A point M is called internal to a certain set G if it belongs to this set together with some of its neighborhood. A point N is called a boundary point for a set G if in any complete neighborhood of it there are points both belonging to G and not belonging to it.

The set of all boundary points of a set G is called the boundary of G.

A set G will be called a region if all its points are internal (open set). A set G with an associated boundary Г is called a closed region. A region is called bounded if it is entirely contained within a circle of sufficiently large radius.

The smallest and largest values ​​of a function in a given area are called the absolute extrema of the function in this area.

Weierstrass's theorem: a function continuous in a bounded and closed area, reaches its minimum and maximum values ​​in this region.

Consequence. The absolute extremum of a function in a given region is achieved either at the critical point of the function belonging to this region, or at To find the largest and smallest values ​​of a function in a closed region G, it is necessary to find all its critical points in this region, calculate the values ​​of the function at these points (including boundary ones) and by comparing the numbers obtained, select the largest and smallest of them.

Example 4.1. Find the absolute extremum of the function (largest and smallest values)
in a triangular region D with vertices
,
,
(Fig. 1).

;
,

that is, point O(0, 0) is a critical point belonging to the region D. z(0,0)=0.

Let's explore the border:

a) OA: y=0
;z(x, 0)=0; z(0, 0)=0; z(1, 0)=0,

b) OB: x=0
z(0,y)=0;

z(0, 0)=0; z(0, 2)=0,
,

c) AB: ; Example 4.2.
.

Find the largest and smallest values ​​of a function in a closed area bounded by the coordinate axes and the straight line

,
,

.

Let's explore the border. Because boundary consists of a segment OA of the Ox axis, a segment OB of the Oy axis and a segment AB, then we determine the largest and smallest values ​​of the function z on each of these segments.

, z(0, 2)=–3, z(0, 0)=5, z(0, 4)=5.

M 3 (5/3,7/3), z(5/3, 7/3)=–10/3.

Among all the found values, select z max =z(4, 0)=13; z naim =z(1, 2)=–4.

5. Conditional extremum. Lagrange multiplier method

Let us consider a problem specific to functions of several variables, when its extremum is sought not over the entire domain of definition, but over a set that satisfies a certain condition.

Let us consider the function
, arguments And which satisfy the condition
, called the coupling equation.

Dot
is called a conditional maximum (minimum) point if there is such a neighborhood of this point that for all points
from this neighborhood satisfying the condition
, the inequality holds
or
.

Figure 2 shows the conditional maximum point
. Obviously, it is not the unconditional extremum point of the function
(in Fig. 2 this is the point
).

The simplest way to find the conditional extremum of a function of two variables is to reduce the problem to finding the extremum of a function of one variable. Let us assume the connection equation
managed to resolve with respect to one of the variables, for example, to express through :
. Substituting the resulting expression into a function of two variables, we get

those. function of one variable. Its extremum will be the conditional extremum of the function
.

Example 5.1. Find the maximum and minimum points of a function
given that
.

Solution. Let us express from the equation
variable via variable and substitute the resulting expression
into a function . We get
or
. This function has a unique minimum at
.
Corresponding function value
. Thus,

– point of conditional extremum (minimum).
In the example considered, the coupling equation

turned out to be linear, so it was easily resolved with respect to one of the variables. However, in more complex cases this cannot be done. To find a conditional extremum in the general case, the Lagrange multiplier method is used.

Consider a function of three variables. This function is called the Lagrange function, and– Lagrange multiplier. The following theorem is true.
Theorem.
If the point
is the conditional extremum point of the function given that
, then there is a value
.

such that point
If the point
is the extremum point of the function

Thus, to find the conditional extremum of the function the last of these equations coincides with the coupling equation. The first two equations of the system can be rewritten in the form, i.e. at the conditional extremum point the function gradients
And
collinear. In Fig. Figure 3 shows the geometric meaning of Lagrange's conditions. Line
dotted, level line
functions
solid. From Fig. it follows that at the conditional extremum point the function level line
touches the line
.

Example 5.2. Find the extremum points of the function
given that
, using the Lagrange multiplier method.

Solution. We compose the Lagrange function. Equating its partial derivatives to zero, we obtain a system of equations:

Her only solution. Thus, the conditional extremum point can only be point (3; 1). It is easy to verify that at this point the function
has a conditional minimum. If the number of variables is more than two, several coupling equations can be considered. Accordingly, in this case there will be several Lagrange multipliers.

The problem of finding a conditional extremum is used in solving such economic problems as finding the optimal allocation of resources, choosing an optimal portfolio of securities, etc.

Joseph Louis Lagrange was born in Turin (Italy) into an Italian-French family. He studied and then taught at the Artillery School. In 1759, on the recommendation of Euler, 23-year-old Lagrange was elected a member of the Berlin Academy of Sciences. In 1766 he already became its president. Frederick II invited Lagrange to Berlin. After the death of Frederick II in 1786, Lagrange moved to Paris. From 1722 he was a member of the Paris Academy of Sciences, in 1795 he was appointed a member of the Bureau of Longitudes, and he took an active part in the creation of the metric system of measures. Circle scientific research Lagrange was unusually wide. They are devoted to mechanics, geometry, mathematical analysis, algebra, number theory, and theoretical astronomy. The main direction of Lagrange's research was the presentation of a wide variety of phenomena in mechanics from a unified point of view. He derived an equation that describes the behavior of any system under the influence of forces. In the field of astronomy, Lagrange did a lot to solve the problem of stability solar system; proved some special cases of stable motion, in particular for small bodies located at the so-called triangular libration points.

Lagrange method─ this is a method for solving a problem conditional optimization, in which the constraints, written as implicit functions, are combined with the objective function in the form of a new equation called Lagrangian.

Let's consider special case common task Not linear programming:

Given the system nonlinear equations (1):

(1) gi(x1,x2,…,xn)=bi (i=1..m),

Find the smallest (or largest) value of the function (2)

(2) f (x1,x2,…,xn),

if there are no conditions for the variables to be non-negative and f(x1,x2,…,xn) and gi(x1,x2,…,xn) are functions that are continuous along with their partial derivatives.

To find a solution to this problem, you can use next method: 1. Enter a set of variables λ1, λ2,…, λm, called Lagrange multipliers, compose the Lagrange function (3)

(3) F(х1,х2,…,хn, λ1,λ2,…,λm) = f(х1,х2,…,хn)+ λi.

2. Find the partial derivatives of the Lagrange function with respect to the variables xi and λi and equate them to zero.

3. Solving the system of equations, find the points at which objective function problem may have an extremum.

4. Among the points that are suspicious not an extremum, find those at which the extremum is reached, and calculate the values ​​of the function at these points .

4. Compare the obtained values ​​of the function f and choose the best one.

According to the production plan, the company needs to produce 180 products. These products can be manufactured in two technological ways. When producing x1 products using method I, the costs are 4*x1+x1^2 rubles, and when producing x2 products using method II, they are 8*x2+x2^2 rubles. Determine how many products should be produced using each method so that the total cost of production is minimal.

Solution: The mathematical formulation of the problem is to determine lowest value functions of two variables:

f = 4*x1+x1^2 +8*x2 +x2^2, provided x1 +x2 = 180.

Let's compose the Lagrange function:

F(x1,x2,λ) = 4*x1+x1^2+8*x2+x2^2+λ*(180-x1-x2).

Let's calculate its partial derivatives with respect to x1, x2, λ and equate them to 0:

Let's move λ to the right sides of the first two equations and equate their left sides, we get 4 + 2*x1 = 8 + 2*x2, or x1 − x2 = 2.

Solving the last equation together with the equation x1 + x2 = 180, we find x1 = 91, x2 = 89, that is, we have obtained a solution that satisfies the conditions:

Let's find the value of the objective function f for these values ​​of the variables:

F(x1, x2) = 17278

This point is suspicious for an extreme point. Using second partial derivatives, we can show that at point (91.89) the function f has a minimum.

Description of the method

Where .

Rationale

The following justification for the Lagrange multiplier method is not a rigorous proof of it. It contains heuristic reasoning to help understand geometric meaning method.

Two-dimensional case

Level lines and curve.

Let it be required to find the extremum of some function of two variables under the condition specified by the equation . We will assume that all functions are continuously differentiable, and this equation defines a smooth curve S on surface . Then the problem reduces to finding the extremum of the function f on the curve S. We will also assume that S does not pass through points where the gradient f turns to 0.

Let's draw function level lines on the plane f(that is, curves). From geometric considerations it is clear that the extremum of the function f on the curve S there can only be points at which the tangents to S and the corresponding level line coincide. Indeed, if the curve S crosses the level line f at a point transversally (that is, at some non-zero angle), then moving along the curve S from a point we can get to the level lines corresponding to a larger value f, and less. Therefore, such a point cannot be an extremum point.

Thus, a necessary condition for an extremum in our case will be the coincidence of the tangents. To write it in analytical form, note that it is equivalent to the parallelism of the gradients of the functions f and ψ at a given point, since the gradient vector is perpendicular to the tangent to the level line. This condition is expressed in the following form:

where λ is a non-zero number that is a Lagrange multiplier.

Let's now consider Lagrange function, depending on and λ:

A necessary condition for its extremum is that the gradient is equal to zero. In accordance with the rules of differentiation, it is written in the form

We have obtained a system whose first two equations are equivalent to the necessary condition local extremum(1), and the third - to the equation . You can find it from it. Moreover, since otherwise the gradient of the function f vanishes at the point , which contradicts our assumptions. It should be noted that the points found in this way may not be the desired points of the conditional extremum - the considered condition is necessary, but not sufficient. Finding a conditional extremum using an auxiliary function L and forms the basis of the Lagrange multiplier method, applied here for the simplest case of two variables. It turns out that the above reasoning can be generalized to the case of an arbitrary number of variables and equations that specify the conditions.

Based on the Lagrange multiplier method, it is possible to prove some sufficient conditions for a conditional extremum, requiring analysis of the second derivatives of the Lagrange function.

Application

• The Lagrange multiplier method is used to solve nonlinear programming problems that arise in many fields (for example, in economics).
• The main method for solving the problem of optimizing the quality of encoding audio and video data at a given average bitrate (distortion optimization - English. Rate-Distortion optimization).

see also

Links

• Zorich V. A. Mathematical analysis. Part 1. - ed. 2nd, rev. and additional - M.: FAZIS, 1997.

Wikimedia Foundation.

2010.

See what “Lagrange Multipliers” are in other dictionaries: Lagrange multipliers - additional factors that transform the objective function of an extremal problem of convex programming (in particular, linear programming) when solving it using one of the classical methods, the method of resolving multipliers... ...

Economic-mathematical dictionary Lagrange multipliers - Additional factors that transform the objective function of an extremal convex programming problem (in particular, linear programming) when solving it using one of the classical methods, the method of resolving multipliers (Lagrange method).... ...

Technical Translator's Guide Mechanics. 1) Lagrange equations of the 1st kind, differential equations of mechanical motion. systems, which are given in projections onto rectangular coordinate axes and contain the so-called. Lagrange multipliers. Obtained by J. Lagrange in 1788. For a holonomic system, ... ...

Physical encyclopedia Ordinary mechanics differential equations 2nd order, describing the movements of mechanical. systems under the influence of forces applied to them. L.u. established by J. Lag range in two forms: L. u. 1st kind, or equations in Cartesian coordinates with... ...

Mathematical Encyclopedia Mechanics. 1) Lagrange equations of the 1st kind, differential equations of mechanical motion. systems, which are given in projections onto rectangular coordinate axes and contain the so-called. Lagrange multipliers. Obtained by J. Lagrange in 1788. For a holonomic system, ... ...

1) in hydromechanics, the equation of fluid (gas) motion in Lagrange variables, which are the coordinates of the medium. Received French scientist J. Lagrange (approx. 1780). From L. u. the law of motion of the medium is determined in the form of dependencies... ...

Lagrange multiplier method, a method for finding the conditional extremum of the function f(x), where, relative to m constraints, i varies from one to m. Contents 1 Description of the method ... Wikipedia A function used in solving problems on the conditional extremum of functions of many variables and functionals. With the help of L. f. are recorded the necessary conditions 2nd order, describing the movements of mechanical. systems under the influence of forces applied to them. L.u. established by J. Lag range in two forms: L. u. 1st kind, or equations in Cartesian coordinates with... ...

optimality in problems on conditional extremum. In this case, it is not necessary to express only variables...

Variables, with the help of which the Lagrange function is constructed when studying problems on a conditional extremum. The use of linear methods and the Lagrange function allows us to obtain the necessary optimality conditions in problems involving a conditional extremum in a uniform way... 2nd order, describing the movements of mechanical. systems under the influence of forces applied to them. L.u. established by J. Lag range in two forms: L. u. 1st kind, or equations in Cartesian coordinates with... ...

1) in hydromechanics, the equations of motion of a fluid medium, written in Lagrange variables, which are the coordinates of particles of the medium. From L. u. the law of motion of particles of the medium is determined in the form of dependences of coordinates on time, and from them... ... Great Soviet Encyclopedia

• Tutorial

Everyone good day. In this article I want to show one of graphic methods construction mathematical models for dynamic systems, which is called bond graph(“bond” - connections, “graph” - graph). In Russian literature, I found descriptions of this method only in Tomsky’s Textbook Polytechnic University, A.V. Voronin “MODELING OF MECHATRONIC SYSTEMS” 2008 Also show classic method through the Lagrange equation of the 2nd kind.

Lagrange method

I will not describe the theory, I will show the stages of calculations with a few comments. Personally, it’s easier for me to learn from examples than to read theory 10 times. It seemed to me that in Russian literature, the explanation of this method, and indeed mathematics or physics in general, is very rich complex formulas, which accordingly requires a serious mathematical background. While studying the Lagrange method (I study at the Polytechnic University of Turin, Italy), I studied Russian literature to compare calculation methods, and it was difficult for me to follow the progress of solving this method. Even remembering the modeling courses at the Kharkov Aviation Institute, the derivation of such methods was very cumbersome, and no one bothered themselves in trying to understand this issue. This is what I decided to write, a manual for constructing mathematical models according to Lagrange, as it turned out it is not at all difficult, it is enough to know how to calculate derivatives with respect to time and partial derivatives. For more complex models, rotation matrices are also added, but there is nothing complicated in them either.

Features of modeling methods:

• Newton-Euler: vector equations based on dynamic equilibrium force And moments
• Lagrange: scalar equations based on state functions associated with kinetic and potential energies
• Bond Count: flow based method power between system elements

Let's start with simple example. Mass with spring and damper. We ignore the force of gravity.

Fig 1. Mass with spring and damper

First of all, we designate:

• initial system coordinates(NSK) or fixed sk R0(i0,j0,k0). Where? You can point your finger at the sky, but by twitching the tips of the neurons in the brain, the idea passes through to place the NSC on the line of movement of the M1 body.
• coordinate systems for each body with mass(we have M1 R1(i1,j1,k1)), the orientation can be arbitrary, but why complicate your life, set it with minimal difference from the NSC
• generalized coordinates q_i(the minimum number of variables that can describe the movement), in this example there is one generalized coordinate, movement only along the j axis

Fig 2. We put down coordinate systems and generalized coordinates

Fig 3. Position and speed of body M1

Then we will find the kinetic (C) and potential (P) energies and the dissipative function (D) for the damper using the formulas:

Fig 4. Complete formula kinetic energy

In our example there is no rotation, the second component is equal to 0.

Fig 5. Calculation of kinetic, potential energy and dissipative function

The Lagrange equation has the following form:

Fig 6. Lagrange Equation and Lagrangian

Delta W_i This is virtual work done by applied forces and moments. Let's find her:

Fig 7. Calculation of virtual work

Where delta q_1 virtual movement.

We substitute everything into the Lagrange equation:

Fig 8. The resulting mass model with spring and damper

This is where Lagrange's method ended. As you can see, it’s not that complicated, but it’s still a very simple example, for which most likely the Newton-Euler method would even be simpler. For more complex systems, where there will be several bodies rotated relative to each other at different angles, the Lagrange method will be easier.

Bond method graph

I’ll show you right away what the model looks like in bond-graph for an example with a mass, a spring and a damper:

Fig 9. Bond-graph masses with spring and damper

Here you will have to tell a little theory, which will be enough to build simple models. If anyone is interested, you can read the book ( Bond Graph Methodology) or ( Voronin A.V. Modeling of mechatronic systems: tutorial. – Tomsk: Tomsk Polytechnic University Publishing House, 2008).

Let us first determine that complex systems consist of several domains. For example, an electric motor consists of electrical and mechanical parts or domains.

bond graph based on the exchange of power between these domains, subsystems. Note that power exchange, of any form, is always determined by two variables ( variable power) with the help of which we can study the interaction of various subsystems within a dynamic system (see table).

As can be seen from the table, the expression of power is almost the same everywhere. In summary, Power- This work " flow - f" on " effort - e».

An effort(English) effort) in the electrical domain this is voltage (e), in the mechanical domain it is force (F) or torque (T), in hydraulics it is pressure (p).

Flow(English) flow) in the electrical domain it is current (i), in the mechanical domain it is speed (v) or angular velocity(omega), in hydraulics – fluid flow or flow rate (Q).

Taking these notations, we obtain an expression for power:

Fig 10. Power formula through power variables

In the bond-graph language, the connection between two subsystems that exchange power is represented by a bond. bond). That's why it's called this method bond-graph or g raf-connections, connected graph. Let's consider block diagram connections in a model with an electric motor (this is not a bond-graph yet):

Fig 11. Block diagram of power flow between domains

If we have a voltage source, then accordingly it generates voltage and transfers it to the motor for winding (this is why the arrow is directed towards the motor), depending on the resistance of the winding, a current appears according to Ohm’s law (directed from the motor to the source). Accordingly, one variable is an input to the subsystem, and the second must be exit from the subsystem. Here the voltage ( effort) – input, current ( flow) - exit.

If you use a current source, how will the diagram change? Right. The current will be directed to the motor, and the voltage to the source. Then the current ( flow) – input, voltage ( effort) - exit.

Let's look at an example in mechanics. Force acting on a mass.

Fig 12. Force applied to mass

The block diagram will be as follows:

Fig 13. Block diagram

In this example, Strength ( effort) – input variable for mass. (Force applied to mass)
According to Newton's second law:

Mass responds with speed:

In this example, if one variable ( force - effort) is entrance into the mechanical domain, then another power variable ( speed - flow) – automatically becomes exit.

To distinguish where the input is and where the output is, a vertical line is used at the end of the arrow (connection) between the elements, this line is called sign of causality or causation (causality). It turns out: applied force is the cause, and speed is the effect. This sign is very important for the correct construction of a system model, since causality is a consequence physical behavior and the exchange of powers of two subsystems, therefore the choice of the location of the causality sign cannot be arbitrary.

Fig 14. Designation of causality

This vertical line shows which subsystem receives the force ( effort) and as a result produce a flow ( flow). In the example with mass it would be like this:

Fig 14. Causal relationship for the force acting on the mass

It is clear from the arrow that the input for mass is - force, and the output is speed. This is done so as not to clutter the diagram with arrows and systematize the construction of the model.

Next important point. Generalized impulse(amount of movement) and moving(energy variables).

Table of power and energy variables in different domains

The table above introduces two additional physical quantities used in the bond-graph method. They're called generalized impulse (R) And generalized movement (q) or energy variables, and they can be obtained by integrating power variables over time:

Fig 15. Relationship between power and energy variables

In the electrical domain :

Based on Faraday's law, voltage at the ends of the conductor is equal to the derivative of the magnetic flux through this conductor.

A Current strength - physical quantity, equal to the ratio of the amount of charge Q passing through some time t cross section conductor, to the value of this period of time.

Mechanical domain:

From Newton's 2nd law, Force– time derivative of impulse

And correspondingly, speed- time derivative of displacement:

Let's summarize:

Basic elements

All elements in dynamic systems can be divided into two-pole and four-pole components.
Let's consider bipolar components:

Sources
There are sources of both effort and flow. Analogy in the electrical domain: source of effort – voltage source, stream source – current source. Causal signs for sources should only be like this.

Fig 16. Causal connections and designation of sources

Component R – dissipative element

Component I – inertial element

Component C – capacitive element

As can be seen from the figures, different elements of the same type R,C,I described by the same equations. ONLY there is a difference for electrical capacitance, you just need to remember it!

Quadrupole components:

Let's look at two components: a transformer and a gyrator.

The last important components in the bond-graph method are the connections. There are two types of nodes:

That's it with the components.

The main steps for establishing causal relationships after constructing a bond-graph:

1. Give causal connections to everyone sources
2. Go through all the nodes and put down causal relationships after point 1
3. For components I assign an input causal relationship (effort is included in this component), for components C assign output causality (effort comes out of this component)
4. Repeat point 2
5. Insert causal connections for R components

This concludes the mini-course on theory. Now we have everything we need to build models.
Let's solve a couple of examples. Let's start with an electrical circuit; it is better to understand the analogy of constructing a bond-graph.

Example 1

Let's start building a bond-graph with a voltage source. Just write Se and put an arrow.

See, everything is simple! Let's look further, R and L are connected in series, which means the same current flows in them, if we speak in power variables - the same flow. Which node has the same flow? The correct answer is 1-node. We connect the source, resistance (component - R) and inductance (component - I) to the 1-node.

Next, we have capacitance and resistance in parallel, which means they have the same voltage or force. 0-node is suitable like no other. We connect the capacitance (component C) and resistance (component R) to the 0-node.

We also connect nodes 1 and 0 to each other. The direction of the arrows is chosen arbitrarily; the direction of the connection only affects the sign in the equations.

You will get the following connection graph:

Now we need to establish causal relationships. Following the instructions for the sequence of their placement, let's start with the source.

1. We have a source of voltage (effort), such a source has only one causality option - output. Let's put it.
2. Next there is component I, let’s see what they recommend. We put
3. We put it down for 1-node. Eat
4. A 0-node must have one input and all output causal connections. We have one day off for now. We are looking for components C or I. We found it. We put
5. Let's list what's left

That's all. Bond graph is built. Hurray, Comrades!

All that remains is to write the equations that describe our system. To do this, create a table with 3 columns. The first will contain all the components of the system, the second will contain the input variable for each element, and the third will contain the output variable for the same component. We have already defined the input and output by causal relationships. So there shouldn't be any problems.

Let's number each connection for ease of recording the levels. We take the equations for each element from the list of components C, R, I.

Having compiled a table, we define the state variables, in this example there are 2 of them, p3 and q5. Next you need to write down the equations of state:

That's it, the model is ready.

Example 2. I would like to immediately apologize for the quality of the photo, the main thing is that you can read

Let's solve another example for a mechanical system, the same one that we solved using the Lagrange method. I will show the solution without comment. Let's check which of these methods is simpler and easier.

In Matbala, both mathematical models with the same parameters were compiled, obtained by the Lagrange method and bond-graph. The result is below: Add tags

The method for determining a conditional extremum begins with constructing an auxiliary Lagrange function, which in the region of feasible solutions reaches a maximum for the same values ​​of variables x 1 , x 2 , ..., x n , which is the same as the objective function z . Let the problem of determining the conditional extremum of the function be solved z = f(X) under restrictions φ i ( x 1 , x 2 , ..., x n ) = 0, i = 1, 2, ..., m , m < n

Let's compose a function

which is called Lagrange function. X , - constant factors ( Lagrange multipliers). Note that Lagrange multipliers can be given an economic meaning. If f(x 1 , x 2 , ..., x n ) - income consistent with the plan X = (x 1 , x 2 , ..., x n ) , and the function φ i (x 1 , x 2 , ..., x n ) - costs of the i-th resource corresponding to this plan, then X , is the price (estimate) of the i-th resource, characterizing the change in the extreme value of the objective function depending on the change in the size of the i-th resource (marginal estimate). L(X) - function n+m variables (x 1 , x 2 , ..., x n , λ 1 , λ 2 , ..., λ n ) . Determining the stationary points of this function leads to solving the system of equations

It's easy to see that . Thus, the task of finding the conditional extremum of the function z = f(X) reduces to finding the local extremum of the function L(X) . If a stationary point is found, then the question of the existence of an extremum in the simplest cases is resolved on the basis of sufficient conditions for the extremum - studying the sign of the second differential d 2 L(X) at a stationary point, provided that the variable increments Δx i - connected by relationships

obtained by differentiating the coupling equations.

Solving a system of nonlinear equations in two unknowns using the Find Solution tool

Settings Finding a solution allows you to find a solution to a system of nonlinear equations with two unknowns:

Where
- nonlinear function of variables x And y ,
- arbitrary constant.

It is known that the couple ( x , y ) is a solution to system of equations (10) if and only if it is a solution to the following equation with two unknowns:

WITH on the other hand, the solution to system (10) is the intersection points of two curves: f ] (x, y) = C And f 2 (x, y) = C 2 on surface XOY.

This leads to a method for finding the roots of the system. nonlinear equations:

Determine (at least approximately) the interval of existence of a solution to the system of equations (10) or equation (11). Here it is necessary to take into account the type of equations included in the system, the domain of definition of each of their equations, etc. Sometimes the selection of an initial approximation of the solution is used;

Tabulate the solution to equation (11) for the variables x and y on the selected interval, or construct graphs of functions f 1 (x, y) = C, and f 2 (x,y) = C 2 (system(10)).

Localize the supposed roots of the system of equations - find several minimum values ​​from the table tabulating the roots of equation (11), or determine the intersection points of the curves included in the system (10).

4. Find the roots for the system of equations (10) using the add-in Finding a solution.