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

Definition 3: Local minimum and maximum points are called extremum points.

Conditional Extremes

When searching for extrema of a function of many variables, problems often arise related to the so-called conditional extreme. This concept can be explained by the example of a function of two variables.

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

It is quite 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: a conditional extremum point may not be a conventional extremum point. Let me explain this 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; it corresponds to the top M hemispheres. If the line L there is a line passing through the points BUT and AT(her equation x+y-1=0), then it is geometrically clear that for the points of this line the maximum value of the function is reached at the point lying in the middle between the points BUT and AT. This is the point of the conditional extremum (maximum) of the function on the given line; it corresponds to the point M 1 on the hemisphere, and it can be seen from the figure that there can be no question 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 extremal values ​​of the function on the boundary of this region, i.e. on some line, and thereby solve the problem for a conditional extremum.

Let us now proceed to the practical search for the points of the conditional extremum of the function Z= f(x, y) provided that the variables x and y are related by the equation (x, y) = 0. This relation will be called the constraint equation. If from the connection equation y can be expressed explicitly in terms of x: y \u003d (x), we get a function of one variable Z \u003d f (x, (x)) \u003d Ф (x).

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

So, in the above example, from the equation of communication 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 conditional extremum problem is solved very simply even when the constraint equation can be represented by parametric equations x=x(t), y=y(t). Substituting the 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 constraint equation has a more complex form and we can neither explicitly express one variable in terms of another, nor replace it with parametric equations, then the problem 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 is the derivative y`, found by 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 constraint 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 as a proportion and introducing a new auxiliary unknown:

(a minus sign is placed in front for convenience). It is easy to pass from these equalities to the following system:

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

which, together with the constraint 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 constraint equation (x, y) = 0, you need to form an auxiliary function

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

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

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

Let us first consider the case of a function of two variables. The conditional extremum of the function $z=f(x,y)$ at the point $M_0(x_0;y_0)$ is the extremum of this function, reached under the condition that the variables $x$ and $y$ in the vicinity of this point satisfy the constraint equation $\ varphi(x,y)=0$.

The name "conditional" extremum is due to the fact that the additional condition $\varphi(x,y)=0$ is imposed on the variables. If it is possible to express one variable in terms of another from the connection equation, then the problem of determining the conditional extremum is reduced to the problem of the usual extremum of a function of one variable. For example, if $y=\psi(x)$ follows from the constraint equation, then substituting $y=\psi(x)$ into $z=f(x,y)$, we get a function of one variable $z=f\left (x,\psi(x)\right)$. In the general case, however, this method is of little use, so a new algorithm is required.

Method of Lagrange multipliers for functions of two variables.

The method of Lagrange multipliers is that to find the conditional extremum, the Lagrange function is composed: $F(x,y)=f(x,y)+\lambda\varphi(x,y)$ (the parameter $\lambda$ is called the Lagrange multiplier ). The necessary extremum conditions are given 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.$$

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 constraint equation we get: $\varphi_(x)^(")dx+\varphi_(y)^(")dy=0$, $dy=-\frac(\varphi_(x)^("))(\varphi_ (y)^("))dx$, so 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:

Elements of the $\left| \begin(array) (cc) F_(xx)^("") & F_(xy)^("") \\ F_(xy)^("") & F_(yy)^("") \end (array) \right|$ 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 $z=f(x,y)$.

Note on the form of the $H$ determinant. 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 changes as follows: if $H > 0$, then the function has a conditional minimum, and for $H< 0$ 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

1. Compose the Lagrange function $F(x,y)=f(x,y)+\lambda\varphi(x,y)$
2. Solve system $ \left \( \begin(aligned) & \frac(\partial F)(\partial x)=0;\\ & \frac(\partial F)(\partial y)=0;\\ & \ varphi(x,y)=0.\end(aligned)\right.$
3. 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 $H$ and find out its sign
   • Taking into account the constraint equation, calculate the sign of $d^2F$

Lagrange multiplier method for functions of n variables

Suppose we have a function of $n$ variables $z=f(x_1,x_2,\ldots,x_n)$ and $m$ constraint 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.$$

It is possible to 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:

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

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

Example #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 value 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 in terms of another from the constraint 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 down the system of equations for determining 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 says 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 $H$ at each of the points.

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

At the 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 point $M_1(1;3)$ the function $z(x,y)=x+3y$ has a conditional maximum, $z_(\max)=z(1;3)=10$.

Similarly, at the 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 open it in a general way. In order not to clutter up the text with details, I will hide this method under a note.

Determinant $H$ notation 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 which 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 also 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 $H$. Find the sign of $d^2F$ at each stationary point:

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

I note that the notation $dx^2$ means exactly $dx$ raised to the second power, i.e. $\left(dx\right)^2$. Hence we have: $dx^2+dy^2>0$, so for $\lambda_1=-\frac(1)(2)$ 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 the point $(-1;-3)$ the function has a conditional minimum, $z_(\min)=-10$. At the point $(1;3)$ the function has a conditional maximum, $z_(\max)=10$

Example #2

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

The first way (the method of Lagrange multipliers)

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.$$

Solving the system, we get: $x_1=0$, $y_1=0$, $\lambda_1=0$ 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 $M_2\left(\frac{10}{9};-\frac{10}{9}\right)$ $H=10 >0$, so at this point the function has a conditional maximum, $z_(\max)=\frac(500)(243)$.

We investigate the nature of the extremum at each of the points by 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 constraint 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 constraint 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.

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

Got points $M_1(0;0)$ and $M_2\left(\frac(10)(9); -\frac(10)(9)\right)$. Further research is known from the course of the differential calculus of functions of one variable. Examining the sign of $u_(xx)^("")$ at each stationary point or checking the sign change of $u_(x)^(")$ at the found points, we obtain the same conclusions as in the first solution. For example, 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 $u(x)$, while $u_(\min)=u(0)=0$ . 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 $u(x)$ under the 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 desired 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 one more example, in which we find out the nature of the extremum by determining the sign of $d^2F$.

Example #3

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

Compose the Lagrange function: $F=5xy-4+\lambda \left(\frac(x^2)(8)+\frac(y^2)(2)-1 \right)$. 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 stipulated in the condition of the problem). 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 $x=2$, $\lambda=-10$. The nature of the extremum at the point $(2;1)$ is determined from 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 $\lambda=-10$, thus 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 for a conditional extremum, there may be several stationary points. In such cases, it is better to represent $d^2F$ in a 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 the 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.

Example

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

This problem can be solved as follows: from the equation
find
X:

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

Geometrically, the problem means the following: on an ellipse obtained by crossing the cylinder
plane
, it is required to find the maximum or minimum value of the applicate (Fig. 9). This problem can be solved as follows: from the equation
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 a segment.

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

We will say that dot
, satisfying the constraint equation, is a point of local conditional maximum (minimum) if there is a neighborhood
such that for any points
, whose coordinates satisfy the constraint equation, the inequality holds.

If from the equation of communication it is possible to find an expression for at, then, 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 problem of finding a conditional extremum has been reduced to finding local extremum points for the Lagrange function. To find the points of a possible extremum, it is necessary to solve a system of 3 equations with three unknowns x, y and.

Then one should use the following sufficient extremum condition.

THEOREM. Let the point be a point of possible extremum for the Lagrange function. We assume that in the vicinity of the point
there are continuous second-order partial derivatives of the functions and . Denote

Then if
, then
─ conditional extremum point of the function
at the constraint equation
meanwhile, if
, then
─ conditional minimum point, if
, then
─ point of conditional maximum.

§eight. Gradient and directional derivative

Let the function
defined in some (open) domain. Consider any point
this area and any directed straight line (axis) passing through this point (Fig. 1). Let
- some other point of this axis,
- the 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 function derivative
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, and ordinary partial derivatives ,can also be thought of as derivatives "with respect to direction".

Suppose now 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
set 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 of the function at a given point.

Example

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

. Function derivative in the direction of the vector :

So,
,
.◄

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

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

Sufficient extremum conditions are used to select extremum points from the set of critical points and are listed below.

Let the function have continuous second partial derivatives at the critical point. If at this point,

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 the 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 confine ourselves to presenting the procedure for studying a function for an extremum. Solving the system of equations, one should find the critical points of the function, and then at each of the critical points calculate the quantities

If all three quantities are positive, then the critical point under consideration is a minimum point; if then the given critical point is a maximum point.

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

To find conditional extremum points, use the Lagrange function

where the number is called the Lagrange multiplier. Solving the 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 gives 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, proceeding from the essence of the problem, it is often possible to establish the nature of the critical point.

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