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The differential equation with separated variables is written as: (1). In this equation, one term depends only on x, and the other depends on y. By integrating this equation term by term, we obtain:
is its general integral.

Example: find the general integral of the equation:
.

Solution: This equation is a differential equation with separated variables. That's why
or
Denote
. Then
is the general integral of the differential equation.

The separable variable equation has the form (2). Equation (2) can easily be reduced to equation (1) by dividing it term by term by
. We get:

is the general integral.

Example: solve the equation .

Solution: transform the left side of the equation: . We divide both sides of the equation by


The solution is the expression:
those.

Homogeneous differential equations. Bernoulli's equations. Linear differential equations of the first order.

The type equation is called homogeneous, if
and
are homogeneous functions of the same order (measurement). Function
is called a homogeneous function of the first order (measurement) if, when multiplying each of its arguments by an arbitrary factor the whole function is multiplied by , i.e.
=
.

The homogeneous equation can be reduced to the form
. With the help of substitution
(
) the homogeneous equation is reduced to an equation with separable variables with respect to the new function .

The first order differential equation is called linear if it can be written in the form
.

Bernoulli method

Equation solution
is sought as a product of two other functions, i.e. using substitution
(
).

Example: integrate the equation
.

We believe
. Then , i.e. . First we solve the equation
=0:


.

Now we solve the equation
those.


. So, common decision this equation is
those.

J. Bernoulli equation

An equation of the form , where
called Bernoulli's equation. This equation is solved using the Bernoulli method.

Homogeneous Second Order Differential Equations with Constant Coefficients

A homogeneous second-order linear differential equation is an equation of the form (1) , where and are constant.

Particular solutions of equation (1) will be sought in the form
, where to- some number. Differentiating this function two times and substituting expressions for
into equation (1), we get m.e. or
(2) (
).

Equation 2 is called the characteristic equation of the differential equation.

When solving the characteristic equation (2), three cases are possible.

Case 1 Roots and equations (2) are real and different:

and

.

Case 2 Roots and equations (2) are real and equal:
. In this case, particular solutions of equation (1) are the functions
and
. Therefore, the general solution of equation (1) has the form
.

Case 3 Roots and equations (2) are complex:
,
. In this case, particular solutions of equation (1) are the functions
and
. Therefore, the general solution of equation (1) has the form

Example. solve the equation
.

Decision: we compose the characteristic equation:
. Then
. The general solution of this equation
.

Extremum of a function of several variables. Conditional extreme.

Extremum of a function of several variables

Definition.Point M (x about ,y about ) is calledmaximum (minimum) point functionsz= f(x, y) if there exists a neighborhood of the point M such that for all points (x, y) from this neighborhood the inequality
(
)

On fig. 1 point AND
- there is a minimum point, and the point AT
- maximum point.

Necessarythe extremum condition is a multidimensional analogue of Fermat's theorem.

Theorem.Let the point
is an extremum point of a differentiable functionz= f(x, y). Then the partial derivatives
and
inthis point are zero.

Points at which the necessary conditions for the extremum of the function are satisfied z= f(x, y), those. partial derivatives z" x and z" y equal to zero are called critical or stationary.

The equality of partial derivatives to zero expresses only a necessary but insufficient condition for the extremum of a function of several variables.

On fig. the so-called saddle point M (x about ,y about ). Partial derivatives
and
are equal to zero, but, obviously, no extremum at the point M(x about ,y about ) no.

Such saddle points are two-dimensional analogs of inflection points for functions of one variable. The challenge is to separate them from the extremum points. In other words, you need to know sufficient extreme condition.

Theorem (sufficient condition for an extremum of a function of two variables).Let the functionz= f(x, y): a) is defined in some neighborhood of the critical point (x about ,y about ), wherein
=0 and
=0 ;

b) has continuous second-order partial derivatives at this point
;

;
Then, if ∆=AC-B 2 >0, then at the point (x about ,y about ) functionz= f(x, y) has an extremum, and if AND<0 - maximum if A>0 - minimum. In the case of ∆=AC-B 2 <0, функция z= f(x, y) has no extremum. If ∆=AC-B 2 =0, then the question of the presence of an extremum remains open.

Investigation of a function of two variables for an extremum it is recommended to carry out the following scheme:

Find Partial Derivatives of Functions z" x and z" y .

Solve a system of equations z" x =0, z" y =0 and find the critical points of the function.

Find second-order partial derivatives, calculate their values ​​at each critical point, and, using a sufficient condition, draw a conclusion about the presence of extrema.

Find extrema (extreme values) of the function.

Example. Find extrema of a function

Decision. 1. Find partial derivatives

2. Critical points of the function are found from the system of equations:

having four solutions (1; 1), (1; -1), (-1; 1) and (-1; -1).

3. We find partial derivatives of the second order:

;
;
, we calculate their values ​​at each critical point and check the fulfillment of the sufficient extremum condition at it.

For example, at the point (1; 1) A= z"(1; 1)= -1; B=0; C= -1. Because ∆ =AC-B 2 = (-1) 2 -0=1 >0 and A=-1<0, then the point (1; 1) is the maximum point.

Similarly, we establish that (-1; -1) is the minimum point, and at points (1; -1) and (-1; 1), in which ∆ =AC-B 2 <0, - экстремума нет. Эти точки являются седловыми.

4. Find the extrema of the function z max = z(l; 1) = 2, z min = z(-l; -1) = -2,

Conditional extreme. Method of Lagrange multipliers.

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

Let the function z = f(x, y), arguments X and at which satisfy the condition g(x, y)= WITH, called connection equation.

Definition.Dot
called a pointconditional maximum (minimum), if there is such a neighborhood of this point that for all points (x, y) from this neighborhood satisfying the conditiong (x, y) = С, the inequality

(
).

On fig. the conditional maximum point is shown
. It is obvious that it is not an unconditional extremum point of the function z = f(x, y) (in the figure this is a 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. Assume the constraint equation g (x, y) = With managed to resolve with respect to one of the variables, for example, to express at through X:
. Substituting the resulting expression into a function of two variables, we obtain z = f(x, y) =
, those. function of one variable. Its extremum will be the conditional extremum of the function z = f(x, y).

Example. X 2 + y 2 given that 3x + 2y = 11.

Decision. We express the variable y from the equation 3x + 2y \u003d 11 in terms of the variable x and substitute the resulting
into a function z. Get z= x 2 +2
or z =
. This function has a single minimum at = 3. Corresponding function value
Thus, (3; 1) is a conditional extremum (minimum) point.

In the considered example, the constraint equation g(x, y) = C 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 the conditional extremum, in the general case, we use method of Lagrange multipliers.

Consider a function of three variables

This function is called Lagrange function, a - Lagrange multiplier. The following theorem is true.

Theorem.If point
is the conditional extremum point of the functionz = f(x, y) given thatg (x, y) = C, then there is a value such that the point
is the extremum point of the functionL{ x, y, ).

Thus, to find the conditional extremum of the function z = f(x, y) given that g(x, y) = C need to find a solution to the system

On fig. the geometric meaning of the Lagrange conditions is shown. Line g(x, y)= C dotted, level line g(x, y) = Q functions z = f(x, y) solid.

From fig. follows that at the conditional extremum point, the level line of the function z= f(x, y) touches the lineg(x, y) = C.

Example. Find the maximum and minimum points of the function z = X 2 + y 2 given that 3x + 2y = 11 using the Lagrange multiplier method.

Decision. Compose the Lagrange function L= x 2 + 2y 2 +

Equating its partial derivatives to zero, we obtain the system of equations

Its only solution (x=3, y=1, =-2). Thus, only the point (3;1) can be a conditional extremum point. It is easy to verify that at this point the function z= f(x, y) has a conditional minimum.

Ordinary differential equations.

The solution of various geometric, physical and engineering problems often leads to equations that relate independent variables that characterize a particular problem with some function of these variables and derivatives of this function of various orders.

As an example, we can consider the simplest case of uniformly accelerated motion of a material point.

It is known that the displacement of a material point during uniformly accelerated motion is a function of time and is expressed by the formula:

In turn, the acceleration a is the time derivative t from speed V, which is also a derivative with respect to time t from moving S. Those.

Then we get:
- the equation relates the function f(t) to the independent variable t and the second-order derivative of the function f(t).

Definition. differential equation called an equation relating independent variables, their functions and derivatives (or differentials) of this function.

Definition. If a differential equation has one independent variable, then it is called ordinary differential equation, if there are two or more independent variables, then such a differential equation is called partial differential equation.

Definition. The highest order of derivatives in an equation is called the order of the differential equation.

Example.

- ordinary differential equation of the 1st order. In general, it is written
.

- ordinary differential equation of the 2nd order. In general, it is written

- differential equation in partial derivatives of the first order.

Definition. General solution differential equation is such a differentiable function y = (x, C), which, when substituted into the original equation instead of an unknown function, turns the equation into an identity.

Properties of the general solution.

1) Because Since the constant C is an arbitrary value, then in general the differential equation has an infinite number of solutions.

2) Under any initial conditions x \u003d x 0, y (x 0) \u003d y 0, there is such a value C \u003d C 0 for which the solution of the differential equation is the function y \u003d  (x, C 0).

Definition. A solution of the form y \u003d  (x, C 0) is called private decision differential equation.

Definition. Cauchy problem(Augustin Louis Cauchy (1789-1857) - French mathematician) is called finding any particular solution to a differential equation of the form y \u003d  (x, C 0) that satisfies the initial conditions y (x 0) \u003d y 0.

Cauchy's theorem. (theorem on the existence and uniqueness of the solution of the differential equation of the 1st order)

If the functionf(x, y) is continuous in some domainDin planeXOYand has a continuous partial derivative in this region
, then whatever the point (x 0 , y 0 ) in areaD, there is only one solution
equations
, defined in some interval containing the point x 0 , accepting at x = x 0 meaning (X 0 ) = y 0 , i.e. there is a unique solution to the differential equation.

Definition. integral differential equation is any equation that does not contain derivatives, for which this differential equation is a consequence.

Example. Find the general solution of the differential equation
.

The general solution of the differential equation is sought by integrating the left and right sides of the equation, which is preliminarily transformed as follows:

Now let's integrate:

is the general solution of the original differential equation.

Suppose some initial conditions are given: x 0 = 1; y 0 = 2, then we have

By substituting the obtained value of the constant into the general solution, we obtain a particular solution for given initial conditions (the solution of the Cauchy problem).

Definition. integral curve the graph y = (x) of the solution of a differential equation on the XOY plane is called.

Definition. special solution of a differential equation is such a solution, at all points of which the Cauchy uniqueness condition is called (cf. Cauchy's theorem.) is not satisfied, i.e. in a neighborhood of some point (x, y) there are at least two integral curves.

The singular solutions do not depend on the constant C.

Special solutions cannot be obtained from the general solution for any values ​​of the constant C. If we construct a family of integral curves for a differential equation, then the special solution will be represented by a line that touches at least one integral curve at each of its points.

Note that not every differential equation has singular solutions.

Example.
Find a special solution if it exists.

This differential equation also has a special solution at= 0. This solution cannot be obtained from the general one, however, when substituting into the original equation, we obtain an identity. opinion that the solution y = 0 can be obtained from the general solution for With 1 = 0 wrong, because C 1 = e C  0.

First order differential equations.

Definition. First order differential equation is the relation connecting the function, its first derivative and the independent variable, i.e. aspect ratio:

If this ratio is converted to the form
then this first-order differential equation will be called the equation, allowed with respect to the derivative.

We represent the function f(x,y) as:
then, substituting into the above equation, we have:

this so-called differential form first order equations.

Equations of the formy ’ = f ( x ).

Let the function f(x) be defined and continuous on some interval

a< x < b. В таком случае все решения данного дифференциального уравнения находятся как
. If the initial conditions x 0 and y 0 are given, then the constant C can be determined.

Separable Variable Equations

Definition. Differential equation
called separable equation if it can be written in the form

.

This equation can also be represented as:

Let's move on to new notation

We get:

After finding the corresponding integrals, a general solution of a differential equation with separable variables is obtained.

If the initial conditions are given, then when they are substituted into the general solution, a constant value C is found, and, accordingly, a particular solution.

Example. Find the general solution of the differential equation:

The integral on the left side is taken by parts (see Fig. Integration by parts.):

this is the general integral of the original differential equation, since the desired function and is not expressed in terms of the independent variable. This is what difference general (private) integral from the general (private) solutions.

To check the correctness of the obtained answer, we differentiate it with respect to the variable x.

- right

Example. Find a solution to a differential equation
provided y(2) = 1.

for y(2) = 1 we get

Total:
or
- private decision;

Examination:
, total

- right.

Example. solve the equation

- common integral

- common decision

Example. solve the equation

Example. solve the equation
provided y(1) = 0.

The integral on the left side will be taken by parts (see Fig. Integration by parts.).

If y(1) = 0, then

So the private integral is:
.

Example. Solve the equation.

To find the integral on the left side of the equation, see Table of basic integrals. item 16. We get the general integral:

Example. solve the equation

Let's transform the given equation:

We have obtained the general integral of this differential equation. If we express the desired function y from this relation, then we obtain the general solution.

Example. solve the equation
.

;
;

Suppose some initial conditions x 0 and y 0 are given. Then:

We get a private solution

Homogeneous equations.

Definition. The function f(x, y) is called homogeneousn– th dimension with respect to their arguments x and y, if for any value of the parameter t (except zero) the identity holds:

Example. Is the function homogeneous?

Thus, the function f(x, y) is homogeneous of the 3rd order.

Definition. Differential equation of the form
called homogeneous, if its right side f(x, y) is a homogeneous function of dimension zero with respect to its arguments.

Any equation of the form is homogeneous if the functions P(x, y) and Q(x, y) are homogeneous functions of the same dimension.

The solution of any homogeneous equation is based on reducing this equation to an equation with separable variables.

Consider the homogeneous equation

Because function f(x, y) is homogeneous of zero dimension, then we can write:

Because the parameter t is generally arbitrary, suppose that . We get:

The right side of the resulting equality actually depends on only one argument
, i.e.

The original differential equation can thus be written as:

thus, we have obtained an equation with separable variables for the unknown function u.

Example. solve the equation
.

We introduce an auxiliary function u.

.

Note that the function introduced by us u is always positive, because otherwise, the original differential equation containing
.

We substitute into the original equation:

Separating variables:

Integrating, we get:

Passing from the auxiliary function back to the y function, we obtain the general solution:

Equations Reducing to Homogeneous.

In addition to the equations described above, there is a class of equations that, with the help of certain substitutions, can be reduced to homogeneous ones.

These are equations of the form
.

If the determinant
then the variables can be separated by substitution

where  and  are solutions of the system of equations

Example. solve the equation

We get

Finding the value of the determinant
.

We solve the system of equations

We apply the substitution in the original equation:

Replacing the variable
when substituting into the expression written above, we have:

Differential equations.

Basic concepts about ordinary differential equations.

Definition 1. Ordinary differential equation n-th order for the function y argument x is called a relation of the form

where F is a given function of its arguments. In the name of this class of mathematical equations, the term "differential" emphasizes that they include derivatives (functions formed as a result of differentiation); the term - "ordinary" says that the desired function depends on only one real argument.

An ordinary differential equation may not explicitly contain an argument x, the desired function and any of its derivatives, but the highest derivative must be included in the equation n- order. For example

a) is a first order equation;

b) is a third order equation.

When writing ordinary differential equations, the notation of derivatives through differentials is often used:

in) is a second order equation;

d) is a first order equation,

forming after division by dx equivalent form of the equation: .

A function is called a solution to an ordinary differential equation if, when substituted into it, it becomes an identity.

For example, the 3rd order equation

Has a solution .

To find by one method or another, for example, selection, one function that satisfies an equation does not mean solving it. To solve an ordinary differential equation means to find Everybody functions that form an identity when substituted into the equation. For equation (1.1), the family of such functions is formed with the help of arbitrary constants and is called the general solution of the ordinary differential equation n th order, and the number of constants coincides with the order of the equation: y(x): In this case, the solution is called the general integral of equation (1.1).

For example, the following expression is a general solution to a differential equation: , and the second term can be written as , since an arbitrary constant divided by 2 can be replaced by a new arbitrary constant .

By setting some admissible values ​​for all arbitrary constants in the general solution or in the general integral, we obtain a certain function that no longer contains arbitrary constants. This function is called a particular solution or a particular integral of equation (1.1). To find the values ​​of arbitrary constants, and hence the particular solution, various additional conditions to equation (1.1) are used. For example, the so-called initial conditions for (1.2) can be given

In the right parts of the initial conditions (1.2), the numerical values ​​of the function and derivatives are given, and the total number of initial conditions is equal to the number of arbitrary constants being determined.

The problem of finding a particular solution to equation (1.1) from initial conditions is called the Cauchy problem.

§ 2. Ordinary differential equations of the 1st order - basic concepts.

Ordinary differential equation of the 1st order ( n=1) has the form: or, if it can be resolved with respect to the derivative: . Common decision y=y(x, C) or the general integral of the 1st order equations contain one arbitrary constant. The only initial condition for the 1st order equation allows you to determine the value of the constant from the general solution or from the general integral. Thus, a particular solution will be found or, which is also the Cauchy problem will be solved. The question of the existence and uniqueness of a solution to the Cauchy problem is one of the central ones in the general theory of ordinary differential equations. For a first-order equation, in particular, the theorem is valid, which is accepted here without proof.

Theorem 2.1. If in an equation a function and its partial derivative are continuous in some region D plane XOY , and a point is given in this region, then there exists and, moreover, a unique solution that satisfies both the equation and the initial condition.

The geometrically general solution of the 1st order equation is a family of curves in the plane XOY, which do not have common points and differ from each other in one parameter - the value of the constant C. These curves are called integral curves for the given equation. The integral curves of the equation have an obvious geometric property: at each point, the tangent of the slope of the tangent to the curve is equal to the value of the right side of the equation at that point: . In other words, the equation is given in the plane XOY field of directions of tangents to integral curves. Comment: It should be noted that for the equation the equation and the so-called equation in symmetric form are given .

First order differential equations with separable variables.

Definition. A differential equation with separable variables is an equation of the form (3.1)

or an equation of the form (3.2)

In order to separate the variables in equation (3.1), i.e. reduce this equation to the so-called equation with separated variables, perform the following actions:

;

Now we need to solve the equation g(y)=0. If it has a real solution y=a, then y=a will also be a solution of equation (3.1).

Equation (3.2) is reduced to an equation with separated variables by dividing by the product:

, which allows us to obtain the general integral of equation (3.2): . (3.3)

The integral curves (3.3) will be supplemented with solutions if such solutions exist.

Solve the equation: .

Separating variables:

.

Integrating, we get

The differential equation with separated variables is written as: (1). In this equation, one term depends only on x, and the other depends on y. By integrating this equation term by term, we obtain:
is its general integral.

Example: find the general integral of the equation:
.

Solution: This equation is a differential equation with separated variables. That's why
or
Denote
. Then
is the general integral of the differential equation.

The separable variable equation has the form (2). Equation (2) can easily be reduced to equation (1) by dividing it term by term by
. We get:

is the general integral.

Example: solve the equation .

Solution: transform the left side of the equation: . We divide both sides of the equation by


The solution is the expression:
those.

Homogeneous differential equations. Bernoulli's equations. Linear differential equations of the first order.

The type equation is called homogeneous, if
and
are homogeneous functions of the same order (measurement). Function
is called a homogeneous function of the first order (measurement) if, when multiplying each of its arguments by an arbitrary factor the whole function is multiplied by , i.e.
=
.

The homogeneous equation can be reduced to the form
. With the help of substitution
(
) the homogeneous equation is reduced to an equation with separable variables with respect to the new function .

The first order differential equation is called linear if it can be written in the form
.

Bernoulli method

Equation solution
is sought as a product of two other functions, i.e. using substitution
(
).

Example: integrate the equation
.

We believe
. Then , i.e. . First we solve the equation
=0:


.

Now we solve the equation
those.


. So the general solution to this equation is
those.

J. Bernoulli equation

An equation of the form , where
called Bernoulli's equation. This equation is solved using the Bernoulli method.

Homogeneous Second Order Differential Equations with Constant Coefficients

A homogeneous second-order linear differential equation is an equation of the form (1) , where and are constant.

Particular solutions of equation (1) will be sought in the form
, where to- some number. Differentiating this function two times and substituting expressions for
into equation (1), we get m.e. or
(2) (
).

Equation 2 is called the characteristic equation of the differential equation.

When solving the characteristic equation (2), three cases are possible.

Case 1 Roots and equations (2) are real and different:

and

.

Case 2 Roots and equations (2) are real and equal:
. In this case, particular solutions of equation (1) are the functions
and
. Therefore, the general solution of equation (1) has the form
.

Case 3 Roots and equations (2) are complex:
,
. In this case, particular solutions of equation (1) are the functions
and
. Therefore, the general solution of equation (1) has the form

Example. solve the equation
.

Decision: we compose the characteristic equation:
. Then
. The general solution of this equation
.

Extremum of a function of several variables. Conditional extreme.

Extremum of a function of several variables

Definition.Point M (x about ,y about ) is calledmaximum (minimum) point functionsz= f(x, y) if there exists a neighborhood of the point M such that for all points (x, y) from this neighborhood the inequality
(
)

On fig. 1 point AND
- there is a minimum point, and the point AT
- maximum point.

Necessarythe extremum condition is a multidimensional analogue of Fermat's theorem.

Theorem.Let the point
is an extremum point of a differentiable functionz= f(x, y). Then the partial derivatives
and
inthis point are zero.

Points at which the necessary conditions for the extremum of the function are satisfied z= f(x, y), those. partial derivatives z" x and z" y equal to zero are called critical or stationary.

The equality of partial derivatives to zero expresses only a necessary but insufficient condition for the extremum of a function of several variables.

On fig. the so-called saddle point M (x about ,y about ). Partial derivatives
and
are equal to zero, but, obviously, no extremum at the point M(x about ,y about ) no.

Such saddle points are two-dimensional analogs of inflection points for functions of one variable. The challenge is to separate them from the extremum points. In other words, you need to know sufficient extreme condition.

Theorem (sufficient condition for an extremum of a function of two variables).Let the functionz= f(x, y): a) is defined in some neighborhood of the critical point (x about ,y about ), wherein
=0 and
=0 ;

b) has continuous second-order partial derivatives at this point
;

;
Then, if ∆=AC-B 2 >0, then at the point (x about ,y about ) functionz= f(x, y) has an extremum, and if AND<0 - maximum if A>0 - minimum. In the case of ∆=AC-B 2 <0, функция z= f(x, y) has no extremum. If ∆=AC-B 2 =0, then the question of the presence of an extremum remains open.

Investigation of a function of two variables for an extremum it is recommended to carry out the following scheme:

Find Partial Derivatives of Functions z" x and z" y .

Solve a system of equations z" x =0, z" y =0 and find the critical points of the function.

Find second-order partial derivatives, calculate their values ​​at each critical point, and, using a sufficient condition, draw a conclusion about the presence of extrema.

Find extrema (extreme values) of the function.

Example. Find extrema of a function

Decision. 1. Find partial derivatives

2. Critical points of the function are found from the system of equations:

having four solutions (1; 1), (1; -1), (-1; 1) and (-1; -1).

3. We find partial derivatives of the second order:

;
;
, we calculate their values ​​at each critical point and check the fulfillment of the sufficient extremum condition at it.

For example, at the point (1; 1) A= z"(1; 1)= -1; B=0; C= -1. Because ∆ =AC-B 2 = (-1) 2 -0=1 >0 and A=-1<0, then the point (1; 1) is the maximum point.

Similarly, we establish that (-1; -1) is the minimum point, and at points (1; -1) and (-1; 1), in which ∆ =AC-B 2 <0, - экстремума нет. Эти точки являются седловыми.

4. Find the extrema of the function z max = z(l; 1) = 2, z min = z(-l; -1) = -2,

Conditional extreme. Method of Lagrange multipliers.

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

Let the function z = f(x, y), arguments X and at which satisfy the condition g(x, y)= WITH, called connection equation.

Definition.Dot
called a pointconditional maximum (minimum), if there is such a neighborhood of this point that for all points (x, y) from this neighborhood satisfying the conditiong (x, y) = С, the inequality

(
).

On fig. the conditional maximum point is shown
. It is obvious that it is not an unconditional extremum point of the function z = f(x, y) (in the figure this is a 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. Assume the constraint equation g (x, y) = With managed to resolve with respect to one of the variables, for example, to express at through X:
. Substituting the resulting expression into a function of two variables, we obtain z = f(x, y) =
, those. function of one variable. Its extremum will be the conditional extremum of the function z = f(x, y).

Example. X 2 + y 2 given that 3x + 2y = 11.

Decision. We express the variable y from the equation 3x + 2y \u003d 11 in terms of the variable x and substitute the resulting
into a function z. Get z= x 2 +2
or z =
. This function has a single minimum at = 3. Corresponding function value
Thus, (3; 1) is a conditional extremum (minimum) point.

In the considered example, the constraint equation g(x, y) = C 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 the conditional extremum, in the general case, we use method of Lagrange multipliers.

Consider a function of three variables

This function is called Lagrange function, a - Lagrange multiplier. The following theorem is true.

Theorem.If point
is the conditional extremum point of the functionz = f(x, y) given thatg (x, y) = C, then there is a value such that the point
is the extremum point of the functionL{ x, y, ).

Thus, to find the conditional extremum of the function z = f(x, y) given that g(x, y) = C need to find a solution to the system

On fig. the geometric meaning of the Lagrange conditions is shown. Line g(x, y)= C dotted, level line g(x, y) = Q functions z = f(x, y) solid.

From fig. follows that at the conditional extremum point, the level line of the function z= f(x, y) touches the lineg(x, y) = C.

Example. Find the maximum and minimum points of the function z = X 2 + y 2 given that 3x + 2y = 11 using the Lagrange multiplier method.

Decision. Compose the Lagrange function L= x 2 + 2y 2 +

Equating its partial derivatives to zero, we obtain the system of equations

Its only solution (x=3, y=1, =-2). Thus, only the point (3;1) can be a conditional extremum point. It is easy to verify that at this point the function z= f(x, y) has a conditional minimum.