Definition: Point x0 is called a point of local maximum (or minimum) of a function if in some neighborhood of point x0 the function takes the greatest (or smallest) value, i.e. for all x from some neighborhood of the point x0 the condition f(x) f(x0) (or f(x) f(x0)) is satisfied.

Local maximum or minimum points are combined common name- points of local extremum of the function.

Note that at local extremum points the function reaches its maximum or lowest value only in some local area. There may be cases when according to the value уmaxуmin.

A necessary sign of the existence of a local extremum of a function

Theorem . If continuous function y = f(x) has a local extremum at the point x0, then at this point the first derivative is either zero or does not exist, i.e. a local extremum occurs at critical points of the first kind.

At local extremum points, either the tangent is parallel to the 0x axis, or there are two tangents (see figure). Note that critical points are a necessary but not sufficient condition for a local extremum. A local extremum occurs only at critical points of the first kind, but not at all critical points a local extremum occurs.

For example: a cubic parabola y = x3 has a critical point x0 = 0, at which the derivative y/(0)=0, but the critical point x0=0 is not an extremum point, but an inflection point at it (see below).

A sufficient sign of the existence of a local extremum of a function

Theorem . If, when the argument passes through a critical point of the first kind from left to right, the first derivative y / (x)

changes sign from “+” to “-”, then the continuous function y(x) at this critical point has a local maximum;

changes sign from “-” to “+”, then the continuous function y(x) has a local minimum at this critical point

does not change sign, then at this critical point there is no local extremum, there is an inflection point here.

For a local maximum, the region of increasing function (y/0) is replaced by the region of decreasing function (y/0). For a local minimum, the region of decreasing function (y/0) is replaced by the region of increasing function (y/0).

Example: Examine the function y = x3 + 9x2 + 15x - 9 for monotonicity, extremum and construct a graph of the function.

Let's find critical points of the first kind by defining the derivative (y/) and equating it to zero: y/ = 3x2 + 18x + 15 =3(x2 + 6x + 5) = 0

Let's decide quadratic trinomial using a discriminant:

x2 + 6x + 5 = 0 (a=1, b=6, c=5) D=, x1k = -5, x2k = -1.

2) Let's divide the number line with critical points into 3 regions and determine the signs of the derivative (y/) in them. Using these signs we will find areas of monotonicity (increasing and decreasing) of functions, and by changing the signs we will determine the points of local extremum (maximum and minimum).

We present the research results in the form of a table, from which the following conclusions can be drawn:

1. On the interval y /(-10) 0 the function increases monotonically (the sign of the derivative y was estimated using the control point x = -10 taken in this interval);
2. On the interval (-5 ; -1) y /(-2) 0 the function decreases monotonically (the sign of the derivative y was estimated using the control point x = -2, taken in this interval);
3. On the interval y /(0) 0, the function increases monotonically (the sign of the derivative y was estimated using the control point x = 0, taken in this interval);
4. When passing through the critical point x1k = -5, the derivative changes sign from “+” to “-”, therefore this point is a local maximum point
(ymax(-5) = (-5)3+9(-5)2 +15(-5)-9=-125 + 225 - 75 - 9 =16);
5. When passing through the critical point x2k = -1, the derivative changes sign from “-” to “+”, therefore this point is a local minimum point
(ymin(-1) = -1 + 9 - 15 - 9 = - 16).

x -5 (-5 ; -1) -1

3) We will construct a graph based on the results of the study using additional calculations of function values at control points:

construct a rectangular coordinate system Oxy;

We show by coordinates the points of maximum (-5; 16) and minimum (-1;-16);

to clarify the graph, we calculate the value of the function at control points, selecting them to the left and right of the maximum and minimum points and inside the average interval, for example: y(-6)=(-6)3 +9(-6)2+15(-6 )-9=9; y(-3)=(-3)3+9(-3)2+15(-3)-9=0;

y(0)= -9 (-6;9); (-3;0) and (0;-9) - calculated control points that we plot to construct the graph;

We show the graph in the form of a curve convex upward at the maximum point and convex downward at the minimum point and passing through the calculated control points.

>>Extrema

Extremum of the function

Definition of extremum

Function y = f(x) is called increasing (decreasing) in a certain interval, if for x 1< x 2 выполняется неравенство (f (x 1) < f (x 2) (f (x 1) >f (x 2)).

If the differentiable function y = f (x) increases (decreases) on an interval, then its derivative on this interval f " (x)> 0

(f"(x)< 0).

Dot x O called local maximum point (minimum) function f (x) if there is a neighborhood of the point x o, for all points of which the inequality f (x) is true≤ f (x o ) (f (x )≥ f (x o )).

The maximum and minimum points are called extremum points, and the values of the function at these points are its extremes.

Extremum points

Necessary conditions for an extremum . If the point x O is the extremum point of the function f (x), then either f " (x o ) = 0, or f(x o ) does not exist. Such points are called critical, and the function itself is defined at the critical point. The extrema of a function should be sought among its critical points.

The first sufficient condition. Let x O - critical point. If f" (x ) when passing through a point x O changes the plus sign to minus, then at the point x o the function has a maximum, otherwise it has a minimum. If, when passing through the critical point, the derivative does not change sign, then at the point x O there is no extreme.

Second sufficient condition. Let the function f(x) have
f" (x ) in the vicinity of the point x O and the second derivative at the point itself x o. If f"(x o) = 0, >0 ( <0), то точка x o is the local minimum (maximum) point of the function f (x). If =0, then you need to either use the first sufficient condition or involve higher ones.

On a segment, the function y = f (x) can reach its minimum or maximum value either at critical points or at the ends of the segment.

Example 3.22.

Solution. Because f " (

Problems of finding the extremum of a function

Example 3.23. a

Solution. x And y y
0 ≤ x≤
> 0, and when x >a /4 S " < 0, значит, в точке x=a /4 функция S имеет максимум. Значение functions kv. units).

Example 3.24. p ≈

Solution. p p
S"
R = 2, H = 16/4 = 4.

Example 3.22.Find the extrema of the function f (x) = 2x 3 - 15x 2 + 36x - 14.

Solution. Because f " (x ) = 6x 2 - 30x +36 = 6(x -2)(x - 3), then the critical points of the function x 1 = 2 and x 2 = 3. Extrema can only be at these points. Since when passing through the point x 1 = 2 the derivative changes sign from plus to minus, then at this point the function has a maximum. When passing through the point x 2 = 3, the derivative changes its sign from minus to plus, so at the point x 2 = 3 the function has a minimum. Having calculated the function values at the points
x 1 = 2 and x 2 = 3, we find the extrema of the function: maximum f (2) = 14 and minimum f (3) = 13.

Example 3.23.It is necessary to build a rectangular area near the stone wall so that it is fenced off on three sides with wire mesh and the fourth side is adjacent to the wall. For this there is a linear meters of mesh. At what aspect ratio will the site have the largest area?

Solution.Let us denote the sides of the platform by x And y. The area of the site is S = xy. Let y- this is the length of the side adjacent to the wall. Then, by condition, the equality 2x + y = a must be satisfied. Therefore y = a - 2x and S = x (a - 2x), where
0 ≤ x≤ a /2 (the length and width of the area cannot be negative). S " = a - 4x, a - 4x = 0 at x = a/4, whence
y = a - 2 × a/4 =a/2. Since x = a /4 is the only critical point; let’s check whether the sign of the derivative changes when passing through this point. At x a /4 S "> 0, and when x >a /4 S " < 0, значит, в точке x=a /4 функция S имеет максимум. Значение functions S(a/4) = a/4(a - a/2) = a 2 /8 (kv. units). Since S is continuous on and its values at the ends S(0) and S(a /2) are equal to zero, then the value found will be highest value functions. Thus, the most favorable aspect ratio of the site under the given conditions of the problem is y = 2x.

Example 3.24.It is required to manufacture a closed cylindrical tank with a capacity of V=16 p ≈ 50 m 3. What should be the dimensions of the tank (radius R and height H) so that the least amount of material is used for its manufacture?

Solution.Square full surface cylinder is equal to S = 2 p R(R+H). We know the volume of the cylinder V = p R 2 Н Þ Н = V/ p R 2 =16 p / p R2 = 16/R2. So S(R) = 2 p (R 2 +16/R). We find the derivative of this function:
S"(R) = 2 p (2R- 16/R 2) = 4 p (R- 8/R 2). S" (R) = 0 at R 3 = 8, therefore,
R = 2, H = 16/4 = 4.

For a function f(x) of many variables, point x is a vector, f'(x) is a vector of first derivatives (gradient) of function f(x), f ′ ′(x) is a symmetric matrix of second partial derivatives (Hessian matrix - Hessian) functions f(x).
For function of many condition variables optimalities are formulated as follows.
A necessary condition for local optimality. Let f(x) be differentiable at the point x * R n . If x * is a local extremum point, then f’(x *) = 0.
As before, points that are solutions to a system of equations are called stationary. The nature of the stationary point x * is associated with the definite sign of the Hessian matrix f′ ′(x).
The sign of matrix A depends on the signs of the quadratic form Q(α)=< α A, α >for all non-zero α∈R n .
Here and further through denotes the scalar product of the vectors x and y. By definition,

A matrix A is positive (non-negative) definite if Q(α)>0 (Q(α)≥0) for all non-zero α∈R n ; negative (non-positive) definite if Q(α)<0 (Q(α)≤0) при всех ненулевых α∈R n ; неопределенной, если Q(α)>0 for some non-zero α∈R n and Q(α)<0 для остальных ненулевых α∈R n .
Sufficient condition for local optimality. Let f(x) be twice differentiable at the point x * R n, and f’(x *)=0, i.e. x * − stationary point. Then, if the matrix f′′(x *) is positive (negative) definite, then x * is a local minimum (maximum) point; if the matrix f′′(x *) is undefined, then x * is a saddle point.
If the matrix f′′(x *) is non-negatively (non-positively) definite, then to determine the nature of the stationary point x * requires the study of higher order derivatives.
To check the sign of a matrix, as a rule, the Sylvester criterion is used. According to this criterion, a symmetric matrix A is positive definite if and only if all its angular minors are positive. In this case, the angular minor of matrix A is the determinant of a matrix constructed from elements of matrix A located at the intersection of rows and columns with the same (and first) numbers. To check the symmetric matrix A for negative definiteness, you need to check the matrix (−A) for positive definiteness.
So, the algorithm for determining local extrema points of a function of many variables is as follows.
1. Find f′(x).
2. The system is being solved

As a result, stationary points x i are calculated.
3. Find f′′(x), set i=1.
4. Find f′′(x i)
5. The angular minors of the matrix f′′(x i) are calculated. If not all angular minors are nonzero, then determining the nature of the stationary point x i requires the study of higher order derivatives. In this case, the transition to step 8 is carried out.
Otherwise, go to step 6.
6. The signs of the angular minors f′′(x i) are analyzed. If f′′(x i) is positive definite, then x i is a local minimum point. In this case, the transition to step 8 is carried out.
Otherwise, go to step 7.
7. The angular minors of the matrix -f′′(x i) are calculated and their signs are analyzed.
If -f′′(x i) − is positive definite, then f′′(x i) is negative definite and x i is a local maximum point.
Otherwise f′′(x i) is undefined and x i is a saddle point.
8. The condition for determining the nature of all stationary points i=N is checked.
If it is fulfilled, then the calculations are completed.
If the condition is not met, then i=i+1 is assumed and the transition to step 4 is carried out.

Example No. 1. Determine the points of local extrema of the function f(x) = x 1 3 – 2x 1 x 2 + x 2 2 – 3x 1 – 2x 2

Since all angular minors are non-zero, the character of x 2 is determined using f′′(x).
Since the matrix f′′(x 2) is positive definite, x 2 is a local minimum point.
Answer: the function f(x) = x 1 3 – 2x 1 x 2 + x 2 2 – 3x 1 – 2x 2 has a local minimum at the point x = (5/3; 8/3).

The function is said to have at the internal point
region D local maximum(minimum), if there is such a neighborhood of the point
, for each point
which holds the inequality

If a function has at a point
local maximum or local minimum, then we say that it has at this point local extremum(or just an extreme).

Theorem (necessary condition for the existence of an extremum). If the differentiable function reaches an extremum at the point
, then each first-order partial derivative of the function at this point it becomes zero.

The points at which all first-order partial derivatives vanish are called stationary points of the function
. The coordinates of these points can be found by solving the system of equations

The necessary condition for the existence of an extremum in the case of a differentiable function can be briefly formulated as follows:

There are cases when at individual points some partial derivatives have infinite values or do not exist (while the rest are equal to zero). Such points are called critical points of the function. These points should also be considered as “suspicious” for an extremum, just like stationary ones.

In the case of a function of two variables necessary condition extremum, namely the equality to zero of the partial derivatives (differential) at the extremum point, has a geometric interpretation: tangent plane to the surface
at the extremum point must be parallel to the plane
.

20. Sufficient conditions for the existence of an extremum

The fulfillment of the necessary condition for the existence of an extremum at some point does not at all guarantee the presence of an extremum there. As an example, we can take the everywhere differentiable function
. Both of its partial derivatives and the function itself vanish at the point
. However, in any neighborhood of this point there are both positive (large
), and negative (smaller
) values of this function. Therefore, at this point, by definition, no extremum is observed. Therefore, it is necessary to know sufficient conditions under which a point suspicious of an extremum is an extremum point of the function under study.

Let's consider the case of a function of two variables. Let's assume that the function
defined, continuous and has continuous partial derivatives up to the second order inclusive in the neighborhood of some point
, which is the stationary point of the function
, that is, satisfies the conditions

,
.

Let us introduce the following notation:

Theorem (sufficient conditions for the existence of an extremum). Let the function
satisfies the above conditions, namely: it is differentiable in some neighborhood of a stationary point
and is twice differentiable at the point itself
. Then if

In case
then the function
at the point
reaches

local maximum at
And

local minimum at
.

In general, for the function
sufficient condition for existence at the point
localminimum(maximum) is positive(negative) certainty of the second differential.

In other words, the following statement is true.

Theorem . If at the point
for function

for any not equal to zero at the same time
, then at this point the function has minimum(similar to maximum, If
).

Example 18.Find local extremum points of a function

Solution. Let's find the partial derivatives of the function and equate them to zero:

Solving this system, we find two possible extremum points:

Let's find the second order partial derivatives for this function:

At the first stationary point, therefore, and
Therefore, additional research is required at this point. Function value
at this point is zero:
Next,

at

A

Therefore, in any neighborhood of the point
function
takes values as large
, and smaller
, and, therefore, at the point
function
, by definition, has no local extremum.

At the second stationary point

therefore, therefore, since
then at the point
the function has a local maximum.

$E \subset \mathbb(R)^(n)$. They say $f$ has local maximum at the point $x_(0) \in E$, if there is a neighborhood $U$ of the point $x_(0)$ such that for all $x \in U$ the inequality $f\left(x\right) \leqslant f is satisfied \left(x_(0)\right)$.

The local maximum is called strict , if the neighborhood $U$ can be chosen so that for all $x \in U$ different from $x_(0)$ there is $f\left(x\right)< f\left(x_{0}\right)$.

Definition
Let $f$ be a real function on the open set $E \subset \mathbb(R)^(n)$. They say $f$ has local minimum at the point $x_(0) \in E$, if there is a neighborhood $U$ of the point $x_(0)$ such that the inequality $f\left(x\right) \geqslant f holds for all $x \in U$ \left(x_(0)\right)$.

A local minimum is called strict if a neighborhood $U$ can be chosen so that for all $x \in U$ different from $x_(0)$ there is $f\left(x\right) > f\left(x_( 0)\right)$.

Local extremum combines the concepts of local minimum and local maximum.

Theorem (necessary condition for the extremum of a differentiable function)
Let $f$ be a real function on the open set $E \subset \mathbb(R)^(n)$. If at the point $x_(0) \in E$ the function $f$ has a local extremum at this point, then $$\text(d)f\left(x_(0)\right)=0.$$ Equal to zero differential is equivalent to the fact that all are equal to zero, i.e. $$\displaystyle\frac(\partial f)(\partial x_(i))\left(x_(0)\right)=0.$$

In the one-dimensional case this is – . Let us denote $\phi \left(t\right) = f \left(x_(0)+th\right)$, where $h$ is an arbitrary vector. The function $\phi$ is defined for values of $t$ that are sufficiently small in absolute value. In addition, with respect to , it is differentiable, and $(\phi)’ \left(t\right) = \text(d)f \left(x_(0)+th\right)h$.
Let $f$ have a local maximum at point x $0$. This means that the function $\phi$ at $t = 0$ has a local maximum and, by Fermat’s theorem, $(\phi)’ \left(0\right)=0$.
So, we got that $df \left(x_(0)\right) = 0$, i.e. function $f$ at point $x_(0)$ is equal to zero on any vector $h$.

Definition
Points at which the differential is zero, i.e. those in which all partial derivatives are equal to zero are called stationary. Critical points functions $f$ are those points at which $f$ is not differentiable or is equal to zero. If the point is stationary, then it does not follow from this that the function has an extremum at this point.

Example 1.
Let $f \left(x,y\right)=x^(3)+y^(3)$. Then $\displaystyle\frac(\partial f)(\partial x) = 3 \cdot x^(2)$,$\displaystyle\frac(\partial f)(\partial y) = 3 \cdot y^(2 )$, so $\left(0,0\right)$ is a stationary point, but the function has no extremum at this point. Indeed, $f \left(0,0\right) = 0$, but it is easy to see that in any neighborhood of the point $\left(0,0\right)$ the function takes both positive and negative values.

Example 2.
The function $f \left(x,y\right) = x^(2) − y^(2)$ has a stationary point at its origin, but it is clear that there is no extremum at this point.

Theorem (sufficient condition for extremum).
Let the function $f$ be twice continuously differentiable on the open set $E \subset \mathbb(R)^(n)$. Let $x_(0) \in E$ be a stationary point and $$\displaystyle Q_(x_(0)) \left(h\right) \equiv \sum_(i=1)^n \sum_(j=1) ^n \frac(\partial^(2) f)(\partial x_(i) \partial x_(j)) \left(x_(0)\right)h^(i)h^(j).$$ Then

if $Q_(x_(0))$ – , then the function $f$ at the point $x_(0)$ has a local extremum, namely, a minimum if the form is positive definite, and a maximum if the form is negative definite;
if the quadratic form $Q_(x_(0))$ is undefined, then the function $f$ at the point $x_(0)$ has no extremum.

Let's use the expansion according to Taylor's formula (12.7 p. 292). Considering that the first order partial derivatives at the point $x_(0)$ are equal to zero, we obtain $$\displaystyle f \left(x_(0)+h\right)−f \left(x_(0)\right) = \ frac(1)(2) \sum_(i=1)^n \sum_(j=1)^n \frac(\partial^(2) f)(\partial x_(i) \partial x_(j)) \left(x_(0)+\theta h\right)h^(i)h^(j),$$ where $0<\theta<1$. Обозначим $\displaystyle a_{ij}=\frac{\partial^{2} f}{\partial x_{i} \partial x_{j}} \left(x_{0}\right)$. В силу теоремы Шварца (12.6 стр. 289-290) , $a_{ij}=a_{ji}$. Обозначим $$\displaystyle \alpha_{ij} \left(h\right)=\frac{\partial^{2} f}{\partial x_{i} \partial x_{j}} \left(x_{0}+\theta h\right)−\frac{\partial^{2} f}{\partial x_{i} \partial x_{j}} \left(x_{0}\right).$$ По предположению, все непрерывны и поэтому $$\lim_{h \rightarrow 0} \alpha_{ij} \left(h\right)=0. \left(1\right)$$ Получаем $$\displaystyle f \left(x_{0}+h\right)−f \left(x_{0}\right)=\frac{1}{2}\left.$$ Обозначим $$\displaystyle \epsilon \left(h\right)=\frac{1}{|h|^{2}}\sum_{i=1}^n \sum_{j=1}^n \alpha_{ij} \left(h\right)h_{i}h_{j}.$$ Тогда $$|\epsilon \left(h\right)| \leq \sum_{i=1}^n \sum_{j=1}^n |\alpha_{ij} \left(h\right)|$$ и, в силу соотношения $\left(1\right)$, имеем $\epsilon \left(h\right) \rightarrow 0$ при $h \rightarrow 0$. Окончательно получаем $$\displaystyle f \left(x_{0}+h\right)−f \left(x_{0}\right)=\frac{1}{2}\left. \left(2\right)$$ Предположим, что $Q_{x_{0}}$ – положительноопределенная форма. Согласно лемме о положительноопределённой квадратичной форме (12.8.1 стр. 295, Лемма 1) , существует такое положительное число $\lambda$, что $Q_{x_{0}} \left(h\right) \geqslant \lambda|h|^{2}$ при любом $h$. Поэтому $$\displaystyle f \left(x_{0}+h\right)−f \left(x_{0}\right) \geq \frac{1}{2}|h|^{2} \left(λ+\epsilon \left(h\right)\right).$$ Так как $\lambda>0$, and $\epsilon \left(h\right) \rightarrow 0$ for $h \rightarrow 0$, then the right-hand side will be positive for any vector $h$ of sufficiently small length.
So, we have come to the conclusion that in some neighborhood of the point $x_(0)$ the inequality $f \left(x\right) >f \left(x_(0)\right)$ holds, if only $x \neq x_ (0)$ (we put $x=x_(0)+h$\right). This means that at the point $x_(0)$ the function has a strict local minimum, and thus the first part of our theorem is proved.
Suppose now that $Q_(x_(0))$ – indefinite form. Then there are vectors $h_(1)$, $h_(2)$ such that $Q_(x_(0)) \left(h_(1)\right)=\lambda_(1)>0$, $Q_ (x_(0)) \left(h_(2)\right)= \lambda_(2)<0$. В соотношении $\left(2\right)$ $h=th_{1}$ $t>$0. Then we get $$f \left(x_(0)+th_(1)\right)−f \left(x_(0)\right) = \frac(1)(2) \left[ t^(2) \ lambda_(1) + t^(2) |h_(1)|^(2) \epsilon \left(th_(1)\right) \right] = \frac(1)(2) t^(2) \ left[ \lambda_(1) + |h_(1)|^(2) \epsilon \left(th_(1)\right) \right].$$ For sufficiently small $t>0$, the right-hand side is positive. This means that in any neighborhood of the point $x_(0)$ the function $f$ takes values $f \left(x\right)$ greater than $f \left(x_(0)\right)$.
Similarly, we find that in any neighborhood of the point $x_(0)$ the function $f$ takes values less than $f \left(x_(0)\right)$. This, together with the previous one, means that at the point $x_(0)$ the function $f$ does not have an extremum.

Let us consider a special case of this theorem for the function $f \left(x,y\right)$ of two variables, defined in some neighborhood of the point $\left(x_(0),y_(0)\right)$ and having continuous partial derivatives of the first and second orders. Assume that $\left(x_(0),y_(0)\right)$ is a stationary point and denote $$\displaystyle a_(11)= \frac(\partial^(2) f)(\partial x ^(2)) \left(x_(0) ,y_(0)\right), a_(12)=\frac(\partial^(2) f)(\partial x \partial y) \left(x_( 0), y_(0)\right), a_(22)=\frac(\partial^(2) f)(\partial y^(2)) \left(x_(0), y_(0)\right ).$$ Then the previous theorem takes the following form.

Theorem
Let $\Delta=a_(11) \cdot a_(22) − a_(12)^2$. Then:

if $\Delta>0$, then the function $f$ has a local extremum at the point $\left(x_(0),y_(0)\right)$, namely, a minimum if $a_(11)>0$ , and maximum if $a_(11)<0$;
if $\Delta<0$, то экстремума в точке $\left(x_{0},y_{0}\right)$ нет. Как и в одномерном случае, при $\Delta=0$ экстремум может быть, а может и не быть.

Examples of problem solving

Algorithm for finding the extremum of a function of many variables:

Finding stationary points;
Find the 2nd order differential at all stationary points
Using the sufficient condition for the extremum of a function of many variables, we consider the 2nd order differential at each stationary point

Investigate the function for extremum $f \left(x,y\right) = x^(3) + 8 \cdot y^(3) + 18 \cdot x — 30 \cdot y$.
Solution
Let's find the 1st order partial derivatives: $$\displaystyle \frac(\partial f)(\partial x)=3 \cdot x^(2) - 6 \cdot y;$$ $$\displaystyle \frac(\partial f)(\partial y)=24 \cdot y^(2) — 6 \cdot x.$$ Let's compose and solve the system: $$\displaystyle \begin(cases)\frac(\partial f)(\partial x) = 0\\\frac(\partial f)(\partial y)= 0\end(cases) \Rightarrow \begin(cases)3 \cdot x^(2) - 6 \cdot y= 0\\24 \cdot y^(2) — 6 \cdot x = 0\end(cases) \Rightarrow \begin(cases)x^(2) — 2 \cdot y= 0\\4 \cdot y^(2) — x = 0 \end(cases)$$ From the 2nd equation we express $x=4 \cdot y^(2)$ - substitute it into the 1st equation: $$\displaystyle \left(4 \cdot y^(2)\right )^(2)-2 \cdot y=0$$ $$16 \cdot y^(4) — 2 \cdot y = 0$$ $$8 \cdot y^(4) — y = 0$$ $$y \left(8 \cdot y^(3) -1\right)=0$$ As a result, 2 stationary points are obtained:
1) $y=0 \Rightarrow x = 0, M_(1) = \left(0, 0\right)$;
2) $\displaystyle 8 \cdot y^(3) -1=0 \Rightarrow y^(3)=\frac(1)(8) \Rightarrow y = \frac(1)(2) \Rightarrow x=1 , M_(2) = \left(\frac(1)(2), 1\right)$
Let's check whether the sufficient condition for an extremum is satisfied:
$$\displaystyle \frac(\partial^(2) f)(\partial x^(2))=6 \cdot x; \frac(\partial^(2) f)(\partial x \partial y)=-6; \frac(\partial^(2) f)(\partial y^(2))=48 \cdot y$$
1) For the point $M_(1)= \left(0,0\right)$:
$$\displaystyle A_(1)=\frac(\partial^(2) f)(\partial x^(2)) \left(0,0\right)=0; B_(1)=\frac(\partial^(2) f)(\partial x \partial y) \left(0,0\right)=-6; C_(1)=\frac(\partial^(2) f)(\partial y^(2)) \left(0,0\right)=0;$$
$A_(1) \cdot B_(1) — C_(1)^(2) = -36<0$ , значит, в точке $M_{1}$ нет экстремума.
2) For point $M_(2)$:
$$\displaystyle A_(2)=\frac(\partial^(2) f)(\partial x^(2)) \left(1,\frac(1)(2)\right)=6; B_(2)=\frac(\partial^(2) f)(\partial x \partial y) \left(1,\frac(1)(2)\right)=-6; C_(2)=\frac(\partial^(2) f)(\partial y^(2)) \left(1,\frac(1)(2)\right)=24;$$
$A_(2) \cdot B_(2) — C_(2)^(2) = 108>0$, which means that at point $M_(2)$ there is an extremum, and since $A_(2)>0$, then this is the minimum.
Answer: The point $\displaystyle M_(2)\left(1,\frac(1)(2)\right)$ is the minimum point of the function $f$.
Investigate the function for the extremum $f=y^(2) + 2 \cdot x \cdot y - 4 \cdot x - 2 \cdot y - 3$.
Solution
Let's find stationary points: $$\displaystyle \frac(\partial f)(\partial x)=2 \cdot y - 4;$$ $$\displaystyle \frac(\partial f)(\partial y)=2 \cdot y + 2 \cdot x — 2.$$
Let's compose and solve the system: $$\displaystyle \begin(cases)\frac(\partial f)(\partial x)= 0\\\frac(\partial f)(\partial y)= 0\end(cases) \ Rightarrow \begin(cases)2 \cdot y - 4= 0\\2 \cdot y + 2 \cdot x - 2 = 0\end(cases) \Rightarrow \begin(cases) y = 2\\y + x = 1\end(cases) \Rightarrow x = -1$$
$M_(0) \left(-1, 2\right)$ is a stationary point.
Let's check whether the sufficient condition for the extremum is met: $$\displaystyle A=\frac(\partial^(2) f)(\partial x^(2)) \left(-1,2\right)=0; B=\frac(\partial^(2) f)(\partial x \partial y) \left(-1,2\right)=2; C=\frac(\partial^(2) f)(\partial y^(2)) \left(-1,2\right)=2;$$
$A \cdot B — C^(2) = -4<0$ , значит, в точке $M_{0}$ нет экстремума.
Answer: there are no extremes.

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Task 1 of 4

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Investigate the function $f$ for extrema: $f=e^(x+y)(x^(2)-2 \cdot y^(2))$

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Does the function $f = 4 + \sqrt((x^(2)+y^(2))^(2))$ have an extremum