In practice, it is quite common to use the derivative to calculate the largest and smallest value of a function. We perform this action when we figure out how to minimize costs, increase profits, calculate the optimal load on production, etc., that is, in cases where we need to determine the optimal value of a parameter. To solve such problems correctly, you need to have a good understanding of what the largest and smallest values ​​of a function are.

Typically we define these values ​​within a certain interval x, which in turn may correspond to the entire domain of the function or part of it. It can be like a segment [a; b ] , and open interval (a ; b), (a ; b ], [ a ; b), infinite interval (a ; b), (a ; b ], [ a ; b) or infinite interval - ∞ ; a , (- ∞ ; a ] , [ a ; + ∞) , (- ∞ ; + ∞) .

In this material we will tell you how to calculate the largest and smallest values ​​of an explicitly defined function with one variable y=f(x) y = f (x) .

Basic definitions

Let's start, as always, with the formulation of basic definitions.

Definition 1

The largest value of the function y = f (x) on a certain interval x is the value m a x y = f (x 0) x ∈ X, which for any value x x ∈ X, x ≠ x 0 makes the inequality f (x) ≤ f (x) valid 0) .

Definition 2

The smallest value of the function y = f (x) on a certain interval x is the value m i n x ∈ X y = f (x 0), which for any value x ∈ X, x ≠ x 0 makes the inequality f(X f (x) ≥ f (x 0) .

These definitions are quite obvious. Even simpler, we can say this: the largest value of a function is its largest value on a known interval at abscissa x 0, and the smallest is the smallest accepted value on the same interval at x 0.

Definition 3

Stationary points are those values ​​of the argument of a function at which its derivative becomes 0.

Why do we need to know what stationary points are? To answer this question, we need to remember Fermat's theorem. It follows from it that a stationary point is the point at which the extremum of the differentiable function is located (i.e., its local minimum or maximum). Consequently, the function will take the smallest or largest value on a certain interval precisely at one of the stationary points.

A function can also take on the largest or smallest value at those points at which the function itself is defined and its first derivative does not exist.

The first question that arises when studying this topic: in all cases can we determine the largest or smallest value of a function on a given interval? No, we cannot do this when the boundaries of a given interval coincide with the boundaries of the definition area, or if we are dealing with an infinite interval. It also happens that a function in a given segment or at infinity will take infinitely small or infinitely large values. In these cases, it is not possible to determine the largest and/or smallest value.

These points will become clearer after being depicted on the graphs:

The first figure shows us a function that takes the largest and smallest values ​​(m a x y and m i n y) at stationary points located on the segment [ - 6 ; 6].

Let us examine in detail the case indicated in the second graph. Let's change the value of the segment to [ 1 ; 6 ] and we find that the maximum value of the function will be achieved at the point with the abscissa at the right boundary of the interval, and the minimum at the stationary point.

In the third figure, the abscissas of the points represent the boundary points of the segment [ - 3 ; 2]. They correspond to the largest and smallest value of a given function.

Now let's look at the fourth picture. In it, the function takes m a x y (the largest value) and m i n y (the smallest value) at stationary points on the open interval (- 6 ; 6).

If we take the interval [ 1 ; 6), then we can say that the smallest value of the function on it will be achieved at a stationary point. The greatest value will be unknown to us. The function could take its maximum value at x equal to 6 if x = 6 belonged to the interval. This is exactly the case shown in graph 5.

In graph 6, this function acquires its smallest value at the right boundary of the interval (- 3; 2 ], and we cannot draw definite conclusions about the largest value.

In Figure 7 we see that the function will have m a x y at a stationary point having an abscissa equal to 1. The function will reach its minimum value at the boundary of the interval on the right side. At minus infinity, the function values ​​will asymptotically approach y = 3.

If we take the interval x ∈ 2 ; + ∞ , then we will see that the given function will take neither the smallest nor the largest value on it. If x tends to 2, then the values ​​of the function will tend to minus infinity, since the straight line x = 2 is a vertical asymptote. If the abscissa tends to plus infinity, then the function values ​​will asymptotically approach y = 3. This is exactly the case shown in Figure 8.

In this paragraph we will present the sequence of actions that need to be performed to find the largest or smallest value of a function on a certain segment.

1. First, let's find the domain of definition of the function. Let's check whether the segment specified in the condition is included in it.
2. Now let's calculate the points contained in this segment at which the first derivative does not exist. Most often they can be found in functions whose argument is written under the modulus sign, or in power functions whose exponent is a fractionally rational number.
3. Next, we will find out which stationary points will fall in the given segment. To do this, you need to calculate the derivative of the function, then equate it to 0 and solve the resulting equation, and then select the appropriate roots. If we don’t get a single stationary point or they don’t fall into the given segment, then we move on to the next step.
4. We determine what values ​​the function will take at given stationary points (if any), or at those points at which the first derivative does not exist (if there are any), or we calculate the values ​​for x = a and x = b.
5. 5. We have a number of function values, from which we now need to select the largest and smallest. These will be the largest and smallest values ​​of the function that we need to find.

Let's see how to correctly apply this algorithm when solving problems.

Example 1

Condition: the function y = x 3 + 4 x 2 is given. Determine its largest and smallest values ​​on the segments [ 1 ; 4 ] and [ - 4 ; - 1 ] .

Solution:

Let's start by finding the domain of definition of a given function. In this case, it will be the set of all real numbers except 0. In other words, D (y) : x ∈ (- ∞ ; 0) ∪ 0 ; + ∞ . Both segments specified in the condition will be inside the definition area.

Now we calculate the derivative of the function according to the rule of fraction differentiation:

y " = x 3 + 4 x 2 " = x 3 + 4 " x 2 - x 3 + 4 x 2 " x 4 = = 3 x 2 x 2 - (x 3 - 4) 2 x x 4 = x 3 - 8 x 3

We learned that the derivative of a function will exist at all points of the segments [ 1 ; 4 ] and [ - 4 ; - 1 ] .

Now we need to determine the stationary points of the function. Let's do this using the equation x 3 - 8 x 3 = 0. It has only one real root, which is 2. It will be a stationary point of the function and will fall into the first segment [1; 4].

Let us calculate the values ​​of the function at the ends of the first segment and at this point, i.e. for x = 1, x = 2 and x = 4:

y (1) = 1 3 + 4 1 2 = 5 y (2) = 2 3 + 4 2 2 = 3 y (4) = 4 3 + 4 4 2 = 4 1 4

We found that the largest value of the function m a x y x ∈ [ 1 ; 4 ] = y (2) = 3 will be achieved at x = 1, and the smallest m i n y x ∈ [ 1 ; 4 ] = y (2) = 3 – at x = 2.

The second segment does not include a single stationary point, so we need to calculate the function values ​​only at the ends of the given segment:

y (- 1) = (- 1) 3 + 4 (- 1) 2 = 3

This means m a x y x ∈ [ - 4 ; - 1 ] = y (- 1) = 3 , m i n y x ∈ [ - 4 ; - 1 ] = y (- 4) = - 3 3 4 .

Answer: For the segment [ 1 ; 4 ] - m a x y x ∈ [ 1 ; 4 ] = y (2) = 3 , m i n y x ∈ [ 1 ; 4 ] = y (2) = 3 , for the segment [ - 4 ; - 1 ] - m a x y x ∈ [ - 4 ; - 1 ] = y (- 1) = 3 , m i n y x ∈ [ - 4 ; - 1 ] = y (- 4) = - 3 3 4 .

See picture:

Before studying this method, we advise you to review how to correctly calculate the one-sided limit and the limit at infinity, as well as learn the basic methods for finding them. To find the largest and/or smallest value of a function on an open or infinite interval, perform the following steps sequentially.

1. First you need to check whether the given interval is a subset of the domain of definition of this function.
2. Let us determine all points that are contained in the required interval and at which the first derivative does not exist. They usually occur for functions where the argument is enclosed in the modulus sign, and for power functions with a fractionally rational exponent. If these points are missing, then you can proceed to the next step.
3. Now let’s determine which stationary points will fall within the given interval. First, we equate the derivative to 0, solve the equation and select suitable roots. If we do not have a single stationary point or they do not fall within the specified interval, then we immediately proceed to further actions. They are determined by the type of interval.

• If the interval is of the form [ a ; b) , then we need to calculate the value of the function at the point x = a and the one-sided limit lim x → b - 0 f (x) .
• If the interval has the form (a; b ], then we need to calculate the value of the function at the point x = b and the one-sided limit lim x → a + 0 f (x).
• If the interval has the form (a ; b), then we need to calculate the one-sided limits lim x → b - 0 f (x) , lim x → a + 0 f (x) .
• If the interval is of the form [ a ; + ∞), then we need to calculate the value at the point x = a and the limit at plus infinity lim x → + ∞ f (x) .
• If the interval looks like (- ∞ ; b ] , we calculate the value at the point x = b and the limit at minus infinity lim x → - ∞ f (x) .
• If - ∞ ; b , then we consider the one-sided limit lim x → b - 0 f (x) and the limit at minus infinity lim x → - ∞ f (x)
• If - ∞; + ∞ , then we consider the limits on minus and plus infinity lim x → + ∞ f (x) , lim x → - ∞ f (x) .

1. At the end, you need to draw a conclusion based on the obtained function values ​​and limits. There are many options available here. So, if the one-sided limit is equal to minus infinity or plus infinity, then it is immediately clear that nothing can be said about the smallest and largest values ​​of the function. Below we will look at one typical example. Detailed descriptions will help you understand what's what. If necessary, you can return to Figures 4 - 8 in the first part of the material.

Example 2

Condition: given function y = 3 e 1 x 2 + x - 6 - 4 . Calculate its largest and smallest value in the intervals - ∞ ; - 4, - ∞; - 3 , (- 3 ; 1 ] , (- 3 ; 2) , [ 1 ; 2) , 2 ; + ∞ , [ 4 ; + ∞) .

Solution

First of all, we find the domain of definition of the function. The denominator of the fraction contains a quadratic trinomial, which should not turn to 0:

x 2 + x - 6 = 0 D = 1 2 - 4 1 (- 6) = 25 x 1 = - 1 - 5 2 = - 3 x 2 = - 1 + 5 2 = 2 ⇒ D (y) : x ∈ (- ∞ ; - 3) ∪ (- 3 ; 2) ∪ (2 ; + ∞)

We have obtained the domain of definition of the function to which all the intervals specified in the condition belong.

Now let's differentiate the function and get:

y" = 3 e 1 x 2 + x - 6 - 4 " = 3 e 1 x 2 + x - 6 " = 3 e 1 x 2 + x - 6 1 x 2 + x - 6 " = = 3 · e 1 x 2 + x - 6 · 1 " · x 2 + x - 6 - 1 · x 2 + x - 6 " (x 2 + x - 6) 2 = - 3 · (2 ​​x + 1) · e 1 x 2 + x - 6 x 2 + x - 6 2

Consequently, derivatives of a function exist throughout its entire domain of definition.

Let's move on to finding stationary points. The derivative of the function becomes 0 at x = - 1 2 . This is a stationary point that lies in the intervals (- 3 ; 1 ] and (- 3 ; 2) .

Let's calculate the value of the function at x = - 4 for the interval (- ∞ ; - 4 ], as well as the limit at minus infinity:

y (- 4) = 3 e 1 (- 4) 2 + (- 4) - 6 - 4 = 3 e 1 6 - 4 ≈ - 0 . 456 lim x → - ∞ 3 e 1 x 2 + x - 6 = 3 e 0 - 4 = - 1

Since 3 e 1 6 - 4 > - 1, it means that m a x y x ∈ (- ∞ ; - 4 ] = y (- 4) = 3 e 1 6 - 4. This does not allow us to uniquely determine the smallest value of the function. We can only conclude that there is a constraint below - 1, since it is to this value that the function approaches asymptotically at minus infinity.

The peculiarity of the second interval is that there is not a single stationary point and not a single strict boundary in it. Consequently, we will not be able to calculate either the largest or smallest value of the function. Having defined the limit at minus infinity and as the argument tends to - 3 on the left side, we get only an interval of values:

lim x → - 3 - 0 3 e 1 x 2 + x - 6 - 4 = lim x → - 3 - 0 3 e 1 (x + 3) (x - 3) - 4 = 3 e 1 (- 3 - 0 + 3) (- 3 - 0 - 2) - 4 = = 3 e 1 (+ 0) - 4 = 3 e + ∞ - 4 = + ∞ lim x → - ∞ 3 e 1 x 2 + x - 6 - 4 = 3 e 0 - 4 = - 1

This means that the function values ​​will be located in the interval - 1; +∞

To find the largest value of the function in the third interval, we determine its value at the stationary point x = - 1 2 if x = 1. We will also need to know the one-sided limit for the case when the argument tends to - 3 on the right side:

y - 1 2 = 3 e 1 - 1 2 2 + - 1 2 - 6 - 4 = 3 e 4 25 - 4 ≈ - 1 . 444 y (1) = 3 e 1 1 2 + 1 - 6 - 4 ≈ - 1 . 644 lim x → - 3 + 0 3 e 1 x 2 + x - 6 - 4 = lim x → - 3 + 0 3 e 1 (x + 3) (x - 2) - 4 = 3 e 1 - 3 + 0 + 3 (- 3 + 0 - 2) - 4 = = 3 e 1 (- 0) - 4 = 3 e - ∞ - 4 = 3 0 - 4 = - 4

It turned out that the function will take the greatest value at a stationary point m a x y x ∈ (3; 1 ] = y - 1 2 = 3 e - 4 25 - 4. As for the smallest value, we cannot determine it. Everything we know , is the presence of a lower limit to - 4 .

For the interval (- 3 ; 2), take the results of the previous calculation and once again calculate what the one-sided limit is equal to when tending to 2 on the left side:

y - 1 2 = 3 e 1 - 1 2 2 + - 1 2 - 6 - 4 = 3 e - 4 25 - 4 ≈ - 1 . 444 lim x → - 3 + 0 3 e 1 x 2 + x - 6 - 4 = - 4 lim x → 2 - 0 3 e 1 x 2 + x - 6 - 4 = lim x → - 3 + 0 3 e 1 (x + 3) (x - 2) - 4 = 3 e 1 (2 - 0 + 3) (2 - 0 - 2) - 4 = = 3 e 1 - 0 - 4 = 3 e - ∞ - 4 = 3 · 0 - 4 = - 4

This means that m a x y x ∈ (- 3 ; 2) = y - 1 2 = 3 e - 4 25 - 4, and the smallest value cannot be determined, and the values ​​of the function are limited from below by the number - 4.

Based on what we got in the two previous calculations, we can say that on the interval [ 1 ; 2) the function will take the largest value at x = 1, but it is impossible to find the smallest.

On the interval (2 ; + ∞) the function will not reach either the largest or the smallest value, i.e. it will take values ​​from the interval - 1 ; + ∞ .

lim x → 2 + 0 3 e 1 x 2 + x - 6 - 4 = lim x → - 3 + 0 3 e 1 (x + 3) (x - 2) - 4 = 3 e 1 (2 + 0 + 3 ) (2 + 0 - 2) - 4 = = 3 e 1 (+ 0) - 4 = 3 e + ∞ - 4 = + ∞ lim x → + ∞ 3 e 1 x 2 + x - 6 - 4 = 3 e 0 - 4 = - 1

Having calculated what the value of the function will be equal to at x = 4, we find out that m a x y x ∈ [ 4 ; + ∞) = y (4) = 3 e 1 14 - 4 , and the given function at plus infinity will asymptotically approach the straight line y = - 1 .

Let's compare what we got in each calculation with the graph of the given function. In the figure, the asymptotes are shown by dotted lines.

That's all we wanted to tell you about finding the largest and smallest values ​​of a function. The sequences of actions that we have given will help you make the necessary calculations as quickly and simply as possible. But remember that it is often useful to first find out at which intervals the function will decrease and at which it will increase, after which you can draw further conclusions. This way you can more accurately determine the largest and smallest values ​​of the function and justify the results obtained.

If you notice an error in the text, please highlight it and press Ctrl+Enter

The standard algorithm for solving such problems involves, after finding the zeros of the function, determining the signs of the derivative on the intervals. Then the calculation of values ​​at the found maximum (or minimum) points and at the boundary of the interval, depending on what question is in the condition.

I advise you to do things a little differently. Why? I wrote about this.

I propose to solve such problems as follows:

1. Find the derivative.
2. Find the zeros of the derivative.
3. Determine which of them belong to this interval.
4. We calculate the values ​​of the function at the boundaries of the interval and points of step 3.
5. We draw a conclusion (answer the question posed).

While solving the presented examples, solving quadratic equations is not discussed in detail; you must be able to do this. They should also know.

Let's look at examples:

77422. Find the largest value of the function y=x 3 –3x+4 on the segment [–2;0].

Let's find the zeros of the derivative:

The point x = –1 belongs to the interval specified in the condition.

We calculate the values ​​of the function at points –2, –1 and 0:

The largest value of the function is 6.

Answer: 6

77425. Find the smallest value of the function y = x 3 – 3x 2 + 2 on the segment.

Let's find the derivative of the given function:

Let's find the zeros of the derivative:

The point x = 2 belongs to the interval specified in the condition.

We calculate the values ​​of the function at points 1, 2 and 4:

The smallest value of the function is –2.

Answer: –2

77426. Find the largest value of the function y = x 3 – 6x 2 on the segment [–3;3].

Let's find the derivative of the given function:

Let's find the zeros of the derivative:

The point x = 0 belongs to the interval specified in the condition.

We calculate the values ​​of the function at points –3, 0 and 3:

The smallest value of the function is 0.

Answer: 0

77429. Find the smallest value of the function y = x 3 – 2x 2 + x +3 on the segment.

Let's find the derivative of the given function:

3x 2 – 4x + 1 = 0

We get the roots: x 1 = 1 x 1 = 1/3.

The interval specified in the condition contains only x = 1.

Let's find the values ​​of the function at points 1 and 4:

We found that the smallest value of the function is 3.

Answer: 3

77430. Find the largest value of the function y = x 3 + 2x 2 + x + 3 on the segment [– 4; –1].

Let's find the derivative of the given function:

Let's find the zeros of the derivative and solve the quadratic equation:

3x 2 + 4x + 1 = 0

Let's get the roots:

The root x = –1 belongs to the interval specified in the condition.

We find the values ​​of the function at points –4, –1, –1/3 and 1:

We found that the largest value of the function is 3.

Answer: 3

77433. Find the smallest value of the function y = x 3 – x 2 – 40x +3 on the segment.

Let's find the derivative of the given function:

Let's find the zeros of the derivative and solve the quadratic equation:

3x 2 – 2x – 40 = 0

Let's get the roots:

The interval specified in the condition contains the root x = 4.

Find the function values ​​at points 0 and 4:

We found that the smallest value of the function is –109.

Answer: –109

Let's consider a way to determine the largest and smallest values ​​of functions without a derivative. This approach can be used if you have big problems with determining the derivative. The principle is simple - we substitute all the integer values ​​from the interval into the function (the fact is that in all such prototypes the answer is an integer).

77437. Find the smallest value of the function y=7+12x–x 3 on the segment [–2;2].

Substitute points from –2 to 2: View solution

77434. Find the largest value of the function y=x 3 + 2x 2 – 4x + 4 on the segment [–2;0].

That's all. Good luck to you!

Sincerely, Alexander Krutitskikh.

P.S: I would be grateful if you tell me about the site on social networks.

A miniature and fairly simple problem of the kind that serves as a life preserver for a floating student. It's mid-July in nature, so it's time to settle down with your laptop on the beach. Early in the morning, the sunbeam of theory began to play, in order to soon focus on practice, which, despite the declared ease, contains shards of glass in the sand. In this regard, I recommend that you conscientiously consider the few examples of this page. To solve practical problems you must be able to find derivatives and understand the material of the article Monotonicity intervals and extrema of the function.

First, briefly about the main thing. In the lesson about continuity of function I gave the definition of continuity at a point and continuity at an interval. The exemplary behavior of a function on a segment is formulated in a similar way. A function is continuous on an interval if:

1) it is continuous on the interval ;
2) continuous at a point right and at the point left.

In the second paragraph we talked about the so-called one-sided continuity functions at a point. There are several approaches to defining it, but I will stick to the line I started earlier:

The function is continuous at the point right, if it is defined at a given point and its right-hand limit coincides with the value of the function at a given point: . It is continuous at the point left, if defined at a given point and its left-hand limit is equal to the value at this point:

Imagine that the green dots are nails with a magic elastic band attached to them:

Mentally take the red line in your hands. Obviously, no matter how far we stretch the graph up and down (along the axis), the function will still remain limited– a fence at the top, a fence at the bottom, and our product grazes in the paddock. Thus, a function continuous on an interval is bounded on it. In the course of mathematical analysis, this seemingly simple fact is stated and strictly proven. Weierstrass's first theorem....Many people are annoyed that elementary statements are tediously substantiated in mathematics, but this has an important meaning. Suppose a certain inhabitant of the terry Middle Ages pulled a graph into the sky beyond the limits of visibility, this was inserted. Before the invention of the telescope, the limited function in space was not at all obvious! Really, how do you know what awaits us over the horizon? After all, the Earth was once considered flat, so today even ordinary teleportation requires proof =)

According to Weierstrass's second theorem, continuous on a segmentthe function reaches its exact upper bound and yours exact bottom edge .

The number is also called the maximum value of the function on the segment and are denoted by , and the number is the minimum value of the function on the segment marked .

In our case:

Note : in theory, recordings are common .

Roughly speaking, the largest value is where the highest point on the graph is, and the smallest value is where the lowest point is.

Important! As already emphasized in the article about extrema of the function, greatest function value And smallest function value – NOT THE SAME, What maximum function And minimum function. So, in the example under consideration, the number is the minimum of the function, but not the minimum value.

By the way, what happens outside the segment? Yes, even a flood, in the context of the problem under consideration, this does not interest us at all. The task only involves finding two numbers and that's it!

Moreover, the solution is purely analytical, therefore no need to make a drawing!

The algorithm lies on the surface and suggests itself from the above figure:

1) Find the values ​​of the function in critical points, which belong to this segment.

Catch another bonus: here there is no need to check the sufficient condition for an extremum, since, as just shown, the presence of a minimum or maximum does not guarantee yet, what is the minimum or maximum value. The demonstration function reaches a maximum and, by the will of fate, the same number is the largest value of the function on the segment. But, of course, such a coincidence does not always occur.

So, in the first step, it is faster and easier to calculate the values ​​of the function at critical points belonging to the segment, without bothering whether there are extrema in them or not.

2) We calculate the values ​​of the function at the ends of the segment.

3) Among the function values ​​found in the 1st and 2nd paragraphs, select the smallest and largest number and write down the answer.

We sit down on the shore of the blue sea and hit the shallow water with our heels:

Example 1

Find the largest and smallest values ​​of a function on a segment

Solution:
1) Let's calculate the values ​​of the function at critical points belonging to this segment:

Let's calculate the value of the function at the second critical point:

2) Let’s calculate the values ​​of the function at the ends of the segment:

3) “Bold” results were obtained with exponents and logarithms, which significantly complicates their comparison. For this reason, let’s arm ourselves with a calculator or Excel and calculate approximate values, not forgetting that:

Now everything is clear.

Answer:

Fractional-rational example for independent solution:

Example 6

Find the maximum and minimum values ​​of a function on a segment

Algorithm for finding the largest and smallest values ​​of a continuous function on a segment:

1) Find all critical points of the function belonging to the segment ;

2) Calculate the values ​​of the function at these points and at the ends of the segment;

3) From the obtained values, select the largest and smallest.

Example 8.1. Find the largest and smallest values ​​of a function
on the segment
.

Solution. 1) Find the critical points of the function.

,




.

On the segment
the denominator does not vanish. Therefore, a fraction is equal to zero if and only if the numerator is equal to zero:




.

Means,
– critical point of the function. It belongs to this segment.

Let's find the value of the function at the critical point:

2) Find the values ​​of the function at the ends of the segment:

, .

3) From the obtained values, select the largest and smallest:

,
.

9. Problems of finding the largest and smallest values ​​of quantities

When solving problems involving calculating the smallest and largest values ​​of quantities, you must first determine for which quantity in the problem you need to find the smallest or largest value. This value will be the function under study. Then one of the quantities on the change of which the application of the function depends should be taken as an independent variable and the function expressed through it. In this case, it is necessary to select as the independent variable the value through which the function under study is expressed most simply. After this, the problem of finding the smallest and largest values ​​of the resulting function in a certain interval of change in the independent variable is solved, which is usually established from the very essence of the problem.

Example 9.1. Find the height of the cone of the largest volume that can be inscribed in a ball of radius .

R decision. Designating the radius of the base, height and volume of the cone, respectively ,And , let's write
. This equality expresses the dependence on two variables And ; let us exclude one of these quantities, namely . To do this, from a right triangle
we derive (using the theorem about the square of a perpendicular dropped from the vertex of a right angle to the hypotenuse):

Figure 6 – Illustration for example 9.1.

or
.

Substituting the value into the formula for the volume of a cone, we get:

.

We see that the volume cone inscribed in a ball of radius , there is a function of the height of this cone . Finding the height at which the inscribed cone has a large volume means finding such , at which the function has a maximum. We are looking for the maximum function:

1)
,

2)
,
,
, where
or
,

3)
.

Substituting instead at first
, and then
, we get:

In the first case we have a minimum (
at
), in the second the desired maximum (since
at
).

Therefore, when
cone inscribed in a ball of radius , has the largest volume.

P Example 9.2. It is required to fence with a wire mesh length 60 m a rectangular area adjacent to the wall of the house (Fig. 7). What should be the length and width of the plot so that it has the largest area?

Solution. Let the width of the plot m, and the area m 2 , Then:

Figure 7 – Illustration for example 9.2.

Values And cannot be negative, so the multiplier
, A
.

Square there is a function , we determine the intervals of its increase and decrease:

.
, and the function increases when
;
, and the function decreases when
. Therefore, the point
is the maximum point. Since this is the only point belonging to the interval
, then at the point
function matters most.

Therefore, the area of ​​the plot is greatest (maximum) if the width
m, and the length m.

Example 9.3. What should be the dimensions of a rectangular room whose area 36 m 2 so that its perimeter is smallest?

Solution. Let the length be m, then the width of the rectangle m, and the perimeter:

.

Perimeter there is a function of length , defined for all positive values :
.

Let us determine the intervals of its increase and decrease:

The sign of the derivative is determined by the sign of the difference
. In between

, and in between

.

Therefore, the point
is the minimum point. Since this is the only point belonging to the interval:
, then at the point
function has the smallest value.

Therefore, the perimeter of a rectangle has the smallest value (minimum) if its length 6 m and width m = 6 m, that is, when it is a square.

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

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

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

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


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

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

Example 1

In a limited closed area

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

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

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

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

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

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

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

II) We explore the border of the region.

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

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

Alternatively, you can do it like this:

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

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

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

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

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

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

The geometric situation is related to the previous point:

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

Let's examine the second end of the segment:

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

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

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

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

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

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

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

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

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

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

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

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

Example 2

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

Example 3

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

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

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

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

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

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

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

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

Example 4

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

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

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

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

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

I) Find stationary points:

The system is an idiot's dream :)

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

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

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

1) If , then

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

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

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

Control:

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

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


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

Check the function yourself.

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

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

To solve it yourself:

Example 5

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

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

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

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

Solutions and answers:

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