Hi there. Let us recall the structure of this course. We Multivariate Functions in One We have seen the basic concepts. Some calculation methods and I have seen some limited applications. Two Functions of Several Variables in ie more applications in this course and We will see some additional advanced methods. Our previous session two In the integral-storey circular coordinates seeing We enrich our knowledge. In the first part of the first course here again very Variables one we saw in applications, concepts, methods We continue to enrich. Here in the first chapter on derivatives Topics will see a selection of applications. On two of the main concepts Our first lesson again wider saw gradient We will see applications. And yet chained derivatives We have seen that the concept of We saw method Multivariate Functions in One. We will further enrich it here more variable functions We will look at the relationship between and Although the same basic concepts we will see much more of a sudden We will encounter a rich structure. Gene remind terms Multivariate Functions in One We saw derivative calculation methods. The most fundamental issue chaining derivatives, total derivative and there was the concept of the directional We had a gradient of differentiation and interests. We finished it here more generally, will do. Now, again related to derivatives applications under We saw two main issues: Taylor series and extreme value. Extreme values ??is actually the best Finding optimization applications. There are four issues here at the limit. There are local extreme values There are always outliers, There are outliers under the restrictions of a There are extreme values ??in the function. Integral extreme values ??in Thanks to this function Find the extreme values ??of this integral will it is called the variational calculus. This is the first issue, the local extreme values In the first part we saw. Now we will see that the other three. As a reminder only variable functions at the local and minimum As values ??were finding. FX it the first derivative of y is equal to zero was going to be a necessary and to find the most critical points smallest and largest values but it was not enough to make. If the second derivative as proficiency increases the little value if the second derivative axis were finding the largest value. Gene context of local endpoints Multivariate Functions in One As we have seen in two variables We found a similar structure in functions. Here too there is a requirement condition but Because the two variables for a single Enough with the requirements univariate condition Our derivative where two kinds of functions for that, because I have a partial derivative both supposed to be zero. This requirement was a condition. Qualifying as a condition how wherein the second derivative plus or the second derivative is negative most of us gives the largest and smallest values still have them, but this is not enough, a second comprising derivatives of these determinants plus is supposed to be valuable. Thus, a bit of qualifying conditions This requires more information more natural because our variable but the same determinants from scratch A second criterion is smaller this time without the need to look smallest or the largest value We understand that this kind We call it point to point in I have seen a lot of them practices. Of course, if the point univariate No anti-in functions. Now again with a reminiscent end do we proceed with the values ??of the local extreme value difference. Absolute values ??of a closed-end terminal region is defined whereas in Near a point of local extreme values is defined in an infinitesimal region. Therefore, this fundamental difference. A local outlier in 're not talking about a region We're talking about points, whereas the absolute value of the absolute end values ??have a closed-terminal region. He has a theory:Continuous the absolute extreme values ??of functions you local extreme values ??occur at the point of or at the boundary point occurs. After knowing this theorem Few things to learn. This method is implemented as follows. First, the local extreme values because we find the absolute end values ??to occur there also may occur knows, but we need to find it. The second theorem says that the absolute end limit values ??at points says may occur. His tip for border border border on extreme values ??are determined. The third step with the local extreme values extreme values ??on the boundary By comparing the largest and smallest Select'll find the absolute extreme values??. This is a very, extraordinary theory Let's consider a simple application. I want to be a function like this. This function is currently in closed regions and finite x is equal to that in the closed region x is equal to minus one with a vertical line, y is equal to y is equal to one-half three divided by two horizontal lines formed in rectangular region We want to find the absolute extreme values??. There is nothing more to do here first We are looking for business-critical point, ie, never thinking that in this function's local extreme values We find here, we find fx, fy'y the We find that the first partial derivatives of the we identify this local spot. This local spot this region of We're looking at whether or not in outside the region because the local point If you are experiencing can not take it as a yardstick. Here indeed from this point zero in a closed a zero in the closed because it is not within the region y is split in two by a whereas two of the three splits this is defined thereof y's goes beyond regional borders. So we will take care of this point. As a second step the border We want to find outliers. Our borders as shown here x is equal to minus one and y is one and x is equal to one divided by two divided by two equals three years consists of four lines identified. These are respectively a to y from x to split in two here two of these three split a y the boundaries of the rectangle in this way if it crosses We choose it because they come and putting a minus y We find in the fourth. These extreme values We need to find. I left it for homework to be done This job is very simple g of a single variable x index ends of this single variable functions We find the value and the end value of course also a minus bir'l whether you're checking. Similarly, the second Y represents a borderline in x is equal to y is in a limit in functions of one variable occurs again y is a function of derivative thereof critical point we find its greatest Do we determine whether it is the smallest all We are at the point in it. Corner points of a subject in There are also four corner points there. We find this corner points are as follows: x minus one-half of a y's A two-point x y's which one-half of y's a point divided by three divided by two, a split of the points x and y's a point that divides the two x three. Four corners of the rectangle. This point values ??in We are also accounts and comparing them at all We find the largest and smallest values??. And that the answers given here I'm waiting for you to do as homework. There's a new theory to be learned here. One important concept absolute end value is defined in a region, and a finite area in a zone to consequently defined here as the the largest and smallest values According to the theorem of local, the largest and the smallest values ??also call border the largest or the smallest We are the values ??obtained by calling. Here can be difficult to calculate, but What is important to understand the concept. Here is an example of the second would say. Because I've given a lot of steps. With two arguments, a gene absolute value of the research. As a difference here 's given function. But the limits given in a circle. BI in the previous four direct It was composed of a rectangle wall. If a circle here. Therefore with a continuous function Here you will be given still more, regardless of what that limit function to find the critical value, the largest and the smallest to set values??. Then find the value of the border. Border means finding values ??y square, for example, y, y here minus the square root of x squared minus nine or that one x minus y in terms of placing the square is minus nine years of functions of one variable are downloading. See if we get here the first two terms, common y times x minus y here Is the start of one frame. y times x squared minus one he thought Although we have created here an extra year. He would remove y and two behind cube minus y squared terms are three years. that's when we organize them as x minus one here too We can easily solve. Already problems so it is possible that designed for. Or maybe the real problem may not be so easy to solve. When we place only y of the function occurs. Therefore, a restriction on the f We find the values ??of the border. Single variables by taking the derivative of it, By differentiating critical to y points are determined. Here's the function of the border here largest and smallest values and these are determined by comparing the The absolute best in this problem the largest and smallest values ??are determined. Gene is an example. Here it might be difficult, of course. This time we want to make three variables. A rectangular region given prism. x is equal to a level of minus two in three dimensions. it is also a level of x equals three. x, and x minus two equals two equals three vertical planes passing through. From here the y equals minus four and y is equal to zero on the horizontal plane. Sorry, this vertical planes. In this, if z is equal to the minus two and z is equal to two horizontal planes passing through ie, six-sided, six-plane surface defined a vertical prism in absolute We want to find the value. Closed in our region, and therefore also the limits to contain. Here again we do this work function of the three variables of the function x According to y, z by the partial take derivatives at the smallest To determine the location of the major points. Then, the largest and the smallest of them It looks at whether the values ??and to calculate the values ??at that point. Here you are in the closed region There are now so critical points. According to Theorem border values we need to look at. Here, of course, because there is a lot of work There are six at the boundary surface. There are 12one edge of the six surfaces. If you think of a rectangular prism Four horizontal surface, the following four horizontal surface, There are those in kenardar. 12units have the edge. There are also eight corners. Accounts not difficult. But one should make them one. There are computer programs ready them by demanding more than you does not require thinking but to recover by this In order to use the software need to understand the issues. Here let's start with an example. Here, a surface and an edge the value of one of the corner will advance by calculating problems. For example, the given function. If we think of x is equal to minus two surfaces, where x minus two we put in place, becomes a function of y and z. As you can see a few variables function has become. We find that the critical values??. Here to opposing it, the values??, z e, w values ??are achieved. Of course, to find the critical value of this because it is a function of y and z to y We take partial derivatives and by z. Two unknowns, two equations here We find the values ??of y and z. This is the y and z values function by placing x minus the value on the two surfaces, We find the critical value. That the remaining 11, There are five more surface sorry. They also need to build on. Similarly, an edge on example. x is equal to minus four minus two years at the side end of the function What is the value we thought, where x is the given function y z function, but we choose the x and y minus two and minus four for the leaving only a function of z remains. In this single-variable functions, because we are moving along a straight line, We are moving along one edge, We find that the critical value. Here the value of z is known. by knowing the value of z, where x and y de at this point we had chosen We can calculate the value. But our account, rectangle because it was not done There are 11more edges of the prism. We need to do for each of them. After finishing this one We need to look to the corner. In the corners, for example one, We have eight corners. How certain corners of the border. on a rectangular prism. Minus four and minus two minus the two corners away wherein the function value are account. We have seven more corner again. Once you find them in by comparison between them We choose the largest and the smallest values??. As you can see, the process more There is nothing that requires new knowledge. The only thing that will be remembered theorem, absolute values ??you local values ??of the region you in knowing that in the border. Gave the following as homework. E, but I myself I do not want to deal with it. There is nothing more I'll learn from. But at least the problem How was complete as he You are now on I'm just saying. Computer is already here The importance emerges. This kind of calculation porterage You make the computer. There are also ready-made software, but In order to identify the problem, To edit this You need to understand the subject. Now today, E, I'm cutting this session here. We will do next There is a second derivative applications. These three values ??under the constraints and Lagrange multiplier method, so that is that some of duymuÅu. The name of this problem, the best of such Please call or optimizasiyo problem. You will see in the next session. Until then, your goodbye.
