Extrema, maximum and minimum values of functions. Label: local extremum

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

The local maximum is called strict , if the neighborhood $U$ can be chosen in such a way 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 an open set $E \subset \mathbb(R)^(n)$. It is said that $f$ has local minimum at the point $x_(0) \in E$ if there exists a neighborhood $U$ of the point $x_(0)$ such that for all $x \in U$ the inequality $f\left(x\right) \geqslant f \left(x_(0)\right)$.

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

A local extremum combines the concepts of a local minimum and a local maximum.

Theorem (necessary condition for extremum of a differentiable function)
Let $f$ be a real function on an 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 as well, then $$\text(d)f\left(x_(0)\right)=0.$$ Equality 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 . Denote $\phi \left(t\right) = f \left(x_(0)+th\right)$, where $h$ is an arbitrary vector. The function $\phi$ is defined for sufficiently small modulo values of $t$. Moreover, 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 x $0$. Hence, 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 the point $x_(0)$ is equal to zero on any vector $h$.

Definition
The points at which the differential is equal to 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 its equal to zero. If the point is stationary, then it does not yet follow 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 the origin of coordinates as a stationary point, but it is clear that there is no extremum at this point.

Theorem (sufficient condition for an extremum).
Let a function $f$ be twice continuously differentiable on an 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 $Q_(x_(0))$ is , then the function $f$ at the point $x_(0)$ has a local extremum, namely, the minimum if the form is positive-definite and the maximum if the form is negative-definite;
if the quadratic form $Q_(x_(0))$ is indefinite, then the function $f$ at the point $x_(0)$ has no extremum.

Let's use the expansion according to the Taylor formula (12.7 p. 292) . Taking into account that the first order partial derivatives at the point $x_(0)$ are equal to zero, we get $$\displaystyle f \left(x_(0)+h\right)−f \left(x_(0)\right) = \ frac(1)(2) \sum_(i=1)^n \sum_(j=1)^n \frac(\partial^(2) f)(\partial x_(i) \partial x_(j)) \left(x_(0)+\theta h\right)h^(i)h^(j),$$ 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 side is positive for any vector $h$ of sufficiently small length.
Thus, we have come to the conclusion that in some neighborhood of the point $x_(0)$ the inequality $f \left(x\right) >f \left(x_(0)\right)$ is satisfied 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 $Q_(x_(0))$ is an 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 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 obtain 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 the function $f$ does not have an extremum at the point $x_(0)$.

Let us consider a particular case of this theorem for a 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:

We find stationary points;
We find the differential of the 2nd order at all stationary points
Using the sufficient condition for the extremum of a function of several variables, we consider the second-order differential at each stationary point

Investigate the function to the extremum $f \left(x,y\right) = x^(3) + 8 \cdot y^(3) + 18 \cdot x — 30 \cdot y$.
Solution
Find partial derivatives of the 1st order: $$\displaystyle \frac(\partial f)(\partial x)=3 \cdot x^(2) - 6 \cdot y;$$ $$\displaystyle \frac(\partial f)(\partial y)=24 \cdot y^(2) — 6 \cdot x.$$ 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 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 us check the fulfillment of the sufficient extremum condition:
$$\displaystyle \frac(\partial^(2) f)(\partial x^(2))=6 \cdot x; \frac(\partial^(2) f)(\partial x \partial y)=-6; \frac(\partial^(2) f)(\partial y^(2))=48 \cdot y$$
1) For 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 $M_(2)$:
$$\displaystyle A_(2)=\frac(\partial^(2) f)(\partial x^(2)) \left(1,\frac(1)(2)\right)=6; B_(2)=\frac(\partial^(2) f)(\partial x \partial y) \left(1,\frac(1)(2)\right)=-6; C_(2)=\frac(\partial^(2) f)(\partial y^(2)) \left(1,\frac(1)(2)\right)=24;$$
$A_(2) \cdot B_(2) — C_(2)^(2) = 108>0$, so there is an extremum at $M_(2)$, 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
Find stationary points: $$\displaystyle \frac(\partial f)(\partial x)=2 \cdot y - 4;$$ $$\displaystyle \frac(\partial f)(\partial y)=2 \cdot y + 2 \cdot x — 2.$$
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 the fulfillment of the sufficient extremum condition: $$\displaystyle A=\frac(\partial^(2) f)(\partial x^(2)) \left(-1,2\right)=0; B=\frac(\partial^(2) f)(\partial x \partial y) \left(-1,2\right)=2; C=\frac(\partial^(2) f)(\partial y^(2)) \left(-1,2\right)=2;$$
$A \cdot B - C^(2) = -4<0$ , значит, в точке $M_{0}$ нет экстремума.
Answer: there are no extrema.

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>> Extremes

Function extremum

Definition of extremum

Function y = f(x) is called increasing (waning) in some interval if for x 1< x 2 выполняется неравенство (f (x 1) < f (x 2) (f (x 1) >f(x2)).

If a differentiable function y \u003d f (x) on a segment increases (decreases), then its derivative on this segment f " (x )> 0

(f"(x)< 0).

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

extremum points

Necessary conditions for an extremum . If point x O is an 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, where 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 the 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 the derivative does not change sign when passing through a critical point, then at the point x O there is no extremum.

The 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 very point x o. If f"(x o) = 0, >0 ( <0), то точка x o is a local minimum (maximum) point of the function f(x). If =0, then one must either use the first sufficient condition, or involve higher ones.

On a segment, the function y \u003d f (x) can reach the smallest or largest value either at critical points or at the ends of the segment.

Example 3.22.

Solution. Because f " (

Tasks for finding the extremum of a function

Example 3.23. a

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

Example 3.24. p ≈

Solution. pp
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) \u003d 6x 2 - 30x +36 \u003d 6 (x -2) (x - 3), then the critical points of the function x 1 \u003d 2 and x 2 \u003d 3. Extreme points can only be at these points. Since when passing through the point x 1 \u003d 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 \u003d 3, the derivative changes sign from minus to plus, therefore, at the point x 2 \u003d 3, the function has a minimum. Calculating the values of the function in 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 with wire mesh on three sides, and adjoins the wall on the fourth side. For this there is a linear meters of the grid. At what aspect ratio will the site have the largest area?

Solution.Denote the sides of the site through x And y. The area of the site is equal to S = xy. Let y is the length of the side adjacent to the wall. Then, by condition, the equality 2x + y = a must hold. Therefore y = a - 2x and S = x (a - 2x), where
0 ≤ x≤ a /2 (the length and width of the pad cannot be negative). S "= a - 4x, a - 4x = 0 for x = a/4, whence
y \u003d a - 2 × a / 4 \u003d a / 2. Because the x = a /4 is the only critical point, let's check whether the sign of the derivative changes when passing through this point. For x a /4 S "> 0, while 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 (sq.. units). Since S is continuous on and its values at the ends of S(0) and S(a /2) are equal to zero, then the value found will be the largest value of the function. 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 make 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) in order to use the least amount of material for its manufacture?

Solution.The total surface area of the cylinder is S = 2 p R(R+H). We know the volume of the cylinder V = p R 2 N Þ N \u003d V / p R 2 \u003d 16 p / p R 2 \u003d 16 / R 2. So S(R) = 2 p (R2+16/R). We find the derivative of this function:
S"(R) \u003d 2 p (2R- 16 / R 2) \u003d 4 p (R- 8 / R 2). S" (R) = 0 for R 3 = 8, therefore,
R = 2, H = 16/4 = 4.

MAXIMUM AND MINIMUM POINTS

points at which it takes the largest or smallest values in the domain of definition; such points are called also points of absolute maximum or absolute minimum. If f is defined on a topological space X, then the point x 0 called point of local maximum (local minimum), if such a point exists x 0, that for the restriction of the function under consideration to this neighborhood, the point x 0 is the absolute maximum (minimum) point. Distinguish points of strict and non-strict maximum (mini m u m a) (both absolute and local). For example, a point called point of a non-strict (strict) local maximum of the function f, if there exists such a neighborhood of the point x 0, which holds for all (respectively, f(x) x0). )/

For functions defined on finite-dimensional domains, in terms of differential calculus, there are conditions and criteria for a given point to be a local maximum (minimum) point. Let the function f be defined in a certain neighborhood of the box x 0 of the real axis. If x 0 - point of non-strict local maximum (minimum) and at this point there exists f"( x0), then it is equal to zero.

If a given function f is differentiable in a neighborhood of a point x 0 , except, perhaps, for this point itself, at which it is continuous, and the derivative f" on each side of the point x0 preserves a constant sign in this neighborhood, then in order to x0 was a point of a strict local maximum (local minimum), it is necessary and sufficient that the derivative changes sign from plus to minus, i.e., that f "(x)> 0 at x<.x0 and f"(x)<0 при x>x0(respectively from minus to plus: f"(X) <0 at x<x0 and f"(x)>0 when x>x 0). However, not for every function differentiable in a neighborhood of a point x 0 , one can speak of a change in the sign of the derivative at this point. . "

If the function f has at the point x 0 t derivatives, moreover, in order to x 0 is a point of strict local maximum, it is necessary and sufficient that τ be even and that f (m) ( x0)<0, и - локального минимума, чтобы m было четно и f (m) (x0)>0.

Let the function f( x 1 ..., x p] is defined in an n-dimensional neighborhood of a point and is differentiable at this point. If x (0) is a non-strict local maximum (minimum) point, then the function f at this point is equal to zero. This condition is equivalent to the equality to zero at this point of all partial derivatives of the 1st order of the function f. If a function has 2nd continuous partial derivatives at x(0) , all its 1st derivatives vanish at x(0) and the 2nd order differential at x(0) is a negative (positive) quadratic shape, then x(0) is a point of strict local maximum (minimum). Conditions are known for M. and M. T. differentiable functions, when certain restrictions are imposed on the changes in the arguments: the constraint equations are satisfied. Necessary and sufficient conditions for the maximum (minimum) of a real function, which has a more complex structure, are studied in special branches of mathematics: for example, in convex analysis, mathematical programming(see also Maximization and function minimization). M. and m.t. functions defined on manifolds are studied in calculus of variations in general, and M. and m.t. for functions defined on function spaces, i.e., for functionals, in variational calculus. There are also various methods of numerical approximate finding of M. and m. t.

Lit.: Il'in V. A., Poznya to E. G., Fundamentals of Mathematical Analysis, 3rd ed., Part 1, M., 1971; KudryavtsevL. L. D. Kudryavtsev.

Mathematical encyclopedia. - M.: Soviet Encyclopedia. I. M. Vinogradov. 1977-1985.

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The change of a function at a certain point and is defined as the limit of the increment of the function to the increment of the argument, which tends to zero. To find it, use the table of derivatives. For example, the derivative of the function y = x3 will be equal to y’ = x2.

Equate this derivative to zero (in this case x2=0).

Find the value of the given variable. These will be the values for which this derivative will be equal to 0. To do this, substitute arbitrary numbers in the expression instead of x, at which the entire expression will become zero. For example:

2-2x2=0
(1-x)(1+x) = 0
x1=1, x2=-1

Apply the obtained values on the coordinate line and calculate the sign of the derivative for each of the obtained ones. Points are marked on the coordinate line, which are taken as the origin. To calculate the value in the intervals, substitute arbitrary values that match the criteria. For example, for the previous function up to the interval -1, you can choose the value -2. For -1 to 1, you can choose 0, and for values greater than 1, choose 2. Substitute these numbers in the derivative and find out the sign of the derivative. In this case, the derivative with x = -2 will be equal to -0.24, i.e. negative and there will be a minus sign on this interval. If x=0, then the value will be equal to 2, and a sign is put on this interval. If x=1, then the derivative will also be equal to -0.24 and a minus is put.

If, when passing through a point on the coordinate line, the derivative changes its sign from minus to plus, then this is a minimum point, and if from plus to minus, then this is a maximum point.

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 existence of an extremum there. As an example, we can take the everywhere differentiable function
. Both 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, there is no extremum. Therefore, it is necessary to know sufficient conditions under which a point suspected of an extremum is an extremum point of the function under study.

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

,
.

Let us introduce the notation:

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

If
then the function
at the point
reaches

local maximum at
And

local minimum at
.

In general, for a function
sufficient condition for existence at a point
localminimum(maximum) is positive(negative) the definiteness 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 maximum, If
).

Example 18.Find local extremum points of a function

Solution. Find the partial derivatives of the function and equate them to zero:

Solving this system, we find two possible extremum points:

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

At the first stationary point , therefore, and
Therefore, further research is required for this point. Function value
at this point is zero:
Further,

at

A

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