (при наличии этих точек), а сами точки изображаются выколотыми при решении строгих неравенств, или обычными при решении нестрогих неравенств.

When the drawing is ready, on it at the second step of the algorithm
- when solving the quadratic inequality a·x 2 +b·x+c>0, the intervals at which the parabola is located above the abscissa axis are determined;
- when solving the inequality a x 2 +b x+c≥0, the intervals are determined at which the parabola is located above the abscissa axis and the abscissas of the intersection points (or the abscissa of the tangent point) are added to them;
- when solving the inequality a x 2 +b x+c<0 находятся промежутки, на которых парабола ниже оси Ox ;
- finally, when solving a quadratic inequality of the form a x 2 +b x + c≤0, there are intervals where the parabola is below the Ox axis and the abscissas of the intersection points (or the abscissa of the tangency point) are added to them;
they constitute the desired solution of the quadratic inequality, and if there are no such intervals and no points of contact, then the original quadratic inequality has no solutions.

It remains only to solve a few quadratic inequalities using this algorithm.

Examples with Solutions

Example.

Solve the inequality .

Decision.

We need to solve a quadratic inequality, we will use the algorithm from the previous paragraph. In the first step, we need to draw a sketch of the graph of the quadratic function . The coefficient at x 2 is 2, it is positive, therefore, the branches of the parabola are directed upwards. Let us also find out whether the parabola with the abscissa axis has common points, for this we calculate the discriminant of the square trinomial . We have . The discriminant turned out to be greater than zero, therefore, the trinomial has two real roots: and , that is, x 1 =−3 and x 2 =1/3.

From this it is clear that the parabola intersects the axis Ox at two points with abscissas −3 and 1/3. We will depict these points in the drawing as ordinary points, since we are solving a non-strict inequality. According to the clarified data, we obtain the following drawing (it fits the first template from the first paragraph of the article):

We pass to the second step of the algorithm. Since we are solving a non-strict quadratic inequality with the ≤ sign, we need to determine the intervals at which the parabola is located below the abscissa axis and add the abscissas of the intersection points to them.

It can be seen from the drawing that the parabola is below the abscissa in the interval (−3, 1/3) and we add the abscissas of the intersection points to it, that is, the numbers −3 and 1/3. As a result, we arrive at the numerical segment [−3, 1/3] . This is the desired solution. It can be written as a double inequality −3≤x≤1/3 .

Answer:

[−3, 1/3] or −3≤x≤1/3 .

Example.

Find a solution to the quadratic inequality −x 2 +16 x−63<0 .

Decision.

As usual, we start with a drawing. The numerical coefficient for the square of the variable is negative, −1, therefore, the branches of the parabola are directed downwards. Let's calculate the discriminant, or better, its fourth part: D"=8 2 −(−1)(−63)=64−63=1. Its value is positive, we calculate the roots of the square trinomial: and , x 1 =7 and x 2 =9. So the parabola intersects the Ox axis at two points with abscissas 7 and 9 (the initial inequality is strict, so we will depict these points with an empty center). Now we can make a schematic drawing:

Since we are solving a strict signed quadratic inequality<, то нас интересуют промежутки, на которых парабола расположена ниже оси абсцисс:

The drawing shows that the solutions to the original quadratic inequality are two intervals (−∞, 7) , (9, +∞) .

Answer:

(−∞, 7)∪(9, +∞) or in another notation x<7 , x>9 .

When solving square inequalities, when the discriminant of a square trinomial on its left side is equal to zero, you need to be careful with the inclusion or exclusion of the abscissa of the tangent point from the answer. It depends on the sign of the inequality: if the inequality is strict, then it is not a solution to the inequality, and if it is non-strict, then it is.

Example.

Does the quadratic inequality 10 x 2 −14 x+4.9≤0 have at least one solution?

Decision.

Let's plot the function y=10 x 2 −14 x+4.9 . Its branches are directed upwards, since the coefficient at x 2 is positive, and it touches the abscissa at the point with the abscissa 0.7, since D "=(−7) 2 −10 4.9=0, whence or 0.7 as a decimal.Schematically, it looks like this:

Since we are solving a quadratic inequality with the sign ≤, then its solution will be the intervals on which the parabola is below the Ox axis, as well as the abscissa of the tangent point. It can be seen from the drawing that there is not a single gap where the parabola would be below the axis Ox, therefore, its solution will be only the abscissa of the point of contact, that is, 0.7.

Answer:

this inequality has a unique solution 0.7 .

Example.

Solve the quadratic inequality –x 2 +8 x−16<0 .

Decision.

We act according to the algorithm for solving quadratic inequalities and start by plotting. The branches of the parabola are directed downwards, since the coefficient at x 2 is negative, −1. Find the discriminant of the square trinomial –x 2 +8 x−16 , we have D'=4 2 −(−1)(−16)=16−16=0 and further x 0 =−4/(−1) , x 0 =4 . So, the parabola touches the Ox axis at the point with the abscissa 4 . Let's make a drawing:

We look at the sign of the original inequality, it is<. Согласно алгоритму, решение неравенства в этом случае составляют все промежутки, на которых парабола расположена строго ниже оси абсцисс.

In our case, these are open rays (−∞, 4) , (4, +∞) . Separately, we note that 4 - the abscissa of the tangent point - is not a solution, since at the tangent point the parabola is not lower than the Ox axis.

Answer:

(−∞, 4)∪(4, +∞) or in other notation x≠4 .

Pay special attention to cases where the discriminant of the square trinomial on the left side of the square inequality is less than zero. There is no need to rush here and say that the inequality has no solutions (we are used to making such a conclusion for quadratic equations with a negative discriminant). The point is that the quadratic inequality for D<0 может иметь решение, которым является множество всех действительных чисел.

Example.

Find the solution to the quadratic inequality 3 x 2 +1>0 .

Decision.

As usual, we start with a drawing. The coefficient a is 3, it is positive, therefore, the branches of the parabola are directed upwards. Calculate the discriminant: D=0 2 −4 3 1=−12 . Since the discriminant is negative, the parabola has no common points with the x-axis. The information obtained is sufficient for a schematic diagram:

We are solving a strict quadratic inequality with > sign. Its solution will be all the intervals where the parabola is above the Ox axis. In our case, the parabola is above the x-axis along its entire length, so the desired solution will be the set of all real numbers.

Ox , and also you need to add the abscissa of the intersection points or the abscissa of the touch point to them. But the drawing clearly shows that there are no such gaps (since the parabola is everywhere below the abscissa axis), as well as there are no intersection points, just as there are no points of contact. Therefore, the original quadratic inequality has no solutions.

Answer:

there are no solutions or in another notation ∅.

Bibliography.

Algebra: textbook for 8 cells. general education institutions / [Yu. N. Makarychev, N. G. Mindyuk, K. I. Neshkov, S. B. Suvorova]; ed. S. A. Telyakovsky. - 16th ed. - M. : Education, 2008. - 271 p. : ill. - ISBN 978-5-09-019243-9.
Algebra: Grade 9: textbook. for general education institutions / [Yu. N. Makarychev, N. G. Mindyuk, K. I. Neshkov, S. B. Suvorova]; ed. S. A. Telyakovsky. - 16th ed. - M. : Education, 2009. - 271 p. : ill. - ISBN 978-5-09-021134-5.
Mordkovich A. G. Algebra. 8th grade. At 2 pm Part 1. A textbook for students of educational institutions / A. G. Mordkovich. - 11th ed., erased. - M.: Mnemozina, 2009. - 215 p.: ill. ISBN 978-5-346-01155-2.
Mordkovich A. G. Algebra. Grade 9 At 2 pm Part 1. Textbook for students of educational institutions / A. G. Mordkovich, P. V. Semenov. - 13th ed., Sr. - M.: Mnemosyne, 2011. - 222 p.: ill. ISBN 978-5-346-01752-3.
Mordkovich A. G. Algebra and beginning of mathematical analysis. Grade 11. At 2 pm Part 1. Textbook for students of educational institutions (profile level) / A. G. Mordkovich, P. V. Semenov. - 2nd ed., erased. - M.: Mnemosyne, 2008. - 287 p.: ill. ISBN 978-5-346-01027-2.

Lesson type:

Type of lesson: Lecture, problem solving lesson.

Duration: 2 hours.

Goals:1) Learn the graphic method.

2) Show the use of the Maple program in solving systems of inequalities using a graphical method.

3) Develop perception and thinking on the topic.

Lesson plan:

Course progress.

Stage 1: The graphical method consists in constructing a set of feasible LLP solutions, and finding a point in this set corresponding to the max / min of the objective function.

Due to the limited possibilities of a visual graphical representation, this method is used only for systems of linear inequalities with two unknowns and systems that can be reduced to this form.

In order to visually demonstrate the graphical method, we will solve the following problem:

Since both the graphs and the area of admissible solutions are in the first quarter. In order to find the boundary points, we solve equations (1)=(2), (1)=(3) and (2)=(3).

As can be seen from the illustration, the polyhedron ABCDE forms an area of feasible solutions.

If the domain of admissible solutions is not closed, then either max(f)=+ ? or min(f)= -?.

2. Now we can proceed to directly finding the maximum of the function f.

Alternately substituting the coordinates of the vertices of the polyhedron into the function f and comparing the values, we find that f(C)=f(4;1)=19 is the maximum of the function.

This approach is quite beneficial for a small number of vertices. But this procedure can be delayed if there are quite a lot of vertices.

In this case, it is more convenient to consider a level line of the form f=a. With a monotonous increase in the number a from -? to +? lines f=a are displaced along the normal vector The normal vector has coordinates (С1;С2), where C1 and C2 are the coefficients of the unknowns in the objective function f=C1?X1+C2?X2+C0.. If there is some point during such a displacement of the level line X is the first common point of the area of feasible solutions (polytope ABCDE) and the level line, then f(X) is the minimum of f on the set ABCDE. If X is the last point of intersection of the level line and the set ABCDE, then f(X) is the maximum on the set of feasible solutions. If for a>-? the line f=a intersects the set of admissible solutions, then min(f)= -?. If this happens when a>+?, then max(f)=+?.

In our example, the line f=a crosses the area ABCDE at the point С(4;1). Since this is the last point of intersection, max(f)=f(C)=f(4;1)=19.

Solve graphically the system of inequalities. Find corner solutions.

x1>=0, x2>=0

>with(plots);

>with(plottools);

> S1:=solve((f1x = X6, f2x = X6), );

Answer: All points Si where i=1..10 for which x and y are positive.

Area bounded by these points: (54/11.2/11) (5/7.60/7) (0.5) (10/3, 10/3)

Stage 3. Each student is given one of 20 options, in which the student is asked to independently solve the inequality using a graphical method, and the rest of the examples as homework.

Lesson №4 Graphical solution of a linear programming problem

Lesson type: lesson learning new material.

Type of lesson: Lecture + problem solving lesson.

Duration: 2 hours.

Goals: 1) Study the graphical solution of the linear programming problem.

2) Learn to use the Maple program when solving a linear programming problem.

2) Develop perception, thinking.

Lesson plan: Stage 1: learning new material.

Stage 2: Development of new material in the Maple mathematical package.

Stage 3: checking the studied material and homework.

Course progress.

The graphical method is quite simple and clear for solving linear programming problems with two variables. It is based on geometric representation of admissible solutions and digital filter of the problem.

Each of the inequalities of the linear programming problem (1.2) defines a certain half-plane on the coordinate plane (Fig. 2.1), and the system of inequalities as a whole defines the intersection of the corresponding planes. The set of intersection points of these half-planes is called domain of feasible solutions(ODR). ODR is always convex figure, i.e. which has the following property: if two points A and B belong to this figure, then the entire segment AB belongs to it. ODR can be graphically represented by a convex polygon, an unlimited convex polygonal area, a segment, a ray, a single point. If the system of constraints of problem (1.2) is inconsistent, then the ODE is an empty set.

All of the above also applies to the case when the system of constraints (1.2) includes equalities, since any equality

can be represented as a system of two inequalities (see Fig. 2.1)

The digital filter at a fixed value defines a straight line on the plane. By changing the values of L, we get a family of parallel lines, called level lines.

This is due to the fact that a change in the value of L will only change the length of the segment cut off by the level line on the axis (initial ordinate), and the slope of the straight line will remain constant (see Fig. 2.1). Therefore, for the solution, it will be enough to construct one of the level lines, arbitrarily choosing the value of L.

The vector with coordinates from the CF coefficients at and is perpendicular to each of the level lines (see Fig. 2.1). The direction of the vector is the same as the direction increasing CF, which is an important point for solving problems. Direction descending The digital filter is opposite to the direction of the vector.

The essence of the graphical method is as follows. In the direction (against the direction) of the vector in the ODR, the search for the optimal point is performed. The optimal point is the point through which the level line passes, corresponding to the largest (smallest) value of the function. The optimal solution is always located on the ODT boundary, for example, at the last vertex of the ODT polygon through which the target line passes, or on its entire side.

When searching for the optimal solution to linear programming problems, the following situations are possible: there is a unique solution to the problem; there is an infinite number of solutions (alternative optium); CF is not limited; the area of feasible solutions is a single point; the problem has no solutions.

Figure 2.1 Geometric interpretation of the constraints and the CF of the problem.

Methodology for solving LP problems by a graphical method

I. In the constraints of problem (1.2), replace the signs of inequalities with signs of exact equalities and construct the corresponding straight lines.

II. Find and shade the half-planes allowed by each of the inequality constraints of problem (1.2). To do this, you need to substitute the coordinates of some point [for example, (0; 0)] into a specific inequality and check the truth of the resulting inequality.

If a true inequality,

then it is necessary to shade the half-plane containing the given point;

otherwise(the inequality is false) it is necessary to shade the half-plane that does not contain the given point.

Since and must be non-negative, their valid values will always be above the axis and to the right of the axis, i.e. in the I quadrant.

Equality constraints allow only those points that lie on the corresponding line. Therefore, it is necessary to highlight such lines on the graph.

III. Define the ODR as a part of the plane that simultaneously belongs to all allowed areas, and select it. In the absence of an SDE, the problem has no solutions.

IV. If the ODS is not an empty set, then it is necessary to construct the target line, i.e. any of the level lines (where L is an arbitrary number, for example, a multiple of and, i.e. convenient for calculations). The method of construction is similar to the construction of direct constraints.

V. Construct a vector that starts at the point (0;0) and ends at the point. If the target line and vector are built correctly, then they will perpendicular.

VI. When searching for the maximum of the digital filter, it is necessary to move the target line in the direction vector, when searching for the minimum of the digital filter - against direction vector. The last top of the ODR in the direction of movement will be the maximum or minimum point of the CF. If there is no such point(s), then we can conclude that unboundedness of the digital filter on the set of plans from above (when searching for a maximum) or from below (when searching for a minimum).

VII. Determine the coordinates of the point max (min) of the digital filter and calculate the value of the digital filter. To calculate the coordinates of the optimal point, it is necessary to solve the system of equations of straight lines at the intersection of which it is located.

Solve a linear programming problem

1. f(x)=2x1+x2 ->extr

x1>=0, x2>=0

>plots((a+b<=3,a+3*b<=5,5*a-b<=5,a+b>=0,a>=0,b>=0), a=-2..5, b=-2..5, optionsfeasible=(color=red),

optionsopen=(color=blue, thickness=2),

optionsclosed=(color=green, thickness=3),

optionsexcluded=(color=yellow));

> with(simplex):

> C:=(x+y<=3, x+3*y <=5, 5*x-y <=5,x+y >=0};

> dp:=setup((x+y<=3, x+3*y <=5, 5*x-y <=5,x+y >=0});

>n:=basis(dp);

W display(C,);

> L:=cterm(C);

W X:=dual(f,C,p);

W f_max:=subs(R,f);

W R1:=minimize(f,C ,NONNEGATIVE);

f_min:=subs(R1,f);

ANSWER: When x 1 =5/4 x 2 =5/4 f_max=15/4; At x 1 =0 x 2 =0 f_min=0;

Lesson #5

Lesson type: lesson control + lesson learning new material. Type of lesson: Lecture.

Duration: 2 hours.

Goals:1) Check and consolidate knowledge on the past material in previous lessons.

2) Learn a new method for solving matrix games.

3) develop memory, mathematical thinking and attention.

Stage 1: check homework in the form of independent work.

Stage 2: give a brief description of the zigzag method

Stage 3: consolidate new material and give homework.

Course progress.

Linear programming methods - numerical methods for solving optimization problems that are reduced to formal models of linear programming.

As is known, any linear programming problem can be reduced to a canonical model for minimizing a linear objective function with linear equality-type constraints. Since the number of variables in a linear programming problem is greater than the number of constraints (n > m), a solution can be obtained by equating (n - m) variables to zero, called free. The remaining m variables, called basic, can be easily determined from the system of equality constraints by the usual methods of linear algebra. If a solution exists, then it is called basic. If the basic solution is admissible, then it is called basic admissible. Geometrically, basic feasible solutions correspond to the vertices (extreme points) of a convex polyhedron, which limits the set of feasible solutions. If a linear programming problem has optimal solutions, then at least one of them is basic.

The above considerations mean that when searching for an optimal solution to a linear programming problem, it suffices to confine ourselves to enumeration of basic admissible solutions. The number of basic solutions is equal to the number of combinations of n variables in m:

C = m n! /nm! * (n - m)!

and can be large enough to enumerate them by direct enumeration in real time. The fact that not all basic solutions are admissible does not change the essence of the problem, since in order to evaluate the admissibility of a basic solution, it must be obtained.

The problem of rational enumeration of basic solutions of a linear programming problem was first solved by J. Dantzig. The simplex method proposed by him is by far the most common general linear programming method. The simplex method implements a directed enumeration of feasible basic solutions along the corresponding extreme points of the convex polyhedron of feasible solutions as an iterative process, where the values of the objective function strictly decrease at each step. The transition between the extreme points is carried out along the edges of the convex polyhedron of feasible solutions in accordance with simple linear-algebraic transformations of the system of constraints. Since the number of extreme points is finite, and the objective function is linear, then by sorting through the extreme points in the direction of decreasing objective function, the simplex method converges to the global minimum in a finite number of steps.

Practice has shown that for most applied problems of linear programming, the simplex method allows finding the optimal solution in a relatively small number of steps compared to the total number of extreme points of an admissible polyhedron. At the same time, it is known that for some linear programming problems with a specially selected form of the admissible region, the use of the simplex method leads to a complete enumeration of the extreme points. This fact to a certain extent stimulated the search for new efficient methods for solving a linear programming problem, based on ideas other than the simplex method, which allow solving any linear programming problem in a finite number of steps, significantly less than the number of extreme points.

Among the polynomial linear programming methods that are invariant to the configuration of the range of permissible values, the most common is the method of L.G. Khachiyan. However, although this method has a polynomial complexity estimate depending on the dimension of the problem, it nevertheless turns out to be non-competitive in comparison with the simplex method. The reason for this is that the dependence of the number of iterations of the simplex method on the dimension of the problem is expressed by a 3rd order polynomial for most practical problems, while in the Khachiyan method, this dependence always has an order of at least 4th. This fact is of decisive importance for practice, where applied problems complex for the simplex method are extremely rare.

It should also be noted that for applied problems of linear programming that are important in a practical sense, special methods have been developed that take into account the specific nature of the constraints of the problem. In particular, for a homogeneous transport problem, special algorithms for choosing the initial basis are used, the most famous of which are the northwest corner method and the approximate Vogel method, and the algorithmic implementation of the simplex method itself is close to the specifics of the problem. To solve the linear assignment problem (choice problem), instead of the simplex method, either the Hungarian algorithm is usually used, based on the interpretation of the problem in terms of graph theory as the problem of finding the maximum weighted perfect matching in a bipartite graph, or the Mack method.

Solve a 3x3 matrix game

f(x)=x 1 +x 2 +x 3

x1>=0, x2>=0, x3>=0

> with(simplex):

> C:=( 0*x+3*y+2*z<=1, 2*x+0*y+1*z <=1, 3*x+0*y+0*z <=1};

W display(C,);

> feasible(C, NONNEGATIVE , "NewC", "Transform");

> S:=dual(f,C,p);

W R:=maximize(f,C ,NONNEGATIVE);

W f_max:=subs(R,f);

W R1:=minimize(S ,NONNEGATIVE);

>G:=p1+p2+p3;

> f_min:=subs(R1,G);

Find the price of the game

> V:=1/f_max;

Finding the optimal strategy for the first player >X:=V*R1;

Finding the optimal strategy for the second player

ANSWER: When X=(3/7, 3/7,1/7) V=9/7; With Y=(3/7.1/7.3/7) V=9/7;

Each student is given one of 20 options, in which the student is asked to independently solve the 2x2 matrix game, and the rest of the examples as homework.