§9 Extremum of a function of two variables. Determining the largest and smallest values ​​of a function of two variables in a closed domain. The largest and smallest value of a function of several variables.

§ Extrema, Maximum and minimum values ​​of functions of several variables - page No. 1/1

§ 8. Extrema. The largest and smallest values ​​of functions of several variables.

1. Extrema of functions of several variables.

Dot
called maximum point functions
, if for any point

inequality holds

.

Likewise point
called minimum point functions
, if for any point
from some neighborhood of a point
inequality holds

.

Notes. 1) According to the definitions, the function
must be defined in some neighborhood of the point
. Those. maximum and minimum points of the function
there can only be internal points of the region
.

2) If there is a neighborhood of a point
, in which for any point
different from
inequality holds

(

), then the point
called strict maximum point (respectively strict minimum point ) functions
. In this regard, the maximum and minimum points defined above are sometimes called non-strict maximum and minimum points.

The concepts of extrema are local in nature: the value of a function at a point
is compared with the function values ​​at fairly close points. In a given area, a function may have no extrema at all, or it may have several minima, several maxima, and even an infinite number of both. Moreover, some minimums may be greater than some of its maximums. Do not confuse the maximum and minimum values ​​of a function with its maximum and minimum values.

Let us find the necessary condition for an extremum. Let, for example,
– maximum point of the function
. Then, by definition, there is a gif" align=absmiddle width="17px" height="18px">-neighborhood of the point
such that
for any point
from this area. In particular,

(1)

Where
,
, And

(2)

Where
,
. But (1) means that a function of one variable
has at point maximum or is on the interval
constant. Hence,

or
- does not exist,

⇒
or
- does not exist.

Similarly, from (2) we obtain that

or
- does not exist.

Thus, the following theorem is valid.

THEOREM 8.1. (necessary conditions for an extremum). If the function
at the point
has an extremum, then at this point either both of its first-order partial derivatives are equal to zero, or at least one of these partial derivatives does not exist.

Geometrically, Theorem 8.1 means that if
– extremum point of the function
, then the tangent plane to the graph of this function at the point is either parallel to the plane
, or does not exist at all. To verify this, it is enough to remember how to find the equation of a tangent plane to a surface (see formula (4.6)).

Points that satisfy the conditions of Theorem 8.1 are called critical points functions
. Just as for a function of one variable, the necessary conditions for an extremum are not sufficient. Those. not every critical point of a function will be its extremum point.

EXAMPLE. Consider the function
. Dot
is critical for this function, since at this point both of its first-order partial derivatives
And
are equal to zero. However, it will not be an extreme point. Really,
, but in any neighborhood of the point
there are points at which the function takes positive values ​​and points at which the function takes negative values. This is easy to verify if you build a graph of the function - a hyperbolic paraboloid.

For a function of two variables, the most convenient sufficient conditions are given by the following theorem.

THEOREM 8.2. (sufficient conditions for the extremum of a function of two variables). Let
– critical point of the function
and in some neighborhood of the point
the function has continuous partial derivatives up to and including the second order. Let's denote

,
,
.

Then 1) if
, then point
is not an extremum point;


If we use Theorem 8.2 to investigate the critical point
failed (i.e. if
or the function has no point in the neighborhood at all
continuous partial derivatives the required order), answer to the question about availability at the point
extremum will give the sign of the function increment at this point.

Indeed, from the definition it follows that if the function
has at point
strict maximum then

for all points
from some neighborhood of a point
, or, otherwise

for all sufficiently small
And
. Likewise, if
is a point of strict minimum, then for all sufficiently small
And
inequality will be satisfied
.

So, to find out whether the critical point is
extremum point, it is necessary to examine the increment of the function at this point. If for all small enough
And
it will preserve the sign, then at the point
the function has a strict extremum (minimum if
, and the maximum if
).

Comment. The rule remains true for a non-strict extremum, but with the amendment that for some values
And
the function increment will be zero
EXAMPLE. Find extrema of functions:

1)
; 2)
.

To study critical points, we apply Theorem 8.2. We have:

,
,
.

Let's explore the point
:

,
,
,

;
.

Therefore, at the point
this function has a minimum, namely
.

Exploring the critical point
:

,
,
,


.

Therefore, the second critical point is not the extremum point of the function.

To study the critical point, we apply Theorem 8.2. We have:

,
,
,

,
,
,

.

Determine the presence or absence of an extremum at a point
using Theorem 8.2 failed.

Let's examine the sign of the function increment at the point
:

If
, That
;

If
, That
.

Because the
does not preserve sign in a neighborhood of a point
, then at this point the function does not have an extremum.

2. The largest and smallest values ​​of the function.

.

Similarly, the value of the function at the point
called the smallest , if for any point
from the region
inequality holds

.

Earlier, we already said that if a function is continuous and the area
– is closed and limited, then the function takes its greatest and smallest values ​​in this area. At the same time, points
And
can lie both inside the area
, and on its border. If the point
(or
) lies inside the region
, then this will be the maximum (minimum) point of the function
, i.e. critical point of a function inside a region
. Therefore, to find the largest and smallest values ​​of the function
in area
need to:
.

Theorem 1.5 Let in a closed region D function specified z=z(x,y) , having continuous partial derivatives of the first order. Border G region D is piecewise smooth (that is, it consists of pieces of “smooth to the touch” curves or straight lines). Then in the area D function z (x,y) reaches its greatest M and the least m values.

No proof.

You can suggest the following location plan M And m .
1. We build a drawing, select all parts of the area boundary D and find all the “corner” points of the border.
2. Find stationary points inside D .
3. Find stationary points on each of the boundaries.
4. We calculate at all stationary and corner points, and then select the largest M and least m meanings.

Example 1.14 Find the greatest M and least m function values z = 4x2-2xy+y2-8x in a closed area D , limited: x = 0, y = 0, 4x+3y=12 .

1. Let's build an area D (Fig. 1.5) on a plane Ohoo .

Corner points: O (0; 0), B (0; 4), A (3; 0) .

Border G region D consists of three parts:

2. Find stationary points inside the region D :

3. Stationary points on the boundaries l 1, l 2, l 3 :

4. We calculate six values:

Examples

Example 1.

This function is defined for all values ​​of the variables x And y , except at the origin, where the denominator goes to zero.

Polynomial x 2 +y 2 is continuous everywhere, and therefore the square root of a continuous function is continuous.

The fraction will be continuous everywhere except at points where the denominator is zero. That is, the function under consideration is continuous on the entire coordinate plane Ohoo , excluding the origin.

Example 2.

Examine the continuity of a function z=tg (x,y) . The tangent is defined and continuous for all finite values ​​of the argument, except for values ​​equal to an odd number of the quantity π /2 , i.e. excluding points where

For every fixed "k" equation (1.11) defines a hyperbola. Therefore, the function under consideration is a continuous function x and y , excluding points lying on curves (1.11).

Example 3.

Find partial derivatives of a function u=z -xy , z > 0 .

Example 4.

Show that function

satisfies the identity:

– this equality is valid for all points M(x;y;z) , except for the point M 0 (a;b;c) .

Let's consider the function z=f(x,y) of two independent variables and establish the geometric meaning of the partial variables z"x =f"x (x,y) And z" y =f" y (x,y) .

In this case, the equation z=f (x,y) there is an equation of some surface (Fig. 1.3). Let's draw a plane y = const . In a section of this surface plane z=f (x,y) you get some line l 1 intersection along which only the quantities change X And z .

Partial derivative z" x (its geometric meaning directly follows from the known geometric meaning of the derivative of a function of one variable) is numerically equal to the tangent of the angle α tilt, relative to the axis Oh , tangent L 1 to the curve l 1 , resulting in a section of the surface z=f (x,y) plane y = const at the point M(x,y,f(xy)): z" x = tanα .

In the section of the surface z=f (x,y) plane X = const you get an intersection line l 2 , along which only the quantities change at And z . Then the partial derivative z" y numerically equal to the tangent of the angle β tilt relative to the axis OU , tangent L 2 to the specified line l 2 intersections at a point M(x,y,f(xy)): z" x = tanβ .

Example 5.

What angle does it make with the axis? Oh tangent to line:

at the point M(2,4,5) ?

We use the geometric meaning of the partial derivative with respect to a variable X (at constant at ):

Example 6.

According to (1.31):

Example 7.

Assuming that the equation

implicitly defines a function

find z" x , z" y .

therefore, according to (1.37), we get the answer.

Example 8.

Explore to the extreme:

1. Find stationary points by solving system (1.41):

that is, four stationary points are found.
2.

by Theorem 1.4 at the point there is a minimum.

Moreover

4. We calculate six values:

From the six values ​​obtained, select the largest and smallest.

Bibliography:

ü Belko I. V., Kuzmich K. K. Higher mathematics for economists. I semester: Express course. – M.: New knowledge, 2002. – 140 p.

ü Gusak A. A.. Mathematical analysis and differential equations. – Mn.: TetraSystems, 1998. – 416 p.

ü Gusak A. A.. Higher mathematics. A textbook for university students in 2 volumes. – Mn., 1998. – 544 p. (1 volume), 448 pp. (2 t.).

ü Kremer N. Sh., Putko B. A., Trishin I. M., Fridman M. N. Higher mathematics for economists: Textbook for universities / Ed. prof. N. Sh. Kremer. – M.: UNITI, 2002. – 471 p.

ü Yablonsky A.I., Kuznetsov A.V., Shilkina E.I. and others. Higher mathematics. General course: Textbook / Under general. ed. S. A. Samal. – Mn.: Vysh. school, 2000. – 351 p.

Highest and Lowest Values

A function bounded in a bounded closed region reaches its maximum and minimum values ​​either at stationary points or at points lying on the boundary of the region.

To find the largest or smallest values ​​of a function you need to:

1. Find stationary points lying inside this area and calculate the value of the function at them.

2. Find the largest (smallest) value of the function on the boundary of the region.

3. Compare all obtained function values: the largest (smallest) will be the largest (smallest) value of the function in this area.

Example 2. Find the largest (smallest) value of the function: in a circle.

Solution.

stationary point; .

2 .The boundary of this closed area is a circle or , where .

The function at the boundary of the region becomes a function of one variable: , where . Let's find the largest and smallest values ​​of this function.

When x=0 ; (0,-3) and (0,3) are critical points.

Let's calculate the values ​​of the function at the ends of the segment

3 . Comparing the values ​​with each other we get,

At points A and B.

At points C and D.

Example 3. Find the largest and smallest values ​​of the function in a closed region defined by the inequality:

Solution. The area is a triangle limited by the coordinate axes and the straight line x+y=1.

1. We find stationary points inside the region:

; ; y = - 1/ 8; x = 1/8.

The stationary point does not belong to the region under consideration, so the z value in it is not calculated.

2 .We study the function on the boundary. Since the boundary consists of three sections described by three different equations, we study the function on each section separately:

A) in section 0A: y=0 - equation 0A, then ; from the equation it is clear that the function increases by 0A from 0 to 1. This means .

b) in section 0B: x=0 - equation 0B, then ; –6y+1=0; - critical point.

V) on the line x+y = 1: y=1-x, then we get the function

Let's calculate the value of the function z at point B(0,1).

3 .Comparing the numbers we get that

On straight AB.

At point B.

Tests for self-control of knowledge.

1 . The extremum of the function is

a) its first order derivatives

b) its equation

c) her schedule

d) its maximum or minimum

2. The extremum of a function of several variables can be achieved:

a) only at points lying inside its domain of definition, at which all first-order partial derivatives are greater than zero

b) only at points lying inside its domain of definition, at which all first-order partial derivatives are less than zero

c) only at points lying inside its domain of definition, at which all first-order partial derivatives are not equal to zero

d) only at points lying inside its domain of definition, at which all first-order partial derivatives are equal to zero

3. A function that is continuous in a limited closed region reaches its maximum and minimum values:

a) at stationary points

b) either at stationary points or at points lying on the boundary of the region

c) at points lying on the boundary of the region

d) at all points

4. Stationary points for a function of several variables are the points:

a) in which all first-order partial derivatives are not equal to zero

b) in which all first-order partial derivatives are greater than zero

c) in which all first-order partial derivatives are equal to zero

d) in which all first-order partial derivatives are less than zero

Lecture 28. Study on extremum of functions of several variables. Conditional extremum of functions of several variables.

Studying functions of many variables to an extremum is a much more complex procedure than a similar procedure for functions of one variable. Therefore, we will limit ourselves to considering this issue at the simplest and most clear example functions of two variables (see Fig. 1). Here M 1(x 1 ; y 1), M 2(x 2 ; y 2), M 3(x 3 ; y 3) are the extremum points of this function. Namely, points M 1 And M 3 – the minimum point of the function, and the point M 2– its maximum point. Figure 1 shows a function with three extremum points, but, naturally, there can be more or less of these points.

Let us define more precisely what the extremum points are for a function of two variables.

Definition. The function has maximum(minimum) at a point, if for any point located in a certain neighborhood - a neighborhood of the point, the following holds: (). - the neighborhood can be represented by a set of points whose coordinates satisfy the condition , where is a positive sufficiently small number.

The maxima and minima of a function are called extremes, A - extreme point.

Let M0(x 0 ; y 0) – a point of any extremum (maximum point or minimum point) of the function. Then it is fair

Theorem 1.

If at the extremum point M0(x 0 ; y 0) there are partial derivatives And , then they are both equal to zero:

2) Let us now consider the function . Because is the extreme value of this function, then the derivative of this function at y = y 0, if it exists, is equal to zero:

The theorem has been proven.

Note that conditions (1) are only necessary extremum conditions at the point M0(x 0 ; y 0) function differentiable at this point. That is, these conditions are not sufficient conditions for what is at the point M0(x 0 ; y 0) the function will have an extremum (maximum or minimum). In other words, period M0(x 0 ; y 0), in which both equalities (1) are satisfied, is only suspicious to the extremum point for the function. The final conclusion about the nature of such a suspicious point for an extremum can be made using the following theorem (we present it without derivation):

Theorem 2.(Sufficient conditions for an extremum)

Let M0(x 0 ; y 0) – such a point from the region D defining a function that the necessary conditions (1) for the extremum of this function are satisfied. That is M0(x 0 ; y 0) – a point suspicious for an extremum. Let's find the numbers at this point

1) If > 0 and > 0 (or С>0 at A=0), That M0(x 0 ; y 0) – minimum point of the function .

2) If > 0 and < 0 (or WITH<0 at A=0), That M0(x 0 ; y 0) – maximum point of the function .

3) If < 0, then point M0(x 0 ; y 0) – not the extremum point of the function .

4) If = 0, then the question remains open - additional research is needed.

Example 1. Let X And at– the quantity of two goods produced; p 1 = 8 rub. And p 2 = 10 rub. – unit price of each of these goods, respectively; C= 0,01(x 2 + xy + y 2) is a function of costs (in rubles) for the production of these goods. Then income R from the sale of goods will be R = 8x+10y(rub.), and profit P will be (in rubles)

P = R – C = 8x+ 10y – 0,01(x 2 +xy+y 2).

Let's find the volumes X And at goods for which profit P will be maximum.

1) First, let's find the values ​​( x;y), suspicious for an extremum for the function P:

2) Now we examine the found suspicious extremum function P point M 0(200; 400). To do this, we find at this point the values ​​determined by expressions (4). Because

and this is true for any ( X; at), and therefore at the point M 0(200; 400), then

Because that's the point M 0(200; 400) – maximum point of the function P. That is, profit P from sales will be maximum at x = 200(units) And y = 400(units) and is equal to 2800 rubles.

Example 2. Find extremum points and extreme values ​​of a function

Solution. This function is a function of two variables, defined for any X And at, that is, on the entire plane howe, and having partial derivatives of the first order at each point:

First we find the points of the plane howe, suspicious for an extremum for this function:

Then, having found the second-order partial derivatives of the function, we write expressions for:

Now calculating the numerical values ​​of these quantities for each of the four points suspicious for an extremum, we obtain the following conclusions about these points:

Dot min.

Dot max.

Not an extreme point.

Not an extreme point.

Now let’s find two extreme (maximum) values ​​of the function that determine the height of the two vertices of the graph of this function:

Determination of the largest and smallest values ​​of a function of two variables in a closed region.

Let's consider the following problem. Let be some continuous function of two variables, considered in a closed domain, where is the interior of the domain, and G– its border (Fig. 8.6).

The fact that a function is continuous in the region means that the graph of this function (surface in space) is a continuous (without discontinuities) surface for all . That is, the concept of continuity of a function of two variables is similar to the concept of continuity of a function of one variable. Like functions of one variable, functions of two variables formed from elementary functions are continuous for all values ​​of their arguments for which they are defined. This also applies to functions of three, four or more variables.

Let's return to Fig. 2. Let us pose the following question: at what points in the region does the function reach its largest and smallest values? z max And z name? And what are these values? Note that this problem is similar to that which was considered for a function of one variable considered on a closed interval [ a; b] axes Oh.

It is obvious that the required points of the region, in which the function reaches its greatest and minimum values, are contained either among the extremum points of this function located inside the region (in the region), or are located somewhere on the boundary G this area. In a closed region such points will certainly exist (Weierstrass theorem). And in an open area (without border G) there may not be such points.

From the above, the following follows: diagram for finding these points, similar to that outlined for functions of one variable.

1. Find all points of the function that are suspicious for an extremum and are located in the area D. These are the points at which both partial derivatives and are equal to zero (or one is zero and the other does not exist; or both do not exist).

2. We find all points of the function that are suspicious for an extremum and are located on the boundary G areas. In this case, we use the boundary equation G.

3. Without examining the suspicious points found in steps 1 and 2 (this is unnecessary), we find the function values ​​at all found suspicious points and select those where z will be the largest and the smallest.

Example 3. Find z max And z name function considered in a closed region, which is a triangular plate with vertices O(0; 0), A(1; 0), B(0; 1)(Fig. 3).

Solution. Let's follow the above diagram.

1. Find inside the triangle (in the area D) points suspicious for an extremum for our function z. To do this, we first find the first order partial derivatives and :

These derivatives exist (they can be calculated) for any (x;y). Consequently, points suspicious for an extremum will only be those for which both of these partial derivatives are equal to zero:

The point obviously belongs to the region D(to the triangle in question). That is, it is a point suspicious for an extremum for a given function z inside the triangle, and she is the only one there.

2. Let us now find points suspicious for an extremum on the border of the triangle.

a) Let’s explore the area first OA borders ( at= 0; £0 X£ 1). In this section - a function of one variable X. Its derivative exists for everyone xÎ . Therefore, its extreme values ​​are a function z may have either at the point where , that is, at the point, or at the ends of the segment OA, that is, at points ABOUT(0; 0) and A(1; 0).

b) Let’s now explore the area OB boundaries of the triangle (there X= 0; £0 at£ 1). In this section the function (0 £ at£ 1) – function of one variable at. Repeating the reasoning of point (a), we come to the conclusion that its extreme values ​​are a function z may have either at the point or at the ends of the segment OB, that is, at points ABOUT(0; 0) and B(0; 1).

c) Finally, we explore the area AB borders. Since on AB(make sure of this) y = - x + 1 (0 £ X£ 1), then there is a function z takes the form: (0 £ X£ 1). Its derivative, therefore, is a function of its extreme values z can reach only at the point where , that is, at the point, or at the ends of the segment AB, that is, at points A And IN.

So, the complete set of points of the function suspicious for extremum
in a triangle OAV is:

; ; ; ; ; ; .

3. Now let’s find the values ​​of the function z at all found suspicious points and select the largest value from these values z max and the smallest value z name:

Thus, z max = 3 and is achieved by the function z in a triangle OAV at two points at once - at its vertices A And IN. And is achieved by the function z in a triangle OAV at its internal point.

Example 4. The city budget has the opportunity to spend no more than 600 million rubles on social housing, while having projects and land plots for 10 five-story buildings with 90 apartments each and 8 nine-story buildings with 120 apartments each. The average estimated cost of one apartment in a five-story building is 400 thousand rubles, and in a nine-story building 500 thousand rubles. How many five-story and how many nine-story buildings should the city build to get the maximum number of apartments?

Solution. Let X– the required number of five-story buildings, y – nine-story, and z – total number of apartments in these buildings:

z = 90x+ 120y

The cost of all apartments in five-story buildings will be 90 × 0.4· X = 36X million rubles, and in nine-story buildings 120 × 0.5 at = 60at million rubles. According to the conditions of the problem we have:

0 £ X£10; £0 at£8; 36 X + 60at£600

These restrictive inequalities are obviously satisfied in the pentagon (Fig. 4). In this closed area you need to find a point M(x;y), for which the function z = 90x+ 120y will take the greatest value z max.

Let us implement the above scheme for solving such problems.

1. Find points inside the pentagon that are suspicious for an extremum for the function z. Because , and these partial derivatives are obviously not equal to zero, then there are no points suspicious for an extremum inside the pentagon.

2. Find points suspicious for extremum on the boundaries of the pentagon. On each of the five segments that make up the boundary of the pentagon, the function z– linear function of the form z = ax + by, and therefore, it reaches its largest and smallest values ​​at the boundaries of the segments. That is, the desired maximum value z max function z reaches at one of the corner points (O; A; M 1; M 2; B). Calculating value z at these points, we get:

z(ABOUT) = 0; z( A) = 960; z( M 1) = 1260; z( M 2) = 1380; z( B) = 900.

Thus z naimbo= 1380 and it is reached at the point M 2(10; 4). That is, the largest number of apartments (1380) will be obtained if 10 five-story buildings and 4 nine-story buildings are built.

Example 5. Prove that of all triangles having a given perimeter 2p, the equilateral triangle has the largest area. M(2p/3, 2p/3), because the remaining points do not satisfy the meaning of the problem: there cannot be a triangle whose side is equal to half the perimeter.

We examine the extremum point M(2p/3, 2p/3):

∂ 2 f/∂x 2 = -2p(p-y); ∂ 2 f/∂x∂y = p(2x+2y-3p); ∂ 2 f/∂y 2 = -2p(p-x);

D=AC-B 2 = ;

D>0, and because A<0 , then at the point under study the function reaches a maximum. So, at a single stationary point the function reaches its maximum, and therefore its greatest value; thus, with x=2p/3, y=2p/3 the function reaches its maximum value. But then z=2p-x-y=2p/3. And because x=y=z, then the triangle is equilateral.

And to solve it you will need minimal knowledge of the topic. Another school year is ending, everyone wants to go on vacation, and in order to bring this moment closer, I will immediately get to the point:

Let's start with the area. The area referred to in the condition is limited closed set of points on a plane. For example, the set of points bounded by a triangle, including the WHOLE triangle (if from borders“prick out” at least one point, then the region will no longer be closed). In practice, there are also areas of rectangular, round and slightly more complex shapes. It should be noted that in the theory of mathematical analysis strict definitions are given limitations, isolation, boundaries, etc., but I think everyone is aware of these concepts on an intuitive level, and now nothing more is needed.

A flat area is standardly denoted by the letter , and, as a rule, is specified analytically - by several equations (not necessarily linear); less often inequalities. Typical verbiage: “closed area bounded by lines.”

An integral part of the task under consideration is the construction of an area in the drawing. How to do it? You need to draw all the listed lines (in this case 3 straight) and analyze what happened. The searched area is usually lightly shaded, and its border is marked with a thick line:


The same area can also be set linear inequalities: , which for some reason are often written as an enumerated list rather than system.
Since the boundary belongs to the region, then all inequalities, of course, lax.

And now the essence of the task. Imagine that the axis comes out straight towards you from the origin. Consider a function that continuous in each area point. The graph of this function represents some surface, and the small happiness is that to solve today’s problem we don’t need to know what this surface looks like. It can be located higher, lower, intersect the plane - all this does not matter. And the following is important: according to Weierstrass's theorems, continuous V limited closed area the function reaches its greatest value (the “highest”) and the least (the “lowest”) values ​​that need to be found. Such values ​​are achieved or V stationary points, belonging to the regionD , or at points that lie on the border of this area. This leads to a simple and transparent solution algorithm:

Example 1

In a limited closed area

Solution: First of all, you need to depict the area in the drawing. Unfortunately, it is technically difficult for me to make an interactive model of the problem, and therefore I will immediately present the final illustration, which shows all the “suspicious” points found during the research. They are usually listed one after the other as they are discovered:

Based on the preamble, the decision can be conveniently divided into two points:

I) Find stationary points. This is a standard action that we performed repeatedly in class. about extrema of several variables:

Found stationary point belongs areas: (mark it on the drawing), which means we should calculate the value of the function at a given point:

- as in the article The largest and smallest values ​​of a function on a segment, I will highlight important results in bold. It is convenient to trace them in a notebook with a pencil.

Pay attention to our second happiness - there is no point in checking sufficient condition for an extremum. Why? Even if at a point the function reaches, for example, local minimum, then this DOES NOT MEAN that the resulting value will be minimal throughout the region (see the beginning of the lesson about unconditional extrema) .

What to do if the stationary point does NOT belong to the area? Almost nothing! It should be noted that and move on to the next point.

II) We explore the border of the region.

Since the border consists of the sides of a triangle, it is convenient to divide the study into 3 subsections. But it’s better not to do it anyhow. From my point of view, it is first more advantageous to consider the segments parallel to the coordinate axes, and first of all, those lying on the axes themselves. To grasp the entire sequence and logic of actions, try to study the ending “in one breath”:

1) Let's deal with the bottom side of the triangle. To do this, substitute directly into the function:

Alternatively, you can do it like this:

Geometrically, this means that the coordinate plane (which is also given by the equation)"carves" out of surfaces a "spatial" parabola, the top of which immediately comes under suspicion. Let's find out where is she located:

– the resulting value “fell” into the area, and it may well turn out that at the point (marked on the drawing) the function reaches the largest or smallest value in the entire region. One way or another, let's do the calculations:

The other “candidates” are, of course, the ends of the segment. Let's calculate the values ​​of the function at points (marked on the drawing):

Here, by the way, you can perform an oral mini-check using a “stripped-down” version:

2) To study the right side of the triangle, substitute it into the function and “put things in order”:

Here we will immediately perform a rough check, “ringing” the already processed end of the segment:
, Great.

The geometric situation is related to the previous point:

– the resulting value also “came into the sphere of our interests,” which means we need to calculate what the function at the appeared point is equal to:

Let's examine the second end of the segment:

Using the function , let's perform a control check:

3) Probably everyone can guess how to explore the remaining side. We substitute it into the function and carry out simplifications:

Ends of the segment have already been researched, but in the draft we still check whether we have found the function correctly :
– coincided with the result of the 1st subparagraph;
– coincided with the result of the 2nd subparagraph.

It remains to find out if there is anything interesting inside the segment:

- There is! Substituting the straight line into the equation, we get the ordinate of this “interestingness”:

We mark a point on the drawing and find the corresponding value of the function:

Let’s check the calculations using the “budget” version :
, order.

And the final step: We CAREFULLY look through all the “bold” numbers, I recommend that beginners even make a single list:

from which we select the largest and smallest values. Answer Let's write down in the style of the problem of finding the largest and smallest values ​​of a function on a segment:

Just in case, I’ll comment once again on the geometric meaning of the result:
– here is the highest point of the surface in the region;
– here is the lowest point of the surface in the area.

In the analyzed task, we identified 7 “suspicious” points, but their number varies from task to task. For a triangular region, the minimum “research set” consists of three points. This happens when the function, for example, specifies plane– it is completely clear that there are no stationary points, and the function can reach its maximum/smallest values ​​only at the vertices of the triangle. But there are only one or two similar examples - usually you have to deal with some surface of 2nd order.

If you solve such tasks a little, then triangles can make your head spin, and that’s why I have prepared unusual examples for you to make it square :))

Example 2

Find the largest and smallest values ​​of a function in a closed area bounded by lines

Example 3

Find the largest and smallest values ​​of a function in a limited closed region.

Pay special attention to the rational order and technique of studying the boundary of the region, as well as to the chain of intermediate checks, which will almost completely avoid computational errors. Generally speaking, you can solve it any way you like, but in some problems, for example, in Example 2, there is every chance of making your life much more difficult. An approximate sample of the final assignments at the end of the lesson.

Let’s systematize the solution algorithm, otherwise with my diligence as a spider, it somehow got lost in the long thread of comments of the 1st example:

– At the first step, we build the area, it is advisable to shade it and highlight the border with a bold line. During the solution, points will appear that need to be marked on the drawing.

– Find stationary points and calculate the values ​​of the function only in those of them that belong to the region. We highlight the resulting values ​​in the text (for example, circle them with a pencil). If a stationary point does NOT belong to the region, then we mark this fact with an icon or verbally. If there are no stationary points at all, then we draw a written conclusion that they are absent. In any case, this point cannot be skipped!

– We are exploring the border of the region. First, it is beneficial to understand the straight lines that are parallel to the coordinate axes (if there are any at all). We also highlight the function values ​​calculated at “suspicious” points. A lot has been said above about the solution technique and something else will be said below - read, re-read, delve into it!

– From the selected numbers, select the largest and smallest values ​​and give the answer. Sometimes it happens that a function reaches such values ​​at several points at once - in this case, all these points should be reflected in the answer. Let, for example, and it turned out that this is the smallest value. Then we write down that

The final examples cover other useful ideas that will come in handy in practice:

Example 4

Find the largest and smallest values ​​of a function in a closed region .

I have retained the author's formulation, in which the area is given in the form of a double inequality. This condition can be written by an equivalent system or in a more traditional form for this problem:

I remind you that with nonlinear we encountered inequalities on, and if you do not understand the geometric meaning of the notation, then please do not delay and clarify the situation right now;-)

Solution, as always, begins with constructing an area that represents a kind of “sole”:

Hmm, sometimes you have to chew not only the granite of science...

I) Find stationary points:

The system is an idiot's dream :)

A stationary point belongs to the region, namely, lies on its boundary.

And so, it’s okay... the lesson went well - this is what it means to drink the right tea =)

II) We explore the border of the region. Without further ado, let's start with the x-axis:

1) If , then

Let's find where the vertex of the parabola is:
– appreciate such moments – you have “hit” right to the point from which everything is already clear. But we still don’t forget about checking:

Let's calculate the values ​​of the function at the ends of the segment:

2) Let’s deal with the lower part of the “sole” “in one sitting” - without any complexes we substitute it into the function, and we will only be interested in the segment:

Control:

This already brings some excitement to the monotonous driving along the knurled track. Let's find critical points:

Let's decide quadratic equation, do you remember anything else about this? ...However, remember, of course, otherwise you wouldn’t be reading these lines =) If in the two previous examples calculations in decimal fractions were convenient (which, by the way, is rare), then here the usual ordinary fractions await us. We find the “X” roots and use the equation to determine the corresponding “game” coordinates of the “candidate” points:


Let's calculate the values ​​of the function at the found points:

Check the function yourself.

Now we carefully study the won trophies and write down answer:

These are “candidates”, these are “candidates”!

To solve it yourself:

Example 5

Find the smallest and largest values ​​of a function in a closed area

An entry with curly braces reads like this: “a set of points such that.”

Sometimes in such examples they use Lagrange multiplier method, but there is unlikely to be a real need to use it. So, for example, if a function with the same area “de” is given, then after substitution into it – with the derivative from no difficulties; Moreover, everything is drawn up in “one line” (with signs) without the need to consider the upper and lower semicircles separately. But, of course, there are also more complex cases, where without the Lagrange function (where, for example, is the same equation of a circle) It’s hard to get by – just as it’s hard to get by without a good rest!

Have a good time everyone and see you soon next season!

Solutions and answers:

Example 2: Solution: Let's depict the area in the drawing: