What happens to the area of a rectangular sheet of paper. “Application of the derivative to solving problems
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Chapter five.
DISAPPEARANCE OF FIGURES. SECTION I
In this and the next chapter we will follow the development of many remarkable geometrical paradoxes. They all begin with cutting the figure into pieces and end with the compilation of these pieces of a new figure. At the same time, it seems that part of the original figure (this may be part of the area of the figure or one of several drawings depicted on it) has disappeared without a trace. When the pieces return to their original places, the disappeared part of the area or the drawing mysteriously reappears.
The geometric nature of these curious disappearances and appearances justifies the inclusion of these paradoxes in the category of mathematical puzzles.
Paradox with lines
All the many paradoxes that we are going to consider here are based on the same principle, which we will call the "principle of covert redistribution." Here is one very old and very elementary paradox, which immediately explains the essence of this principle.
Draw ten vertical lines of the same length on a rectangular sheet of paper and draw a diagonal with a dotted line, as shown in Fig. fifty.
Let's look at the segments of these lines above and below the diagonal; it is easy to see that the length of the first decreases, and the second increases accordingly.
Let's cut the rectangle along the dotted line and move the lower part to the left down, as shown in Fig. 51.
By counting the number of vertical lines, you will find that there are now nine. Which line disappeared and where? Move the left side back to its original position and the line that disappeared will reappear.
But which line fell into place and where did it come from?
At first, these questions seem puzzling, but after a little reflection it becomes clear that no single line disappears or appears. What happens is that these eight increments are exactly equal to the length of each of the original lines.
Perhaps the essence of the paradox will come out even more clearly if it is illustrated on pebbles.
Take five piles of pebbles, four pebbles in a pile. Let's move one pebble from the second pile to the first, two pebbles from the third to the second, three from the fourth to the third, and finally all four pebbles from the fifth to the fourth. Rice. 52 explains our actions.
After such a shift, it turns out that there were only four piles. It is impossible to answer the question which pile disappeared, since the pebbles were redistributed so that each of the four piles had a pebble added. Exactly the same thing happens in the line paradox. When parts of the sheet are shifted diagonally, the segments of the cut lines are redistributed and each resulting line becomes slightly longer than the original.
Facial Disappearance
Let's move on to a description of ways in which the paradox with lines can be made more interesting and entertaining. This can be achieved, for example, by replacing the disappearance and appearance of lines by the same disappearance and appearance of plane figures. Images of pencils, cigarettes, bricks, high-crowned hats, glasses of water, and other vertically extended objects, the nature of the image of which remains the same before and after the shift, are especially suitable here. With some artistic ingenuity, you can take more complex objects. Look, for example, at the disappearing face in fig. 53.
By shifting the lower band on the top of the picture to the left, all the hats remain unaffected, but one face disappears completely! (see bottom of picture). It is pointless to ask what kind of face, because when shifting, four faces are divided into two parts. These parts are then redistributed, with each face receiving several additional features: one, for example, a longer nose, another a more elongated chin, etc. However, these small redistributions are ingeniously hidden, and the disappearance of the entire face is, of course, much more striking than the disappearance of a piece of the line.
"Vanishing Warrior"
In this puzzle, the paradox with lines is given a circular shape and straight line segments are replaced by figures of 13 warriors (Fig. 54).
At the same time, a large arrow points to the northeast of the S.V. the arrow will point to the northwest NW, there will be 12 warriors in the figure (Fig. 55).
When the circle is rotated in the opposite direction to the position when the large arrow will again stand on the NE, the disappeared warrior will appear again.
If fig. 54 take a closer look, you can see that the two warriors in the lower left part of the picture are located in a special way: they are opposite each other, while all the rest are placed in a chain. These two figures correspond to the extreme lines in the line segment paradox. Based on the requirements of the drawing, each of these figures should be missing a part of the leg, and in order to make this defect less noticeable in the turned position of the wheel, it would be better to depict them side by side.
We also note that the warriors are depicted in the figure with much more ingenuity than it might seem at first glance. So, for example, in order for the figures to remain in a vertical position in all places of the globe, it is necessary in one case to have the right leg instead of the left, and in the other, on the contrary, instead of the right leg, the left.
Lost Bunny
The paradox of vertical lines can obviously be shown on more complex objects, such as human faces, animal figures, etc. In fig. 56 shows one option.
When, after cutting along a thick line, rectangles A and B are swapped, one rabbit disappears, leaving an Easter egg in its place. If instead of swapping rectangles A and B, cut the right half of the drawing along the dotted line and swap the right parts, the number of rabbits increases to 12, but one rabbit loses its ears and other funny details appear.
Chapter six.
DISAPPEARANCE OF FIGURES. SECTION I I
The checkerboard paradox
Closely related to the paradoxes discussed in the previous chapter is another class of paradoxes in which the "principle of covert redistribution" explains the mysterious disappearance or appearance of squares. One of the oldest and most simple examples paradoxes of this kind are shown in fig. 57.
The chessboard is cut obliquely, as shown in the left half of the figure, and then part B is shifted down to the left, as shown in the right half of the figure. If the triangle protruding in the upper right corner is cut off with scissors and placed in an empty space that looks like a triangle in the lower left corner of the picture, then a rectangle of 7x9 square units will be obtained.
The original area was 64 square units, but now it is 63. Where did the one missing square unit go?
The answer is that our diagonal line runs slightly below the lower left corner of the square located in the upper right corner of the board.
Due to this, the cut off triangle has a height equal not to 1, but to 1 1/7. And thus, the height is not 9, but 9 1/7 units. The 1/7 unit height increase is almost imperceptible, but when taken into account, it results in a required rectangle area of 64 square units.
The paradox becomes even more striking if, instead of a chessboard, we take just a square sheet of paper without cells, since in our case, careful examination reveals an inaccurate closure of cells along the cut line.
The connection of our paradox with the paradox of vertical lines, discussed in the previous chapter, becomes clear if we follow the cells at the cut line. When moving upward along the cut line, it is found that parts of the cut cells above the line (they are darkened in the figure) gradually decrease, and gradually increase below the line. There were fifteen shaded squares on the chessboard, but on the rectangle obtained after rearranging the pieces, there were only fourteen of them. The apparent disappearance of one darkened cell is simply another form of the paradox discussed above. When we cut and then shuffle the little triangle, we are effectively cutting piece A of the chessboard into two pieces, which are then swapped along the diagonal.
For the puzzle, only the cells adjacent to the cut line are important, while the rest do not matter, playing the role of decoration. However, their presence changes the nature of the paradox. Instead of the disappearance of one of several small cells (or a somewhat more complex figure, say, a playing card, a human face, etc., which could be drawn inside each cell), we are faced here with a change in the area of a large geometric figure.
The square paradox
Here is another paradox with the area. By changing the position of parts A and C, as shown in Fig. 58, it is possible to convert a rectangle of 30 square units into two smaller rectangles with a total area of 32 square units, thus obtaining a "gain" of two square units. As in the previous paradox, only the cells adjacent to the cut line play a role here. The rest are just for decoration.
In this paradox, there are two essentially different ways of cutting the figure into pieces.
You can start with a large 3x10 unit rectangle (upper part of Fig. 58), carefully drawing a diagonal in it, then two smaller rectangles (bottom part of Fig. 58) will be 1/5 units shorter than their apparent dimensions.
But you can also start with a figure made up of two neatly drawn smaller rectangles of 2x6 and 4x5 units; then the segments connecting point X with point Y and point Y with point Z will not form a straight line. And only because the obtuse angle formed by them with the apex at the point Y is very close to the developed one, the broken line XYZ seems to be a straight line. Therefore, a figure made up of parts of small rectangles will not actually be a rectangle, since these parts will overlap slightly along the diagonal. The chessboard paradox, just like most of other paradoxes that we are going to consider in this chapter can also be presented in two ways. In one of them, the paradox is obtained due to a slight decrease or increase in the height (or width) of the figures, in the other - due to an increase or loss of area along the diagonal, caused either by the overlapping of the figures, as in the case just considered, or by the appearance of empty spaces, with which we'll meet soon.
By changing the size of the figures and the slope of the diagonal, this paradox can be given a very different design. You can achieve a loss or gain in area of 1 square unit or 2, 3, 4, 5 units, etc.
Square variant
In one elegant version, the original rectangles of 3x8 and 5x8 units, when placed next to each other, form a regular 8x8 chessboard. These rectangles are cut into pieces, which, after redistribution, form a new large rectangle with an apparent area increase of one square unit (Fig. 59).
The essence of the paradox is as follows. With a careful construction of a drawing of a square, a strict diagonal of a large rectangle does not work. Instead, a diamond-shaped figure appears, so elongated that its sides seem to almost merge. On the other hand, if you carefully draw the diagonal of a large rectangle; the height of the top of the two rectangles that make up the square will be slightly larger than it should be, and the bottom rectangle will be slightly wider. Note that the inaccurate closure of the parts of the figure in the second method of cutting is more striking than the inaccuracies along the diagonal in the first; so the first way is preferable. As in the previously encountered examples, inside the cells, dissected by the diagonal, you can draw circles, physiognomies, or some kind of figures; when rearranging constituent parts the rectangles of these figures will become one more or less.
Fibonacci numbers
It turns out that the lengths of the sides of the four parts that make up the figures (Fig. 59 and 60) are members of the Fibonacci series, that is, a series of numbers starting with two units: 1, 1, each of which, starting from the third, is the sum of two previous ones. Our series looks like 1, 1, 2, 3, 5, 8, 13, 21, 34…
The arrangement of the parts, from which the square was cut, in the form of a rectangle illustrates one of the properties of the Fibonacci series, namely the following: when squaring any member of this series, the product of two adjacent members of the series plus or minus one is obtained. In our example, the side of the square is 8, and the area is 64. The 8 in the Fibonacci series is located between 5 and 13. Since the numbers 5 and 13 become the lengths of the sides of the rectangle, its area should be equal to 65, which gives an increase in area by one unit.
Thanks to this property of the series, it is possible to construct a square whose side is any Fibonacci number greater than one, and then cut it in accordance with the two previous numbers of this series.
If, for example, we take a square of 13x13 units, then its three sides should be divided into segments of length 5 and 8 units, and then cut, as shown in Fig. 60. The area of this square is 169 square units. The sides of the rectangle formed by the parts of the squares will be 21 and 8, giving an area of 168 square units. Here, due to the overlapping of parts along the diagonal, one square unit is not added, but lost.
If we take a square with a side of 5, then there will also be a loss of one square unit. It is also possible to formulate a general rule: taking for the side of the square some number from the “first” subsequence of Fibonacci numbers (3, 8 ...) located through one and composing a rectangle from the parts of this square, we will get a clearance along its diagonal and, as a result, an apparent increase in area for one unit. Taking some number from the “second” subsequence (2, 5, 13…) as the side of the square, we get overlapping areas along the diagonal of the rectangle and the loss of one square unit of area.
You can build a paradox even on a square with a side of two units. But then there is such an obvious overlap in the 3x1 rectangle that the paradox effect is completely lost.
Using other Fibonacci series for paradox, you can get countless options. So, for example, squares based on a row of 2, 4, 6, 10, 16, 26, etc. result in a loss or gain of 4 square units. The magnitude of these losses or gains can be found by calculating for a given series the difference between the square of any of its terms and the product of its two adjacent terms on the left and right. The row 3, 4, 7, 11, 18, 29, etc. gives an increase or loss of five square units. T. de Moulidar gave a drawing of a square based on the series 1, 4, 5, 9, 14, etc. The side of this square is taken equal to 9, and after converting it into a rectangle, 11 square units are lost. Row 2, 5, 7, 12, 19 ... also gives a loss or gain of 11 square units. In both cases, the overlaps (or gaps) along the diagonal are so large that they can be seen immediately.
Denoting any three consecutive Fibonacci numbers by A, B and C, and by X - loss or gain in area, we get the following two formulas:
A + B = C
B 2 \u003d AC ± X
If we substitute instead of X the desired gain or loss, and instead of B a number that is taken as the length of the side of the square, then we can construct quadratic equation, from which there are two other Fibonacci numbers, although these, of course, will not necessarily be rational numbers. It turns out, for example, that by dividing a square into figures with rational side lengths, one cannot obtain an increase or loss of two or three square units. By using irrational numbers this can of course be achieved. Thus, the Fibonacci series 2 1/2 , 2 2 1/2 , 3 2 1/2 , 5 2 1/2 gives an increase or loss of two square units, and the series 3 1/2 , 2 3 1/ 2 , 3 3 1/2 , 5 3 1/2 results in a gain or loss of three square units.
Rectangle variant
There are many ways in which a rectangle can be cut into a small number of pieces and then folded into another rectangle of larger or smaller area. On fig. 61 shows a paradox, also based on the Fibonacci series.
Like the square case just discussed, choosing some Fibonacci number from the "second" subsequence as the width of the first rectangle (13 in this case) results in an increase in the area of the second rectangle by one square unit.
If we take some Fibonacci number from the “additional” subsequence as the width of the first rectangle, then the area in the second rectangle will decrease by one unit. Losses and gains in area are explained by small overlaps or gaps along the diagonal section of the second rectangle. Another version of such a rectangle, shown in Fig. 62, when constructing the second rectangle, results in an increase in area by two square units.
If the shaded part of the area of the second rectangle is placed over the unshaded part, the two diagonal cuts merge into one large diagonal. Now rearranging parts A and B (as in Fig. 61), we get a second rectangle of a larger area.
Another version of the paradox
When summing the areas of the parts, the permutation of triangles B and C in the upper part of Fig. 63 results in an apparent loss of one square unit.
As the reader will note, this is due to the areas of the shaded parts: on the top of the figure there are 15 shaded squares, on the bottom - 16. Replacing the shaded pieces with two figures of a special kind covering them, we arrive at a new, striking form of paradox. Now we have a rectangle in front of us, which can be cut into 5 parts, and then, swapping them, make up a new rectangle, and, despite the fact that its linear dimensions remain the same, a hole with an area of one square unit appears inside (Fig. 64).
The possibility of converting one figure into another, of the same external dimensions, but with a hole inside the perimeter, is based on the following. If you take the point X exactly three units from the base and five units from the side of the rectangle, then the diagonal will not pass through it. However, the polyline connecting the point X with the opposite vertices of the rectangle will deviate so little from the diagonal that it will be almost imperceptible.
After interchanging triangles B and C in the lower half of the drawing, the parts of the figure will slightly overlap along the diagonal.
On the other hand, if in the upper part of the figure we consider the line connecting the opposite vertices of the rectangle as an exactly drawn diagonal, then the XW line will be slightly longer than three units. And as a consequence of this, the second rectangle will be slightly higher than it seems. In the first case, the missing unit of area can be considered as distributed from corner to corner and forming an overlap along the diagonals. In the second case, the missing square is distributed over the width of the rectangle. As we already know from the previous one, all paradoxes of this kind can be attributed to one of these two construction options. In both cases, the inaccuracies of the figures are so slight that they are completely invisible.
The most elegant form of this paradox is the squares, which, after redistribution of parts and the formation of a hole, remain squares.
Such squares are known in countless variants and with holes of any number of square units. Some of the most interesting of them are shown in Fig. 65 and 66.
Can be pointed to a simple formula, relating the size of the hole to the proportions of the large triangle. The three sizes that will be discussed, we will denote by A, B to C (Fig. 67).
The area of the hole in square units is equal to the difference between the product of A and C and the nearest multiple of size B. So, in the last example, the product of A and C is 25. The nearest multiple of size B to 25 is 24, so the hole is one square unit. This rule applies regardless of whether the real diagonal is drawn or the point X in fig. 67 is applied neatly at the intersection of the lines of the square grid.
If the diagonal, as it should be, is drawn as a strictly straight line, or if the point X is taken exactly at one of the vertices of the square grid, then no paradox is obtained. In these cases, the formula gives a hole of zero square units, denoting, of course, that there is no hole at all.
Triangle variant
Let's return to the first example of a paradox (see Fig. 64). Note that the large triangle A does not change its position, while the other parts move. Since this triangle does not play significant role in a paradox, it can be discarded altogether, leaving only the right triangle, cut into four parts. These parts can then be redistributed, thus obtaining right triangle with a hole (Fig. 68), as if equal to the original.
Composing two such right-angled triangles with legs, you can build many variants of isosceles triangles, similar to the one shown in Fig. 69.
As in the paradoxes discussed earlier, these triangles can be built in two ways: either draw their sides strictly rectilinearly, then the point X will not fall on the intersection of the lines of the square grid, or place the point X exactly at the intersection, then the sides will be slightly convex or concave. The latter method seems to mask inaccuracies in the drawing better. The paradox will seem even more surprising if lines of a square grid are applied to the parts that make up the triangle, thereby emphasizing that the parts were made with the necessary accuracy.
By giving our isosceles triangles different sizes, we can gain or lose any even number of square units.
Some typical examples are given in Fig. 70, 71 and 72.
Making two bases isosceles triangle any of these types, you can build a variety of options for a rhombic view; however, they add nothing essentially new to our paradox.
Four Piece Squares
All the types of paradoxes with a change in area that we have considered so far are closely related to each other in terms of the method of construction. However, there are paradoxes obtained by completely different methods. You can, for example, cut a square into four parts of the same shape and size (fig. 73), and then compose them in a new way, as shown in fig. 74. This results in a square, the dimensions of which seem to have not changed and at the same time with a hole in the middle.
Similarly, you can cut a rectangle with any ratio of side lengths. It is curious that point A, at which the two intersect, seem unchanged and at the same time with a hole in the middle.
Similarly, you can cut a rectangle with any ratio of side lengths. Curiously, point A, at which two mutually perpendicular cut lines intersect, can be located anywhere inside the rectangle. In each case, when the parts are redistributed, a hole appears, and its size depends on the angle formed by the cut lines with the sides of the rectangle.
This paradox is comparatively simple, but it loses a lot because even a superficial study shows that the sides of the second rectangle should be slightly larger than the sides of the first.
A more complicated way of cutting a square into four parts, which results in an inner hole, is shown in Fig. 75.
It is based on the chessboard paradox that opens this chapter. Note that when redistributing parts, two of them need to be flipped reverse side up. Note also that when part A is discarded, we get a right-angled triangle made up of three parts, inside which a hole can be formed.
Three piece squares
Is there a way to cut a square into three pieces that can be re-arranged to make a square with a hole inside? The answer will be positive. One elegant solution is based on the application of the paradox discussed in the previous chapter.
Instead of placing the pictures in a special way in ledges, and making the cut straight (horizontally), the pictures are placed on one straight line, and the cut is made in ledges. The result is amazing: not only does the picture disappear, but a hole appears at the place of its disappearance.
Two piece squares
Is it possible to do the same with two parts?
I do not think that in this case it is possible by any method to obtain an internal hole in a square by imperceptibly increasing its height or width. However, it has been shown that the hole-in-a-square paradox can be built on the same principle as the vanishing warrior paradox. In this case, instead of placing the figures in a spiral or step, they are placed strictly in a circle, while the cut is made spiral or stepped; in the latter case, it has the form of a gear wheel with teeth of various sizes. When this wheel is rotated, one figure disappears and a hole appears in its place.
The fixed and rotating parts are neatly fitted to each other only in the position where the hole appears. In the initial position, small gaps are visible at each tooth if the incision was stepped, or one continuous circular gap with a spiral incision.
If the original rectangle is not a square, it can be cut into two parts, and then a hole can be obtained inside with very little noticeable change in its external dimensions. On fig. 76 shows one option.
Both parts are identical both in shape and in size. The easiest way to demonstrate this paradox is as follows: cut out pieces of cardboard, fold them into a rectangle without a hole, put them on a piece of paper and trace around the perimeter with a pencil. Folding the parts in a different way now, you can see that they still do not go beyond the drawn line, although a hole has formed in the middle of the rectangle.
To our two parts we can, of course, add a third, made in the form of a strip, which, when applied to one of the sides of the rectangle, turns it into a square; thus we get another way of cutting a square into three parts, giving an inner hole.
Curvilinear and 3D variants
The examples we have given clearly show that the field of paradoxes with a change in area is just beginning to be developed. Are there any curvilinear shapes, such as circles or ellipses, that can be cut into pieces and then reassembled in such a way that, without noticeable distortion of the figure, internal holes are obtained?
Are there three-dimensional figures that are specific to three dimensions, i.e., are not a trivial consequence of two-dimensional figures? After all, it is clear that any flat figure that we have met in this chapter can be “added a dimension” by simply cutting it out of fairly thick cardboard, the height of which is equal to the “length of the third dimension”).
Is it possible to cut a cube or, say, a pyramid not very in a complicated way into parts so that, composing them in a new way, get noticeable voids inside?
The answer will be this: if you do not limit the number of parts, then it is not difficult at all to indicate such spatial figures. This is clear enough in the case of a cube.
Here an inner void can be obtained, however the question of the smallest number parts with which this can be achieved is more complex. It can certainly be made from six parts; it is possible that this can be achieved with a smaller number.
Such a cube can be effectively demonstrated as follows: take it out of a box made exactly according to the cube, disassemble it into parts, revealing a ball inside, put the parts back into a solid cube and show that it (without the ball) still fills the box densely. We will conjecture that there must be many such figures, both flat and spatial, and also distinguished by simplicity and elegance of form. Future explorers of this curious field will have the pleasure of discovering them.
Example 1 . From a wire 20 cm long, it is necessary to make a rectangle of the largest area. Find its dimensions.
Solution: Let's denote one side of the rectangle through x cm, then the second will be (10-x) cm, the area S (x) \u003d (10-x) * x \u003d 10x-x 2;
S / (x)=10-2x; S/(x)=0; x=5;
According to the condition of the problem x (0; 10)
Find the sign of the derivative on the interval (0;5) and on the interval (5;10). The derivative changes sign from “+” to “-”. Hence: x \u003d 5 maximum point, S (5) \u003d 25 cm 2 - highest value. Therefore, one side of the rectangle is 5cm, the second is 10x=10-5=5cm;
Example 2 A plot with an area of 2400m 2 must be divided into two rectangular sections so that the length of the fence is the smallest. Find the sizes of the plots.
Solution: Let's denote one side of the site through x m, then the second will be m, the length of the fence P (x) \u003d 3x +;
P / (x) \u003d 3-; P / (x) \u003d 0; 3x 2 \u003d 4800; x 2 \u003d 1600; x=40. We take only a positive value according to the condition of the problem.
By the condition of the problem x (0; )
Find the sign of the derivative on the interval (0;40) and on the interval (40; ?). The derivative changes sign from “-” to “+”. Hence x=40 is the minimum point, therefore P(40)=240m smallest value, so one side is 40m, the other = 60m.
Example 3 A rectangular area with one side adjacent to the building. With a given perimeter of 1 m, it is necessary to enclose the site so that the area is the largest.
Solution:
Let's denote one side of the rectangular section through x m, then the second will be (-2x) m, the area S (x) \u003d (-2x) x \u003d x -2x 2;
S / (x) \u003d -4x; S/(x)=0; -4x; x = ;
By the condition of the problem x (0; )
Find the sign of the derivative on the interval (0; ) and on the interval ( ; ). The derivative changes sign from “+” to “-”. Hence x = maximum point. Therefore, one side of the site \u003d m, the second -2x \u003d m;
Example 4 From a rectangular sheet of cardboard with sides of 80 cm and 50 cm, you need to make a rectangular box, cutting out squares along the edges and bending the resulting edges. How tall should the box be to maximize its volume?
Solution: We denote the height of the box (this is the side of the cut out square) through x m, then one side of the base will be (80-2x) cm, the second (50-2x) cm, volume V (x) \u003d x (80-2x) (50-2x) \u003d 4x 3 -260x 2 + 4000x;
V / (x) \u003d 12x 2 -520x + 4000; V / (x)=0; 12x 2 -520x+4000=0; x 1 =10; x 2 =
According to the condition of the problem x (0; 25); x 1 (0; 25), x 2 (0; 25)
Find the sign of the derivative on the interval (0; 10) and on the interval (10; 25). The derivative changes sign from “+” to “-”. Hence x = 10 is the maximum point. Therefore, the height of the box = 10cm.
Example 5 A rectangular area with one side adjacent to the building. With a given perimeter of 20 m, it is necessary to fence the site so that the area is the largest.
Solution:
Let's denote one side of the rectangle through x m, then the second will be (20 -2x) m, the area S (x) \u003d (20-2x) x \u003d 20x -2x 2;
S / (x) \u003d 20 -4x; S/(x)=0; 20 -4x = 0; x = =5;
According to the condition of the problem x (0; 10)
Find the sign of the derivative on the interval (0; 5) and on the interval (5; 10). The derivative changes sign from “+” to “-”. Hence x = 5 is the maximum point. Therefore, one side of the site \u003d 5m, the second 20 -2x \u003d 10m;
Example 6 . To reduce the friction of the liquid against the walls and bottom of the channel, the area wetted by it must be made as small as possible. It is required to find the dimensions of an open rectangular channel with a cross-sectional area of 4.5 m 2, at which the wetted area will be the smallest.
Solution:
We denote the depth of the ditch through x m, then the width will be m, P (x) \u003d 2x +;
P / (x)=2-; P / (x) \u003d 0; 2x 2 \u003d 4.5; x=1.5. We take only a positive value according to the condition of the problem.
By the condition of the problem x (0; )
Find the sign of the derivative on the interval (0; 1.5) and on the interval (1.5; ?). The derivative changes sign from “-” to “+”. Hence, x=1.5 is the minimum point, therefore, P(1.5)=6m is the smallest value, which means that one side of the ditch is 1.5m, the second = 3m.
Example 7 A rectangular area with one side adjacent to the building. With a given perimeter of 200m, it is necessary to fence the site so that the area is the largest.
Sections: Maths
The purpose of the lesson:
- Generalization and systematization of the acquired knowledge.
- Expansion of students' ideas about solving problems of finding the largest and smallest values.
During the classes
1 stage of the lesson
Teacher introduction: every person from time to time finds himself in a situation where you need to find the best way solving any task.
For example: process engineers try to organize production in such a way as to get as many products as possible, designers want to plan devices in such a way on spaceship so that the mass of the device is the smallest, etc.
We can say that the problems of finding the largest and smallest values have practical applications.
In proof of my words, I want to cite from the story of L.N. Tolstoy "How Much Land Does a Man Need" about the peasant Pakhom, who bought land from the Bashkirs.
- What will be the price? Pahom says.
- We have one price: 1000 r. per day.
Pahom didn't understand.
- What is this measure - a day? How many tithes will it have?
“We don’t know how to count this,” he says. And we sell in a day; how much you get around a day, then it’s yours, and the price is 1000 rubles.
Pahom was surprised.
“But this,” he says, “there will be a lot to go around the earth in a day.
The foreman laughed.
“All yours,” he says. - Only one agreement: if you don’t come back on the day to the place from which you start, your money is gone.
- But how, - says Pahom, - to mark where I will pass?
- And we will stand in the place where you choose; we will stand, and you go, make a circle, and take a scraper with you and, where necessary, notice, at the corners of the hole, put a swarm; then we’ll go from hole to hole with a plow. Take whatever circle you want, just before sunset, come to the place from which you started. What you get around is all yours.
The figure that Pakhom turned out is shown in the figure. What is this figure? (Rectangular trapezoid)
Question: What do you think, did Pahom get the largest area? (taking into account that the plots are usually in the form of a quadrilateral)? We will find out in today's lesson.
To solve this problem, we need to remember what steps are involved in solving extreme problems?
- The task is translated into the language of the function.
- Analysis tools look for the largest or smallest value.
- Find out what practical meaning the result has.
Task #1 (decided by the whole class)
The perimeter of the rectangle is 120 cm. What length should the sides of the rectangle have in order for the area to be the largest.
We return to the task with which we started the lesson. Did Pakhom get the largest area (given that plots are usually quadrilateral)? We discuss with the students what the largest area Pahom could get.
2 stage of the lesson
There is an explanation of the written tasks on the board in advance (there are two of them).
Task #1
Find under what conditions the consumption of tin for the manufacture of cylindrical cans of a given capacity will be the smallest.
I draw the attention of the guys that hundreds of millions of cans are produced in our country and the saved consumption of tinplate by at least 1% will allow additional production of millions of cans.
Task #2
The boats are located at a distance of 3 km from the nearest point A of the shore. At point B, located at a distance of 5 km from A, there is a fire. The boatman wants to help, so he needs to get there as soon as possible. The boat is moving at a speed of 4 km/h and the passenger is moving at 5 km/h. At what point on the coast should the boatman land?
3 stage of the lesson
Work in groups with subsequent protection of tasks.
Task #1
One of the faces cuboid- square. The sum of the lengths of the edges coming out of one vertex of the parallelepiped is 12. Find its largest possible volume.
Task #2
To mount the equipment, a 240 dm 3 base in the form of a rectangular parallelepiped is required. The base of the stand to be built into the floor is a rectangle. The length of the rectangle is three times the width. The rear longer wall of the stand will be built into the workshop wall. When installing the stand, its walls that are not built into the floor or into the wall are interconnected by welding. Determine the dimensions of the stand that will give the shortest overall length of the weld.
Task #3
A beam with a rectangular section of the largest area is cut out of a round log. Find the dimensions of the beam section if the radius of the log section is 30 cm.
Task #4
From a rectangular sheet of cardboard with sides of 80 cm and 50 cm, you need to make a rectangular box, cutting out squares along the edges and bending the resulting edges. How tall should the box be to maximize its volume. Find this volume.
4 Stage of the lesson
Solving tasks for evaluation by choice.
Task #1
From a wire 80 cm long, it is necessary to make a rectangle of the largest area. Find its dimensions.
Task #2
The sum of the lengths of the edges of the correct triangular prism equals 18√3. Find the largest possible volume of such a prism.
Task #3
The diagonal of a rectangular parallelepiped, one of the side faces of which is a square, is equal to 2√3. Find the largest possible volume of such a parallelepiped.
Stage 5 lesson
Khrestina Nadezhda Mikhailovna, teacher for developmental work with children, NOU DOD "DRC" Wonderland ", Ryazan [email protected]
Application of TRIZ elements in mathematics lessons
Annotation. The article discusses the use of elements of the structure of a creative lesson in an innovative pedagogical system NFTMTRIZ. The author proposes methodical development math lesson in grade 5, which demonstrated how to develop students' creativity within the framework of school curriculum. Keywords: universal learning activities, creative thinking, system-activity approach, creative lesson, reflection.
Mathematics is a science that is vital for everyone. From the very young age, the child is surrounded by the world of numbers, shapes, etc. And at the same time, this world is very complex and multifaceted. Many children, faced with difficulties in learning the material, lose interest in the subject and “ignorance” accumulates like a snowball. Therefore, the teacher faces a problem: not only to teach, but also to instill interest, which means to give the child the tools for self-learning new knowledge (universal learning activities). The task of the teacher is to make the lesson interesting, exciting, using a variety of teaching methods, to systematically develop creative thinking, the ability to work with a problem and solve it, draw conclusions, look for new original approaches, see the beauty of the results. The message for this is the Federal State Educational Standard (FSES) of the main general education dated December 17, 2010. It is based on a systematic activity approach, with the value of a free and responsible personality of the student. The standard dictates that we move away from the classroom system of Jan Amos Comenius, in which the teacher is the “narrator” and the students are the “narrators.” New lesson types, such as: “ brainstorm", dispute, project activity, will help the child in a constantly changing world. What results should the teacher get as a result of his work? training and knowledge, conscious choice of profession; form communicative competence; the ability to set goals, look for ways to achieve them, master the basics of self-control, etc. Also, the student must have sufficient knowledge and competencies, be able to be responsible for their actions and their consequences, respect the law, be a free and responsible, tolerant citizen. The advancement of science and technology leads to increase in the number of inventions and new professions, the student must be prepared for the ever-changing demands of the labor market. The foregoing allows us to conclude that in order to achieve all these results, the teacher must not only transfer knowledge, he must “teach him to learn.” The teacher, going to the lesson, must understand that subject results are no longer the only main ones, he also needs to form personal and meta-subject results. The very wording of the results has changed, since the child must now master the methods of action, i.e. universal learning activities, which are the meta-subject results. Only a set of universal actions will make it possible to form the student's ability to learn as a system. It makes it possible to visually trace how and at what stage certain universal educational actions are formed. To achieve the goals, the use of elements of a creative pedagogical system of continuous formation can help the teacher. creative thinking(NPTM), which has the tools of decision theory inventive problems(TRIZ). This allows students to develop creative imagination and fantasy, systemic and dialectical thinking. The use of the structure of a creative lesson at school allows you to make the lesson brighter, less stressful for the child, keep the child focused throughout the lesson, and most importantly, not provide him with ready-made knowledge, but give him the opportunity to get it himself. Also important issue is a partial transition from closed-type tasks to open-type tasks. Open-type tasks that affect the everyday experience of students make students think already when reading the condition, since it is insufficient, “blurry”, and may contain an excess of information. A variety of solution methods leads to the destruction of psychological inertia - the habit of standard actions in a familiar situation or the desire to think and act in accordance with accumulated experience. A set of possible answers helps to teach the child reflection and self-esteem. One cannot speak of a complete rejection of closed tasks. They are good in small quantities when you just need to get your hands on a particular formula or property. But explaining new material cannot be without a problem. After all, the first question after reading the topic in the lesson in the head of the children is: “Why do I need this?” or “Where will I need this?” All of the above is given to us by the NFTW system – the continuous formation of creative thinking and development creativity children. I present a lesson in mathematics grade 5, with elements of the structure of a creative lesson in the innovative pedagogical system NFTMTRIZ. Units of area »Lesson type: Lesson learning new material. Lesson objectives: 1. Subject: to form students' idea of the area of the figure, to establish links between the units of measurement of the area, to acquaint students with the formulas for the area of a rectangle and a square.2. Personal: to form the ability to determine the methods of action within the proposed conditions and requirements, to adjust their actions in accordance with the changing situation.3. Metasubject: to form the ability to see math problem in the context problem situation, in the surrounding life.Planned results:
students will get an idea of the area of \u200b\u200bfigures and its properties, learn to establish relationships between units of measurement of the area, apply the formulas for the area of \u200b\u200ba rectangle and square; they will receive inability to analyze, compare, generalize, draw conclusions; students will develop cognitive interest through game moments of the "little miracle"; communication skills work in a group and pairs. Textbook: A.G. Merzlyak, V.B. Polonsky, M.S. Yakir. Mathematics Grade 5. Textbook for students of educational institutions. 2014.
Stages of the lesson Tasks of the stage Teacher activity Student activity lesson, organization of children's attention. Focus with playing bones: first, in a transparent case, 1 large bone, after hitting the cover of the case, 8 small ones appear in it. - How did this happen? - What did we do in the last lesson? - Today we will continue working with rectangles. Included in the business rhythm of the lesson.
The guys try to solve the focus. They activate the knowledge of the last lesson.
Personal: self-determination. Regulatory: self-organization. Communicative: planning educational cooperation with a teacher and peers. Cognitive: research skills. The neighbors are in strife. The owner of the blue area, in order to get to his garden, must pass through the neighbor's red area. What to do? Entrance to the sites
Fig. 1 We know from experience that equal plots of land have equal areas. – What conclusion can we draw? Problem. A man decided to paint the floor in his country house. But the floor has an unusual shape. But he does not know how much paint is needed, on a can of paint it says 100g per 1m2. The area of the smaller figure is 12m2, the area of the larger one is 20m2. What should I do?
Put forward versions of the settlement of the dispute. Together with the teacher, they choose the right one: the blue one needs to take a piece of the red land, and in return give it an equal size.
They conclude: equal figures have equal areas. The guys put forward versions, together we choose the correct one: you need to add the areas of two figures and find the paint consumption. The students themselves derive the second property: The area of \u200b\u200bthe figure is equal to the sum of the areas of the figures of which it consists. Personal: self-determination. Regulatory: development of regulation learning activities.Communicative: the ability to work in a team, to hear and respect the opinions of others, the ability to defend one's position. Cognitive: research skills. Development of creative thinking.
Fig. 2 Heuristic conversation with the elements of the trial and error method. There is a ruler, a compass, a protractor on the teacher's table. We talked about the area, but how can we measure it? Let's measure the area of our board. - What do we have for measuring segments? - What do we have for measuring angles? We conclude: for the unit of measurement of the area, we select a square whose side is equal to single segment. What do we call such a square? To measure the area, you need to calculate how many unit squares fit in it?
The guys go through all the possible tools, come to the conclusion that they are not enough.
– Ruler, unit segment – Protractor, unit angle. the topic of the lesson: "The area of the rectangle." 3. Psychological unloading. Give students the opportunity to change the type of activity. Tasks for the development of creative abilities. Orientation in space. 1. A pair of horses ran 20 km. How many kilometers did each horse run? (20 km) 2. There were 4 rabbits in the cage. Four guys each bought one of these rabbits and one rabbit remained in the cage. How could this happen? (One rabbit was bought together with a cage) 3. There are two coins in two wallets, moreover, in one wallet there are twice as many coins as in the other. How can this be? (One wallet lies inside another) The class is divided into groups of 6 people, in groups the captain is selected by the teacher, who, after discussing the problem, chooses the correct answer. 1 minute is given for discussion.
Personal: self-determination. Regulatory: development of regulation of educational activities. Communicative: interaction with partners in joint activities. Cognitive: research skills. Development of creative thinking.
4. Two sons and two fathers ate 3 eggs. How many eggs did each one eat? (One egg each). Toy: "Touch the neighbor's right ear on the left with the elbow of your left hand" 4. Puzzle.
Present a system of increasingly complex puzzles embodied in real objects. Independent solution of tasks. 1. How many centimeters are: 1 dm, 5m 3dm, 12dm 5cm; 2. How many meters are: 1 km, 4km 16 m, 800 cm. 40 km. In how many hours will it pass 24 km at the same speed?
Right answers.
Fig. 3 Only answers are written in notebooks, then they exchange notebooks with a neighbor on their desk and check with each other. At the end, the correct answers appear on the screen. Personal: sense formation. Regulatory: self-regulation of emotional and functional states, self-organization. Communicative: the ability to work in pairs. Development of creative thinking.
5.Intellectual warm-up.Develop logical thinking and creativity. 1. The side of a rectangular sheet of paper has an integer length (in centimeters), and the sheet area is 12 cm2. How many squares with an area of 4 cm2 can be cut out of this rectangle? 2. The following figure is displayed on the board through the projector. How to divide the resulting figure into two figures with equal areas with one straight cut. One student at the blackboard, the rest work from their place. Personal: meaning formation, the ability to complete work. Regulatory: self-organization. research activities.6. Content.
It contains the program material of the training course and ensures the formation of systemic thinking and the development of creative abilities. Was it hard for us to calculate the area using a square? If we need to calculate the area of the stadium, let's go and try? If one side of the board is 2m and the other side is 1m, the board is rectangular, then it can be divided into 2x1 unit squares. Therefore, what is the area of the board? If a and b are adjacent sides of the rectangle expressed in the same units. How to find the area of such a rectangle?
Problem.–How to find the area of a regular quadrilateral in which all sides and angles are equal?
New units of area are introduced: ar (weave), hectare.1 a = 10 m * 10 m = 100 m2
1 ha = 100 m* 100 m = 10000 m2
What measurements require such large units of area?
S= a b We write down the formula in a notebook. Students discuss the problem in groups previously formed in a psychological warm-up, the only one group becomes experts (having listened to the versions put forward, they process them and offer one that they think is correct). There is a discussion of the solution to the problem. Then in notebooks we write down the resulting formula for the area of the square S = a 2
–For area measurement land plots, villages, stadiums, etc. Personal: self-determination. Regulatory: development of the regulation of educational activities. Communicative: the ability to work in a team, hear and respect the opinions of others, the ability to defend one's position. Cognitive: research skills. Development of creative thinking.
7. Computer intellectual warm-up. Provide motivation and development of thinking. Establish the correctness and awareness of the study of the topic.
Test on a computer. The teacher controls the number of errors. Fig. 5 (the figure is under the table)
Students work on a computer in pairs, pass a test. Personal: self-determination. Regulatory: development of regulation of educational activities. Summary. Homework. Summing up the lesson. Provide feedback in the lesson. The teacher offers to clap who liked the lesson and stomp if they find this lesson boring. – What new did you learn in the lesson?
Homework. Given a square with a side of 8 cm. Find its area. Using multi-colored pieces, explain and then refute my hypothesis: 8 * 8 = 65 Fig. 6 Students evaluate the lesson, their actions in the lesson, the actions of their peers.
-The formula for the area of a rectangle, square, area unit. At home, students conduct an experiment with parts of a square.
Such calculations are obtained because a gap is formed between the parts when assembling the rectangle. Personal: self-development of moral consciousness and orientation of students in the field of moral and ethical relations. Regulatory: development of the regulation of educational activities. Communicative: the ability to express one’s thoughts with sufficient completeness and accuracy.
Links to sources1.Federal State educational standard basic general education. Federal Law of the Russian Federation of December 17, 2010 No. No. 1897FZ.2.M.M.Zinovkin. NFTMTRIZ: creative education of the XXI century. Moscow, 2007. –313s.
"Application of the derivative to problem solving"
(Grade 10)
The methodological system of the teacher's activity on this lesson involves the formation of the ability of students to independently plan and carry out in stages research work. The student has the right to consult with the teacher, discuss, receive advice or tips from the teacher in order to help the child understand the variety of solutions and determine the right one.
The lesson discusses the theoretical material, the class is divided into groups to ensure the diversity of the methods of reasoning they offer, followed by the selection of the most appropriate of them.
Along with independent activity, it is advisable to use differentiated tasks of different levels in the lesson and evaluate them accordingly.
The analysis of the results of the performance of these tasks by students, in addition to information about their assimilation, gives the teacher a picture of the main difficulties of the students, their main gaps, which helps to identify the main ways to solve problems.
The purpose of the lesson: mastering skills independently in a complex to apply knowledge, skills and abilities, to carry out their transfer to new conditions using the research method.
Tasks:
Educational and cognitive: consolidation, systematization and generalization of knowledge and skills related to mastering the concept of "the largest and smallest value of a function"; practical application of the formed skills and abilities.
Developing: development of skills to work independently, express thoughts clearly, conduct self-assessment of educational activities in the classroom.
Communicative: the ability to participate in discussions, listen and hear.
During the classes
Organizing time
1. Every person from time to time finds himself in a situation where it is necessary to find the best way to solve a problem, and mathematics becomes a means of solving problems of organizing production, searching for optimal solutions. An important condition for increasing the efficiency of production and improving the quality of products is the widespread introduction of mathematical methods in technology.
Repetition
Among the tasks of mathematics important role assign tasks to extremes, i.e. tasks to find the largest and smallest values, the best, the most profitable, the most economical. Representatives of various specialties have to deal with such problems: process engineers try to organize production in such a way that as many products as possible are obtained, designers want to plan the device on a spacecraft in such a way that the mass of the device is the smallest, economists try to plan the attachment of factories to sources of raw materials in such a way to keep transport costs to a minimum. We can say that the problems of finding the smallest and largest values have a great practical application. Today in the lesson we will deal with such problems.
Consolidation of the studied material
2. Two “strong” students are called to the board to solve tasks (10 min.).
1st student: Given a tank without a lid in the form of a rectangular parallelepiped, at the base of which lies a square and whose volume is 108 cm 3 . At what dimensions of the tank will the least amount of material be used to make it?
Solution: Let us denote the side of the base through x cm, express the height of the parallelepiped. Find the sign of the derivative on the intervals. The derivative changes sign from "-" to "+". Hence x=6 is the minimum point, therefore, S(6)=108 cm 2 is the smallest value. So, the side of the base is 6 cm, the height is 12 cm.
2nd student: A rectangle of the largest area is inscribed in a circle with a radius of 30 cm. Find its dimensions.
Solution: Let us denote one side of the rectangle as x cm, then we will express the area of the rectangle. Find the sign of the derivative on the interval (0;30) and on the interval (30;60). The derivative changes sign from "+" to "-". Hence x=30 is the maximum point. Therefore, one side of the rectangle is 30, the other is 30.
3.At this time, youA mutual check is carried out on the topic “Application of the derivative” (1 point is given for each correct answer). Each student answers and, for verification, passes his answer to a neighbor in his desk.
Questions are written on a portable board, only the answer is given:
A function is called increasing on a given interval if...
A function is said to be decreasing on a given interval if...
The point x 0 is called the minimum point if ...
The point x 0 is called the maximum point if ...
Stationary points of a function are called points ...
Write general form tangent equations
The physical meaning of the derivative
Drawing conclusions
4. The class is divided into groups. Groups perform tasks to find the minimum and maximum of the function.
5. The word "strong" students is given. Students in the class check their solutions (10 min.).
6. Issued tasks of choice for each group (10 min.).
1 group.
To mark "3"
For the function f (x) \u003d x 2 * (6-x), find the smallest value on the segment.
Solution: f (x) \u003d x 2 * (6-x) \u003d 6x 2 + x 3; f / (x) \u003d 12x-3x 2; f / (x)=0; 12x-3x 2 \u003d 0; x 1 =0; x 2 =4;
f(0)=0; f(6)=0; f(4)=32-max.
To mark "4"
From a wire 20 cm long, it is necessary to make a rectangle of the largest area. Find its dimensions.
Solution: Let's denote one side of the rectangle through x cm, then the second will be (10-x) cm, the area S (x) \u003d (10-x) * x \u003d 10x-x 2; S / (x)=10-2x; S/(x)=0; x=5. By the condition of the problem x (0; 10). Find the sign of the derivative on the interval (0;5) and on the interval (5;10). The derivative changes sign from "+" to "-". Hence: x=5 - maximum point, S(5)=25 cm 2 - the largest value. Therefore, one side of the rectangle is 5 cm, the other side is 10x=10-5=5 cm.
To mark "5"
A plot of 2400 m 2 must be divided into two rectangular sections so that the length of the fence is the smallest. Find the sizes of the plots.
Solution: Let's denote one side of the site through x m, write down the length of the fence and find the derivative P / (x) = 0; 3x 2 \u003d 4800; x 2 \u003d 1600; x=40. We take only a positive value according to the condition of the problem.
Find the sign of the derivative on the interval (0;40) and on the interval (40;?). The derivative changes sign from "-" to "+". Hence x=40 is the minimum point, therefore P(40)=240 is the smallest value, which means that one side is 40 m, the other is 60 m.
2 group.
To mark "3"
For the function f (x) \u003d x 2 + (16-x) 2, find the smallest value on the segment.
Solution: f / (x)=2x-2(16-x)x=4x-32; f / (x)=0; 4x-32=0; x=8; f(0)=256; f(16)=256; f(8)=128-min.
To mark "4"
A rectangular area with one side adjacent to the building. Given the dimensions of the perimeter in m, it is necessary to enclose the site so that the area is the largest.
To mark "5"
From a rectangular sheet of cardboard with sides of 80 cm and 50 cm, you need to make a rectangular box, cutting out squares along the edges and bending the resulting edges. How tall should the box be to maximize its volume?
We denote the height of the box (this is the side of the cut out square) through x m, then one side of the base will be (80-2x) cm, the second - (50-2x) cm, volume V (x) \u003d x (80-2x) (50-2x ) \u003d 4x 3, 260x 2 + 4000x; V / (x) \u003d 12x 2 -520x + 4000; V / (x)=0; 12x 2 -520x+4000=0.
According to the condition of the problem x (0; 25); x 1 (0; 25), x 2 (0; 25).
Find the sign of the derivative on the interval (0;10) and on the interval (10;25). The derivative changes sign from "+" to "-". Hence x=10 is the maximum point. Therefore, the height of the box = 10 cm.
3rd group.
To mark "3"
For the function f (x) \u003d x * (60-x), find the largest value on the segment.
Solution: f (x) \u003d x * (60-x) \u003d 60x-x 2; f / (x)=60-2x; f / (x)=0; 60-2x=0; x=30; f(0)=0; f(60)=0; f(30)=900-max.
To mark "4"
A rectangular area with one side adjacent to the building. With a given perimeter of 20 m, it is necessary to fence the site so that the area is the largest.
We denote one side of the rectangle through x m, then the second will be (20-2x) m, the area S (x) \u003d (20-2x) x \u003d 20x-2x 2; S / (x)=20-4x; S/(x)=0; 20-4x=0; x=5. By the condition of the problem x € (0;10). Find the sign of the derivative on the interval (0;5) and on the interval (5;10). The derivative changes sign from "+" to "-". Hence x=5 is the maximum point. Therefore, one side of the site = 5 m, the second - 20-2 * 5 = 10 m.
To mark "5"
To reduce the friction of the liquid against the walls and bottom of the channel, the area wetted by it must be made as small as possible. It is required to find the dimensions of an open rectangular channel with a cross-sectional area of 4.5 m 2, at which the wetted area will be the smallest.
We denote the depth of the ditch through x m, P / (x) = 0; 2x 2 \u003d 4.5; x=1.5. We take only a positive value according to the condition of the problem. Find the sign of the derivative on the interval (0;1.5) and on the interval (1.5;?). The derivative changes sign from "-" to "+". Hence x=1.5 is the minimum point, therefore, P(1.5)=6 m is the smallest value, which means that one side of the ditch is 1.5 m, the second is 3 m.
4 group.
To mark "3"
For the function f (x) \u003d x 2 (18-x), find the largest value on the segment.
f (x) \u003d x 2 (18-x) \u003d 18x 2 -x 3; f / (x) \u003d (18x 2 -x 3) /; f / (x)=0; 36x-3x 2 \u003d 0; x 1 =0; x 2 =12 f(0)=0; f(18)=0; f(12)=864-max.
At the mark "4".
A rectangular area with one side adjacent to the building. With a given perimeter of 200 m, it is necessary to fence the site so that the area is the largest.
Let's denote one side of the rectangular section through x m, then the second will be (200-2x) m, the area S (x) \u003d (200-2x) x \u003d 200x-2x 2; S / (x)=200-4x; S/(x)=0; 200-4x=0; x=200/4=50. By the condition of the problem x (0; 100). Find the sign of the derivative on the interval (0;50) and on the interval (50;100). The derivative changes sign from "+" to "-". Hence x=50 is the maximum point. Therefore, one side of the site = 50 m, the second - 200-2x = 100 m.
To mark "5"
It is required to make an open box in the form of a rectangular parallelepiped with a square base, with the smallest volume, if 300 cm 2 can be spent on its manufacture.
We denote one side of the base through x cm and express the volume, then V / (x) \u003d 0 300-3x 2 \u003d 0; x 2 =100; x=10. We take only a positive value according to the condition of the problem.
Find the sign of the derivative on the interval (0;10) and on the interval (10;0). The derivative changes sign from "-" to "+". Hence x \u003d 10 - the minimum point, therefore, V (10) \u003d 500 cm 3 - the smallest value, which means that the side of the base is 10 cm, the height is 50 cm.
Questions for the class
7. Delegates from the groups explain the solution of the selected problems (10 min.).
8. Taking into account the points in the warm-up and work in groups, marks are given for the lesson.
Summing up the lesson
Homework
Solving a problem one point higher; students who completed the task on "5" are exempted from homework.
The analysis of the results of the fulfillment of these tasks by students, in addition to information about their assimilation, gives the teacher a picture of the main difficulties of the students, their main gaps, which helps to outline the main ways to eliminate them.
FOMKINA TATYANA FYODOROVNA |
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BUSINESS CARD |
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Job title | Teacher of Russian language and literature |
Place of work | Municipal educational institution"Average comprehensive school No. 9 "of the city of Orenburg |
Work experience in the position | |
Competitive score | |
The theme of pedagogical experience | Formation of linguistic competence of students on the basis of an activity-system approach in teaching the Russian language according to the EMC S.I. Lvova |
The essence of the methodological system of the teacher, reflecting the leading ideas of experience | The essence of the teacher's methodological system is in the organization of educational activity as a movement from a question of a linguistic nature (allowing students to pay attention to the meaningful linguistic essence of a particular spelling) to a method of action (based on a rule, referring to a dictionary), and then to the result (free operating rules in the course of writing or using a spelling dictionary). |
Work on dissemination of own experience, presentation of the methodological system at various levels (forms, intellectual products) | Experience of Fomkina T.F. summarized in 2009 at the level of the MO MOU "Secondary School No. 9" and approved by the Methodological Council. In 2009 and 2010 represented among the teachers of the city of Orenburg at the municipal level. Tatyana Fedorovna spoke at district methodological associations on the following issues: “The use of ICT in the lessons of the Russian language and literature as a means of forming linguistic competence”, “An activity approach to building educational standards”. |
The effectiveness of the implementation of the methodological system | Formation of sustainable positive motivation and increasing students' interest in the subject; Positive dynamics in relation of students to the teacher, lessons of the Russian language and literature, development of students' ability to predictive activities and activation of cognitive processes; Significant increase in quality creative works, essays, which is confirmed by the results final exams: in 2007, according to the results of the GIA, the academic performance was 100%, the number of those who completed the tasks for "4" and "5" - 87%; in 2008 by USE results academic performance - 100%, the number of those who completed the tasks for "4" and "5" - 92%, the highest score - 87; in 2009, according to the results of the Unified State Examination, the academic performance was 100%, the number of those who completed the tasks for “4” and “5” was 58%, the highest score was 96; Increase in the number of students participating in scientific and practical conferences, competitions, olympiads: X district scientific and practical conference of students "You are an Orenburger" (III place), XV city conference of students "Intellectuals of the XXI century" (diploma for "Diversified research of the family"), All-Russian Correspondence Competition "Knowledge and Creativity", 2010 (III place, laureate), regional intramural-correspondence competition "Fatherland", 2009 (III place), VI International Olympiad in Fundamentals of Sciences, 2010 (I and II degree diplomas), International game-competition "Russian Bear Cub", 2010 (15th place in the region). Monitoring educational activities shows high level the level of education of students Fomkina Tatyana Fedorovna: Russian language - 69% (2009), literature - 77% (2009). |
MATERIALS FROM WORK EXPERIENCE
Lesson in learning new knowledge
with multi-level differentiation of training
"NOT with nouns"
(grade 5)
The presented summary of the lesson was compiled in accordance with the “Russian Language Program for Grades 5-6” by S.I. Lvova (M.; "Mnemosyne", 2008). The lesson is aimed at the formation of linguistic, language and speech competence of students. The material included in the lesson is educational, developing, educating.
Lesson objectives:
1) develop communication skills: formulate a question and give an answer to a grammatical topic; to carry out speech interaction in a mobile group; create your own texts on a given topic;
2) to form linguistic and linguistic competence: to know the spelling rule NOT with a noun be able to use the algorithm to apply this rule in practice; repeat spelling « NOT with a verb" , noun rule;
3) to cultivate a careful attitude to the word as the spiritual value of the people.
Equipment: multimedia equipment, video presentation, reference cards, test, research task files.
During the classes
Organizing time
Dear colleagues! Yes, yes, colleagues. I didn't call you guys that by accident. Today we will be engaged in a common cause: solve linguistic problems, discover the secrets of spelling words. After all, according to Leo Tolstoy, “The word is a great thing ... With a word you can serve love, but with a word you can serve enmity and hatred” (an epigraph to the lesson).
Linguistic warm-up "Yes - no"
Here is a word skill that will help you cope with the linguistic warm-up, which is called "Yes - no." The rules of this warm-up are as follows: I guessed the rule, and you will try to guess it by asking leading questions, which should be formulated in such a way that I can answer with the words “yes” or “no”. I will evaluate your answers today with tokens. Ask me questions.
The students ask the teacher questions. For example:
1. Did we learn this rule in 5th grade? (Yes)
2. Is this a spelling rule? (Not)
3. Is this a rule about parts of speech? (Yes)
4. Is this a noun rule? (Yes)
- Well done! Guessed!
Knowledge update
Now let's remember what a noun is. But we will tell about it in a chain, passing the baton to each other, like athletes in competitions. Anyone who wants to can use the answer helper cards. I will evaluate your answers with tokens ( student responses).
Did a great job! We need knowledge of the rule about a noun in order to be able to distinguish nouns from other parts of speech.
We will test this skill by executing oral distributive dictation.
Read the words carefully (When you click the mouse on the projector screen, the image fades).
But what is it? What happened to the image? Guys, there is a mistake!
Catch her! (Reception "Catch the mistake")
"Indignant" should be written together. Why?
This is a verb that is not used without NOT.
(mouse click)
Exercise: divide the words into two groups according to parts of speech. (Students do the task)
1. What parts of speech did you come across? (nouns and verbs)
2. Name the nouns.
3. Name the verbs.
4. How to spell NOT with a verb?
goal setting
So, knowing the rule about nouns and about spelling NOT with a verb will help us deal with new theme which sounds like this: "NOT with nouns".Write it down in a notebook.
I recorded the course of our thoughts in "Thinkingsheet", which consists of three columns: “I know”, “I want to know”, “I learned (a)”.
In the graph "I know" given the rule on which we will rely today. This rule is about writing NOT with a verb .
In the graph "I want to know" the question of the day was formulated: "Find out when NOT is written together with a noun, and when - separately."
In the graph "I found out" we will write down the answer to this question.
But first let's do vocabulary work.
Guys, who are they? ignoramus and ignorant? What kind of people do we call that? (Student answers)
Write these words in your notebook and lexical meanings. Now make up phrases or sentences with them (optional).
Learning new material
Why do you guys think the words "ignorant" and "ignorant" are spelled together? (Because they are not used without NOT)Report
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