The largest and smallest values of the function concepts mathematical analysis. The value taken by a function at some point of the set on which this function is defined is called the largest (smallest) value on this set if the function does not have a larger (smaller) value at any other point in the set. N. and n. h. f. in comparison with its values at all sufficiently close points are called extrema (respectively, maxima and minima) of the function. N. and n. h. f., given on a segment, can be achieved either at points where the derivative is equal to zero, or at points where it does not exist, or at the ends of the segment. A continuous function given on a segment necessarily reaches its maximum and minimum values; if a continuous function is considered on an interval (that is, a segment with excluded ends), then among its values on this interval there may not be a maximum or a minimum. For example, the function at = x, given on the interval , reaches the largest and smallest values, respectively, at x= 1 and x= 0 (i.e., at the ends of the segment); if we consider this function on the interval (0; 1), then among its values on this interval there is neither the largest nor the smallest, since for each x0 there is always a point of this interval lying to the right (to the left) x0, and such that the value of the function at this point will be greater (respectively, less) than at the point x0. Similar statements are valid for functions of several variables. See also Extreme.
Big soviet encyclopedia. - M.: Soviet Encyclopedia. 1969-1978 .
See what "The largest and smallest values of a function" are in other dictionaries:
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Concepts of mathematical analysis. The value taken by the function at some point of the set on which this function is given is called the largest (smallest) on this set, if at no other point the function has a larger (smaller) ... ... encyclopedic Dictionary
The concepts of mathematics. analysis. The value taken by the function at a particular point of the set, pa rum this function is given, called. largest (smallest) on this set, if at no other point does the function have a larger (smaller) value ... Natural science. encyclopedic Dictionary
MAXIMUM AND MINIMUM FUNCTION- respectively, the largest and smallest values of the function in comparison with its values at all sufficiently close points. High and low points are called extreme points... Great Polytechnic Encyclopedia
The largest and, accordingly, the smallest values of a function that takes real values. The point of the domain of definition of the function in question, in which it takes a maximum or minimum, is called. respectively the maximum point or the minimum point ... ... Mathematical Encyclopedia
A ternary function in the theory of functional systems and ternary logic is a function of type, where is a ternary set, and is a non-negative integer, which is called the arity or locality of the function. The elements of the set are digital ... ... Wikipedia
Representation of Boolean functions by normal forms (see Normal forms of Boolean functions). the simplest with respect to some measure of complexity. Usually, the complexity of a normal form is understood as the number of letters in it. In this case, the simplest form is called ... ... Mathematical Encyclopedia
A function that receives infinitesimal increments as the argument increments infinitesimally. A single-valued function f (x) is called continuous for the value of the argument x0, if for all values of the argument x that differ sufficiently little from x0 ... Great Soviet Encyclopedia
- (Latin maximum and minimum, literally the largest and smallest) (Math.), the largest and smallest values of a function compared to its values \u200b\u200bin sufficiently close points. In the figure, the function y \u003d f (x) has a maximum at the points x1 and x3, and at the point x2 ... ... encyclopedic Dictionary
- (from the Latin maximum and minimum, the largest and smallest) (mathematical), the largest and smallest values of a function compared to its values \u200b\u200bin sufficiently close points. High and low points are called extreme points... Modern Encyclopedia
And to solve it, you need minimal knowledge of the topic. The next academic year, everyone wants to go on vacation, and to bring this moment closer, I immediately get down to business:
Let's start with the area. The area referred to in the condition is limited closed set of points in the plane. For example, a set of points bounded by a triangle, including the ENTIRE triangle (if from borders“Poke out” at least one point, then the area 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 more is not needed now.
The flat area is standardly denoted by the letter , and, as a rule, is given analytically - by several equations (not necessarily linear); less often inequalities. A typical verbal turnover: "closed arealimited by lines".
An integral part of the task under consideration is the construction of the area on the drawing. How to do it? It is necessary to draw all the lines listed (in this case 3 straight) and analyze what happened. The desired area is usually lightly hatched, and its border is highlighted with a bold line: 
The same area can be set linear inequalities: , which for some reason are more often written as an enumeration list, and not system.
Since the boundary belongs to the region, then all inequalities, of course, non-strict.
And now the crux of the matter. Imagine that the axis goes straight to you from the origin of coordinates. Consider a function that continuous in each area point. The graph of this function is surface, and the small happiness is that in order to solve today's problem, we do not need to know what this surface looks like at all. It can be located above, below, cross the plane - all this is not important. And the following is important: according to Weierstrass theorems, continuous in limited closed area, the function reaches its maximum (of the "highest") and least (of the "lowest") values to be found. These values are achieved or in stationary points, belonging to the regionD , or at points that lie on the boundary of this region. From which follows a simple and transparent solution algorithm:
Example 1
In a limited enclosed area
Decision: First of all, you need to depict the area on the drawing. Unfortunately, it is technically difficult for me to make an interactive model of the problem, and therefore I will immediately give the final illustration, which shows all the "suspicious" points found during the study. Usually they are put down one after another as they are found: 
Based on the preamble, the decision can be conveniently divided into two points:
I) Let's find stationary points. This is a standard action that we have repeatedly performed in the lesson. about extrema of several variables:
Found stationary point belongs areas: (mark it on the drawing), which means that 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 the important results in bold. In a notebook, it is convenient to circle them with a pencil.
Pay attention to our second happiness - there is no point in checking sufficient condition for an extremum. Why? Even if at the 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 extremes) .
What if the stationary point does NOT belong to the area? Almost nothing! It should be noted that and go to the next paragraph.
II) We investigate the border of the region.
Since the border consists of the sides of a triangle, it is convenient to divide the study into 3 subparagraphs. But it is better to do it not anyhow. From my point of view, at first it is more advantageous to consider the segments parallel to the coordinate axes, and first of all, those lying on the axes themselves. To catch the whole sequence and logic of actions, try to study the ending "in one breath":
1) Let's deal with the lower side of the triangle. To do this, we substitute directly into the function:
Alternatively, you can do it like this: 
Geometrically, this means that coordinate plane (which is also given by the equation)"cut out" from surfaces"spatial" parabola, the top of which immediately falls under suspicion. Let's find out where is she:
- the resulting value "hit" in the area, and it may well be that at the point (mark on the drawing) the function reaches the largest or smallest value in the entire area. Anyway, let's do the calculations:
Other "candidates" are, of course, the ends of the segment. Calculate the values of the function at points 
(mark on the drawing):
Here, by the way, you can perform an oral mini-check on the "stripped down" version: 
2) To study the right side of the triangle, we substitute it into the function and “put things in order there”:
Here we immediately perform a rough check, “ringing” the already processed end of the segment:
, perfect.
The geometric situation is related to the previous point:
- the resulting value also “entered the scope of our interests”, which means that we need to calculate what the function is equal to at the appeared point:
Let's examine the second end of the segment:
Using the function 
, let's check: 
3) Everyone probably knows how to explore the remaining side. We substitute into the function and carry out simplifications:
Line ends 
have already been investigated, but on the draft we still check whether we 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 something interesting inside the segment :
- there is! Substituting a 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 control the calculations according to the "budget" version 
:
, order.
And the final step: CAREFULLY look through all the "fat" numbers, I recommend even beginners to compose single list:
from which we choose the largest and smallest values. Answer write in the style of the problem of finding the largest and smallest values of the function on the segment:
I'll comment again just in case. geometric sense result:
- here is the most high point surfaces in the area ;
- here is the lowest point of the surface in the area.
In the analyzed problem, we found 7 “suspicious” points, but their number varies from task to task. For a triangular region, the minimum "exploration set" consists of three points. This happens when the function, for example, sets plane- it is quite clear that there are no stationary points, and the function can reach the maximum / minimum values only at the vertices of the triangle. But there are no such examples once, twice - usually you have to deal with some kind of surface of the 2nd order.
If you solve such tasks a little, then triangles can make your head spin, and therefore 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 bounded closed area.
Special attention pay attention to the rational order and technique of studying the boundary of the area, as well as to the chain of intermediate checks, which will almost completely avoid computational errors. Generally speaking, you can solve it as you like, but in some problems, for example, in the same Example 2, there is every chance to significantly complicate your life. An approximate example of finishing assignments at the end of the lesson.
We systematize the solution algorithm, otherwise, with my diligence of a spider, it somehow got lost in a long thread of comments of the 1st example:
- At the first step, we build an area, it is desirable to shade it, and highlight the border with a thick line. During the solution, points will appear that need to be put on the drawing.
– Find stationary points and calculate the values of the function only in those, which belong to the area . The obtained values are highlighted in the text (for example, circled with a pencil). If the stationary point does NOT belong to the area, 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 item cannot be skipped!
– Exploring the border area. First, it is advantageous to deal with straight lines that are parallel to the coordinate axes (if there are any). The function values calculated at "suspicious" points are also highlighted. A lot has been said about the solution technique above and something else will be said below - read, re-read, delve into!
- From the selected numbers, select the largest and smallest values \u200b\u200band give an answer. Sometimes it happens that the 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 that
The final examples are devoted to other useful ideas that will come in handy in practice:
Example 4
Find the largest and smallest values of a function in a closed area 
.
I have kept the author's formulation, in which the area is given as a double inequality. This condition can be written in an equivalent system or in a more traditional form for this problem: 
I remind you that with non-linear we encountered inequalities on , and if you do not understand the geometric meaning of the entry, then please do not delay and clarify the situation right now ;-)
Decision, as always, begins with the construction of the area, which is a kind of "sole": 
Hmm, sometimes you have to gnaw not only the granite of science ....
I) Find stationary points: 
Idiot's dream system :) 
The stationary point belongs to the region, namely, lies on its boundary.
And so, it’s nothing ... fun lesson went - that’s what it means to drink the right tea =)
II) We investigate the border of the region. Without further ado, let's start with the x-axis:
1) If , then
Find where the top of the parabola is:
- Appreciate such moments - "hit" right to the point, from which everything is already clear. But don't forget to check: 
Let's calculate the values of the function at the ends of the segment: 
2) C bottom Let's figure out the "soles" "in one sitting" - without any complexes we substitute into the function, moreover, we will be interested only in the segment:
The control:
Now this is already bringing some revival to the monotonous ride on a knurled track. Let's find the critical points:
We decide quadratic equation do you remember this one? ... However, remember, of course, otherwise you would not have read these lines =) If in the two previous examples calculations in decimal fractions were convenient (which, by the way, is rare), then here we are waiting for the usual common fractions. We find the “x” roots and, using the equation, 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:
Here are the "candidates", so the "candidates"!
For a standalone solution:
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 the real need to use it is unlikely to arise. So, for example, if a function with the same area "de" is given, then after substitution into it - with a derivative of no difficulties; moreover, everything is drawn up in a “one line” (with signs) without the need to consider the upper and lower semicircles separately. But, of course, there are more complicated cases, where without the Lagrange function (where , for example, is the same circle equation) it's hard to get by - how hard it is to get by without a good rest!
All the best to pass the session and see you soon next season!
Solutions and answers:
Example 2: Decision: draw the area on the drawing:
In this article I will talk about algorithm for finding the largest and smallest value function, minimum and maximum points.
From theory, we will definitely need derivative table and differentiation rules. It's all in this board:
Algorithm for finding the largest and smallest values.
I find it easier to explain with a concrete example. Consider:
Example: Find the largest value of the function y=x^5+20x^3–65x on the segment [–4;0].
Step 1. We take the derivative.
Y" = (x^5+20x^3–65x)" = 5x^4 + 20*3x^2 - 65 = 5x^4 + 60x^2 - 65
Step 2 Finding extremum points.
extremum point we name such points at which the function reaches its maximum or minimum value.
To find the extremum points, it is necessary to equate the derivative of the function to zero (y "= 0)
5x^4 + 60x^2 - 65 = 0
Now let's solve this quadratic equation and the found roots are our extremum points.
I solve such equations by replacing t = x^2, then 5t^2 + 60t - 65 = 0.
Reduce the equation by 5, we get: t^2 + 12t - 13 = 0
D = 12^2 - 4*1*(-13) = 196
T_(1) = (-12 + sqrt(196))/2 = (-12 + 14)/2 = 1
T_(2) = (-12 - sqrt(196))/2 = (-12 - 14)/2 = -13
We make the reverse substitution x^2 = t:
X_(1 and 2) = ±sqrt(1) = ±1
x_(3 and 4) = ±sqrt(-13) (we exclude, there cannot be negative numbers under the root, unless of course we are talking about complex numbers)
Total: x_(1) = 1 and x_(2) = -1 - these are our extremum points.
Step 3 Determine the largest and smallest value.
Substitution method.
In the condition, we were given the segment [b][–4;0]. The point x=1 is not included in this segment. So we don't consider it. But in addition to the point x=-1, we also need to consider the left and right borders of our segment, that is, the points -4 and 0. To do this, we substitute all these three points into the original function. Notice the original one is the one given in the condition (y=x^5+20x^3–65x), some start substituting into the derivative...
Y(-1) = (-1)^5 + 20*(-1)^3 - 65*(-1) = -1 - 20 + 65 = [b]44
y(0) = (0)^5 + 20*(0)^3 - 65*(0) = 0
y(-4) = (-4)^5 + 20*(-4)^3 - 65*(-4) = -1024 - 1280 + 260 = -2044
This means that the maximum value of the function is [b]44 and it is reached at the points [b]-1, which is called the maximum point of the function on the segment [-4; 0].
We decided and got an answer, we are great, you can relax. But stop! Don't you think that counting y(-4) is somehow too complicated? In conditions of limited time, it is better to use another method, I call it like this:
Through intervals of constancy.
These gaps are found for the derivative of the function, that is, for our biquadratic equation.
I do it in the following way. I draw a directional line. I set the points: -4, -1, 0, 1. Despite the fact that 1 is not included in the given segment, it should still be noted in order to correctly determine the intervals of constancy. Let's take some number many times greater than 1, let's say 100, mentally substitute it into our biquadratic equation 5(100)^4 + 60(100)^2 - 65. Even without counting anything, it becomes obvious that at the point 100 the function has plus sign. This means that for intervals from 1 to 100 it has a plus sign. When passing through 1 (we go from right to left), the function will change sign to minus. When passing through the point 0, the function will retain its sign, since this is only the boundary of the segment, and not the root of the equation. When passing through -1, the function will again change sign to plus.
From theory, we know that where the derivative of the function is (and we drew this for it) changes sign from plus to minus (point -1 in our case) function reaches its local maximum (y(-1)=44 as calculated earlier) on this segment (this is logically very clear, the function has ceased to increase, since it reached its maximum and began to decrease).
Accordingly, where the derivative of the function changes sign from minus to plus, achieved local minimum of a function. Yes, yes, we also found the local minimum point, which is 1, and y(1) is the minimum value of the function on the interval, let's say from -1 to +∞. Please note that this is only a LOCAL MINIMUM, that is, a minimum on a certain segment. Since the actual (global) minimum function will reach somewhere there, in -∞.
In my opinion, the first method is simpler theoretically, and the second one is simpler in terms of arithmetic operations, but much more difficult in terms of theory. After all, sometimes there are cases when the function does not change sign when passing through the root of the equation, and indeed you can get confused with these local, global maxima and minima, although you will have to master this well anyway if you plan to enter technical university(and why otherwise hand over profile exam and solve this problem). But practice and only practice will teach you how to solve such problems once and for all. And you can train on our website. Here .
If you have any questions, or something is not clear, be sure to ask. I will be happy to answer you, and make changes, additions to the article. Remember we are making this site together!
From a practical point of view, the most interesting is the use of the derivative to find the largest and smallest value of a function. What is it connected with? Maximizing profits, minimizing costs, determining the optimal load of equipment... In other words, in many areas of life, one has to solve the problem of optimizing some parameters. And this is the problem of finding the largest and smallest values of the function.
It should be noted that the largest and smallest value of a function is usually sought on some interval X , which is either the entire domain of the function or part of the domain. The interval X itself can be a line segment, an open interval 
, an infinite interval .
In this article, we will talk about finding the largest and smallest values of an explicitly given function of one variable y=f(x) .
Page navigation.
The largest and smallest value of a function - definitions, illustrations.
Let us briefly dwell on the main definitions.
The largest value of the function 
, which for any 
the inequality is true.
The smallest value of the function y=f(x) on the interval X is called such a value 
, which for any the inequality is true.
These definitions are intuitive: the largest (smallest) value of a function is the largest (smallest) value accepted on the interval under consideration with the abscissa.
Stationary points are the values of the argument at which the derivative of the function vanishes.
Why do we need stationary points when finding the largest and smallest values? The answer to this question is given by Fermat's theorem. It follows from this theorem that if a differentiable function has an extremum (local minimum or local maximum) at some point, then this point is stationary. Thus, the function often takes its maximum (smallest) value on the interval X at one of the stationary points from this interval.
Also, a function can often take on the largest and smallest values at points where the first derivative of this function does not exist, and the function itself is defined.
Let's immediately answer one of the most common questions on this topic: "Is it always possible to determine the largest (smallest) value of a function"? No not always. Sometimes the boundaries of the interval X coincide with the boundaries of the domain of the function, or the interval X is infinite. And some functions at infinity and on the boundaries of the domain of definition can take both infinitely large and infinitely small values. In these cases, nothing can be said about the largest and smallest value of the function.
For clarity, we give a graphic illustration. Look at the pictures - and much will become clear.
On the segment
In the first figure, the function takes the largest (max y ) and smallest (min y ) values at stationary points inside the segment [-6;6] .
Consider the case shown in the second figure. Change the segment to . In this example, the smallest value of the function is achieved at a stationary point, and the largest - at a point with an abscissa corresponding to the right boundary of the interval.
In figure No. 3, the boundary points of the segment [-3; 2] are the abscissas of the points corresponding to the largest and smallest value of the function.
In the open range
In the fourth figure, the function takes the largest (max y ) and smallest (min y ) values at stationary points within the open interval (-6;6) .
On the interval , no conclusions can be drawn about the largest value.
At infinity
In the example shown in the seventh figure, the function takes the largest value (max y ) at a stationary point with the abscissa x=1 , and the smallest value (min y ) is reached at the right boundary of the interval. At minus infinity, the values of the function asymptotically approach y=3 .
On the interval, the function does not reach either the smallest or the largest value. As x=2 tends to the right, the function values tend to minus infinity (the straight line x=2 is a vertical asymptote), and as the abscissa tends to plus infinity, the function values asymptotically approach y=3 . A graphic illustration of this example is shown in Figure 8.
Algorithm for finding the largest and smallest values of a continuous function on the segment .
We write an algorithm that allows us to find the largest and smallest value of a function on a segment.
- We find the domain of the function and check if it contains the entire segment .
- We find all points at which the first derivative does not exist and which are contained in the segment (usually such points occur in functions with an argument under the module sign and in power functions with a fractional rational exponent). If there are no such points, then go to the next point.
- We determine all stationary points that fall into the segment. To do this, we equate it to zero, solve the resulting equation and choose the appropriate roots. If there are no stationary points or none of them fall into the segment, then go to the next step.
- We calculate the values of the function at the selected stationary points (if any), at points where the first derivative does not exist (if any), and also at x=a and x=b .
- From the obtained values of the function, we select the largest and smallest - they will be the desired maximum and smallest values of the function, respectively.
Let's analyze the algorithm when solving an example for finding the largest and smallest values of a function on a segment.
Example.
Find the largest and smallest value of a function
- on the segment;
- on the interval [-4;-1] .
Decision.
The domain of the function is the entire set real numbers, except for zero, that is, . Both segments fall within the domain of definition.
We find the derivative of the function with respect to: 
Obviously, the derivative of the function exists at all points of the segments and [-4;-1] .
Stationary points are determined from the equation . The only real root is x=2 . This stationary point falls into the first segment.
For the first case, we calculate the values of the function at the ends of the segment and at a stationary point, that is, for x=1 , x=2 and x=4 : 
Therefore, the largest value of the function 
is reached at x=1 , and the smallest value 
– at x=2 .
For the second case, we calculate the values of the function only at the ends of the segment [-4;-1] (since it does not contain a single stationary point): 
Decision.
Let's start with the scope of the function. Square trinomial the denominator of a fraction must not vanish: 
It is easy to check that all intervals from the condition of the problem belong to the domain of the function.
Let's differentiate the function: 
Obviously, the derivative exists on the entire domain of the function.
Let's find stationary points. The derivative vanishes at . This stationary point falls within the intervals (-3;1] and (-3;2) .
And now you can compare the results obtained at each point with the graph of the function. The blue dotted lines indicate the asymptotes.
This can end with finding the largest and smallest value of the function. The algorithms discussed in this article allow you to get results with a minimum of actions. However, it can be useful to first determine the intervals of increase and decrease of the function and only after that draw conclusions about the largest and smallest value of the function on any interval. This gives a clearer picture and a rigorous justification of the results.
Let the function $z=f(x,y)$ be defined and continuous in some bounded closed domain $D$. Let the given function have finite partial derivatives of the first order in this region (with the possible exception of a finite number of points). To find the largest and smallest values of a function of two variables in a given closed region, three steps of a simple algorithm are required.
Algorithm for finding the largest and smallest values of the function $z=f(x,y)$ in the closed domain $D$.
- Find the critical points of the function $z=f(x,y)$ that belong to the region $D$. Compute function values at critical points.
- Investigate the behavior of the function $z=f(x,y)$ on the boundary of the region $D$ by finding the points of possible maximum and minimum values. Calculate the function values at the obtained points.
- From the function values obtained in the previous two paragraphs, choose the largest and smallest.
What are critical points? show/hide
Under critical points imply points where both first-order partial derivatives are equal to zero (i.e. $\frac(\partial z)(\partial x)=0$ and $\frac(\partial z)(\partial y)=0 $) or at least one partial derivative does not exist.
Often the points at which the first-order partial derivatives are equal to zero are called stationary points. Thus, stationary points are a subset of critical points.
Example #1
Find the maximum and minimum values of the function $z=x^2+2xy-y^2-4x$ in the closed region bounded by the lines $x=3$, $y=0$ and $y=x+1$.
We will follow the above, but first we will deal with the drawing of a given area, which we will denote by the letter $D$. We are given the equations of three straight lines, which limit this area. The straight line $x=3$ passes through the point $(3;0)$ parallel to the y-axis (axis Oy). The straight line $y=0$ is the equation of the abscissa axis (Ox axis). Well, to construct a straight line $y=x+1$ let's find two points through which we draw this straight line. You can, of course, substitute a couple of arbitrary values instead of $x$. For example, substituting $x=10$, we get: $y=x+1=10+1=11$. We have found the point $(10;11)$ lying on the line $y=x+1$. However, it is better to find those points where the line $y=x+1$ intersects with the lines $x=3$ and $y=0$. Why is it better? Because we will lay down a couple of birds with one stone: we will get two points for constructing the straight line $y=x+1$ and at the same time find out at what points this straight line intersects other lines that bound the given area. The line $y=x+1$ intersects the line $x=3$ at the point $(3;4)$, and the line $y=0$ - at the point $(-1;0)$. In order not to clutter up the course of the solution with auxiliary explanations, I will put the question of obtaining these two points in a note.
How were the points $(3;4)$ and $(-1;0)$ obtained? show/hide
Let's start from the point of intersection of the lines $y=x+1$ and $x=3$. The coordinates of the desired point belong to both the first and second lines, so to find unknown coordinates, you need to solve the system of equations:
$$ \left \( \begin(aligned) & y=x+1;\\ & x=3. \end(aligned) \right. $$
The solution of such a system is trivial: substituting $x=3$ into the first equation we will have: $y=3+1=4$. The point $(3;4)$ is the desired intersection point of the lines $y=x+1$ and $x=3$.
Now let's find the point of intersection of the lines $y=x+1$ and $y=0$. Again, we compose and solve the system of equations:
$$ \left \( \begin(aligned) & y=x+1;\\ & y=0. \end(aligned) \right. $$
Substituting $y=0$ into the first equation, we get: $0=x+1$, $x=-1$. The point $(-1;0)$ is the desired intersection point of the lines $y=x+1$ and $y=0$ (abscissa axis).
Everything is ready to build a drawing that will look like this:
The question of the note seems obvious, because everything can be seen from the figure. However, it is worth remembering that the drawing cannot serve as evidence. The figure is just an illustration for clarity.
Our area was set using the equations of lines that limit it. It's obvious that these lines define a triangle, don't they? Or not quite obvious? Or maybe we are given a different area, bounded by the same lines:
Of course, the condition says that the area is closed, so the picture shown is wrong. But to avoid such ambiguities, it is better to define regions by inequalities. We are interested in the part of the plane located under the line $y=x+1$? Ok, so $y ≤ x+1$. Our area should be located above the line $y=0$? Great, so $y ≥ 0$. By the way, the last two inequalities are easily combined into one: $0 ≤ y ≤ x+1$.
$$ \left \( \begin(aligned) & 0 ≤ y ≤ x+1;\\ & x ≤ 3. \end(aligned) \right. $$
These inequalities define the domain $D$, and define it uniquely, without any ambiguities. But how does this help us in the question at the beginning of the footnote? It will also help :) We need to check if the point $M_1(1;1)$ belongs to the region $D$. Let us substitute $x=1$ and $y=1$ into the system of inequalities that define this region. If both inequalities are satisfied, then the point lies inside the region. If at least one of the inequalities is not satisfied, then the point does not belong to the region. So:
$$ \left \( \begin(aligned) & 0 ≤ 1 ≤ 1+1;\\ & 1 ≤ 3. \end(aligned) \right. \;\; \left \( \begin(aligned) & 0 ≤ 1 ≤ 2;\\ & 1 ≤ 3. \end(aligned) \right.$$
Both inequalities are true. The point $M_1(1;1)$ belongs to the region $D$.
Now it is the turn to investigate the behavior of the function on the boundary of the domain, i.e. go to. Let's start with the straight line $y=0$.
The straight line $y=0$ (abscissa axis) limits the region $D$ under the condition $-1 ≤ x ≤ 3$. Substitute $y=0$ into given function$z(x,y)=x^2+2xy-y^2-4x$. The resulting substitution function of one variable $x$ will be denoted as $f_1(x)$:
$$ f_1(x)=z(x,0)=x^2+2x\cdot 0-0^2-4x=x^2-4x. $$
Now for the function $f_1(x)$ we need to find the largest and smallest values on the interval $-1 ≤ x ≤ 3$. Find the derivative of this function and equate it to zero:
$$ f_(1)^(")(x)=2x-4;\\ 2x-4=0; \; x=2. $$
The value $x=2$ belongs to the segment $-1 ≤ x ≤ 3$, so we also add $M_2(2;0)$ to the list of points. In addition, we calculate the values of the function $z$ at the ends of the segment $-1 ≤ x ≤ 3$, i.e. at the points $M_3(-1;0)$ and $M_4(3;0)$. By the way, if the point $M_2$ did not belong to the segment under consideration, then, of course, there would be no need to calculate the value of the function $z$ in it.
So, let's calculate the values of the function $z$ at the points $M_2$, $M_3$, $M_4$. You can, of course, substitute the coordinates of these points in the original expression $z=x^2+2xy-y^2-4x$. For example, for the point $M_2$ we get:
$$z_2=z(M_2)=2^2+2\cdot 2\cdot 0-0^2-4\cdot 2=-4.$$
However, the calculations can be simplified a bit. To do this, it is worth remembering that on the segment $M_3M_4$ we have $z(x,y)=f_1(x)$. I'll spell it out in detail:
\begin(aligned) & z_2=z(M_2)=z(2,0)=f_1(2)=2^2-4\cdot 2=-4;\\ & z_3=z(M_3)=z(- 1,0)=f_1(-1)=(-1)^2-4\cdot (-1)=5;\\ & z_4=z(M_4)=z(3,0)=f_1(3)= 3^2-4\cdot 3=-3. \end(aligned)
Of course, there is usually no need for such detailed entries, and in the future we will begin to write down all calculations in a shorter way:
$$z_2=f_1(2)=2^2-4\cdot 2=-4;\; z_3=f_1(-1)=(-1)^2-4\cdot (-1)=5;\; z_4=f_1(3)=3^2-4\cdot 3=-3.$$
Now let's turn to the straight line $x=3$. This line bounds the domain $D$ under the condition $0 ≤ y ≤ 4$. Substitute $x=3$ into the given function $z$. As a result of such a substitution, we get the function $f_2(y)$:
$$ f_2(y)=z(3,y)=3^2+2\cdot 3\cdot y-y^2-4\cdot 3=-y^2+6y-3. $$
For the function $f_2(y)$, you need to find the largest and smallest values on the interval $0 ≤ y ≤ 4$. Find the derivative of this function and equate it to zero:
$$ f_(2)^(")(y)=-2y+6;\\ -2y+6=0; \; y=3. $$
The value $y=3$ belongs to the segment $0 ≤ y ≤ 4$, so we add $M_5(3;3)$ to the points found earlier. In addition, it is necessary to calculate the value of the function $z$ at the points at the ends of the segment $0 ≤ y ≤ 4$, i.e. at the points $M_4(3;0)$ and $M_6(3;4)$. At the point $M_4(3;0)$ we have already calculated the value of $z$. Let us calculate the value of the function $z$ at the points $M_5$ and $M_6$. Let me remind you that on the segment $M_4M_6$ we have $z(x,y)=f_2(y)$, therefore:
\begin(aligned) & z_5=f_2(3)=-3^2+6\cdot 3-3=6; &z_6=f_2(4)=-4^2+6\cdot 4-3=5. \end(aligned)
And, finally, consider the last boundary of $D$, i.e. line $y=x+1$. This line bounds the region $D$ under the condition $-1 ≤ x ≤ 3$. Substituting $y=x+1$ into the function $z$, we will have:
$$ f_3(x)=z(x,x+1)=x^2+2x\cdot (x+1)-(x+1)^2-4x=2x^2-4x-1. $$
Once again we have a function of one variable $x$. And again, you need to find the largest and smallest values of this function on the segment $-1 ≤ x ≤ 3$. Find the derivative of the function $f_(3)(x)$ and equate it to zero:
$$ f_(3)^(")(x)=4x-4;\\ 4x-4=0; \; x=1. $$
The value $x=1$ belongs to the interval $-1 ≤ x ≤ 3$. If $x=1$, then $y=x+1=2$. Let's add $M_7(1;2)$ to the list of points and find out what the value of the function $z$ is at this point. The points at the ends of the segment $-1 ≤ x ≤ 3$, i.e. points $M_3(-1;0)$ and $M_6(3;4)$ were considered earlier, we have already found the value of the function in them.
$$z_7=f_3(1)=2\cdot 1^2-4\cdot 1-1=-3.$$
The second step of the solution is completed. We got seven values:
$$z_1=-2;\;z_2=-4;\;z_3=5;\;z_4=-3;\;z_5=6;\;z_6=5;\;z_7=-3.$$
Let's turn to. Choosing the largest and smallest values from those numbers that were obtained in the third paragraph, we will have:
$$z_(min)=-4; \; z_(max)=6.$$
The problem is solved, it remains only to write down the answer.
Answer: $z_(min)=-4; \; z_(max)=6$.
Example #2
Find the largest and smallest values of the function $z=x^2+y^2-12x+16y$ in the region $x^2+y^2 ≤ 25$.
Let's build a drawing first. The equation $x^2+y^2=25$ (this is the boundary line of the given area) defines a circle with a center at the origin (i.e. at the point $(0;0)$) and a radius of 5. The inequality $x^2 +y^2 ≤ 25$ satisfy all points inside and on the mentioned circle.
We will act on. Let's find partial derivatives and find out the critical points.
$$ \frac(\partial z)(\partial x)=2x-12; \frac(\partial z)(\partial y)=2y+16. $$
There are no points at which the found partial derivatives do not exist. Let us find out at what points both partial derivatives are simultaneously equal to zero, i.e. find stationary points.
$$ \left \( \begin(aligned) & 2x-12=0;\\ & 2y+16=0. \end(aligned) \right. \;\; \left \( \begin(aligned) & x =6;\\ & y=-8.\end(aligned) \right.$$
We got stationary point$(6;-8)$. However, the found point does not belong to the region $D$. This is easy to show without even resorting to drawing. Let's check if the inequality $x^2+y^2 ≤ 25$, which defines our domain $D$, holds. If $x=6$, $y=-8$, then $x^2+y^2=36+64=100$, i.e. the inequality $x^2+y^2 ≤ 25$ is not satisfied. Conclusion: the point $(6;-8)$ does not belong to the region $D$.
Thus, there are no critical points inside $D$. Let's move on, to. We need to investigate the behavior of the function on the boundary of the given area, i.e. on the circle $x^2+y^2=25$. You can, of course, express $y$ in terms of $x$, and then substitute the resulting expression into our function $z$. From the circle equation we get: $y=\sqrt(25-x^2)$ or $y=-\sqrt(25-x^2)$. Substituting, for example, $y=\sqrt(25-x^2)$ into the given function, we will have:
$$ z=x^2+y^2-12x+16y=x^2+25-x^2-12x+16\sqrt(25-x^2)=25-12x+16\sqrt(25-x ^2); \;\; -5≤ x ≤ 5. $$
The further solution will be completely identical to the study of the behavior of the function on the boundary of the region in the previous example No. 1. However, it seems to me more reasonable in this situation to apply the Lagrange method. We are only interested in the first part of this method. After applying the first part of the Lagrange method, we get the points at which we examine the function $z$ for the minimum and maximum values.
We compose the Lagrange function:
$$ F=z(x,y)+\lambda\cdot(x^2+y^2-25)=x^2+y^2-12x+16y+\lambda\cdot (x^2+y^2 -25). $$
We find the partial derivatives of the Lagrange function and compose the corresponding system of equations:
$$ F_(x)^(")=2x-12+2\lambda x; \;\; F_(y)^(")=2y+16+2\lambda y.\\ \left \( \begin (aligned) & 2x-12+2\lambda x=0;\\ & 2y+16+2\lambda y=0;\\ & x^2+y^2-25=0.\end(aligned) \ right. \;\; \left \( \begin(aligned) & x+\lambda x=6;\\ & y+\lambda y=-8;\\ & x^2+y^2=25. \end( aligned)\right.$$
To solve this system, let's immediately indicate that $\lambda\neq -1$. Why $\lambda\neq -1$? Let's try to substitute $\lambda=-1$ into the first equation:
$$ x+(-1)\cdot x=6; \; x-x=6; \; 0=6. $$
The resulting contradiction $0=6$ says that the value $\lambda=-1$ is invalid. Output: $\lambda\neq -1$. Let's express $x$ and $y$ in terms of $\lambda$:
\begin(aligned) & x+\lambda x=6;\; x(1+\lambda)=6;\; x=\frac(6)(1+\lambda). \\ & y+\lambda y=-8;\; y(1+\lambda)=-8;\; y=\frac(-8)(1+\lambda). \end(aligned)
I believe that it becomes obvious here why we specifically stipulated the $\lambda\neq -1$ condition. This was done to fit the expression $1+\lambda$ into the denominators without interference. That is, to be sure that the denominator is $1+\lambda\neq 0$.
Let us substitute the obtained expressions for $x$ and $y$ into the third equation of the system, i.e. in $x^2+y^2=25$:
$$ \left(\frac(6)(1+\lambda) \right)^2+\left(\frac(-8)(1+\lambda) \right)^2=25;\\ \frac( 36)((1+\lambda)^2)+\frac(64)((1+\lambda)^2)=25;\\ \frac(100)((1+\lambda)^2)=25 ; \; (1+\lambda)^2=4. $$
It follows from the resulting equality that $1+\lambda=2$ or $1+\lambda=-2$. Hence, we have two values of the parameter $\lambda$, namely: $\lambda_1=1$, $\lambda_2=-3$. Accordingly, we get two pairs of values $x$ and $y$:
\begin(aligned) & x_1=\frac(6)(1+\lambda_1)=\frac(6)(2)=3; \; y_1=\frac(-8)(1+\lambda_1)=\frac(-8)(2)=-4. \\ & x_2=\frac(6)(1+\lambda_2)=\frac(6)(-2)=-3; \; y_2=\frac(-8)(1+\lambda_2)=\frac(-8)(-2)=4. \end(aligned)
So, we got two points of a possible conditional extremum, i.e. $M_1(3;-4)$ and $M_2(-3;4)$. Find the values of the function $z$ at the points $M_1$ and $M_2$:
\begin(aligned) & z_1=z(M_1)=3^2+(-4)^2-12\cdot 3+16\cdot (-4)=-75; \\ & z_2=z(M_2)=(-3)^2+4^2-12\cdot(-3)+16\cdot 4=125. \end(aligned)
We should choose the largest and smallest values from those that we obtained in the first and second steps. But in this case, the choice is small :) We have:
$$z_(min)=-75; \; z_(max)=125. $$
Answer: $z_(min)=-75; \; z_(max)=125$.