Sufficient condition for the extremum of a function of two variables

1. Let the function be continuously differentiable in some neighborhood of the point and have continuous partial derivatives of the second order (pure and mixed).

2. Let us denote by the second-order determinant

extremum variable lecture function

Theorem

If the point with coordinates is a stationary point for the function, then:

A) At it is a point of local extremum and, at a local maximum, it is a local minimum;

C) at the point is not a local extremum point;

C) if, maybe both.

Proof

Let us write the Taylor formula for the function, limiting ourselves to two terms:

Since, according to the conditions of the theorem, the point is stationary, the second-order partial derivatives are equal to zero, i.e. And. Then

Let's denote

Then the increment of the function will take the form:

Due to the continuity of second-order partial derivatives (pure and mixed), according to the conditions of the theorem at a point, we can write:

Where or; ,

1. Let and, i.e. or.

2. Multiply the increment of the function and divide by, we get:

3. Let's add the expression in curly brackets to the full square of the sum:

4. The expression in curly braces is non-negative, since

5. Therefore, if a means and, then and, therefore, according to the definition, the point is a local minimum point.

6. If a means and, then, according to the definition, the point with coordinates is a point of local maximum.

2. Consider the quadratic trinomial, its discriminant, .

3. If, then there are points such that the polynomial

4. We write the total increment of the function at a point in accordance with the expression obtained in I as:

5. Due to the continuity of second-order partial derivatives, according to the conditions of the theorem at a point, we can write that

Therefore, there is a neighborhood of a point such that, for any point, the quadratic trinomial is greater than zero:

6. Consider the neighborhood of a point.

Let's choose any value, so period. Assuming that in the formula for the increment of the function

What do we get:

7. Since, then.

8. Arguing similarly for the root, we find that in any -neighborhood of a point there is a point for which, therefore, in the neighborhood of the point does not preserve sign, therefore there is no extremum at the point.

Conditional extremum of a function of two variables

When finding extrema of a function of two variables, problems often arise related to the so-called conditional extremum. This concept can be explained using the example of a function of two variables.

Let a function and a line L be given on the 0xy plane. The task is to find a point P (x, y) on line L at which the value of the function is the largest or smallest compared to the values of this function at points on line L located near point P. Such points P are called conditional extremum points functions on line L. Unlike the usual extremum point, the value of the function at the conditional extremum point is compared with the values of the function not at all points of its neighborhood, but only at those that lie on the line L.

It is absolutely clear that the point of ordinary extremum (they also say unconditional extremum) is also the point of conditional extremum for any line passing through this point. The converse, of course, is not true: the conditional extremum point may not be the ordinary extremum point. Let us illustrate this with an example.

Example No. 1. The graph of the function is the upper hemisphere (Fig. 2).

Rice. 2.

This function has a maximum at the origin; it corresponds to the vertex M of the hemisphere. If line L is a straight line passing through points A and B (its equation), then it is geometrically clear that for the points of this line the greatest value of the function is achieved at the point lying in the middle between points A and B. This is the point of conditional extremum (maximum) functions on this line; it corresponds to point M 1 on the hemisphere, and from the figure it is clear that there can be no talk of any ordinary extremum here.

Note that in the final part of the problem of finding the largest and smallest values of a function in a closed region, we have to find the extreme values of the function on the boundary of this region, i.e. on some line, and thereby solve the conditional extremum problem.

Definition 1. They say that where has at a point satisfying the equation a conditional or relative maximum (minimum): if for any point satisfying the equation the inequality

Definition 2. An equation of the form is called a constraint equation.

Theorem

If the functions and are continuously differentiable in the neighborhood of a point, and the partial derivative, and the point is a conditional extremum point of the function with respect to the constraint equation, then the second-order determinant is equal to zero:

Proof

1. Since, according to the conditions of the theorem, the partial derivative and the value of the function, then in a certain rectangle

implicit function defined

A complex function of two variables at a point will have a local extremum, therefore, or.

2. Indeed, according to the invariance property of the first order differential formula

3. The connection equation can be represented in this form, which means

4. Multiply equation (2) by, and (3) by and add them

Therefore, when

arbitrary. etc.

Consequence

The search for conditional extremum points of a function of two variables in practice is carried out by solving a system of equations

So, in the above example No. 1 from the connection equation we have. From here it is easy to check what reaches a maximum at. But then from the communication equation. We obtain point P, found geometrically.

Example No. 2. Find the conditional extremum points of the function relative to the coupling equation.

Let's find the partial derivatives of the given function and the coupling equation:

Let's create a second-order determinant:

Let's write a system of equations to find conditional extremum points:

This means that there are four points of the conditional extremum of the function with coordinates: .

Example No. 3. Find the extremum points of the function.

Equating the partial derivatives to zero: , we find one stationary point - the origin. Here,. Consequently, the point (0, 0) is not an extremum point. The equation is the equation of a hyperbolic paraboloid (Fig. 3) from the figure it can be seen that the point (0, 0) is not an extremum point.

Rice. 3.

The largest and smallest value of a function in a closed region

1. Let the function be defined and continuous in a bounded closed domain D.

2. Let the function have finite partial derivatives in this region, except for individual points of the region.

3. In accordance with Weierstrass’s theorem, in this region there is a point at which the function takes on the largest and smallest values.

4. If these points are internal points of the region D, then obviously they will have a maximum or a minimum.

5. In this case, the points of interest to us are among the suspicious points at the extremum.

6. However, the function can also take on the largest or smallest value at the boundary of region D.

7. In order to find the largest (smallest) value of a function in region D, you need to find all internal points suspicious for an extremum, calculate the value of the function in them, then compare with the value of the function at the boundary points of the region, and the largest of all found values will be largest in closed region D.

8. The method for finding a local maximum or minimum was discussed earlier in section 1.2. and 1.3.

9. It remains to consider the method of finding the largest and smallest values of the function at the boundary of the region.

10. In the case of a function of two variables, the area is usually limited by a curve or several curves.

11. Along such a curve (or several curves), the variables and either depend on one another, or both depend on one parameter.

12. Thus, at the boundary the function turns out to depend on one variable.

13. The method of finding the largest value of a function of one variable was discussed earlier.

14. Let the boundary of region D be given by parametric equations:

Then on this curve the function of two variables will be a complex function of the parameter: . For such a function, the largest and smallest values are determined using the method for determining the largest and smallest values for a function of one variable.

Conditional extremum.

Extrema of a function of several variables

Least squares method.

Local extremum of the FNP

Let the function be given And= f(P), РÎDÌR n and let point P 0 ( A 1 , A 2 , ..., a p) –internal point of set D.

Definition 9.4.

1) Point P 0 is called maximum point functions And= f(P), if there is a neighborhood of this point U(P 0) М D such that for any point P( X 1 , X 2 , ..., x n)О U(P 0) , Р¹Р 0 , the condition is satisfied f(P) £ f(P 0) . Meaning f(P 0) function at the maximum point is called maximum of the function and is designated f(P0) = max f(P) .

2) Point P 0 is called minimum point functions And= f(P), if there is a neighborhood of this point U(P 0)Ì D such that for any point P( X 1 , X 2 , ..., x n)ОU(P 0), Р¹Р 0 , the condition is satisfied f(P)³ f(P 0) . Meaning f(P 0) function at the minimum point is called minimum function and is designated f(P 0) = min f(P).

The minimum and maximum points of a function are called extrema points, the values of the function at the extrema points are called extrema of the function.

As follows from the definition, the inequalities f(P) £ f(P 0) , f(P)³ f(P 0) must be satisfied only in a certain neighborhood of the point P 0, and not in the entire domain of definition of the function, which means that the function can have several extrema of the same type (several minima, several maxima). Therefore, the extrema defined above are called local(local) extremes.

Theorem 9.1. (necessary condition for the extremum of the FNP)

If the function And= f(X 1 , X 2 , ..., x n) has an extremum at the point P 0 , then its first-order partial derivatives at this point are either equal to zero or do not exist.

Proof. Let at point P 0 ( A 1 , A 2 , ..., a p) function And= f(P) has an extremum, for example, a maximum. Let's fix the arguments X 2 , ..., x n, putting X 2 =A 2 ,..., x n = a p. Then And= f(P) = f 1 ((X 1 , A 2 , ..., a p) is a function of one variable X 1. Since this function has X 1 = A 1 extremum (maximum), then f 1 ¢=0or does not exist when X 1 =A 1 (a necessary condition for the existence of an extremum of a function of one variable). But, that means or does not exist at point P 0 - the extremum point. Similarly, we can consider partial derivatives with respect to other variables. CTD.

Points in the domain of a function at which first-order partial derivatives are equal to zero or do not exist are called critical points this function.

As follows from Theorem 9.1, the extremum points of the FNP should be sought among the critical points of the function. But, as for a function of one variable, not every critical point is an extremum point.

Theorem 9.2. (sufficient condition for the extremum of the FNP)

Let P 0 be the critical point of the function And= f(P) and is the second order differential of this function. Then

a) if d 2 u(P 0) > 0 at , then P 0 is a point minimum functions And= f(P);

b) if d 2 u(P0)< 0 при , то Р 0 – точка maximum functions And= f(P);

c) if d 2 u(P 0) is not defined by sign, then P 0 is not an extremum point;

We will consider this theorem without proof.

Note that the theorem does not consider the case when d 2 u(P 0) = 0 or does not exist. This means that the question of the presence of an extremum at the point P 0 under such conditions remains open - additional research is needed, for example, a study of the increment of the function at this point.

In more detailed mathematics courses it is proven that, in particular for the function z = f(x,y) of two variables, the second order differential of which is a sum of the form

the study of the presence of an extremum at the critical point P 0 can be simplified.

Let's denote , , . Let's compose a determinant

Turns out:

d 2 z> 0 at point P 0, i.e. P 0 – minimum point, if A(P 0) > 0 and D(P 0) > 0;

d 2 z < 0 в точке Р 0 , т.е. Р 0 – точка максимума, если A(P0)< 0 , а D(Р 0) > 0;

if D(P 0)< 0, то d 2 z in the vicinity of point P 0 it changes sign and there is no extremum at point P 0;

if D(Р 0) = 0, then additional studies of the function in the vicinity of the critical point Р 0 are also required.

Thus, for the function z = f(x,y) of two variables we have the following algorithm (let’s call it “algorithm D”) for finding an extremum:

1) Find the domain of definition D( f) functions.

2) Find critical points, i.e. points from D( f), for which and are equal to zero or do not exist.

3) At each critical point P 0, check the sufficient conditions for the extremum. To do this, find , where , , and calculate D(P 0) and A(P 0).Then:

if D(P 0) >0, then at point P 0 there is an extremum, and if A(P 0) > 0 – then this is the minimum, and if A(P 0)< 0 – максимум;

if D(P 0)< 0, то в точке Р 0 нет экстремума;

If D(P 0) = 0, then additional research is needed.

4) At the found extremum points, calculate the value of the function.

Example 1.

Find the extremum of the function z = x 3 + 8y 3 – 3xy .

Solution. The domain of definition of this function is the entire coordinate plane. Let's find critical points.

, , Þ P 0 (0,0) , .

Let us check whether the sufficient conditions for the extremum are met. We'll find

6X, = -3, = 48at And = 288xy – 9.

Then D(P 0) = 288×0×0 – 9 = -9< 0 , значит, в точке Р 0 экстремума нет.

D(Р 1) = 36-9>0 – at point Р 1 there is an extremum, and since A(P 1) = 3 >0, then this extremum is a minimum. So min z=z(P 1) = .

Example 2.

Find the extremum of the function .

Solution: D( f) =R 2 . Critical points: ; does not exist when at= 0, which means P 0 (0,0) is the critical point of this function.

2, = 0, = , = , but D(P 0) is not defined, so studying its sign is impossible.

For the same reason, it is impossible to apply Theorem 9.2 directly - d 2 z does not exist at this point.

Let's consider the increment of the function f(x, y) at point P 0 . If D f =f(P) – f(P 0)>0 "P, then P 0 is the minimum point, but if D f < 0, то Р 0 – точка максимума.

In our case we have

D f = f(x, y) – f(0, 0) = f(0+D x,0+D y) – f(0, 0) = .

At D x= 0.1 and D y= -0.008 we get D f = 0,01 – 0,2 < 0, а при Dx= 0.1 and D y= 0.001 D f= 0.01 + 0.1 > 0, i.e. in the vicinity of point P 0 neither condition D is satisfied f <0 (т.е. f(x, y) < f(0, 0) and therefore P 0 is not a maximum point), nor condition D f>0 (i.e. f(x, y) > f(0, 0) and then P 0 is not a minimum point). This means, by definition of an extremum, this function has no extrema.

Conditional extremum.

The considered extremum of the function is called unconditional, since no restrictions (conditions) are imposed on the function arguments.

Definition 9.2. Extremum of the function And = f(X 1 , X 2 , ... , x n), found under the condition that its arguments X 1 , X 2 , ... , x n satisfy the equations j 1 ( X 1 , X 2 , ... , x n) = 0, …, j T(X 1 , X 2 , ... , x n) = 0, where P ( X 1 , X 2 , ... , x n) О D( f), called conditional extremum .

Equations j k(X 1 , X 2 , ... , x n) = 0 , k = 1, 2,..., m, are called connection equations.

Let's look at the functions z = f(x,y) two variables. If the connection equation is one, i.e. , then finding a conditional extremum means that the extremum is sought not in the entire domain of definition of the function, but on some curve lying in D( f) (i.e., it is not the highest or lowest points of the surface that are sought z = f(x,y), and the highest or lowest points among the points of intersection of this surface with the cylinder, Fig. 5).

Conditional extremum of a function z = f(x,y) of two variables can be found in the following way( elimination method). From the equation, express one of the variables as a function of another (for example, write ) and, substituting this value of the variable into the function, write the latter as a function of one variable (in the case considered ). Find the extremum of the resulting function of one variable.

Let the function z - /(x, y) be defined in some domain D and let Mo(xo, Vo) be an interior point of this domain. Definition. If there is a number such that for all satisfying the conditions the inequality is true, then the point Mo(xo, y) is called the local maximum point of the function f(x, y); if for all Dx, Du, satisfying the conditions | then the point Mo(xo,yo) is called a thin local minimum. In other words, the point M0(x0, y0) is a point of maximum or minimum of the function /(x, y) if there is a 6-neighborhood of the point A/o(x0, y0) such that at all points M(x, y) of this in the neighborhood, the increment of the function maintains its sign. Examples. 1. For the function point - minimum point (Fig. 17). 2. For the function, point 0(0,0) is the maximum point (Fig. 18). 3. For a function, point 0(0,0) is a local maximum point. 4 Indeed, there is a neighborhood of the point 0(0, 0), for example, a circle of radius j (see Fig. 19), at any point of which, different from the point 0(0,0), the value of the function /(x,y) less than 1 = We will consider only points of strict maximum and minimum of functions when strict inequality or strict inequality is satisfied for all points M(x) y) from some punctured 6-neighborhood of the point Mq. The value of a function at the maximum point is called the maximum, and the value of the function at the minimum point is called the minimum of this function. The maximum and minimum points of a function are called the extremum points of the function, and the maximums and minimums of the function themselves are called its extrema. Theorem 11 (necessary condition for an extremum). If a function is an extremum of a function of several variables. The concept of an extremum of a function of several variables. Necessary and sufficient conditions for an extremum Conditional extremum The largest and smallest values of continuous functions have an extremum at the point then at this point each partial derivative and u either vanishes or does not exist. Let the function z = f(x) y) have an extremum at the point M0(x0, yо). Let's give the variable y the value oo. Then the function z = /(x, y) will be a function of one variable x\ Since at x = xo it has an extremum (maximum or minimum, Fig. 20), then its derivative with respect to x = “o, | (*o,l>)" Equal to zero or does not exist. Similarly, we are convinced that) is either equal to zero or does not exist. The points at which = 0 and χ = 0 or do not exist are called critical points of the function z = Dx, y). The points at which $£ = φ = 0 are also called stationary points of the function. Theorem 11 expresses only the necessary conditions for the extremum, which are not sufficient. Example Function Fig. 18 Fig. 20 immt derivatives which turn to zero at. But this function is thin on the imvat of the strum. Indeed, the function is equal to zero at the point 0(0,0) and takes positive and negative values at points M(x,y), arbitrarily close to the point 0(0,0). For it, so at points at points (0, y) for arbitrarily small Point 0(0,0) of the indicated type is called a minimax point (Fig. 21). Sufficient conditions for an extremum of a function of two variables are expressed by the following theorem. Theorem 12 (sufficient conditions for an extremum in two variables). Let the point Mo(xo»Yo) be a stationary point of the function f(x, y), and in some neighborhood of the point /, including the point Mo itself, the function f(z, y) has continuous partial derivatives up to the second order inclusive. Then". at the point Mo(xo, V0) the function /(xo, y) does not have an extremum if D(xo, yo)< 0. Если же то в точке Мо(жо> The extremum of the function f(x, y) may or may not exist. In this case, further research is required. m Let us limit ourselves to proving statements 1) and 2) of the theorem. Let us write the second-order Taylor formula for the function /(i, y): where. According to the condition, it is clear that the sign of the increment D/ is determined by the sign of the trinomial on the right side of (1), i.e., the sign of the second differential d2f. Let's denote it for brevity. Then equality (l) can be written as follows: Let at the point MQ(so, V0) we have... Since, by condition, the second-order partial derivatives of the function f(s, y) are continuous, then inequality (3) will also hold at some neighborhood of the point M0(s0,yo). If the condition is satisfied (at the point А/0, and by virtue of continuity, the derivative /,z(s,y) will retain its sign in a certain neighborhood of the point Af0. In the region where А Ф 0, we have. It is clear from this that if ЛС - В2 > 0 in some neighborhood of the point M0(x0) y0), then the sign of the trinomial AAx2 -I- 2BAxAy + CDy2 coincides with the sign of A at the point (so, V0) (as well as with the sign of C, since for AC - B2 > 0 A and C cannot have different signs). Since the sign of the sum AAs2 + 2BAxAy + CAy2 at the point (s0 + $ Ax, y0 + 0 Dyn) determines the sign of the difference, we come to the following conclusion: if for the function /(s,y) at the stationary point (s0, V0) condition, then for sufficiently small || inequality will be satisfied. Thus, at the point (sq, V0) the function /(s, y) has a maximum. If the condition is satisfied at the stationary point (s0, y0), then for all sufficiently small |Dr| and |Du| the inequality is true, which means that at the point (so,yo) the function /(s, y) has a minimum. Examples. 1. Investigate the function for an extremum 4 Using the necessary conditions for an extremum, we look for stationary points of the function. To do this, we find the partial derivatives u and equate them to zero. We obtain a system of equations from where - a stationary point. Let us now use Theorem 12. We have This means that there is an extremum at point Ml. Because this is the minimum. If we transform the function r into form, it is easy to see that the right side (“) will be minimal when is the absolute minimum of this function. 2. Examine the function for an extremum. We find stationary points of the function, for which we compose a system of equations. Hence, so that the point is stationary. Since, by virtue of Theorem 12, there is no extremum at point M. * 3. Investigate the extremum of the function. Find the stationary points of the function. From the system of equations we obtain that, so the point is stationary. Next we have that Theorem 12 does not answer the question about the presence or absence of an extremum. Let's do it this way. For a function about all points different from the point so, by definition, and the point A/o(0,0) the function r has an absolute minimum. By similar calculations we establish that the function has a maximum at the point, but the function does not have an extremum at the point. Let a function of n independent variables be differentiable at a point. Point Mo is called a stationary point of the function if Theorem 13 (up to sufficient conditions for an extremum). Let the function be defined and have continuous partial derivatives of the second order in some neighborhood of the fine Mt(xi..., which is a stationary fine function if the quadratic form (the second differential of the function f in the fine is positive definite (negative definite), the minimum point (respectively, fine maximum) of the function f is fine. If the quadratic form (4) is sign-alternating, then there is no extremum in the fine LG0. In order to determine whether the quadratic form (4) is positive or negative definite, you can use, for example, the Sylvester criterion for positive (negative). ) the certainty of the quadratic form. 15.2 Conditional extrema Until now, we have been looking for local extrema of a function in the entire domain of its definition, when the arguments of the function are not bound by any additional conditions. Such extrema are called unconditional. However, problems of finding so-called conditional extrema are often encountered. Let the function z = /(x, y) be defined in the domain D. Let us assume that a curve L is given in this domain, and we need to find the extrema of the function f(x> y) only among those of its values that correspond to the points of the curve L. The same extrema are called conditional extrema of the function z = f(x) y) on the curve L. Definition They say that at a point lying on the curve L, the function /(x, y) has a conditional maximum (minimum) if the inequality is satisfied at all points M (s, y) y) curve L, belonging to some neighborhood of the point M0(x0, V0) and different from the point M0 (If the curve L is given by an equation, then the problem is to find the conditional extremum of the function r - f(x,y) on the curve! can be formulated as follows: find the extrema of the function x = /(z, y) in the region D, provided that Thus, when finding the conditional extrema of the function z = y), the arguments of wildebeest can no longer be considered as independent variables: they are related to each other by the relation y ) = 0, which is called the coupling equation. To clarify the distinction between unconditional and conditional extremum, let’s look at an example, the unconditional maximum of a function (Fig. 23) is equal to one and is achieved at point (0,0). It corresponds to point M - the vertex of the pvvboloid. Let us add the connection equation y = j. Then the conditional maximum will obviously be equal to it. It is reached at the point (o,|), and it corresponds to the vertex Afj of the ball, which is the line of intersection of the ball with the plane y = j. In the case of an unconditional mvximum, we have a mvximum applicate among all vpplicvt of the surface * = 1 - l;2 ~ y1; summvv conditional - only among all the points of the pvraboloidv, corresponding to the point * of the straight line y = j not of the xOy plane. One of the methods for finding the conditional extremum of a function in the presence and connection is as follows. Let the connection equation y) - O define y as a unique differentiable function of the argument x: Substituting a function instead of y into the function, we obtain a function of one argument in which the connection condition is already taken into account. The (unconditional) extremum of the function is the desired conditional extremum. Example. Find the extremum of a function under the condition Extremum of a function of several variables The concept of extremum of a function of several variables. Necessary and sufficient conditions for an extremum Conditional extremum The largest and smallest values of continuous functions A From the connection equation (2") we find y = 1-x. Substituting this value y into (V), we obtain a function of one argument x: Let us examine it for the extremum: whence x = 1 is a critical point, so it provides a conditional minimum of the function g (Fig. 24). Let us indicate another way to solve the problem of a conditional extremum, called the Lagrange multiplier method. Let there be a point of a conditional extremum of a function in the presence of a connection. Let us assume that the connection equation defines a unique continuously differentiable function in a certain neighborhood of the point xx. Assuming that we obtain that the derivative with respect to x of the function /(r, ip(x)) at the point xq must be equal to zero or, which is equivalent to this, the differential of must be equal to zero. f(x, y) at the point Mo" O) From the connection equation we have (5) Multiplying the last equality by an as yet undetermined numerical factor A and adding term by term with equality (4), we will have (we assume that). Then, due to the arbitrariness of dx, we obtain Equalities (6) and (7) express the necessary conditions for an unconditional extremum at the point of the function, which is called the Lagrange function. Thus, the conditional extremum point of the function /(x, y), if, is necessarily a stationary point of the Lagrange function where A is a certain numerical coefficient. From here we obtain a rule for finding conditional extrema: in order to find points that can be points of the conventional extremum of a function in the presence of a connection, 1) we compose the Lagrange function, 2) by equating the derivatives and μ of this function to zero and adding the connection equation to the resulting equations, we obtain a system of three equations from which we find the values of A and the coordinates x, y of possible extremum points. The question of the existence and nature of the conditional extremum is resolved on the basis of studying the sign of the second differential of the Lagrange function for the considered system of values x0, V0, A, obtained from (8) provided that If, then at the point (x0, V0) the function /(x, y ) has a conditional maximum; if d2F > 0 - then a conditional minimum. In particular, if at a stationary point (xo, J/o) the determinant D for the function F(x, y) is positive, then at the point (®o, V0) there is a conditional maximum of the function f(x, y), if and conditional minimum of the function /(x, y), if Example. Let us turn again to the conditions of the previous example: find the extremum of the function under the condition that x + y = 1. We will solve the problem using the Lagrange multiplier method. The Lagrange function in this case has the form To find stationary points, we compose a system. From the first two equations of the system, we obtain that x = y. Then from the third equation of the system (connection equation) we find that x - y = j are the coordinates of the possible extremum point. In this case (it is indicated that A = -1. Thus, the Lagrange function. is the conditional minimum point of the function * = x2 + y2 under the condition There is no unconditional extremum for the Lagrange function. P(x, y) does not yet mean the absence of a conditional extremum for the function /(x, y) in the presence of a connection Example. Find the extremum of a function under the condition y 4 We compose the Lagrange function and write out a system for determining A and the coordinates of possible extremum points: From the first two equations we obtain x + y = 0 and we come to the system from where x = y = A = 0. Thus, the corresponding Lagrange function has the form At the point (0,0), the function F(x, y; 0) does not have an unconditional extremum, but there is a conditional extremum of the function r = xy when y = x. Indeed, in this case r = x2. From here it is clear that at the point (0,0) there is a conditional minimum "The method of Lagrange multipliers is transferred to the case of functions of any number of arguments. Let us look for the extremum of the function in the presence of connection equations. Let us compose the Lagrange function where. A|, Az,..., A„, are indefinite constant factors. Equating to zero all first-order partial derivatives of the function F and adding connection equations (9) to the resulting equations, we obtain a system of n + m equations, from which we determine Ab A3|..., At and coordinates x\) x2). » xn of possible points of conditional extremum. The question of whether the points found using the Lagrange method are indeed points of a conditional extremum can often be resolved on the basis of considerations of a physical or geometric nature. 15.3. The largest and smallest values of continuous functions Let it be necessary to find the largest (smallest) value of a function z = /(x, y), continuous in some closed limited domain D. According to Theorem 3, in this region there is a point (xo, V0) at which the function takes the largest (smallest) value. If the point (xo, y0) lies inside the domain D, then the function / has a maximum (minimum) in it, so in this case the point of interest to us is contained among the critical points of the function /(x, y). However, the function /(x, y) can reach its greatest (smallest) value at the boundary of the region. Therefore, in order to find the largest (smallest) value taken by the function z = /(x, y) in a limited closed area 2), you need to find all the maxima (minimum) of the function achieved inside this area, as well as the largest (smallest) value of the function in border of this area. The largest (smallest) of all these numbers will be the desired largest (smallest) value of the function z = /(x,y) in region 27. Let us show how this is done in the case of a differentiable function. Prmmr. Find the largest and smallest values of the function of region 4. We find the critical points of the function inside region D. To do this, we compose a system of equations. From here we obtain x = y « 0, so that point 0 (0,0) is the critical point of the function x. Since Let us now find the largest and smallest values of the function on the boundary Г of the region D. On part of the boundary we have that y = 0 is a critical point, and since = then at this point the function z = 1 + y2 has a minimum equal to one. At the ends of the segment Г", at points (, we have. Using symmetry considerations, we obtain the same results for other parts of the boundary. We finally obtain: the smallest value of the function z = x2+y2 in the region "B is equal to zero and it is achieved at the internal point 0( 0, 0) region, and the maximum value of this function, equal to two, is achieved at four points of the boundary (Fig. 25) Fig. 25 Exercises Find the domain of definition of the functions: Construct the level lines of the functions: 9 Find the level surfaces of the functions of three independent variables: Calculate the limits functions: Find partial derivatives of functions and their total differentials: Find derivatives of complex functions: 3 Find J. Extremum of a function of several variables Concept of extremum of a function of several variables Necessary and sufficient conditions for an extremum Conditional extremum The largest and smallest values of continuous functions 34. Using the formula for the derivative of a complex function. two variables, find and functions: 35. Using the formula for the derivative of a complex function of two variables, find |J and functions: Find jj functions given implicitly: 40. Find the slope of the tangent curve at the point of its intersection with the line x = 3. 41. Find the points at which the tangent of the x curve is parallel to the Ox axis. . In the following problems, find and T: Write the equations of the tangent plane and the normal of the surface: 49. Write the equations of the tangent planes of the surface x2 + 2y2 + 3z2 = 21, parallel to the plane x + 4y + 6z = 0. Find the first three or four terms of the expansion using the Taylor formula : 50. y in the vicinity of the point (0, 0). Using the definition of an extremum of a function, examine the following functions for extremum:). Using sufficient conditions for the extremum of a function of two variables, examine the extremum of the function: 84. Find the largest and smallest values of the function z = x2 - y2 in a closed circle 85. Find the largest and smallest values of the function * = x2y(4-x-y) in a triangle bounded by straight lines x = 0, y = 0, x + y = b. 88. Determine the dimensions of a rectangular open pool that has the smallest surface, provided that its volume is equal to V. 87. Find the dimensions of a rectangular parallelepiped that has the maximum volume given the total surface 5. Answers 1. and | A square formed by line segments x including its sides. 3. Family of concentric rings 2= 0,1,2,... .4. The entire plane except for the points on the straight lines. Part of the plane located above the parabola y = -x?. 8. Points of the circle x. The entire plane except for straight lines x The radical expression is non-negative in two cases j * ^ or j x ^ ^ which is equivalent to an infinite series of inequalities, respectively. The domain of definition is shaded squares (Fig. 26); l which is equivalent to an infinite series The function is defined in points. a) Straight lines parallel to straight line x b) concentric circles with the center at the origin. 10. a) parabolas y) parabolas y a) parabolas b) hyperbolas | .Planes xc. 13. Prim - single-cavity hyperboloids of rotation around the Oz axis; when and are two-sheet hyperboloids of rotation around the Oz axis, both families of surfaces are separated by a cone; There is no limit, b) 0. 18. Let us set y = kxt then z lim z = -2, so the given function at the point (0,0) has no limit. 19. a) Point (0,0); b) point (0,0). 20. a) Break line - circle x2 + y2 = 1; b) the break line is the straight line y = x. 21. a) Break lines - coordinate axes Ox and Oy; b) 0 (empty set). 22. All points (m, n), where and n are integers

Necessary and sufficient conditions for extremum of functions of two variables. A point is called a minimum (maximum) point of a function if in a certain neighborhood of the point the function is defined and satisfies the inequality (respectively, the maximum and minimum points are called extremum points of the function.

A necessary condition for an extremum. If at an extremum point a function has first partial derivatives, then they vanish at this point. It follows that to find the extremum points of such a function, one must solve a system of equations. Points whose coordinates satisfy this system are called critical points of the function. Among them there may be maximum points, minimum points, and also points that are not extremum points.

Sufficient extremum conditions are used to identify extremum points from a set of critical points and are listed below.

Let the function have continuous second partial derivatives at the critical point. If at this point it is true

condition then it is a minimum point at and a maximum point at If at a critical point then it is not an extremum point. In this case, a more subtle study of the nature of the critical point is required, which in this case may or may not be an extremum point.

Extrema of functions of three variables. In the case of a function of three variables, the definitions of extremum points repeat verbatim the corresponding definitions for a function of two variables. We limit ourselves to presenting the procedure for studying a function for an extremum. When solving a system of equations, one should find the critical points of the function, and then at each of the critical points calculate the values

If all three quantities are positive, then the critical point in question is the minimum point; if then this critical point is a maximum point.

Conditional extremum of a function of two variables. A point is called a conditional minimum (maximum) point of a function provided that there is a neighborhood of the point at which the function is defined and in which (respectively) for all points whose coordinates satisfy the equation

To find conditional extremum points, use the Lagrange function

where the number is called the Lagrange multiplier. Solving a system of three equations

find the critical points of the Lagrange function (as well as the value of the auxiliary factor A). At these critical points there may be a conditional extremum. The above system provides only necessary conditions for an extremum, but not sufficient ones: it can be satisfied by the coordinates of points that are not points of a conditional extremum. However, based on the essence of the problem, it is often possible to determine the nature of the critical point.

Conditional extremum of a function of several variables. Let us consider the function of variables under the condition that they are related by the equations

Example

Find the extremum of the function provided that X And at are related by the relation: . Geometrically, the problem means the following: on an ellipse
plane
.

This problem can be solved this way: from the equation
we find
X:

provided that
, reduced to the problem of finding the extremum of a function of one variable on the interval
.

Geometrically, the problem means the following: on an ellipse , obtained by crossing the cylinder
plane
, you need to find the maximum or minimum value of the applicate (Fig.9). This problem can be solved this way: from the equation
we find
. Substituting the found value of y into the equation of the plane, we obtain a function of one variable X:

Thus, the problem of finding the extremum of the function
provided that
, reduced to the problem of finding the extremum of a function of one variable on an interval.

So, the problem of finding a conditional extremum– this is the problem of finding the extremum of the objective function
, provided that the variables X And at subject to restriction
, called connection equation.

Let's say that dot
, satisfying the coupling equation, is the point of local conditional maximum (minimum), if there is a neighborhood
such that for any points
, whose coordinates satisfy the connection equation, the inequality is satisfied.

If from the coupling equation one can find an expression for at, then, by substituting this expression into the original function, we turn the latter into a complex function of one variable X.

The general method for solving the conditional extremum problem is Lagrange multiplier method. Let's create an auxiliary function, where ─ some number. This function is called Lagrange function, A ─ Lagrange multiplier. Thus, the task of finding a conditional extremum has been reduced to finding local extremum points for the Lagrange function. To find possible extremum points, you need to solve a system of 3 equations with three unknowns x, y And.

Then you should use the following sufficient condition for an extremum.

THEOREM. Let the point be a possible extremum point for the Lagrange function. Let us assume that in the vicinity of the point
there are continuous second-order partial derivatives of the functions And . Let's denote

Then if
, That
─ conditional extremum point of the function
with the coupling equation
in this case, if
, That
─ conditional minimum point, if
, That
─ conditional maximum point.

§8. Gradient and directional derivative

Let the function
defined in some (open) region. Consider any point
this area and any directed straight line (axis) , passing through this point (Fig. 1). Let
- some other point on this axis,
– length of the segment between
And
, taken with a plus sign, if the direction
coincides with the direction of the axis , and with a minus sign if their directions are opposite.

Let
approaches indefinitely
. Limit

called derivative of a function
in the direction (or along the axis ) and is denoted as follows:

This derivative characterizes the “rate of change” of the function at the point
in the direction . In particular, the ordinary partial derivatives ,can also be thought of as derivatives "with respect to direction".

Let us now assume that the function
has continuous partial derivatives in the region under consideration. Let the axis forms angles with the coordinate axes
And . Under the assumptions made, the directional derivative exists and is expressed by the formula

If the vector
given by its coordinates
, then the derivative of the function
in the direction of the vector
can be calculated using the formula:

Vector with coordinates
called gradient vector functions
at the point
. The gradient vector indicates the direction of the fastest increase in the function at a given point.