Definition1: A function is said to have a local maximum at a point if there is a neighborhood of the point such that for any point M with coordinates (x, y) inequality holds: . In this case, i.e., the increment of the function< 0.

Definition2: A function is said to have a local minimum at a point if there is a neighborhood of the point such that for any point M with coordinates (x, y) inequality holds: . In this case, i.e., the increment of the function > 0.

Definition 3: The points of local minimum and maximum are called extremum points.

Conditional Extremes

When finding extrema of a function of many 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 be given L on surface 0xy. The task is to get on the line L find such a point P(x, y), in which the value of a function is the largest or smallest compared to the values of this function at points on the line L, located near the point P. Such points P are called conditional extremum points functions on line L. In contrast to 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 the usual extremum (they also say unconditional extremum) is also a conditional extremum point 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 me explain what I said with a simple example. The graph of the function is the upper hemisphere (Appendix 3 (Fig. 3)).

This function has a maximum at the origin; the vertex corresponds to it M hemispheres. If the line L there is a line passing through the points A And IN(her equation x+y-1=0), 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 the points A And IN. This is the point of conditional extremum (maximum) of the function 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.

Let us now proceed to the practical search for the conditional extremum points of the function Z= f(x, y) provided that the variables x and y are related by the equation (x, y) = 0. We will call this relation the connection equation. If from the coupling equation y can be expressed explicitly in terms of x: y=(x), we obtain a function of one variable Z= f(x, (x)) = Ф(x).

Having found the value x at which this function reaches an extremum, and then determined from the connection equation the corresponding y values, we obtain the desired points of the conditional extremum.

So, in the above example, from the relation equation x+y-1=0 we have y=1-x. From here

It is easy to check that z reaches its maximum at x = 0.5; but then from the connection equation y = 0.5, and we get exactly the point P, found from geometric considerations.

The problem of a conditional extremum is very easily solved even when the connection equation can be represented by parametric equations x=x(t), y=y(t). Substituting expressions for x and y into this function, we again come to the problem of finding the extremum of a function of one variable.

If the coupling equation has a more complex form and we are unable to either explicitly express one variable in terms of another or replace it with parametric equations, then the task of finding a conditional extremum becomes more difficult. We will continue to assume that in the expression of the function z= f(x, y) the variable (x, y) = 0. The total derivative of the function z= f(x, y) is equal to:

Where the derivative y` is found using the rule of differentiation of the implicit function. At the points of the conditional extremum, the found total derivative must be equal to zero; this gives one equation relating x and y. Since they must also satisfy the coupling equation, we get a system of two equations with two unknowns

Let's transform this system to a much more convenient one by writing the first equation in the form of a proportion and introducing a new auxiliary unknown:

(the minus sign in front is for convenience). From these equalities it is easy to move to the following system:

f` x =(x,y)+` x (x,y)=0, f` y (x,y)+` y (x,y)=0 (*),

which, together with the connection equation (x, y) = 0, forms a system of three equations with unknowns x, y and.

These equations (*) are easiest to remember using the following rule: in order to find points that can be points of the conditional extremum of the function

Z= f(x, y) with the connection equation (x, y) = 0, you need to form an auxiliary function

F(x,y)=f(x,y)+(x,y)

Where is some constant, and create equations to find the extremum points of this function.

The indicated system of equations provides, as a rule, only the necessary conditions, i.e. not every pair of values x and y that satisfies this system is necessarily a conditional extremum point. I will not give sufficient conditions for the conditional extremum points; very often the specific content of the problem itself suggests what the found point is. The described technique for solving problems on a conditional extremum is called the Lagrange multiplier method.

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 partial derivatives of the second order of 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
towards (or along the axis ) and is denoted as follows:

This derivative characterizes the “rate of change” of the function at the point
towards . 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.

Example

Given a function, point A(1, 1) and vector
. Find: 1)grad z at point A; 2) derivative at point A in the direction of the vector .

Partial derivatives of a given function at a point
:

;
.

Then the gradient vector of the function at this point is:
. The gradient vector can also be written using vector decomposition And :

. Derivative of a function in the direction of the vector :

So,
,
.◄

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 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.

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 of coordinates. 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 of 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 on 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.

First, let's consider the case of a function of two variables. The conditional extremum of a function $z=f(x,y)$ at the point $M_0(x_0;y_0)$ is the extremum of this function, achieved under the condition that the variables $x$ and $y$ in the vicinity of this point satisfy the connection equation $\ varphi (x,y)=0$.

The name “conditional” extremum is due to the fact that an additional condition $\varphi(x,y)=0$ is imposed on the variables. If one variable can be expressed from the connection equation through another, then the problem of determining the conditional extremum is reduced to the problem of determining the usual extremum of a function of one variable. For example, if the connection equation implies $y=\psi(x)$, then substituting $y=\psi(x)$ into $z=f(x,y)$, we obtain a function of one variable $z=f\left (x,\psi(x)\right)$. In the general case, however, this method is of little use, so the introduction of a new algorithm is required.

Lagrange multiplier method for functions of two variables.

The Lagrange multiplier method consists of constructing a Lagrange function to find a conditional extremum: $F(x,y)=f(x,y)+\lambda\varphi(x,y)$ (the $\lambda$ parameter is called the Lagrange multiplier ). The necessary conditions for the extremum are specified by a system of equations from which the stationary points are determined:

$$ \left \( \begin(aligned) & \frac(\partial F)(\partial x)=0;\\ & \frac(\partial F)(\partial y)=0;\\ & \varphi (x,y)=0. \end(aligned) \right.

A sufficient condition from which one can determine the nature of the extremum is the sign $d^2 F=F_(xx)^("")dx^2+2F_(xy)^("")dxdy+F_(yy)^("" )dy^2$. If at a stationary point $d^2F > 0$, then the function $z=f(x,y)$ has a conditional minimum at this point, but if $d^2F< 0$, то условный максимум.

There is another way to determine the nature of the extremum. From the coupling equation we obtain: $\varphi_(x)^(")dx+\varphi_(y)^(")dy=0$, $dy=-\frac(\varphi_(x)^("))(\varphi_ (y)^("))dx$, therefore at any stationary point we have:

$$d^2 F=F_(xx)^("")dx^2+2F_(xy)^("")dxdy+F_(yy)^("")dy^2=F_(xx)^( "")dx^2+2F_(xy)^("")dx\left(-\frac(\varphi_(x)^("))(\varphi_(y)^("))dx\right)+ F_(yy)^("")\left(-\frac(\varphi_(x)^("))(\varphi_(y)^("))dx\right)^2=\\ =-\frac (dx^2)(\left(\varphi_(y)^(") \right)^2)\cdot\left(-(\varphi_(y)^("))^2 F_(xx)^(" ")+2\varphi_(x)^(")\varphi_(y)^(")F_(xy)^("")-(\varphi_(x)^("))^2 F_(yy)^ ("") \right)$$

The second factor (located in brackets) can be represented in this form:

The elements of the determinant $\left| are highlighted in red. \begin(array) (cc) F_(xx)^("") & F_(xy)^("") \\ F_(xy)^("") & F_(yy)^("") \end (array)\right|$, which is the Hessian of the Lagrange function. If $H > 0$, then $d^2F< 0$, что указывает на условный максимум. Аналогично, при $H < 0$ имеем $d^2F >0$, i.e. we have a conditional minimum of the function $z=f(x,y)$.

A note regarding the notation of the determinant $H$. show\hide

$$ H=-\left|\begin(array) (ccc) 0 & \varphi_(x)^(") & \varphi_(y)^(")\\ \varphi_(x)^(") & F_ (xx)^("") & F_(xy)^("") \\ \varphi_(y)^(") & F_(xy)^("") & F_(yy)^("") \ end(array) \right| $$

In this situation, the rule formulated above will change as follows: if $H > 0$, then the function has a conditional minimum, and if $H< 0$ получим условный максимум функции $z=f(x,y)$. При решении задач следует учитывать такие нюансы.

Algorithm for studying a function of two variables for a conditional extremum

Compose the Lagrange function $F(x,y)=f(x,y)+\lambda\varphi(x,y)$
Solve the system $ \left \( \begin(aligned) & \frac(\partial F)(\partial x)=0;\\ & \frac(\partial F)(\partial y)=0;\\ & \ varphi (x,y)=0. \end(aligned) \right.$
Determine the nature of the extremum at each of the stationary points found in the previous paragraph. To do this, use any of the following methods:
- Compose the determinant of $H$ and find out its sign
- Taking into account the coupling equation, calculate the sign of $d^2F$

Lagrange multiplier method for functions of n variables

Let's say we have a function of $n$ variables $z=f(x_1,x_2,\ldots,x_n)$ and $m$ coupling equations ($n > m$):

$$\varphi_1(x_1,x_2,\ldots,x_n)=0; \; \varphi_2(x_1,x_2,\ldots,x_n)=0,\ldots,\varphi_m(x_1,x_2,\ldots,x_n)=0.$$

Denoting the Lagrange multipliers as $\lambda_1,\lambda_2,\ldots,\lambda_m$, we compose the Lagrange function:

$$F(x_1,x_2,\ldots,x_n,\lambda_1,\lambda_2,\ldots,\lambda_m)=f+\lambda_1\varphi_1+\lambda_2\varphi_2+\ldots+\lambda_m\varphi_m$$

The necessary conditions for the presence of a conditional extremum are given by a system of equations from which the coordinates of stationary points and the values of the Lagrange multipliers are found:

$$\left\(\begin(aligned) & \frac(\partial F)(\partial x_i)=0; (i=\overline(1,n))\\ & \varphi_j=0; (j=\ overline(1,m)) \end(aligned) \right.$$

You can find out whether a function has a conditional minimum or a conditional maximum at the found point, as before, using the sign $d^2F$. If at the found point $d^2F > 0$, then the function has a conditional minimum, but if $d^2F< 0$, - то условный максимум. Можно пойти иным путем, рассмотрев следующую матрицу:

Determinant of the matrix $\left| \begin(array) (ccccc) \frac(\partial^2F)(\partial x_(1)^(2)) & \frac(\partial^2F)(\partial x_(1)\partial x_(2) ) & \frac(\partial^2F)(\partial x_(1)\partial x_(3)) &\ldots & \frac(\partial^2F)(\partial x_(1)\partial x_(n)) \\ \frac(\partial^2F)(\partial x_(2)\partial x_1) & \frac(\partial^2F)(\partial x_(2)^(2)) & \frac(\partial^2F )(\partial x_(2)\partial x_(3)) &\ldots & \frac(\partial^2F)(\partial x_(2)\partial x_(n))\\ \frac(\partial^2F )(\partial x_(3) \partial x_(1)) & \frac(\partial^2F)(\partial x_(3)\partial x_(2)) & \frac(\partial^2F)(\partial x_(3)^(2)) &\ldots & \frac(\partial^2F)(\partial x_(3)\partial x_(n))\\ \ldots & \ldots & \ldots &\ldots & \ ldots\\ \frac(\partial^2F)(\partial x_(n)\partial x_(1)) & \frac(\partial^2F)(\partial x_(n)\partial x_(2)) & \ frac(\partial^2F)(\partial x_(n)\partial x_(3)) &\ldots & \frac(\partial^2F)(\partial x_(n)^(2))\\ \end( array) \right|$, highlighted in red in the matrix $L$, is the Hessian of the Lagrange function. We use the following rule:

If the signs of the angular minors $H_(2m+1),\; H_(2m+2),\ldots,H_(m+n)$ matrices $L$ coincide with the sign of $(-1)^m$, then the stationary point under study is the conditional minimum point of the function $z=f(x_1,x_2 ,x_3,\ldots,x_n)$.
If the signs of the angular minors $H_(2m+1),\; H_(2m+2),\ldots,H_(m+n)$ alternate, and the sign of the minor $H_(2m+1)$ coincides with the sign of the number $(-1)^(m+1)$, then the stationary the point is the conditional maximum point of the function $z=f(x_1,x_2,x_3,\ldots,x_n)$.

Example No. 1

Find the conditional extremum of the function $z(x,y)=x+3y$ under the condition $x^2+y^2=10$.

The geometric interpretation of this problem is as follows: it is required to find the largest and smallest values of the applicate of the plane $z=x+3y$ for the points of its intersection with the cylinder $x^2+y^2=10$.

It is somewhat difficult to express one variable through another from the coupling equation and substitute it into the function $z(x,y)=x+3y$, so we will use the Lagrange method.

Denoting $\varphi(x,y)=x^2+y^2-10$, we compose the Lagrange function:

$$ F(x,y)=z(x,y)+\lambda \varphi(x,y)=x+3y+\lambda(x^2+y^2-10);\\ \frac(\partial F)(\partial x)=1+2\lambda x; \frac(\partial F)(\partial y)=3+2\lambda y. $$

Let us write a system of equations to determine the stationary points of the Lagrange function:

$$ \left \( \begin(aligned) & 1+2\lambda x=0;\\ & 3+2\lambda y=0;\\ & x^2+y^2-10=0. \end (aligned)\right.$$

If we assume $\lambda=0$, then the first equation becomes: $1=0$. The resulting contradiction indicates that $\lambda\neq 0$. Under the condition $\lambda\neq 0$, from the first and second equations we have: $x=-\frac(1)(2\lambda)$, $y=-\frac(3)(2\lambda)$. Substituting the obtained values into the third equation, we get:

$$ \left(-\frac(1)(2\lambda) \right)^2+\left(-\frac(3)(2\lambda) \right)^2-10=0;\\ \frac (1)(4\lambda^2)+\frac(9)(4\lambda^2)=10; \lambda^2=\frac(1)(4); \left[ \begin(aligned) & \lambda_1=-\frac(1)(2);\\ & \lambda_2=\frac(1)(2). \end(aligned) \right.\\ \begin(aligned) & \lambda_1=-\frac(1)(2); \; x_1=-\frac(1)(2\lambda_1)=1; \; y_1=-\frac(3)(2\lambda_1)=3;\\ & \lambda_2=\frac(1)(2); \; x_2=-\frac(1)(2\lambda_2)=-1; \; y_2=-\frac(3)(2\lambda_2)=-3.\end(aligned) $$

So, the system has two solutions: $x_1=1;\; y_1=3;\; \lambda_1=-\frac(1)(2)$ and $x_2=-1;\; y_2=-3;\; \lambda_2=\frac(1)(2)$. Let us find out the nature of the extremum at each stationary point: $M_1(1;3)$ and $M_2(-1;-3)$. To do this, we calculate the determinant of $H$ at each point.

$$ \varphi_(x)^(")=2x;\; \varphi_(y)^(")=2y;\; F_(xx)^("")=2\lambda;\; F_(xy)^("")=0;\; F_(yy)^("")=2\lambda.\\ H=\left| \begin(array) (ccc) 0 & \varphi_(x)^(") & \varphi_(y)^(")\\ \varphi_(x)^(") & F_(xx)^("") & F_(xy)^("") \\ \varphi_(y)^(") & F_(xy)^("") & F_(yy)^("") \end(array) \right|= \left| \begin(array) (ccc) 0 & 2x & 2y\\ 2x & 2\lambda & 0 \\ 2y & 0 & 2\lambda \end(array) \right|= 8\cdot\left| \begin(array) (ccc) 0 & x & y\\ x & \lambda & 0 \\ y & 0 & \lambda \end(array) \right| $$

At point $M_1(1;3)$ we get: $H=8\cdot\left| \begin(array) (ccc) 0 & x & y\\ x & \lambda & 0 \\ y & 0 & \lambda \end(array) \right|= 8\cdot\left| \begin(array) (ccc) 0 & 1 & 3\\ 1 & -1/2 & 0 \\ 3 & 0 & -1/2 \end(array) \right|=40 > 0$, so at the point The $M_1(1;3)$ function $z(x,y)=x+3y$ has a conditional maximum, $z_(\max)=z(1;3)=10$.

Similarly, at point $M_2(-1,-3)$ we find: $H=8\cdot\left| \begin(array) (ccc) 0 & x & y\\ x & \lambda & 0 \\ y & 0 & \lambda \end(array) \right|= 8\cdot\left| \begin(array) (ccc) 0 & -1 & -3\\ -1 & 1/2 & 0 \\ -3 & 0 & 1/2 \end(array) \right|=-40$. Since $H< 0$, то в точке $M_2(-1;-3)$ имеем условный минимум функции $z(x,y)=x+3y$, а именно: $z_{\min}=z(-1;-3)=-10$.

I note that instead of calculating the value of the determinant $H$ at each point, it is much more convenient to expand it in general form. In order not to clutter the text with details, I will hide this method under a note.

Writing the determinant $H$ in general form. show\hide

$$ H=8\cdot\left|\begin(array)(ccc)0&x&y\\x&\lambda&0\\y&0&\lambda\end(array)\right| =8\cdot\left(-\lambda(y^2)-\lambda(x^2)\right) =-8\lambda\cdot\left(y^2+x^2\right). $$

In principle, it is already obvious what sign $H$ has. Since none of the points $M_1$ or $M_2$ coincides with the origin, then $y^2+x^2>0$. Therefore, the sign of $H$ is opposite to the sign of $\lambda$. You can complete the calculations:

$$ \begin(aligned) &H(M_1)=-8\cdot\left(-\frac(1)(2)\right)\cdot\left(3^2+1^2\right)=40;\ \ &H(M_2)=-8\cdot\frac(1)(2)\cdot\left((-3)^2+(-1)^2\right)=-40. \end(aligned) $$

The question about the nature of the extremum at the stationary points $M_1(1;3)$ and $M_2(-1;-3)$ can be solved without using the determinant $H$. Let's find the sign of $d^2F$ at each stationary point:

$$ d^2 F=F_(xx)^("")dx^2+2F_(xy)^("")dxdy+F_(yy)^("")dy^2=2\lambda \left( dx^2+dy^2\right) $$

Let me note that the notation $dx^2$ means exactly $dx$ raised to the second power, i.e. $\left(dx \right)^2$. Hence we have: $dx^2+dy^2>0$, therefore, with $\lambda_1=-\frac(1)(2)$ we get $d^2F< 0$. Следовательно, функция имеет в точке $M_1(1;3)$ условный максимум. Аналогично, в точке $M_2(-1;-3)$ получим условный минимум функции $z(x,y)=x+3y$. Отметим, что для определения знака $d^2F$ не пришлось учитывать связь между $dx$ и $dy$, ибо знак $d^2F$ очевиден без дополнительных преобразований. В следующем примере для определения знака $d^2F$ уже будет необходимо учесть связь между $dx$ и $dy$.

Answer: at point $(-1;-3)$ the function has a conditional minimum, $z_(\min)=-10$. At point $(1;3)$ the function has a conditional maximum, $z_(\max)=10$

Example No. 2

Find the conditional extremum of the function $z(x,y)=3y^3+4x^2-xy$ under the condition $x+y=0$.

First method (Lagrange multiplier method)

Denoting $\varphi(x,y)=x+y$, we compose the Lagrange function: $F(x,y)=z(x,y)+\lambda \varphi(x,y)=3y^3+4x^2 -xy+\lambda(x+y)$.

$$ \frac(\partial F)(\partial x)=8x-y+\lambda; \; \frac(\partial F)(\partial y)=9y^2-x+\lambda.\\ \left \( \begin(aligned) & 8x-y+\lambda=0;\\ & 9y^2-x+\ lambda=0; \\ & x+y=0. \end(aligned) \right.

Having solved the system, we obtain: $x_1=0$, $y_1=0$, $\lambda_1=0$ and $x_2=\frac(10)(9)$, $y_2=-\frac(10)(9)$ , $\lambda_2=-10$. We have two stationary points: $M_1(0;0)$ and $M_2 \left(\frac(10)(9);-\frac(10)(9) \right)$. Let us find out the nature of the extremum at each stationary point using the determinant $H$.

$$H=\left| \begin(array) (ccc) 0 & \varphi_(x)^(") & \varphi_(y)^(")\\ \varphi_(x)^(") & F_(xx)^("") & F_(xy)^("") \\ \varphi_(y)^(") & F_(xy)^("") & F_(yy)^("") \end(array) \right|= \left| \begin(array) (ccc) 0 & 1 & 1\\ 1 & 8 & -1 \\ 1 & -1 & 18y \end(array) \right|=-10-18y $$

At point $M_1(0;0)$ $H=-10-18\cdot 0=-10< 0$, поэтому $M_1(0;0)$ есть точка условного минимума функции $z(x,y)=3y^3+4x^2-xy$, $z_{\min}=0$. В точке $M_2\left(\frac{10}{9};-\frac{10}{9}\right)$ $H=10 >0$, therefore at this point the function has a conditional maximum, $z_(\max)=\frac(500)(243)$.

We investigate the nature of the extremum at each point using a different method, based on the sign of $d^2F$:

$$ d^2 F=F_(xx)^("")dx^2+2F_(xy)^("")dxdy+F_(yy)^("")dy^2=8dx^2-2dxdy+ 18ydy^2 $$

From the connection equation $x+y=0$ we have: $d(x+y)=0$, $dx+dy=0$, $dy=-dx$.

$$ d^2 F=8dx^2-2dxdy+18ydy^2=8dx^2-2dx(-dx)+18y(-dx)^2=(10+18y)dx^2 $$

Since $ d^2F \Bigr|_(M_1)=10 dx^2 > 0$, then $M_1(0;0)$ is the conditional minimum point of the function $z(x,y)=3y^3+4x^ 2-xy$. Similarly, $d^2F \Bigr|_(M_2)=-10 dx^2< 0$, т.е. $M_2\left(\frac{10}{9}; -\frac{10}{9} \right)$ - точка условного максимума.

Second way

From the connection equation $x+y=0$ we get: $y=-x$. Substituting $y=-x$ into the function $z(x,y)=3y^3+4x^2-xy$, we obtain some function of the variable $x$. Let's denote this function as $u(x)$:

$$ u(x)=z(x,-x)=3\cdot(-x)^3+4x^2-x\cdot(-x)=-3x^3+5x^2. $$

Thus, we reduced the problem of finding the conditional extremum of a function of two variables to the problem of determining the extremum of a function of one variable.

$$ u_(x)^(")=-9x^2+10x;\\ -9x^2+10x=0; \; x\cdot(-9x+10)=0;\\ x_1=0; \ ; y_1=-x_1=0;\\ x_2=\frac(10)(9); \; y_2=-x_2=-\frac(10)(9).

We obtained points $M_1(0;0)$ and $M_2\left(\frac(10)(9); -\frac(10)(9)\right)$. Further research is known from the course of differential calculus of functions of one variable. By examining the sign of $u_(xx)^("")$ at each stationary point or checking the change in the sign of $u_(x)^(")$ at the found points, we obtain the same conclusions as when solving the first method. For example, we will check sign $u_(xx)^("")$:

$$u_(xx)^("")=-18x+10;\\ u_(xx)^("")(M_1)=10;\;u_(xx)^("")(M_2)=- 10.$$

Since $u_(xx)^("")(M_1)>0$, then $M_1$ is the minimum point of the function $u(x)$, and $u_(\min)=u(0)=0$ . Since $u_(xx)^("")(M_2)<0$, то $M_2$ - точка максимума функции $u(x)$, причём $u_{\max}=u\left(\frac{10}{9}\right)=\frac{500}{243}$.

The values of the function $u(x)$ for a given connection condition coincide with the values of the function $z(x,y)$, i.e. the found extrema of the function $u(x)$ are the sought conditional extrema of the function $z(x,y)$.

Answer: at the point $(0;0)$ the function has a conditional minimum, $z_(\min)=0$. At the point $\left(\frac(10)(9); -\frac(10)(9) \right)$ the function has a conditional maximum, $z_(\max)=\frac(500)(243)$.

Let's consider another example in which we will clarify the nature of the extremum by determining the sign of $d^2F$.

Example No. 3

Find the largest and smallest values of the function $z=5xy-4$ if the variables $x$ and $y$ are positive and satisfy the coupling equation $\frac(x^2)(8)+\frac(y^2)(2) -1=0$.

Let's compose the Lagrange function: $F=5xy-4+\lambda \left(\frac(x^2)(8)+\frac(y^2)(2)-1 \right)$. Let's find the stationary points of the Lagrange function:

$$ F_(x)^(")=5y+\frac(\lambda x)(4); \; F_(y)^(")=5x+\lambda y.\\ \left \( \begin(aligned) & 5y+\frac(\lambda x)(4)=0;\\ & 5x+\lambda y=0;\\ & \frac(x^2)(8)+\frac(y^2)(2)- 1=0;\\ & x > 0; \; y > 0. \end(aligned) \right.

All further transformations are carried out taking into account $x > 0; \; y > 0$ (this is specified in the problem statement). From the second equation we express $\lambda=-\frac(5x)(y)$ and substitute the found value into the first equation: $5y-\frac(5x)(y)\cdot \frac(x)(4)=0$ , $4y^2-x^2=0$, $x=2y$. Substituting $x=2y$ into the third equation, we get: $\frac(4y^2)(8)+\frac(y^2)(2)-1=0$, $y^2=1$, $y =1$.

Since $y=1$, then $x=2$, $\lambda=-10$. We determine the nature of the extremum at the point $(2;1)$ based on the sign of $d^2F$.

$$ F_(xx)^("")=\frac(\lambda)(4); \; F_(xy)^("")=5; \; F_(yy)^("")=\lambda. $$

Since $\frac(x^2)(8)+\frac(y^2)(2)-1=0$, then:

$$ d\left(\frac(x^2)(8)+\frac(y^2)(2)-1\right)=0; \; d\left(\frac(x^2)(8) \right)+d\left(\frac(y^2)(2) \right)=0; \; \frac(x)(4)dx+ydy=0; \; dy=-\frac(xdx)(4y). $$

In principle, here you can immediately substitute the coordinates of the stationary point $x=2$, $y=1$ and the parameter $\lambda=-10$, obtaining:

$$ F_(xx)^("")=\frac(-5)(2); \; F_(xy)^("")=-10; \; dy=-\frac(dx)(2).\\ d^2 F=F_(xx)^("")dx^2+2F_(xy)^("")dxdy+F_(yy)^(" ")dy^2=-\frac(5)(2)dx^2+10dx\cdot \left(-\frac(dx)(2) \right)-10\cdot \left(-\frac(dx) (2) \right)^2=\\ =-\frac(5)(2)dx^2-5dx^2-\frac(5)(2)dx^2=-10dx^2. $$

However, in other problems on a conditional extremum there may be several stationary points. In such cases, it is better to represent $d^2F$ in general form, and then substitute the coordinates of each of the found stationary points into the resulting expression:

$$ d^2 F=F_(xx)^("")dx^2+2F_(xy)^("")dxdy+F_(yy)^("")dy^2=\frac(\lambda) (4)dx^2+10\cdot dx\cdot \frac(-xdx)(4y) +\lambda\cdot \left(-\frac(xdx)(4y) \right)^2=\\ =\frac (\lambda)(4)dx^2-\frac(5x)(2y)dx^2+\lambda \cdot \frac(x^2dx^2)(16y^2)=\left(\frac(\lambda )(4)-\frac(5x)(2y)+\frac(\lambda \cdot x^2)(16y^2) \right)\cdot dx^2 $$

Substituting $x=2$, $y=1$, $\lambda=-10$, we get:

$$ d^2 F=\left(\frac(-10)(4)-\frac(10)(2)-\frac(10 \cdot 4)(16) \right)\cdot dx^2=- 10dx^2. $$

Since $d^2F=-10\cdot dx^2< 0$, то точка $(2;1)$ есть точкой условного максимума функции $z=5xy-4$, причём $z_{\max}=10-4=6$.

Answer: at point $(2;1)$ the function has a conditional maximum, $z_(\max)=6$.

In the next part we will consider the application of the Lagrange method for functions of a larger number of variables.

Conditional extremum.

Extrema of a function of several variables

Least square 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 p)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 p)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 p) 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 p, putting X 2 =A 2 ,..., x p = 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

and 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(P 1) = 36-9>0 – at point P 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 p), found under the condition that its arguments X 1 , X 2 , ... , x p satisfy the equations j 1 ( X 1 , X 2 , ... , x p) = 0, …, j T(X 1 , X 2 , ... , x p) = 0, where P ( X 1 , X 2 , ... , x p) О D( f), called conditional extremum .

Equations j k(X 1 , X 2 , ... , x p) = 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.