Lesson type:

Lesson type: 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 the graphical method.

3) Develop perception and thinking on this topic.

Lesson plan:

Progress of the lesson.

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

Due to disabilities clear graphical representation this method applies only to systems linear inequalities with two unknowns and systems that can be reduced to a given form.

In order to clearly demonstrate the graphical method, let’s solve the following problem:

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

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

As can be seen from the illustration, the polyhedron ABCDE forms a region of feasible solutions.

If the region of feasible 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.

By 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 with a small number of vertices. But this procedure may take a long time 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 monotonic increase in the number a from -? to +? straight lines f=a are shifted along the normal vector. The normal vector has coordinates (C1;C2), where C1 and C2 are coefficients for the unknowns in the objective function f=C1?X1+C2?X2+C0.. If with such a movement of the level line there is a certain point X is the first common point of the domain 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 ABCDE set, then f(X) is the maximum on the set of feasible solutions. If for a>-? the straight line f=a intersects the set of feasible solutions, then min(f)= -?. If this happens for a>+?, then max(f)=+?.

In our example, the straight line f=a intersects the region ABCDE at point C(4;1). Since this is the last intersection point, max(f)=f(C)=f(4;1)=19.

Solve the system of inequalities graphically. 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 limited 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 remaining examples are given as homework.

Lesson No. 4 Graphical solution of a linear programming problem

Lesson type: lesson of learning new material.

Lesson type: 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 and thinking.

Lesson plan: Stage 1: learning new material.

Stage 2: Working on new material in the Maple mathematical package.

Stage 3: checking the studied material and homework.

Progress of the lesson.

The graphical method is quite simple and intuitive for solving linear programming problems with two variables. It is based on geometric presentation of feasible solutions and TFs 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 area of ​​feasible solutions(ODR). ODR always represents convex figure, i.e. having 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, or one point. If the system of constraints in problem (1.2) is inconsistent, the ODS is an empty set.

All of the above also applies to the case when the system of restrictions (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 changing the value of L will entail a change only in the length of the segment cut off by the level line on the axis (initial ordinate), and the angular coefficient of the straight line will remain constant (see Fig. 2.1). Therefore, to solve it, 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 coincides with the direction increasing TF, which is an important point for solving problems. Direction descending The CF 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 optimal point is searched. 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 boundary of the ODD, for example, at the last vertex of the ODD polygon through which the target line will pass, or on its entire side.

When searching for an optimal solution to linear programming problems, it is possible following situations: exists the only solution tasks; there is an infinite number of solutions (alternative); TF is not limited; the region of feasible solutions is a single point; the problem has no solutions.

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

Technique for solving LP problems using the graphical method

I. In the constraints of problem (1.2), replace the inequality signs with exact equality signs 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 a point [for example, (0;0)] into a specific inequality and check the truth of the resulting inequality.

If the inequality is true,

That it is necessary to shade the half-plane containing this point;

otherwise(the inequality is false) we must shade the half-plane that does not contain the given point.

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

Equality constraints allow only those points that lie on the corresponding line. Therefore, it is necessary to highlight such straight 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 ODD, the problem has no solutions.

IV. If the ODR is not an empty set, then you need to construct the target line, i.e. any of the level lines (where L is an arbitrary number, for example, a multiple and, i.e., convenient for calculations). The construction method 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 constructed correctly, then they will be perpendicular.

VI. When searching for the maximum CF, you need to move the target line in the direction vector, when searching for the minimum CF - against the direction vector. The last top of the ODR in the direction of movement will be the point of maximum or minimum of the CF. If such a point(s) does not exist, then we can conclude that unlimited TF on many 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 a system of equations of the 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);

Sh display(C,);

> L:=cterm(C);

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

Sh f_max:=subs(R,f);

Sh 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 No. 5. Solving matrix games using linear programming methods and the simplex method

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

Duration: 2 hours.

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

2) Learn a new method for solving matrix games.

3) develop memory, mathematical thinking and attention.

Stage 1: check your homework as independent work.

Stage 2: give a brief description of the zigzag method

Stage 3: consolidate new material and assign homework.

Progress of the lesson.

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

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), it is possible to obtain a solution by setting (n - m) variables, called free. The remaining m variables, called basic, can be easily determined from a system of equality constraints using 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 permissible. Geometrically, the basic feasible solutions correspond to the vertices (extreme points) of a convex polyhedron, which bounds 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 is sufficient to limit ourselves to enumerating basic feasible solutions. The number of basic solutions is equal to the number of combinations of n variables in m:

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

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

The problem of rational enumeration of basis solutions to a linear programming problem was first solved by J. Danzig. The simplex method he proposed is still the most widespread general method linear programming. The simplex method implements a directed search of admissible basic solutions along the corresponding extreme points of a convex polyhedron of admissible solutions in the form of an iterative process, where at each step the values ​​of the objective function strictly decrease. The transition between extreme points is carried out along the edges of a convex polyhedron of admissible solutions in accordance with simple linear algebraic transformations of the system of restrictions. Since the number of extreme points is finite, and the objective function is linear, then by searching through the extreme points in the direction of decreasing the objective function, the simplex method converges to a global minimum in a finite number of steps.

Practice has shown that for most applied linear programming problems, the simplex method allows one to find the optimal solution in a relatively small number of steps compared to the total number of extreme points of the 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 effective methods solutions to linear programming problems, 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 polynomial linear programming methods invariant to the domain configuration acceptable values, the most common is the L.G. method. 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 compared to 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 Khachiyan’s method, this dependence always has an order of at least four. This fact has crucial for practice, where applied problems difficult for the simplex method are extremely rare.

It should also be noted that for applied linear programming problems that are important in a practical sense, special methods, taking into account the specific nature of the constraints of the problem. In particular, for a homogeneous transport problem, special algorithms for selecting 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 (selection 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 weight perfect matching in a bipartite graph, or Mack's 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};

Sh display(C,);

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

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

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

Sh f_max:=subs(R,f);

Sh R1:=minimize(S ,NONNEGATIVE);

> G:=p1+p2+p3;

> f_min:=subs(R1,G);

Let's find the price of the game

> V:=1/f_max;

Let's find the optimal strategy of the first player > X:=V*R1;

Let's find the optimal strategy of the second player

ANSWER: When X=(3/7, 3/7,1/7) V=9/7; When 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 a 2x2 matrix game, and the remaining examples as homework.

Entry level

Solving equations, inequalities, systems using function graphs. Visual guide (2019)

Many tasks that we are used to calculating purely algebraically can be solved much easier and faster; using function graphs will help us with this. You say “how so?” draw something, and what to draw? Believe me, sometimes it is more convenient and easier. Shall we get started? Let's start with the equations!

Graphical solution of equations

Graphical solution of linear equations

As you already know, the graph of a linear equation is a straight line, hence the name of this type. Linear equations are quite easy to solve algebraically - we transfer all the unknowns to one side of the equation, everything we know to the other and voila! We found the root. Now I'll show you how to do it graphically.

So you have the equation:

How to solve it?
Option 1, and the most common one is to move the unknowns to one side and the knowns to the other, we get:

Now let's build. What did you get?

What do you think is the root of our equation? That's right, the coordinate of the intersection point of the graphs is:

Our answer is

That's all the wisdom graphic solution. As you can easily check, the root of our equation is a number!

As I said above, this is the most common option, close to an algebraic solution, but you can solve it in another way. To consider an alternative solution, let's return to our equation:

This time we will not move anything from side to side, but will construct the graphs directly, as they are now:

Built? Let's see!

What is the solution this time? That's right. The same thing - the coordinate of the intersection point of the graphs:

And, again, our answer is.

As you can see, with linear equations everything is extremely simple. It's time to look at something more complex... For example, graphical solution of quadratic equations.

Graphical solution of quadratic equations

So, now let's start solving the quadratic equation. Let's say you need to find the roots of this equation:

Of course, you can now start counting through the discriminant, or according to Vieta’s theorem, but many people make mistakes when multiplying or squaring, especially if the example is with large numbers, and, as you know, you won’t have a calculator for the exam... Therefore, let’s try to relax a little and draw while solving this equation.

You can find solutions to this equation graphically in various ways. Let's consider various options, and you can choose which one you like best.

Method 1. Directly

We simply build a parabola using this equation:

To do this quickly, I'll give you one little hint: It is convenient to start the construction by determining the vertex of the parabola. The following formulas will help determine the coordinates of the vertex of a parabola:

You will say “Stop! The formula for is very similar to the formula for finding the discriminant,” yes, it is, and this is a huge disadvantage of “directly” constructing a parabola to find its roots. However, let's count to the end, and then I'll show you how to do it much (much!) easier!

Did you count? What coordinates did you get for the vertex of the parabola? Let's figure it out together:

Exactly the same answer? Well done! And now we already know the coordinates of the vertex, but to construct a parabola we need more... points. How many minimum points do you think we need? Right, .

You know that a parabola is symmetrical about its vertex, for example:

Accordingly, we need two more points on the left or right branch parabolas, and in the future we will symmetrically reflect these points on the opposite side:

Let's return to our parabola. For our case, period. We need two more points, so we can take positive ones, or we can take negative ones? Which points are most convenient for you? It’s more convenient for me to work with positive ones, so I’ll calculate at and.

Now we have three points, we can easily construct our parabola by reflecting the last two points relative to its vertex:

What do you think is the solution to the equation? That's right, points at which, that is, and. Because.

And if we say that, it means that it must also be equal, or.

Just? We have finished solving the equation with you in a complex graphical way, or there will be more!

Of course, you can check our answer algebraically - you can calculate the roots using Vieta's theorem or Discriminant. What did you get? The same? You see! Now let's look at a very simple graphic solution, I'm sure you'll really like it!

Method 2. Divided into several functions

Let’s take our same equation: , but we’ll write it a little differently, namely:

Can we write it like this? We can, since the transformation is equivalent. Let's look further.

Let's construct two functions separately:

1. - the graph is a simple parabola, which you can easily construct even without defining the vertex using formulas and drawing up a table to determine other points.
2. - the graph is a straight line, which you can just as easily construct by estimating the values ​​in your head without even resorting to a calculator.

Built? Let's compare with what I got:

Do you think that in in this case are the roots of the equation? Right! The coordinates obtained by the intersection of two graphs and, that is:

Accordingly, the solution to this equation is:

What do you say? Agree, this method of solution is much easier than the previous one and even easier than looking for roots through a discriminant! If so, try solving the following equation using this method:

What did you get? Let's compare our graphs:

The graphs show that the answers are:

Did you manage? Well done! Now let's look at the equations a little more complicated, namely, solving mixed equations, that is, equations containing functions of different types.

Graphical solution of mixed equations

Now let's try to solve the following:

Of course, we can bring everything to common denominator, find the roots of the resulting equation, not forgetting to take into account the ODZ, but again, we will try to solve it graphically, as we did in all previous cases.

This time let's build the following 2 graphs:

1. - the graph is a hyperbola
2. - the graph is a straight line, which you can easily construct by estimating the values ​​in your head without even resorting to a calculator.

Realized it? Now start building.

Here's what I got:

Looking at this picture, tell me what are the roots of our equation?

That's right, and. Here's the confirmation:

Try plugging our roots into the equation. Did it work?

That's right! Agree, solving such equations graphically is a pleasure!

Try to solve the equation graphically yourself:

I'll give you a hint: move part of the equation to right side, so that on both sides there are the simplest functions to construct. Did you get the hint? Take action!

Now let's see what you got:

Respectively:

1. - cubic parabola.
2. - ordinary straight line.

Well, let's build:

As you wrote down long ago, the root of this equation is - .

Having decided this large number examples, I'm sure you realized how easily and quickly you can solve equations graphically. It's time to figure out how to solve systems in this way.

Graphic solution of systems

Graphically solving systems is essentially no different from graphically solving equations. We will also build two graphs, and their intersection points will be the roots of this system. One graph is one equation, the second graph is another equation. Everything is extremely simple!

Let's start with the simplest thing - solving systems of linear equations.

Solving systems of linear equations

Let's say we have the following system:

First, let's transform it so that on the left there is everything that is connected with, and on the right - everything that is connected with. In other words, let’s write these equations as a function in our usual form:

Now we just build two straight lines. What is the solution in our case? Right! The point of their intersection! And here you need to be very, very careful! Think about it, why? Let me give you a hint: we are dealing with a system: in the system there is both, and... Got the hint?

That's right! When solving a system, we must look at both coordinates, and not just as when solving equations! Another important point- write them down correctly and not confuse where we have the meaning and where the meaning is! Did you write it down? Now let's compare everything in order:

And the answers: and. Do a check - substitute the found roots into the system and make sure whether we solved it correctly graphically?

Solving systems of nonlinear equations

What if, instead of one straight line, we have quadratic equation? It's okay! You just build a parabola instead of a straight line! Don't believe me? Try solving the following system:

What's our next step? That’s right, write it down so that it’s convenient for us to build graphs:

And now it’s all a matter of small things - build it quickly and here’s your solution! We are building:

Did the graphs turn out the same? Now mark the solutions of the system in the figure and correctly write down the identified answers!

Did you do everything? Compare with my notes:

Is everything right? Well done! You are already cracking these types of tasks like nuts! If so, let’s give you a more complicated system:

What are we doing? Right! We write the system so that it is convenient to build:

I’ll give you a little hint, since the system looks very complicated! When constructing graphs, build them “more”, and most importantly, do not be surprised by the number of intersection points.

So, let's go! Exhaled? Now start building!

So how? Beautiful? How many intersection points did you get? I have three! Let's compare our graphs:

Also? Now carefully write down all the solutions of our system:

Now look at the system again:

Can you imagine that you solved this in just 15 minutes? Agree, mathematics is still simple, especially when looking at an expression you are not afraid to make a mistake, but just take it and solve it! You're great!

Graphical solution of inequalities

Graphical solution of linear inequalities

After last example You can handle everything! Now exhale - compared to the previous sections, this one will be very, very easy!

We will start, as usual, with a graphical solution to a linear inequality. For example, this one:

First, let's carry out the simplest transformations - open the brackets of perfect squares and present similar terms:

The inequality is not strict, therefore it is not included in the interval, and the solution will be all points that are to the right, since more, more, and so on:

Answer:

That's it! Easily? Let's solve a simple inequality with two variables:

Let's draw a function in the coordinate system.

Did you get such a schedule? Now let’s look carefully at what inequality we have there? Less? This means we paint over everything that is to the left of our straight line. What if there were more? That's right, then we would paint over everything that is to the right of our straight line. It's simple.

All solutions to this inequality are “shaded out” orange. That's it, the inequality with two variables is solved. This means that the coordinates of any point from the shaded area are the solutions.

Graphical solution of quadratic inequalities

Now we will understand how to graphically solve quadratic inequalities.

But before we get down to business, let's review some material regarding the quadratic function.

What is the discriminant responsible for? That’s right, for the position of the graph relative to the axis (if you don’t remember this, then definitely read the theory about quadratic functions).

In any case, here's a little reminder for you:

Now that we have refreshed all the material in our memory, let's get down to business - solve the inequality graphically.

I’ll tell you right away that there are two options for solving it.

Option 1

We write our parabola as a function:

Using the formulas, we determine the coordinates of the vertex of the parabola (exactly the same as when solving quadratic equations):

Did you count? What did you get?

Now let's take two more different points and calculate for them:

Let's start building one branch of the parabola:

We symmetrically reflect our points onto another branch of the parabola:

Now let's return to our inequality.

We need it to be less than zero, respectively:

Since in our inequality the sign is strictly less than, we exclude the end points - “puncture out”.

Answer:

Long way, right? Now I will show you a simpler version of the graphical solution using the example of the same inequality:

Option 2

We return to our inequality and mark the intervals we need:

Agree, it's much faster.

Let us now write down the answer:

Let's consider another solution that simplifies the algebraic part, but the main thing is not to get confused.

Multiply the left and right sides by:

Try to solve the following quadratic inequality yourself in any way you like: .

Did you manage?

Look how my graph turned out:

Answer: .

Graphical solution of mixed inequalities

Now let's move on to more complex inequalities!

How do you like this:

It's creepy, isn't it? Honestly, I have no idea how to solve this algebraically... But it’s not necessary. Graphically there is nothing complicated about this! The eyes are afraid, but the hands are doing!

The first thing we will start with is by constructing two graphs:

I won’t write out a table for each one - I’m sure you can do it perfectly on your own (wow, there are so many examples to solve!).

Did you paint it? Now build two graphs.

Let's compare our drawings?

Is it the same with you? Great! Now let's arrange the intersection points and use color to determine which graph we should have larger in theory, that is. Look what happened in the end:

Now let’s just look at where our selected graph is higher than the graph? Feel free to take a pencil and paint over this area! She will be the solution to our complex inequality!

At what intervals along the axis is we located higher than? Right, . This is the answer!

Well, now you can handle any equation, any system, and even more so any inequality!

BRIEFLY ABOUT THE MAIN THINGS

Algorithm for solving equations using function graphs:

1. Let's express it through
2. Let's define the function type
3. Let's build graphs of the resulting functions
4. Let's find the intersection points of the graphs
5. Let’s write the answer correctly (taking into account the ODZ and inequality signs)
6. Let's check the answer (substitute the roots into the equation or system)

For more information about constructing function graphs, see the topic “”.

The graphical method is one of the main methods for solving quadratic inequalities. In the article we will present an algorithm for using the graphical method, and then consider special cases using examples.

The essence of the graphical method

The method is applicable to solving any inequalities, not only quadratic ones. Its essence is this: the right and left sides of the inequality are considered as two separate functions y = f (x) and y = g (x), their graphs are plotted in a rectangular coordinate system and look at which of the graphs is located above the other, and on which intervals. The intervals are estimated as follows:

Definition 1

• solutions to the inequality f (x) > g (x) are intervals where the graph of the function f is higher than the graph of the function g;
• solutions to the inequality f (x) ≥ g (x) are intervals where the graph of the function f is not lower than the graph of the function g;
• solutions to the inequality f(x)< g (x) являются интервалы, где график функции f ниже графика функции g ;
• solutions to the inequality f (x) ≤ g (x) are intervals where the graph of the function f is not higher than the graph of the function g;
• The abscissas of the intersection points of the graphs of the functions f and g are solutions to the equation f (x) = g (x).

Let's look at the above algorithm using an example. To do this, take the quadratic inequality a x 2 + b x + c< 0 (≤ , >, ≥) and derive two functions from it. Left side the inequality will correspond to y = a · x 2 + b · x + c (in this case f (x) = a · x 2 + b · x + c), and the right one is y = 0 (in this case g (x) = 0) .

The graph of the first function is a parabola, the second is a straight line, which coincides with the x-axis Ox. Let's analyze the position of the parabola relative to the O x axis. To do this, let's make a schematic drawing.

The branches of the parabola are directed upward. It intersects the O x axis at points x 1 And x 2. Coefficient a in this case is positive, since it is it that is responsible for the direction of the branches of the parabola. The discriminant is positive, indicating that the quadratic trinomial has two roots a x 2 + b x + c. We denote the roots of the trinomial as x 1 And x 2, and it was accepted that x 1< x 2 , since a point with an abscissa is depicted on the O x axis x 1 to the left of the abscissa point x 2.

The parts of the parabola located above the O x axis will be denoted in red, below - in blue. This will allow us to make the drawing more visual.

Let's select the spaces that correspond to these parts and mark them in the figure with fields of a certain color.

We marked in red the intervals (− ∞, x 1) and (x 2, + ∞), on them the parabola is above the O x axis. They are a · x 2 + b · x + c > 0. We marked in blue the interval (x 1 , x 2) , which is the solution to the inequality a · x 2 + b · x + c< 0 . Числа x 1 и x 2 будут отвечать равенству a · x 2 + b · x + c = 0 .

Let's make a brief summary of the solution. For a > 0 and D = b 2 − 4 a c > 0 (or D " = D 4 > 0 for an even coefficient b) we get:

• decision quadratic inequality a · x 2 + b · x + c > 0 is (− ∞ , x 1) ∪ (x 2 , + ∞) or in another notation x< x 1 , x >x 2 ;
• the solution to the quadratic inequality a · x 2 + b · x + c ≥ 0 is (− ∞ , x 1 ] ∪ [ x 2 , + ∞) or in another notation x ≤ x 1 , x ≥ x 2 ;
• solving the quadratic inequality a x 2 + b x + c< 0 является (x 1 , x 2) или в другой записи x 1 < x < x 2 ;
• the solution to the quadratic inequality a x 2 + b x + c ≤ 0 is [ x 1 , x 2 ] or in another notation x 1 ≤ x ≤ x 2 ,

where x 1 and x 2 are the roots of the quadratic trinomial a x 2 + b x + c, and x 1< x 2 .

In this figure, the parabola touches the O x axis only at one point, which is designated as x 0 a > 0. D=0, therefore, the square trinomial has one root x 0.

The parabola is located above the O x axis completely, with the exception of the point of tangency of the coordinate axis. Let's color the intervals (− ∞ , x 0) , (x 0 , ∞) .

Let's write down the results. At a > 0 And D=0:

• solving the quadratic inequality a x 2 + b x + c > 0 is (− ∞ , x 0) ∪ (x 0 , + ∞) or in another notation x ≠ x 0;
• solving 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 has no solutions (there are no intervals at which the parabola is located below the axis O x);
• quadratic inequality a x 2 + b x + c ≤ 0 has a unique solution x = x 0(it is given by the point of contact),

Where x 0- root of the square trinomial a x 2 + b x + c.

Let's consider the third case, when the branches of the parabola are directed upward and do not touch the axis O x. The branches of the parabola are directed upward, which means that a > 0. The square trinomial has no real roots because D< 0 .

There are no intervals on the graph at which the parabola would be below the x-axis. We will take this into account when choosing a color for our drawing.

It turns out that when a > 0 And D< 0 solving quadratic inequalities 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 And a x 2 + b x + c ≤ 0 have no solutions.

We have three options left to consider when the branches of the parabola are directed downward. There is no need to dwell on these three options in detail, since when we multiply both sides of the inequality by − 1, we obtain an equivalent inequality with a positive coefficient for x 2.

Consideration previous section article prepared us to perceive the algorithm for solving inequalities using graphic method. To carry out calculations, we will need to use a drawing each time, which will depict the coordinate line O x and the parabola that corresponds quadratic function y = a x 2 + b x + c. In most cases, we will not depict the O y axis, since it is not needed for calculations and will only overload the drawing.

To construct a parabola, we will need to know two things:

Definition 2

• the direction of the branches, which is determined by the value of the coefficient a;
• the presence of points of intersection of the parabola and the abscissa axis, which are determined by the value of the discriminant of the quadratic trinomial a · x 2 + b · x + c .

We will denote the points of intersection and tangency in the usual way when solving non-strict inequalities and empty when solving strict ones.

Having a completed drawing allows you to move on to the next step of the solution. It involves determining the intervals at which the parabola is located above or below the O x axis. The intervals and points of intersection are the solution to the quadratic inequality. If there are no points of intersection or tangency and there are no intervals, then it is considered that the inequality specified in the conditions of the problem has no solutions.

Now let's solve several quadratic inequalities using the above algorithm.

Example 1

It is necessary to solve the inequality 2 x 2 + 5 1 3 x - 2 graphically.

Solution

Let's draw a graph of the quadratic function y = 2 · x 2 + 5 1 3 · x - 2 . Coefficient at x 2 positive because it is equal 2 . This means that the branches of the parabola will be directed upward.

Let us calculate the discriminant of the quadratic trinomial 2 x 2 + 5 1 3 x - 2 in order to find out whether the parabola has common points with the abscissa axis. We get:

D = 5 1 3 2 - 4 2 (- 2) = 400 9

As we see, D is greater than zero, therefore, we have two intersection points: x 1 = - 5 1 3 - 400 9 2 2 and x 2 = - 5 1 3 + 400 9 2 2, that is, x 1 = − 3 And x 2 = 1 3.

We solve a non-strict inequality, therefore we put ordinary points on the graph. Let's draw a parabola. As you can see, the drawing has the same appearance as in the first template we considered.

Our inequality has the sign ≤. Therefore, we need to highlight the intervals on the graph where the parabola is located below the O x axis and add intersection points to them.

The interval we need is 3, 1 3. We add intersection points to it and get a numerical segment − 3, 1 3. This is the solution to our problem. You can write the answer in the form double inequality: − 3 ≤ x ≤ 1 3 .

Answer:− 3 , 1 3 or − 3 ≤ x ≤ 1 3 .

Example 2

− x 2 + 16 x − 63< 0 graphical method.

Solution

The square of the variable has a negative numerical coefficient, so the branches of the parabola will be directed downward. Let's calculate the fourth part of the discriminant D " = 8 2 − (− 1) · (− 63) = 64 − 63 = 1. This result tells us that there will be two points of intersection.

Let's calculate the roots of the quadratic trinomial: x 1 = - 8 + 1 - 1 and x 2 = - 8 - 1 - 1, x 1 = 7 and x 2 = 9.

It turns out that the parabola intersects the x-axis at the points 7 And 9 . Let's mark these points on the graph as empty, since we are working with strict inequality. After this, draw a parabola that intersects the O x axis at the marked points.

We will be interested in the intervals at which the parabola is located below the O x axis. Let's mark these intervals in blue.

We get the answer: the solution to the inequality is the intervals (− ∞, 7) , (9, + ∞) .

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

In cases where the discriminant of a quadratic trinomial is zero, it is necessary to carefully consider whether to include the abscissa of the tangent points in the answer. In order to accept the right decision, it is necessary to take into account the inequality sign. In strict inequalities, the point of tangency of the x-axis is not a solution to the inequality, but in non-strict ones it is.

Example 3

Solve quadratic inequality 10 x 2 − 14 x + 4, 9 ≤ 0 graphical method.

Solution

The branches of the parabola in this case will be directed upward. It will touch the O x axis at point 0, 7, since

Let's plot the function y = 10 x 2 − 14 x + 4, 9. Its branches are directed upward, since the coefficient at x 2 positive, and it touches the x-axis at the x-axis point 0 , 7 , because D " = (− 7) 2 − 10 4, 9 = 0, from where x 0 = 7 10 or 0 , 7 .

Let's put a point and draw a parabola.

We solve a non-strict inequality with a sign ≤. Hence. We will be interested in the intervals at which the parabola is located below the x-axis and the point of tangency. There are no intervals in the figure that would satisfy our conditions. There is only a point of contact 0, 7. This is the solution we are looking for.

Answer: The inequality has only one solution 0, 7.

Example 4

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

Solution

The branches of the parabola are directed downwards. The discriminant is zero. Intersection point x 0 = 4.

We mark the point of tangency on the x-axis and draw a parabola.

We are dealing with severe inequality. Consequently, we are interested in the intervals at which the parabola is located below the O x axis. Let's mark them in blue.

The point with abscissa 4 is not a solution, since the parabola at it is not located below the O x axis. Consequently, we get two intervals (− ∞ , 4) , (4 , + ∞) .

Answer: (− ∞ , 4) ∪ (4 , + ∞) or in another notation x ≠ 4 .

Not always with negative value discriminant inequality will have no solutions. There are cases when the solution is the set of all real numbers.

Example 5

Solve the quadratic inequality 3 x 2 + 1 > 0 graphically.

Solution

Coefficient a is positive. The discriminant is negative. The branches of the parabola will be directed upward. There are no points of intersection of the parabola with the O x axis. Let's look at the drawing.

We work with strict inequality, which has a > sign. This means that we are interested in the intervals at which the parabola is located above the x-axis. This is exactly the case when the answer is the set of all real numbers.

Answer:(− ∞, + ∞) or so x ∈ R.

Example 6

It is necessary to find a solution to the inequality − 2 x 2 − 7 x − 12 ≥ 0 graphically.

Solution

The branches of the parabola are directed downwards. The discriminant is negative, therefore, there are no common points between the parabola and the x-axis. Let's look at the drawing.

We are working with a non-strict inequality with the sign ≥, therefore, the intervals in which the parabola is located above the x-axis are of interest to us. Judging by the graph, there are no such gaps. This means that the inequality given in the problem conditions has no solutions.

Answer: No solutions.

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

Due to the limited possibilities of 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 clearly demonstrate the graphical method, let’s solve the following problem:

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

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

As can be seen from the illustration, the polyhedron ABCDE forms a region of feasible solutions.

If the region of feasible 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.

By 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 with a small number of vertices. But this procedure can take a long time 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 monotonic increase in the number a from -? to +? straight lines f=a are shifted along the normal vector. If, with such a movement of the level line, there is a certain point X - the first common point of the region 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 ABCDE set, then f(X) is the maximum on the set of feasible solutions. If for a>-? the straight line f=a intersects the set of feasible solutions, then min(f)= -?. If this happens for a>+?, then max(f)=+?.

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 needs to be maximized.

Let's answer the question: what pairs of numbers ( x; y) are solutions to the system of inequalities, that is, 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 to one linear inequality with two unknowns.
Solving a linear inequality with two unknowns means determining all pairs of unknown values ​​for which the inequality holds.
For example, inequality 3 x – 5y≥ 42 satisfy pairs ( x , y) : (100, 2); (3, –10), etc. The task is to find all such pairs.
Let's 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, let us take a point with coordinate x = x 0 ; then a point lying on a line and having an abscissa x 0, has an ordinate

Let for certainty a< 0, b>0, c>0. All points with abscissa x 0 lying above P(for example, dot M), have y M>y 0 , and all points below the point P, with abscissa x 0 , have y N<y 0 . Since 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 side - points for which ax + by< c.

Figure 1

The inequality sign in the half-plane depends on the numbers a, b , c.
It follows from this next way graphical solution of systems of linear inequalities in two variables. To solve the system you need:

1. For each inequality, write the equation corresponding to this inequality.
2. Construct straight lines that 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 an arbitrary point that does not lie on a line and 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 can be a finite number or an infinite number of solutions. The area can be a closed polygon or unbounded.

Let's look at three relevant examples.

Example 1. Solve the system graphically:
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's construct straight lines given by these equations.

Figure 2

Let us define the half-planes defined by the inequalities. Let's take an arbitrary point, let (0; 0). Let's 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, –2 x – 2y+ 5≥ 0, and we were asked where –2 x – 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 2. Find graphically solutions to the system of inequalities:

Figure 3
1. Let's write out 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 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), IN(0; 1), WITH(6; –2).

Let's consider another example in which the resulting solution domain of the system is not limited.