AI Assistant
Transcript
00:07:290Paolo Guiotto: Okay, I left yesterday to try…
00:12:410Paolo Guiotto: to do this exercise 296, that is an exercise on finding stationary points.
00:22:430Paolo Guiotto: Which is just an algebraic exercise, but let's do some of these, because
00:30:840Paolo Guiotto: I want to show you some…
00:32:880Paolo Guiotto: features of solving this problem. So… but there is written also denied unclassified. You don't know what this means, we will see later. So determine for the moment
00:42:910Paolo Guiotto: stationary… Stationary… points.
00:49:190Paolo Guiotto: Of course.
00:50:400Paolo Guiotto: if any.
00:53:520Paolo Guiotto: off.
00:55:340Paolo Guiotto: We did the number one… no, number two, let's do number one.
01:01:490Paolo Guiotto: FXY… equal.
01:04:890Paolo Guiotto: XY times X plus 1.
01:09:980Paolo Guiotto: So, point XY.
01:14:450Paolo Guiotto: ease.
01:15:950Paolo Guiotto: Stationary.
01:17:490Paolo Guiotto: point.
01:19:50Paolo Guiotto: 4.
01:20:390Paolo Guiotto: F… If the gradient of F at point XY
01:27:30Paolo Guiotto: Is… the gradient is a vector, so is the vector zero?
01:32:690Paolo Guiotto: Now, the gradient is made of the partial derivative, so…
01:37:60Paolo Guiotto: Gradient F is the vector made of the derivative with respect to X of F, derivative with respect to Y of F.
01:45:20Paolo Guiotto: We notice that the derivative with respect to X of X of F is…
01:50:360Paolo Guiotto: But perhaps we may write this,
01:53:70Paolo Guiotto: In an expanded form, so it's like X squared Y plus X times Y.
01:59:530Paolo Guiotto: So we have derivative with respect to X is 2X times Y plus Y. This is… But…
02:08:350Paolo Guiotto: Derivative with respect to X, and the derivative with respect to Y,
02:12:210Paolo Guiotto: is the same, so it is X squared plus X.
02:17:610Paolo Guiotto: Okay? These are the two derivatives.
02:20:370Paolo Guiotto: Well, notice that, PR.
02:26:640Paolo Guiotto: We… Night… a notice.
02:34:650Paolo Guiotto: that… These partial derivatives, dx, F, E,
02:40:100Paolo Guiotto: YF are clearly well-defined and continuous in the interior plane R2.
02:47:500Paolo Guiotto: belong to… continuous functions on R2. So, according to the differentiability Faster.
02:57:450Paolo Guiotto: This says that the function f is differentiable on R2.
03:05:240Paolo Guiotto: But it is something which is not required. But since differentiability is important.
03:10:760Paolo Guiotto: I want to use these examples also to show you how, in concrete cases, we discuss differentiability without any problem. Now.
03:23:240Paolo Guiotto: XY is a stationary point.
03:29:680Paolo Guiotto: If and only if gradient equals zero.
03:33:740Paolo Guiotto: So this means that the two components of the gradient together must be zero, not one equals zero or the other. So you get a system.
03:42:380Paolo Guiotto: Well, there are two equations. The derivative with respect to X of F must be 0, and the derivative with respect to Y must be 0.
03:52:90Paolo Guiotto: Usually, this is… usually… this one will be always a system of two equations, in this case, being two unknown. If you have a function of three variables, we will see later, this will be a system of three equations in three unknown.
04:07:260Paolo Guiotto: Okay? So the size of the system depends on the number of variables, because there is the gradient, and the gradient depends on the… has number of components equal to the number of variables.
04:18:760Paolo Guiotto: Now, the first equation is, 2XY plus Y.
04:27:810Paolo Guiotto: equals 0, and the second equation is X squared plus X equals 0.
04:35:470Paolo Guiotto: A second point that we should observe here is that, normally, these are systems of non-linear equations.
04:44:460Paolo Guiotto: If we have a 2x2 system of linear equations, so something AX plus B equals something CX plus ED equals something else.
04:55:950Paolo Guiotto: This is standard, no? It's linear algebra. You have seen, normally, this is non-standard, so there is not a general way to solve these systems.
05:07:10Paolo Guiotto: Okay?
05:08:920Paolo Guiotto: So, this means that you have time by time to find out the way.
05:15:950Paolo Guiotto: The unique general thing we could say is that since these equations are something equals zero, if possible, you should try to factorize the equations, because in that case, you have alternatives.
05:30:650Paolo Guiotto: Well, actually, in this… and then, you have to do a choice on which equation you should work first.
05:38:70Paolo Guiotto: Well, let's say that the last of the technique will be, I find the X on an equation, and they plug into the other equation to reduce the number of X. That's… almost never works, okay, in these cases. Well, if I look at these two equations.
05:53:130Paolo Guiotto: They are both, easy, in some sense.
05:56:770Paolo Guiotto: I could say, well, the second equation seems easier because there is only X, okay? So, we could start from the second equation. The second equation, if I factorize, is a second degree equation if you want, but you can also write as X times X plus 1 equals 0. And you leave the… if you want, well, let's do the factorization also for the first equation, which is Y times…
06:19:170Paolo Guiotto: 2X plus 1. Of course, the order of the equation is irrelevant. You write first what you prefer, okay?
06:27:470Paolo Guiotto: But the important thing here is that you don't have to miss spaces in the solution, because this will mean that you will lose solutions.
06:36:130Paolo Guiotto: Okay, so for example, let's take… let's choose now one of the two equations, for example, this one.
06:42:230Paolo Guiotto: Maybe later I will show you the solution by choosing the second one, to see that it comes the same.
06:48:410Paolo Guiotto: Now, this one gives two possibilities, either X equals 0, or X equals minus 1.
06:57:240Paolo Guiotto: So what I do with this?
06:59:680Paolo Guiotto: Now, I have an alternative. This means that either… X is 0.
07:06:270Paolo Guiotto: or… X is minus 1, and since X is a specific value, I plug this into the second line.
07:14:840Paolo Guiotto: In this, case, X equals 0, the second line with X equals 0 becomes Y equals 0.
07:22:110Paolo Guiotto: Why?
07:23:700Paolo Guiotto: If X is minus 1, the second line becomes 2X is minus 2 plus 1 minus 1, so minus Y equals 0.
07:33:10Paolo Guiotto: Now, you have two subsystems, that means the solution either verifies the first, or it verifies the second. But the first is already the solution. You have X equals 0, Y equals 0, and for the second, it's basically the solution, because minus Y equals 0 means y equals zero. So X equals minus 1, Y equals 0.
07:52:670Paolo Guiotto: So, conclusion is… the… stationary points… R…
08:03:580Paolo Guiotto: In the first case, we get 0.00, in the second, 0 minus 1.
08:08:470Paolo Guiotto: No.
08:09:790Paolo Guiotto: Wrong Minus 1, 0.
08:15:160Paolo Guiotto: And that's it. The exercise is over.
08:18:390Paolo Guiotto: I could have done an alternative solution at this stage here, so I will do the alternative solution for you now, only once.
08:26:430Paolo Guiotto: Alternative.
08:28:160Paolo Guiotto: solution. This because, of course, you need just to find one way, not all possible ways, okay? One solution is enough, you know, not right, all possible solutions.
08:39:110Paolo Guiotto: So, at this point here, I have the system X times X plus 1 equals 0, and the second line becomes Y times 2X plus 1 equals
08:53:660Paolo Guiotto: Zero.
08:56:180Paolo Guiotto: So… from this, I now choose this second line.
09:04:150Paolo Guiotto: This second line yields an alternative, you see?
09:08:200Paolo Guiotto: That product is 0 if, and only if, one of the two factors is 0. So, Y equals 0, or 2X plus 1 equals 0.
09:20:60Paolo Guiotto: So this means that I have Y equals 0, first case, or 2X plus 1 equals 0, X equals minus 1 half, second case.
09:30:110Paolo Guiotto: In the first case, what happens to the first equation? Nothing, because there is no Y, so it remains X times X plus 1 equals 0.
09:39:960Paolo Guiotto: In the second case, what happens to the first equation? Now it changes, because X is minus 1 half. The first equation becomes minus 1 half times minus 1 half plus one one half equals 0.
09:53:40Paolo Guiotto: So what is the conclusion here?
10:04:580Paolo Guiotto: If you see this, what would you say?
10:08:250Paolo Guiotto: impossible. This no solution
10:13:660Paolo Guiotto: Okay? So you forget, that's over. We continue only with the first one, and the first one, now we have to look at the second equation. And the second equation yields this alternative, X equals 0, or
10:27:350Paolo Guiotto: Y equals 0 and X equals minus 1. As you can see, at the end, we get the same solutions, of course.
10:34:200Paolo Guiotto: So we get solution X equals 0, Y equals 0, solution X equals minus 1, Y equals 0. Same solutions.
10:46:460Paolo Guiotto: Okay, let's do… For example, number 3.
10:54:260Paolo Guiotto: Here we have this function fxy equal to… X cubed plus Y cubed plus 2X square…
11:05:850Paolo Guiotto: plus 2Y square plus X plus Y. It's very…
11:10:610Paolo Guiotto: stupid exercise, this one. However, we do.
11:15:680Paolo Guiotto: So, here…
11:18:70Paolo Guiotto: The gradient of F is the vector made of the two partial derivatives with respect to X and with respect to Y.
11:26:410Paolo Guiotto: As you can see, the function…
11:28:130Paolo Guiotto: depends on X and Y in the same way.
11:31:260Paolo Guiotto: So the derivative with respect to X is the, the re… X squared… for AXA… plus 1.
11:45:130Paolo Guiotto: And the same is for… the Y derivative, 3Y squared plus 4Y,
11:52:560Paolo Guiotto: Last one. It's a stupid exercise because we have exactly the same thing. So, this is equal to vector 0 if and only if
12:01:410Paolo Guiotto: We have these two equations, 3X squared plus 4X. Well, maybe there is something interesting here, however.
12:09:950Paolo Guiotto: to 3Y squared plus 4Y plus 1. Now, we saw that, yeah, there is something interesting to be noticed here.
12:18:840Paolo Guiotto: There is no difficulty in solving the two equations, because they are the same equation, so we may notice that if I say 3T squared plus 40T plus 1 equals 0. Now, you have to look at the delta, the discriminant of this, which is…
12:35:520Paolo Guiotto: 4 square minus 4 times 3 times 1, so it is 16 minus 12.
12:44:30Paolo Guiotto: Equals 4 positive, so we have roots.
12:47:440Paolo Guiotto: And they are T1 real roots. Minus 4 plus minus the root of delta, which is 4, divided by… divided by 6. So, minus 4 plus minus 2 divided by 6, so it means that
13:05:590Paolo Guiotto: Minus 4 plus 2 is minus 2, divided by 6 is minus 1 third.
13:11:570Paolo Guiotto: Okay, minus 6 over 6 is minus 1. So the equation has these two solutions.
13:17:150Paolo Guiotto: So this means that, here, continuing with the system.
13:22:690Paolo Guiotto: Well, if we look at the equation for X, this is X equal minus 1 third or…
13:33:320Paolo Guiotto: or minus 1, and the same is for Y, minus 1 third, and minus 1. Now, I want to know from you how many stationary points are there.
13:47:590Paolo Guiotto: I want the number. Are you able to tell me the number of stationary points?
13:53:480Paolo Guiotto: There are 2 stationary points, 3 stationary points, 4 stationary points, 4…
14:02:260Paolo Guiotto: Because here, you see, X can be either minus 1 third or minus 1. Y can be either minus 1 third or minus 1. There is no other condition. So you must combine all them, so at least that the point XY can be either you take the value minus 1 third for X, and you take minus 1 third for Y.
14:21:830Paolo Guiotto: Or, you take minus 1 third for X, and you take minus 1 for Y. Or, you take minus 1 for X and minus 1 third for Y, and finally, you can take also minus 1 and minus 1 for both.
14:36:550Paolo Guiotto: Okay? So, you have to always to be, careful with Combining the solutions of this.
14:46:330Paolo Guiotto: Now, let's see the number 5, which is an example with the… Functions of three variables.
14:53:570Paolo Guiotto: Fxyz is this.
14:57:140Paolo Guiotto: It's a bit more complicated.
14:59:260Paolo Guiotto: is mostly in the interpretation of the final result. X cubed minus 3X minus Y squared times Z squared plus Z cubed.
15:18:280Paolo Guiotto: So here, the gradient of F is the vector made of the partial derivatives. So F, there are three partial derivatives here, because there are three variables, so DXF,
15:30:720Paolo Guiotto: DYF… BZF.
15:34:630Paolo Guiotto: Well, let's compute in line here. DXF is
15:39:610Paolo Guiotto: Now, derivative with respect to X. I look at that expression as a function of X. I forget of all of the others, they are constant. So, for example, you see that I have Z squared times, in the parentheses, I have 3X squared minus 3, and then 0 for the Y squared, and then plus Z cubed 0, okay?
16:00:470Paolo Guiotto: DYF
16:02:240Paolo Guiotto: DYF, I have, again, Z squared, and in the parentheses, there is only this part here that contains Y, so I have minus 2Y.
16:12:260Paolo Guiotto: So it's 1 minus 2YZ squared.
16:17:930Paolo Guiotto: Then, DZF, this is more complex, because we have that coefficient, X cubed minus 3x minus Y squared, which is a number, a constant for Z. So I keep… then I have the derivative of Z squared, which is 2Z.
16:33:510Paolo Guiotto: plus 3Z squared.
16:39:670Paolo Guiotto: Incidentally, we noticed that… We… Notice.
16:47:150Paolo Guiotto: That, all three partial derivatives, the XF, the YF,
16:53:620Paolo Guiotto: DZ, F, are continuous functions on the full
16:59:280Paolo Guiotto: space R3, which is the natural domain for this function.
17:04:40Paolo Guiotto: And therefore, according to the differentiability test.
17:11:550Paolo Guiotto: the function f is differentiable on R3.
17:18:480Paolo Guiotto: So, as you can see, differentiability is just one line. We don't have to do long calculations.
17:25:460Paolo Guiotto: Of course, this does not mean that the examples we have seen last time are… were stupid or useless calculations, because in those cases.
17:37:330Paolo Guiotto: we had complicated situations. The exercises asked to check differentiability in a point where I have a definition that is different, so there are lots of technical problems here, but that was just to see how the definition works, okay?
17:55:40Paolo Guiotto: In this case, we have, the problem is different, we have to find the stationary points, so…
18:01:240Paolo Guiotto: So, XYZ.
18:07:90Paolo Guiotto: Is a stationary point.
18:11:20Paolo Guiotto: 4.
18:12:230Paolo Guiotto: F, if and only if it solves this system made of the three derivatives, DXF equals 0, DYF equals 0, and DZF equal to 0.
18:26:460Paolo Guiotto: Let's write.
18:28:480Paolo Guiotto: Now, DX is Z squared times, I can factorize that 3, so 3.
18:35:90Paolo Guiotto: Z squared times… X squared minus 1.
18:40:800Paolo Guiotto: Equals zero.
18:42:860Paolo Guiotto: Second equation is, easier, minus 2YZ square equals 0.
18:48:530Paolo Guiotto: Third equation is more complex. I could factorize a Z, for example, Z times… 2…
18:56:960Paolo Guiotto: X cubed minus 3X minus Y squared.
19:03:490Paolo Guiotto: plus 3Z, all this equals…
19:08:660Paolo Guiotto: Now we have to solve this. As you can see, this system is highly nonlinear. There are no linear equations in this thing. There are no equations with the one unit variable, so this is the general equation.
19:21:210Paolo Guiotto: Okay? The previous example was a very, very, very particular situation, okay?
19:27:400Paolo Guiotto: This is a more… more likely how it should be. So, first of all, we could, for example, take out constants like this.
19:35:820Paolo Guiotto: But be careful. Another general, care should be the following. You do not simplify by variables, because value equals zero can be
19:46:850Paolo Guiotto: part of the game. So, now we have to choose one of these three questions to start solving the system. The first, the second, or the third.
19:56:510Paolo Guiotto: I should try to look at the simplest equation, not at the more difficult.
20:01:640Paolo Guiotto: In fact, the last one seems more complicated, so I should say, let's take either the first or the second.
20:08:80Paolo Guiotto: In this case, oh, okay, we can also simplify the 2, the minus here, equals 0.
20:14:720Paolo Guiotto: I can divide by factors which are different from zero. I can't divide by variables that can take value equals zero, so I cannot simplify by Y. That's an error.
20:26:580Paolo Guiotto: Okay?
20:28:330Paolo Guiotto: Be careful, because these stupid things are those who change the solution.
20:34:540Paolo Guiotto: And changing the solution here means that you find the wrong stationary points, and since stationary points are used to solve minimum-maximum problems, you are producing a completely different solution of the problem, which is probably wrong, okay?
20:49:970Paolo Guiotto: Now, I would say that among the first or the second is indifferent, because they are all factorized. Products equals zero, so… but let's say that most of you will think that the second one is easier, so let's take the second one.
21:05:540Paolo Guiotto: So this… what kind of alternative is… well, either Y is 0, so I have a first alternative, Y equals 0, or a second alternative is Z square equals 0. Z square equals 0 means Z equals 0. There's no other possibility, so I can also write this equation.
21:23:880Paolo Guiotto: Okay, now let's rewrite the other equations by plugging the information we have, now, in these subsystems. So the first one, take the case y equals 0, the first equation remains z squared times X squared minus 1 equals 0.
21:40:620Paolo Guiotto: While I get a little bit of simplification in the last one, Z times 2X cubed minus 3x, you see that the term Y squared disappears, plus 3Z equal to 0.
21:57:250Paolo Guiotto: The second alternative is Z equals 0. Now, you see that if I plug Z equals 0 into the first equation, it becomes 0 equals 0.
22:05:820Paolo Guiotto: What does it mean, this identity? Also, let's do in the last one. In the last line, there is this factor, Z, you see, that becomes also… the last one becomes Z equals 0.
22:16:350Paolo Guiotto: Now, what, what do these, equation, equations mean?
22:24:760Paolo Guiotto: What does it mean, zero equals zero?
22:29:450Paolo Guiotto: It means that it is an identity always verified, so it's true, so it doesn't tell anything here, so I can cancel this true, because they do not restrict anything. So, my alternative is now between these two systems, and in particular, the second one produces just Z equals 0.
22:48:30Paolo Guiotto: What does it mean? I cannot do anything more than this, you see, now there is no other equation. So, what kind of solutions
22:57:150Paolo Guiotto: produces these… Z equals 0.
23:01:30Paolo Guiotto: So you have to think that we are solving four points
23:05:440Paolo Guiotto: XYZ. We are looking for points XYZ, such that this system here is verified, which is equivalent to this alternative, either first or second.
23:21:850Paolo Guiotto: R, so it means that all you put together solutions. So what are the solutions of second, Z equals 0?
23:32:300Paolo Guiotto: XYZ0. So this gives points X, Y, 0, not condition on X, not condition on Y, so it means all XY in R.
23:43:530Paolo Guiotto: So, if you want to see these points, we are in space R3, this is the X, this is the Y, and this is the Z axis.
23:53:560Paolo Guiotto: Points of type XYZ0 are the plane XY.
23:57:590Paolo Guiotto: are down here.
23:59:220Paolo Guiotto: So you see, it's plenty of points of this type.
24:02:420Paolo Guiotto: It is possible.
24:03:930Paolo Guiotto: Now, let's continue with the first.
24:06:570Paolo Guiotto: The first, we say that it does not, yes, it does a little bit simplify the circle, the third equation.
24:15:60Paolo Guiotto: But now what we do? Okay, the first one is Y equals 0, we cannot do anything better than this. We work on the second one, which is definitely better than the third one.
24:24:100Paolo Guiotto: So, this second one produces, again, an alternative. What is the alternative? So, I already have Y equals 0,
24:32:390Paolo Guiotto: or still keep Y equals 0. So the alternative is either this factor is 0, so Z square is equal to 0, or this factor is 0. X squared minus 1 equals 0. You see that?
24:50:240Paolo Guiotto: Okay? Now, you have to put these information in the third line. So this means Z equals 0, and when I put in the third line, I get 0 equals 0. So this is
25:03:110Paolo Guiotto: basically completed, so the system is Y equals 0, Z equals 0, and 0 equals 0, which is a useless identity. So this yields what points?
25:15:240Paolo Guiotto: X00 with X not conditioned, so, arbitrary.
25:21:700Paolo Guiotto: As you can see, these solutions are already included into these ones.
25:27:940Paolo Guiotto: So, basically, I have no new solution from this, okay?
25:32:310Paolo Guiotto: So… These are… I'll rate it.
25:45:480Paolo Guiotto: So, I can forget, because they do not add new solutions. About the second one, let's look. Y equals 0, X squared equals 1,
25:55:110Paolo Guiotto: And what about the last line? So, since the last one contains X cubed X, well, for the moment, we keep this.
26:03:970Paolo Guiotto: Because there is a little bit of mess here, so I just copy.
26:10:460Paolo Guiotto: So 2X cubed minus 3X.
26:14:580Paolo Guiotto: minus 3X plus 3Z equals 0. So now it remains only this part here to be completed. So let's say that X, Y equals 0,
26:29:870Paolo Guiotto: That condition, X squared minus 1 equals 0, means x squared equals 1. That yields two possibilities, X equals 1, X equals minus 1. So let's say, either X is equal to plus 1, or a new alternative.
26:43:280Paolo Guiotto: Y equals 0, X equals minus 1, and now we plug this into the third line.
26:49:960Paolo Guiotto: In the third line, there is no Y, we already simplified, but there is X. So when I put X equal 1, I get Z times… that X cubed minus 3x becomes 1 minus 3, so minus 2.
27:03:980Paolo Guiotto: times 2 minus 4, so minus 4 plus 3Z equal to 0.
27:11:640Paolo Guiotto: R.
27:12:880Paolo Guiotto: When X is minus 1, X cubed minus 3x is minus 1 plus 3, so it is plus 2, that makes Z times 4 plus 3Z equals 0.
27:28:90Paolo Guiotto: It is not yet over, because now we have a new sub-alternative. Until you finish all the possibilities, you have to continue. So this one means that…
27:39:650Paolo Guiotto: Y equals 0.
27:41:630Paolo Guiotto: X equals 1, then the subalternative is Z equals 0, or…
27:47:780Paolo Guiotto: Y equals 0. X equals 1, and 3Z minus 4 equals 0, so Z equals 4… Duh.
27:58:900Paolo Guiotto: Okay?
28:02:00Paolo Guiotto: Then, here, we continue, because these are on the same plane, Y equals 0.
28:07:170Paolo Guiotto: X equals minus 1, Z equals 0 for the last line, or Y equals 0. X equals minus 1, and Z equal minus 4.
28:20:200Paolo Guiotto: Duh.
28:22:500Paolo Guiotto: Okay, now we have the information about points. This is point 100. It's not a new point, because it is already in the class XYZ0.
28:35:700Paolo Guiotto: So this is .0… no, sorry.
28:40:00Paolo Guiotto: 1… 0, 4 thirds, and that's new.
28:46:60Paolo Guiotto: You see? Because so far, we have these points, with the third coordinate equals zero. So that's a new point. Let's emphasize this.
28:56:90Paolo Guiotto: Then here we have minus 100, this is not a new point, and this is minus 1, 0, minus 4 thirds, and this is a new point.
29:07:930Paolo Guiotto: So we can conclude now, because there is no other, calculation to do for the system, so conclusion.
29:19:50Paolo Guiotto: D.
29:20:100Paolo Guiotto: stationary points, of F. R.
29:25:980Paolo Guiotto: So we said all points of the plane XY, so points XYZ0 for every X and Y real.
29:34:250Paolo Guiotto: And we have also these two points, 1, 0, 4 thirds, and minus, so we may say plus minus D, okay?
29:45:220Paolo Guiotto: And that's all.
29:50:950Paolo Guiotto: Okay?
29:52:630Paolo Guiotto: Any questions?
29:57:70Paolo Guiotto: No.
29:59:580Paolo Guiotto: Okay, so… Stationary points. Why stationary points?
30:05:700Paolo Guiotto: are so important. That's because of the theorem WIST.
30:10:240Paolo Guiotto: We have, written the statement yesterday, the Fermat theorem, that says… When you have… a function F.
30:22:200Paolo Guiotto: That is differentiable, then you have to check differentiability.
30:26:800Paolo Guiotto: Even at the end, the theorem says that the minimum-maximum point must verify gradient f equals 0, so it must be a stationary point.
30:36:660Paolo Guiotto: Okay, so you will apply this to determine stationary points, to have candidates, minimum, maximum points. You must know that this is valid when the function f
30:50:130Paolo Guiotto: Oh.
30:53:350Paolo Guiotto: When the func… when the function f is differentiable, okay?
31:01:690Paolo Guiotto: So differentiability must be checked. So that work we have done here in the exercise is not just to waste time. It's something that is needed, because when we will apply the Fermat theorem, we will need to know the differentiability.
31:17:640Paolo Guiotto: However, the Fermat theorem says You have a differentiable function.
31:22:610Paolo Guiotto: on a domain
31:24:370Paolo Guiotto: And it says if a point is a minimum or a maximum in the interior, not just a generic point of D, which is minimum-maximum.
31:34:720Paolo Guiotto: then gradient F at that point is 0, so that point is a stationary point.
31:40:720Paolo Guiotto: Now, I want to give a little proof of this factor. I will simplify a bit the proof, which is in the notes, so…
31:50:880Paolo Guiotto: Let's return on this, because this is one of the few results we can prove here, so let's prove it.
31:58:810Paolo Guiotto: So it says that we have a function f of array X defined on D, with values in our…
32:09:630Paolo Guiotto: F, differentiable.
32:12:490Paolo Guiotto: on B.
32:15:420Paolo Guiotto: a point, let's call it X star, in the interior, of the… B, local, mean… or maximum.
32:31:260Paolo Guiotto: point.
32:33:60Paolo Guiotto: Then, the gradient of F at point X star Must be equal to zero.
32:42:110Paolo Guiotto: So, in other words, X star.
32:46:320Paolo Guiotto: In interior of the… local.
32:51:240Paolo Guiotto: Minimum, or maximum.
32:55:240Paolo Guiotto: implies… X star… Stationary point.
33:04:660Paolo Guiotto: But, and this is something we already noticed, the vice versa is non-truth. So it's not an if-and-all if.
33:13:600Paolo Guiotto: So this means that… what is the interest? If it is an if and only if, we may say, okay, you know what? To determine minimum-maximum, I solve the equation gradient F equals zero, because this will tell me that, what are the minimum maximum points. That's wrong.
33:31:470Paolo Guiotto: First, the point must be in the interior.
33:35:10Paolo Guiotto: Okay.
33:36:530Paolo Guiotto: So, if it is not in the interior, that conclusion can be false.
33:41:170Paolo Guiotto: Let's see this remark one. We already seen it in dimension 1, but I want to put this also with the…
33:48:190Paolo Guiotto: two-dimensional example.
33:50:360Paolo Guiotto: take this function, FXY, is X squared plus Y squared.
33:57:940Paolo Guiotto: on domain D, which is the unitary bowl, so it is the set where X squared plus Y squared
34:06:850Paolo Guiotto: is, well, less or equal than 1.
34:14:800Paolo Guiotto: You can have an idea of how this function is made, because if,
34:22:00Paolo Guiotto: We will never plot graphs of these functions here, but however, in this case, it is easy, we can do.
34:31:469Paolo Guiotto: So in the plane XY, I have my domain, which is a disk in the plane. So this is the domain.
34:41:739Paolo Guiotto: It is all the distance.
34:44:429Paolo Guiotto: Now, the function at every point of that domain gives you X squared plus Y squared, which is the distance to the origin.
34:51:350Paolo Guiotto: Now, if you want to have an idea on how this function is made, look at, for example, what happens if you put one of the two coordinates equal 0. So if you take sections along one of the axes, for example, FX0. FX0 is X squared. That means that when I look at the function along the x-axis, I see a parabola.
35:10:540Paolo Guiotto: So if you want to see the picture, it should be like that.
35:16:40Paolo Guiotto: And if you look the same along the y-axis, F, F0, Y, you get exactly the same type of function, Y squared. And it is the same parabola, so it is something like this.
35:32:290Paolo Guiotto: Now, imagine to rotate this, you get a sort of surface, and this should be the graph of this function.
35:44:190Paolo Guiotto: And this surface is called the paraboloid. It's a parabola in three dimensions.
35:51:280Paolo Guiotto: Now, if you look at this figure, it is clear that when you evaluate the function f on these points, the points of the edge of the disk, the green points, these are the points where X squared plus Y squared is equal to 1, no?
36:11:190Paolo Guiotto: At those points, the function f here, FXY, since FXY is just X squared plus Y squared, that value will be constantly equal to 1, no? So all these points are sent into this circle here.
36:29:250Paolo Guiotto: with Z equals 1 for all points.
36:32:410Paolo Guiotto: So it is clear that all these points are maximum points for the function f, because all the other values are less than 1, no?
36:40:300Paolo Guiotto: IFA.
36:42:650Paolo Guiotto: So if X squared plus Y squared is strictly less than 1, I'm talking about points which are inside the disk, so not on the edge, but inside, the value of F at those points
36:58:770Paolo Guiotto: FXY, since it is equal to X squared plus Y squared, it is less than 1, which is the value
37:06:270Paolo Guiotto: Taken at… on the edge points.
37:09:250Paolo Guiotto: On the green points. So, on the green points, you get value 1. On the other points in the interior of the disk, the value is less than 1. So, it is clear that
37:21:720Paolo Guiotto: D.
37:24:380Paolo Guiotto: Maximum.
37:26:180Paolo Guiotto: points… 4.
37:29:630Paolo Guiotto: F… R… Points, X, Y,
37:35:410Paolo Guiotto: Such that they are on this circle. XX squared plus Y squared is equal to 1.
37:42:640Paolo Guiotto: Now, we'll do a second figure, which is a figure on the plane XY only, so on the domain.
37:49:500Paolo Guiotto: The domain is this disk centered in the origin, and radius 1, so this is 1, this is 1. So the blue points are those in the inside of the disk.
38:04:260Paolo Guiotto: And the green points are those of this circle.
38:09:660Paolo Guiotto: So the green points are those for which X squared plus Y squared is 1, and the blue points are those for which X squared plus Y squared is less than 1.
38:24:790Paolo Guiotto: Okay? So the green points are the maximum points for this function.
38:29:470Paolo Guiotto: Now, notice what happens to the gradient.
38:32:140Paolo Guiotto: The gradient of F is… so the F, function f is written there, is X squared plus y squared is 2x
38:40:800Paolo Guiotto: to Y?
38:42:880Paolo Guiotto: And when this gradient is 0, Only if point XY is
38:51:460Paolo Guiotto: is the origin. So, the unique stationary point is this one.
38:59:950Paolo Guiotto: stationary bond.
39:02:40Paolo Guiotto: So this function has a stationary point. In this domain, it is D00.
39:06:970Paolo Guiotto: But in particular, this means that none of the green points is a stationary point.
39:13:430Paolo Guiotto: And if you look at the gradient, it says the unique possibility to have greater than equal zero is that the coordinates of the point are zero. For none of these green points, the gradient will be zero.
39:25:430Paolo Guiotto: So they are all maximum points, but none of them has gradient equals 0.
39:31:690Paolo Guiotto: That's wrong. Fermat theorem says that gradients should be zero.
39:36:70Paolo Guiotto: Nope.
39:37:460Paolo Guiotto: Fermat theorem says that gradient is 0 when
39:41:920Paolo Guiotto: This happens when you are in the interior. And in fact, if you look at this function, you have a minimum. You see what is the minimum point.
39:52:870Paolo Guiotto: is 0, 0. Now, because the function is positive, at 0, it takes value 0, so 0, 0 is the minimum point, no?
40:01:80Paolo Guiotto: We also noticed that… so… so, gradient F is different from 0, in… each… Off.
40:13:340Paolo Guiotto: max points.
40:16:500Paolo Guiotto: And this is because they are not in the interior. So, Fermat theorem does not apply. While…
40:26:510Paolo Guiotto: Gradient F is equal to 0, at 0, 0.
40:32:380Paolo Guiotto: which is in the interior of this domain. Now, you see that 00 is well inside, that belongs to the interior.
40:39:470Paolo Guiotto: of the… And… 0, 0.
40:45:610Paolo Guiotto: is… the… Unique, minimum.
40:50:640Paolo Guiotto: fine.
40:52:680Paolo Guiotto: So, be careful, because this Vermat theorem says gradient F equals zero at min-max points when they are inside the domain, well inside, so they are in the interior. Otherwise, the gradient might not be equal to zero.
41:10:470Paolo Guiotto: So the equation gradient F equals 0 is important to determine possible mean-max points.
41:19:10Paolo Guiotto: provided they are in the interior. Otherwise, you will never, probably, find other min-max points. Look at this example. If you use the equation gradient F equals 0 here, you would find as a unit point for which gradient f equals 0, the .00.
41:38:580Paolo Guiotto: So, you would list exactly all the green points, which are maximum points.
41:44:770Paolo Guiotto: You see? So the equation gradient F equals 0 does not capture exactly all mean max points.
41:52:540Paolo Guiotto: It can capture only those which are inside, in the interior of the device.
42:00:920Paolo Guiotto: So this is an… and now I want also to show another example.
42:06:530Paolo Guiotto: Moreover, So, well, let's say… So… this… Example… shows.
42:25:370Paolo Guiotto: that.
42:27:870Paolo Guiotto: Gradient F.
42:29:940Paolo Guiotto: X star.
42:31:630Paolo Guiotto: equals zero at… Mean or Max.
42:39:370Paolo Guiotto: points.
42:41:620Paolo Guiotto: only… if… X star belongs into the interior of the domain.
42:50:440Paolo Guiotto: A second remark is that equation gradient f equals 0 might be verified at points which are not minimum and neither maximum points.
43:01:30Paolo Guiotto: Okay? So you remind those points with the original tangent, but the function is not. And now I want to show an example with a function of two variables.
43:11:870Paolo Guiotto: So let's take this other example. We were doing the maths, yeah, forget the March 2.
43:21:510Paolo Guiotto: remarked to…
43:23:10Paolo Guiotto: Take now this example, which is a slight modification of the previous one. FXY equals X squared, but now minus Y squared.
43:34:390Paolo Guiotto: Now, look at this function. On domain, let's take the same domain as before. Domain D is still a set of points XY, where X squared plus Y squared is less or equal than 1.
43:50:490Paolo Guiotto: No? The disk.
43:52:460Paolo Guiotto: Now, again, let's try to do a figure to understand what's going on here.
44:00:10Paolo Guiotto: So this is the plane XY, on which I have my domain, which is this disk.
44:06:330Paolo Guiotto: I hope you understand the geometry of figures. I'm not particularly good in drawing figures, but this is a plane list. So this, for the moment, is the plane XY.
44:18:720Paolo Guiotto: Now, if we look at sections here to see what is F, for example, FX0 is X squared, so as before, a parabola, so we may have an idea like this.
44:31:780Paolo Guiotto: for the function F, so let's now put also the z-axis.
44:36:520Paolo Guiotto: But when we look at the section along the y-axis, we see now minus Y squared. It's a negative parabola. So, in the y-axis, it's done like that.
44:50:690Paolo Guiotto: Now…
44:52:680Paolo Guiotto: It's hard to visualize this, but it looks like, you know, the saddle of a horse, no? How it's made? Like that. On a direction, you are like a parabola going up. On another direction, it is like a parabola going down.
45:06:670Paolo Guiotto: So, in fact, we call these kind of points, there is a definition for them, we will see later. This is a saddle point.
45:17:370Paolo Guiotto: No? And in this case, you may notice that here, the gradient of Earth
45:24:480Paolo Guiotto: is, again, it's very similar to the previous one. Now it is 2X minus 2Y, so it is equal to vector 0, again, if and only if both X and Y are 0. So there is a unit point where gradient is 0, and that point is the origin, so that point.
45:42:660Paolo Guiotto: But that point is not a maximum, neither a minimum, because if you look at the function along the x-axis, you would see a minimum.
45:50:90Paolo Guiotto: Along the y-axis, you would see a maximum, and you can have that the function is both a mean and a maximum, which should be constant, but is not constant, of course, this function. You can also notice that, for example, if you evaluate the function along the line Y equal X, FXX, this would be identical equals 0. So in this case, along this line, you would see…
46:13:520Paolo Guiotto: A flat diagram like that.
46:16:30Paolo Guiotto: So, it's… it's a bit complex. I should use the calculator to show you the shape of this surface, but if you have in mind a saddle, a horse saddle, it would… it will work.
46:29:610Paolo Guiotto: But… 0, 0 is… knots.
46:35:990Paolo Guiotto: Mean… nor Max.
46:42:260Paolo Guiotto: So, this, the second remark, the moral is… This… shows.
46:53:10Paolo Guiotto: Even if this point is well decided, even if, Even… if,
47:02:690Paolo Guiotto: 0 is in the interior of this domain. Now, here we are in the interior. And nonetheless, the gradient F equals zero
47:12:520Paolo Guiotto: does not necessarily say that you have found a minimum or a maximum. This shows that equations V.
47:22:420Paolo Guiotto: equation.
47:24:350Paolo Guiotto: gradient F equal 0.
47:28:280Paolo Guiotto: does not…
47:33:250Paolo Guiotto: Do you mean?
47:38:160Paolo Guiotto: does not always.
47:41:720Paolo Guiotto: determine mean… Max.
47:45:870Paolo Guiotto: bonds.
47:48:100Paolo Guiotto: So, it's a very weak condition somehow, okay? But that's the unique one, basically, we have to find out minimum-maximum points, and so we have to move
48:00:110Paolo Guiotto: try to… to give a value to this condition. Otherwise, there is no other condition to determine mean multiples.
48:09:680Paolo Guiotto: Okay, do you want to take a shot back?
48:20:250Paolo Guiotto: Have I… Yeah, fortunately.
48:24:700Paolo Guiotto: I was thinking that I forgot to, to, to… Truly.
48:34:130Paolo Guiotto: So, let's come to the, proof of, St. Mark, you're in?
48:57:320Paolo Guiotto: Okay, let's, we can… but let's do this, let's do a simplified proof, assuming, for simplicity, that F is just a function of two variables.
49:09:800Paolo Guiotto: You can repeat the argument for a generic number of functions.
49:15:190Paolo Guiotto: for simplicity.
49:23:930Paolo Guiotto: let F be a function of two variables, X, Y,
49:30:170Paolo Guiotto: let's call, X star.
49:34:60Paolo Guiotto: Y star.
49:36:190Paolo Guiotto: in the interior of D.
49:40:560Paolo Guiotto: B, local.
49:44:50Paolo Guiotto: Minimum.
49:45:580Paolo Guiotto: or a maximum. It's the same for the proof. We assume that it is a minimum.
49:51:880Paolo Guiotto: or a mat, huh?
49:56:590Paolo Guiotto: So, this means I do this first figure, which is a figure in the domain. So, the domain is in R2.
50:06:710Paolo Guiotto: It's something like this, domain D. There is a point somewhere, this point X star, Y star.
50:15:830Paolo Guiotto: Where the function has a local minimum.
50:19:650Paolo Guiotto: Now, what does it mean, a local minimum? I refresh you, we have seen the definition last time.
50:25:760Paolo Guiotto: Simply, local minimum or maximum means that you have a minimum, not for the entire domain, but only for a little part of domain around your point, okay?
50:38:710Paolo Guiotto: Now, here, there is an important remark. Since the point is dentilo.
50:43:990Paolo Guiotto: I know that. There is a ball that tells me that the point is entirely contained into the domain.
50:51:460Paolo Guiotto: And there is another ball where the pointer has a minimum.
50:55:950Paolo Guiotto: Now, this second boiler, I can always assume that it has a radius smaller than this red baller, so I can always assume that the minimum is on a small neighborhood of that point inferiorly contained in the domain. Otherwise, if it is bigger, you take the red ball.
51:16:10Paolo Guiotto: If the radius of the green ball were this one, you just can say, if it is a minimum on a big ball, it would be a minimum on the small ball.
51:24:860Paolo Guiotto: Not vice versa. So, the point is that since X star
51:31:340Paolo Guiotto: Y star is in the interior of D, and it is a local minimum, I can say that there exists a ball centered at that point, X star. Y star?
51:42:870Paolo Guiotto: with some radius r, which is interiorly contained in the domain. This is… this comes from
51:52:20Paolo Guiotto: The fact that we are in the interior.
51:55:300Paolo Guiotto: And…
51:56:930Paolo Guiotto: Such that, in that ball, F has a minimum, or a maximum. Here, we do the case for the minimum.
52:04:330Paolo Guiotto: So F of X star Y star, is less or equal than F of XY for any other point XY of the ball.
52:16:280Paolo Guiotto: centered at X star, Y star.
52:19:770Paolo Guiotto: Outrageous.
52:22:490Paolo Guiotto: So now, I do a second figure, we forget of whatever is outside of this ball, I draw in the bowl.
52:31:700Paolo Guiotto: So, our word is now the ball. You have a point.
52:36:270Paolo Guiotto: X star. Y star?
52:39:390Paolo Guiotto: We have a bowl, we do a little bit bigger for convenience.
52:43:930Paolo Guiotto: A green ball.
52:46:810Paolo Guiotto: This is… of course, the ball is together with the interior, okay? Maybe I can color with some light green.
52:57:450Paolo Guiotto: Yeah, it's not so light. However…
53:07:760Paolo Guiotto: Okay, now, this is, in green is,
53:11:240Paolo Guiotto: the ball centered at the minimum point, X star, Y star.
53:20:50Paolo Guiotto: Now, the trick consists in the following. Since at any other point, for example, here the function is bigger, if I now look along two particular directions, which is… one is this one, parallel to the x-axis.
53:36:160Paolo Guiotto: And the other is this one, parallel to the Y-axis.
53:40:600Paolo Guiotto: So these points here are what?
53:44:80Paolo Guiotto: Well, for this point, I have…
53:47:160Paolo Guiotto: different typesa, but the same ordinate. So this is a point of type X star Y,
53:55:250Paolo Guiotto: Not exactly the opposite. XY star.
53:59:450Paolo Guiotto: And this point here is a point with the same aptisa, external, and different ordinance.
54:06:830Paolo Guiotto: When these points are in the green ball, I know that, in particular, F of X star
54:14:70Paolo Guiotto: Y star will be less or equal than F of XY star.
54:24:410Paolo Guiotto: for all the possible X, you see that these X are from… this is X star, down here.
54:33:760Paolo Guiotto: And this will be X star plus radius, and this will be X star minus radius.
54:40:110Paolo Guiotto: Okay, so for all X, in the interval from X star minus radius to X star plus radius.
54:51:580Paolo Guiotto: And similarly, in vertical, FX star.
54:56:760Paolo Guiotto: Y star, is less or equal than F of X star Y,
55:03:650Paolo Guiotto: Similarly, for every Y, that is from Y star minus radius to Y star plus radius. So for every Y in the interval, Y star minus radius, Y star plus radius.
55:19:190Paolo Guiotto: But now… Look at these two auxiliary functions, so let's call them…
55:26:860Paolo Guiotto: Let, phi of X be the function.
55:32:590Paolo Guiotto: F of XY star. So, as we are looking at the section, basically, along the, not x-axis, but a line parallel to the x-axis.
55:45:540Paolo Guiotto: This is a function of one single variable, X. Y star is fixed, you see?
55:51:370Paolo Guiotto: So this is for X, which is in the interval X star minus R, X star plus R.
55:59:870Paolo Guiotto: What is written here is that phi at point X star, which is exactly FX star Y star, you see, when you plug X equals X star, you get F
56:13:140Paolo Guiotto: X star, Y star. That's the minimum value. We know that it is less or equal than F of XY star, which is phi of X.
56:24:880Paolo Guiotto: So what are we saying here is that, look only at the two extremes. So F of X star is less or equal than phi of x.
56:36:750Paolo Guiotto: For which X? For every X here.
56:41:150Paolo Guiotto: So it means that this function has a minimum at point X star.
56:46:00Paolo Guiotto: No? It is like if I am looking at this function on the interval X star minus R.
56:52:510Paolo Guiotto: to X star plus r, here in the middle, there is X star. My function at point X star is smaller than all the other values.
57:04:710Paolo Guiotto: And here, we apply what you know since the last year. That point X star is in the interior of that interval, so the derivative of phi at point X star must be equal to zero.
57:18:440Paolo Guiotto: So this is, first, the, calculus.
57:25:540Paolo Guiotto: firma, PRM.
57:31:510Paolo Guiotto: So in first-year calculus, you have seen that this… the one-dimensional version of this theorem.
57:37:490Paolo Guiotto: And what is the derivative of this phi?
57:41:380Paolo Guiotto: But… Phi Prima?
57:45:310Paolo Guiotto: at point X star is what? But remember, the definition is the limit when H goes to 0 of phi of X star.
57:57:610Paolo Guiotto: Too many stars here.
58:01:570Paolo Guiotto: X star plus H minus phi of X star
58:07:360Paolo Guiotto: divided by H. Now, plug into this the definition of phi, you get the limit.
58:13:00Paolo Guiotto: for H going to 0.
58:15:120Paolo Guiotto: What is phi of X is this. It's F of XY star. So phi of X star plus H is FX star plus HY star.
58:28:170Paolo Guiotto: minus… that's FX star, Y star, divided by H.
58:35:60Paolo Guiotto: And what is this?
58:37:990Paolo Guiotto: That's exactly the partial derivative with respect to X at point X star, Y star.
58:46:640Paolo Guiotto: So, since this one must be zero, the other one must be zero.
58:50:730Paolo Guiotto: So the conclusion is… D, X, F, X star.
58:58:80Paolo Guiotto: Y star equal zero.
59:01:570Paolo Guiotto: In the same way, you prove the other partial derivative, similarly.
59:08:340Paolo Guiotto: I don't repeat the calculation. Maybe you do, no?
59:13:320Paolo Guiotto: do… peak days.
59:17:820Paolo Guiotto: So you are forced to understand what we have done. Similarly, you get the derivative with respect to Y of F, at that point will be also 0.
59:28:410Paolo Guiotto: And so you have that these are the two components, in this case, of the gradient, and the conclusion is that the gradient of F at point X star, Y star, equals 0.
59:40:300Paolo Guiotto: So this finishes the growth.
59:43:150Paolo Guiotto: Now, I suggest you to do the… the other case for the partial derivative with respect to Y, but also,
59:50:520Paolo Guiotto: similar. You can repeat again the argument. Here, we assume that the point was a minimum. Do the proof when the point is a maximum. So, in such a way, you have… you will, you will enter better into this,
00:06:990Paolo Guiotto: Okay, now, it's the moment to… apply this.
00:13:730Paolo Guiotto: Okay, so, let's start,
00:19:130Paolo Guiotto: showing how to use this. So, the main…
00:28:300Paolo Guiotto: Application… Jeez.
00:32:610Paolo Guiotto: now, to… the… Search.
00:41:580Paolo Guiotto: Off.
00:42:830Paolo Guiotto: mean… remarks… points.
00:48:250Paolo Guiotto: of F on sum.
00:52:420Paolo Guiotto: domain.
00:56:250Paolo Guiotto: Well, I repeat, it is not clear yet how to use directly this tool, because we have seen there are many exceptions. This is saying…
01:08:210Paolo Guiotto: If you know that you have a mean-max point, if you know that you have.
01:14:140Paolo Guiotto: So if you don't know that you have, it's a problem. So suppose that you know that you have a new mapped product. And this point is in the interior, then it must verify the equation gradient F equals 0.
01:28:80Paolo Guiotto: Okay? But, if it is not in the interior, maybe the gradient is not zero.
01:34:760Paolo Guiotto: So, the equation gradient F equals 0 missed points.
01:40:300Paolo Guiotto: Okay? Second, we know that this equation could take points that are not even minimum and maximum. Now, this was the second example.
01:49:840Paolo Guiotto: So, how can we use, this,
01:53:700Paolo Guiotto: very, apparently, partial result to find… to solve an optimization problem. So let's see how this works on an example. This is the example
02:04:670Paolo Guiotto: I will take the example in 237, which is classified with one star. This means a very simple calculation, so we don't have to focus on calculation, but you have to focus on method. So here we have a function F, XY,
02:21:230Paolo Guiotto: defined in this way. It is X plus Y minus X, Why?
02:27:620Paolo Guiotto: D domain D is this, the set of points XY, Such that,
02:34:980Paolo Guiotto: There are conditions X greater or equal than zero.
02:39:180Paolo Guiotto: Y greater or equal than 0 and less or equal than 3 minus X.
02:45:680Paolo Guiotto: And the problem is determined Determine the minimum
02:54:660Paolo Guiotto: And the maximum, of course, if any, on the… of this function n.
03:02:120Paolo Guiotto: Well, since this is the first time we attack a problem of this nature.
03:07:850Paolo Guiotto: let's use whatever we can use. For example, let's first have an idea of what is the domain, which is… which means we draw the domain, even if it is not necessarily needed, okay, for the solution.
03:20:960Paolo Guiotto: What is the domain? The domain is in the plane XY.
03:28:820Paolo Guiotto: And, it is made of points. XY belongs to D,
03:35:340Paolo Guiotto: If and only if we have… do you see the comma means…
03:40:300Paolo Guiotto: that the first condition, together with the second, etc, means an intersection, okay? So, X must be greater or equal than zero, and
03:50:480Paolo Guiotto: If you want, we can split this into two conditions, no? Y greater or equal than zero, and Y less or equal than 3 minus X. Let's see each of them, what, what consequences are.
04:04:330Paolo Guiotto: The first says X must be greater or equal than zero.
04:08:340Paolo Guiotto: It is always a condition on points XY. So, points X, Y, whose X is positive. So, these are points which are on the right of the Y axis. So, if you want, we discard these points.
04:23:720Paolo Guiotto: I color in black points that are thrown away. Here is X is negative, okay? So X equals 0 would be the y-axis.
04:33:480Paolo Guiotto: X positive would be at the right of the Y axis.
04:37:630Paolo Guiotto: Then, the second one says Y greater than 0, which is similar. Now, we are above the x-axis, so we discard also this part down here.
04:47:230Paolo Guiotto: The third one says Y less or equal than 3 minus X.
04:51:650Paolo Guiotto: So Y equal to 3 minus X is a line, and it is contained, so, let's,
05:00:550Paolo Guiotto: I will give a color at the end to the domain.
05:04:160Paolo Guiotto: For the mom, let's use blue.
05:06:230Paolo Guiotto: So this is a line Y equal to 3 minus X is a straight line.
05:12:360Paolo Guiotto: No? With the negative slope, at X equals 0, we are with y equals 3. At X equals 3, Y is 0.
05:22:210Paolo Guiotto: So the line passes through these two points. It's an infinite line, straight line.
05:29:150Paolo Guiotto: These points are points where Y is equal, so they are in the domain, but because for the domain, we have less or equal.
05:37:620Paolo Guiotto: But for less or equal, it means that Y must be below 3 minus X, so the line… what is below that line? So it means that we discard what is above, so the black points are all here.
05:52:850Paolo Guiotto: Now, I finished the… so what remains is the domain. So let me be clear, because there are the equal, we have to take this part of the segment here, where Y is equal to 3 minus X.
06:06:390Paolo Guiotto: So, also, Y equals 0 is this, and X equals 0 is this. The part which is inside, plus, of course, whatever is
06:16:620Paolo Guiotto: inside this triangle. So that red thing is the domain D.
06:21:750Paolo Guiotto: Okay? So now we have an idea on where we are.
06:26:270Paolo Guiotto: Optimizing this function.
06:29:10Paolo Guiotto: Okay, now…
06:31:560Paolo Guiotto: Let's now pass to the problem of determining these points. I return once again here, because, as you can see, the… let's say, the…
06:42:150Paolo Guiotto: Logical statement is not easy to be understood, because it says if you have a mean-max point, which is in the interior, so you must know that this is this point.
06:53:730Paolo Guiotto: then gradient F equals 0.
06:56:260Paolo Guiotto: So, if I start saying, let's look for points where greater than 30 plus 0, I should risk. First, there are no points, I miss minimum, maximum points.
07:06:20Paolo Guiotto: Because, if they are not in the interior, I don't see through that equation. Second, I don't know if there are points that will verify that equation, even if they are minimum-maximum points. So, how can I use this?
07:20:980Paolo Guiotto: Now, the idea is that, if you notice, this function is a nice function.
07:25:780Paolo Guiotto: at least continuous.
07:27:920Paolo Guiotto: So… we can say that F…
07:32:230Paolo Guiotto: is clearly a continuous function on the full plane R2 is a polynomial, and in particular, it will be continuous on the domain D.
07:43:200Paolo Guiotto: This domain D… But this figure suggests that this domain is bounded.
07:49:600Paolo Guiotto: And since it is defined by weak inequalities, should be also closed.
07:54:850Paolo Guiotto: So… D.
07:57:30Paolo Guiotto: is closed.
08:01:960Paolo Guiotto: Because…
08:06:270Paolo Guiotto: defiled.
08:09:230Paolo Guiotto: Bye.
08:10:690Paolo Guiotto: Large.
08:13:230Paolo Guiotto: inequalities, involving…
08:20:450Paolo Guiotto: continuous.
08:22:279Paolo Guiotto: functions.
08:25:200Paolo Guiotto: No? This is evident.
08:28:390Paolo Guiotto: and bounded.
08:34:29Paolo Guiotto: What?
08:34:920Paolo Guiotto: Since we have done the figure, from the figure, unless it is wrong, of course, we understand that X is between 0 and at maximum 3, and also for Y, we have this bound.
08:49:240Paolo Guiotto: No?
08:50:380Paolo Guiotto: Otherwise, we should prove the boundary if we don't have the feed, you know?
08:54:850Paolo Guiotto: Okay. So, in particular, this means that D is compact.
09:03:160Paolo Guiotto: And this means that we have a continuous function and a compact set, the minimum and the maximum both exist, and that's because of Feisser's theorem.
09:13:210Paolo Guiotto: So, there exists… Both.
09:16:960Paolo Guiotto: Mean?
09:19:160Paolo Guiotto: and maximum.
09:22:380Paolo Guiotto: 4.
09:23:810Paolo Guiotto: S.
09:25:69Paolo Guiotto: on the… Now we know they exist.
09:29:100Paolo Guiotto: So, let's say that, the, the premise of the theorem, suppose that you have a point, it's now, true, no? You have that point, you have a minimum, you have a maximum, because of the Beister's theorem, so you know that they are.
09:46:350Paolo Guiotto: And therefore, now you determine to determine… Minimum.
09:55:30Paolo Guiotto: and maximum point… points.
09:59:330Paolo Guiotto: We… Use… Verma. PRM?
10:07:920Paolo Guiotto: In the following way.
10:17:770Paolo Guiotto: Now, I can say, let's take a minimum or a maximum.
10:22:410Paolo Guiotto: Can I say that it is gradient F at that point equal to zero?
10:27:80Paolo Guiotto: Not necessarily. Depends. If the point is in the interior, yes. Otherwise, no. So I can say this, that we have an alternative.
10:35:500Paolo Guiotto: If, XY is a minimum.
10:41:630Paolo Guiotto: Or a maximum point.
10:43:980Paolo Guiotto: We can have only two possibilities. Dan, The first is either…
10:52:980Paolo Guiotto: The point, XY, belongs to the interior of D.
10:58:410Paolo Guiotto: In this figure is the triangle without the edge.
11:04:10Paolo Guiotto: So what is inside this triangle?
11:07:200Paolo Guiotto: And in this case, Fermat theorem applies, and if it is in the interior, then gradient F must be equal to zero.
11:19:330Paolo Guiotto: So, that point is one of the stationary points of F, if any, in the interior of the
11:27:800Paolo Guiotto: So, since, well… Here, I should say, to be precise, Provided.
11:36:600Paolo Guiotto: Because Fermat theorem asks that the function be differentiable,
11:41:60Paolo Guiotto: Provided F is differentiable.
11:45:250Paolo Guiotto: on P.
11:46:700Paolo Guiotto: But we will easily, check this.
11:51:650Paolo Guiotto: So, let's continue this, language, the second, alternative.
11:56:880Paolo Guiotto: Now, we have that the derivatives with respect to X and respect to Y are F is a…
12:06:290Paolo Guiotto: X plus Y minus XY. The derivative with respect to X is 1 minus Y, right?
12:13:370Paolo Guiotto: So, 1 minus 1.
12:15:760Paolo Guiotto: Derivative with respect to X with respect to Y is 1 minus X.
12:21:930Paolo Guiotto: 1 minus X. So, as you can see, they are both continuous functions.
12:26:910Paolo Guiotto: on the full plan R2, so in particular, only
12:31:320Paolo Guiotto: So this means that F is… differentiable.
12:36:280Paolo Guiotto: on the… so the technical requirements to apply the firm up here is okay.
12:43:640Paolo Guiotto: Banged.
12:44:840Paolo Guiotto: The point… the gradient of F at point XY is 0, so point XY is a stationary point. If and of if, now we have the system, very simple system, 1 minus Y equals 0, 1 minus X equals 0, from which we get them.
13:03:890Paolo Guiotto: X equals 1, Y equals 1. So we found one point.
13:12:90Paolo Guiotto: So the conclusion is that, that's chorally narrow, then XY Must be a point.
13:19:840Paolo Guiotto: 1-1.
13:21:580Paolo Guiotto: So if you forget the… the passages, this says either this min-max point is in the interior, and then it is 1-1,
13:32:750Paolo Guiotto: Okay, now let's see if this is in the individual, because maybe it's not even in the domain. So where is it at this point? 11 is about here, so 1…
13:43:50Paolo Guiotto: Because the absence and the ordinates range from 0 to 3, so it is about here. So it is, in fact, in the domain, so it must be considered.
13:51:460Paolo Guiotto: So that's the point 1-1.
13:55:600Paolo Guiotto: Okay, so either it is in the interior, and then it is 1-1, or…
14:02:600Paolo Guiotto: It is not in the interior.
14:05:850Paolo Guiotto: XY… belongs to D, but not in the interior of D.
14:13:00Paolo Guiotto: So, you know what these,
14:14:990Paolo Guiotto: the points of D which are not in the interior of D,
14:18:590Paolo Guiotto: It is the boundary of me.
14:21:440Paolo Guiotto: So, it must be… the search now must be on this part here.
14:28:260Paolo Guiotto: This is the boundary of the…
14:31:600Paolo Guiotto: And for those points, you cannot say gradient F equals 0, because gradient F is equals 0 only when you are inside.
14:38:860Paolo Guiotto: So, the gradient at this stage is something useless.
14:43:850Paolo Guiotto: And so, in this… in this case, at this… at this point of the… of our course, we don't have any special tool, we have to do by hand, how we say, no?
14:54:500Paolo Guiotto: or XY belongs to boundary of D. So let's write what is the boundary of D in this case. Boundary of D, I now just draw the boundary, so you will see only
15:05:340Paolo Guiotto: Visa.
15:08:410Paolo Guiotto: So that in red is the boundary of the… not the inside.
15:13:330Paolo Guiotto: So the boundary of D, as you can see, is made of 3 segments, no? Segment 1, 2, and 3. Let's give names to these three. Let's say set 1, set 2, and set 3.
15:30:910Paolo Guiotto: Let's try to describe these sets. So this is made of set 1, union, set 2, union, set 3.
15:39:700Paolo Guiotto: Where?
15:41:280Paolo Guiotto: Wow.
15:42:430Paolo Guiotto: How is made at Set 1?
15:45:30Paolo Guiotto: Set 1 is made of points here, what shape they have.
15:51:70Paolo Guiotto: They are points X0,
15:56:610Paolo Guiotto: with X between 0 and… 3. Exactly, no? That's 3, that's C. No?
16:03:780Paolo Guiotto: S2 is similar, these are points 0Y with Y.
16:13:310Paolo Guiotto: And now, what about S3? So, S2 is here, and what about the point which is here, S3?
16:25:460Paolo Guiotto: be mindless.
16:26:280Paolo Guiotto: Yeah, you remind that this is part of the line Y equals 3 minus X. So DY is not arbitrary. If you know X, you will know Y, and you could all do the vice versa. So we can say, they are point with abshesa X, and if X is the abshesa, the Y is this one.
16:45:320Paolo Guiotto: 3 minus X, so 3 minus X.
16:48:600Paolo Guiotto: Now, we have to identify the range for X.
16:52:950Paolo Guiotto: You can see the figure, and you can imagine what is the range.
17:00:770Paolo Guiotto: X is the abscess, so you vary from this to this, so speed from 0 to 3.
17:07:470Paolo Guiotto: Now, we have these three sets.
17:10:30Paolo Guiotto: Okay, so we said either XY is in the interior, then it is necessarily 0.11, or it is on the boundary
17:19:180Paolo Guiotto: So, it must be in one of these three segments, okay? So, it must be either in S1, or in S2, or in S3.
17:27:400Paolo Guiotto: So let's say this.
17:29:90Paolo Guiotto: So… XY… belongs to S1, or XY belongs to S2, Or, XY belongs to S3.
17:45:00Paolo Guiotto: In this case, The XY is made of type X0, with X between 0 and 3.
17:54:410Paolo Guiotto: In this case, we have that it is a .0Y with Y.
17:59:520Paolo Guiotto: A, between 0 and 3.
18:04:40Paolo Guiotto: And in this case, it is a point X3 minus X, with X between 0 and 3.
18:12:550Paolo Guiotto: As you can see, there are lots of points here, no, because they are segments.
18:16:510Paolo Guiotto: So it is one of them, but which one? How can I determine which one of them is
18:22:640Paolo Guiotto: My minimum, maximum point.
18:24:760Paolo Guiotto: Well, in this case, I can say, okay, let's evaluate the function on each point, and let's see what happens. If I evaluate F on a pointer X0,
18:36:880Paolo Guiotto: For example, reminded function was X plus Y minus X times Y. So Y is 0. It is X,
18:45:160Paolo Guiotto: plus Y, 0, minus X times Y. Y is 0, so X times Y is 0. So it is X. You see?
18:52:910Paolo Guiotto: Remind X plus Y minus XY.
18:57:750Paolo Guiotto: So, if the function is X, when X is from 0 to 3, where do you have minimum, where do you have maximum?
19:07:920Paolo Guiotto: Yeah.
19:09:250Paolo Guiotto: So, here I get that my point can be only, for the minimum.
19:15:440Paolo Guiotto: 0.00 for the maximum, 0.30. Cannot be 0.10, because 1.0 is not even a minimum, and not even a maximum for F on that part of domain. So, how can be a minimum maximum for the full domain, you see?
19:31:660Paolo Guiotto: So… In this case, we can say that 0.x0 Ease.
19:40:00Paolo Guiotto: 0, 0.
19:42:00Paolo Guiotto: for MIN, And, 3-0, 4… Max.
19:49:690Paolo Guiotto: Now, let's do the same on the other two cases.
19:52:660Paolo Guiotto: So, here.
19:54:720Paolo Guiotto: Well, let's do down here. F of 0Y, well, it's exactly the same, because the function is identical in 0 and Y. It is Y.
20:05:160Paolo Guiotto: So, when this quantity is minimum, when this quantity is maximum.
20:09:620Paolo Guiotto: Again, I, I say that, it is zero-zero.
20:14:560Paolo Guiotto: for minimum, And 03 for maximum.
20:22:730Paolo Guiotto: Finally, we have the last type of points, which are a little bit more complicated.
20:27:540Paolo Guiotto: So, F of X3 minus X, in this case, we have… the function is X plus Y, so X plus 3 minus X,
20:37:460Paolo Guiotto: minus X times Y. X times 3 minus X.
20:42:130Paolo Guiotto: So, you see that we get 3, these 2 disappear, and then a plus X times X minus 3.
20:51:750Paolo Guiotto: So, this is how the function is when we evaluate the function along points of that diagonal segment.
21:00:470Paolo Guiotto: Now, here I have to understand when this quantity is minimum, when this quantity is maximum.
21:09:400Paolo Guiotto: let's… here, there is a simple thing, because this is a parabola, no? This… this part here, as a function of X,
21:19:400Paolo Guiotto: for X going from 0 to 3 is a parabola that has value 0 here and here, and it is symmetric, so the midpoint will be the maximum. But let's see a more general method. Let's call this function, I don't know, phi of X.
21:36:440Paolo Guiotto: So, if you want to determine the maximum of this function, you do what?
21:40:850Paolo Guiotto: You can see that as a one variable function, you compute the derivative, you have a phi prime of x equal, so the derivative of 3 is 0, then you have x squared, which is 2X minus 3x minus 3.
21:53:860Paolo Guiotto: Okay? So you say that this is greater or equal than zero, so function increasing. If and only if X, is greater…
22:05:730Paolo Guiotto: There's something wrong.
22:17:750Paolo Guiotto: Yes, because the parabola is not this one. The parabola is the other way.
22:24:500Paolo Guiotto: So, the midpoint will be… but however, let's assume that we have not. So, X greater or equal than 3 halves.
22:33:30Paolo Guiotto: So, the function is increasing after 3AF, decreasing before. So, we have a minimum at 3AF, So, fee.
22:43:500Paolo Guiotto: Well, Vax.
22:44:860Paolo Guiotto: Jeez.
22:46:90Paolo Guiotto: minimum, At X equals 3 half.
22:52:110Paolo Guiotto: And the maximum is at x equals 03.
22:56:280Paolo Guiotto: P of X.
22:58:160Paolo Guiotto: is maximum.
23:01:720Paolo Guiotto: X, X equals 0 and 3.
23:05:600Paolo Guiotto: What is phi of X? Phi of X is F at X3 minus X, so it is F at this point on the segment, S3.
23:16:530Paolo Guiotto: So, this says that the function f is a minimum.
23:21:990Paolo Guiotto: at point… at value x equals 3 half, so the point is the point with abs is at 3 half, ordinate 3 minus 3 half, and this is the point 3 half 3 half.
23:39:590Paolo Guiotto: While for X equals 0, I get points.
23:43:660Paolo Guiotto: 03.
23:46:660Paolo Guiotto: And for X equals 3, I get 0.30.
23:51:60Paolo Guiotto: Okay, now we have to draw the conclusion, because we have not yet determined what is the minimum-maximum point. Let's give a look to the argument.
24:02:350Paolo Guiotto: So we say, We know that the minimum-maximum point exists.
24:07:180Paolo Guiotto: If it is in the interior, it must be 1-1. If it is not in the interior, it must be in one of these three pieces of the bundle.
24:16:30Paolo Guiotto: If it is on the first
24:17:930Paolo Guiotto: bound… piece of the boundary, what we call this S1.
24:22:290Paolo Guiotto: Then…
24:23:810Paolo Guiotto: It must be 00 if it is a minimum, or it must be 3-0 if it is a maximum.
24:30:280Paolo Guiotto: On the second… if it is on the second part of the boundary as 2, then we did use that it is 00 for minimum, or 03 for maximum.
24:41:780Paolo Guiotto: If it is on the third component of the boundary, S3, Then we discovered that
24:47:840Paolo Guiotto: It is a minimum at 0.353.5. It is maximum at 0330.
24:54:300Paolo Guiotto: So now we conclude.
24:57:400Paolo Guiotto: Conclusion.
25:00:130Paolo Guiotto: Because we have not… are we able now to respond to the question, which one is the minimum point, which one is the maximum point? Not yet.
25:09:220Paolo Guiotto: So, we can say that.
25:11:480Paolo Guiotto: if XY Is, let's start with the minimums.
25:17:460Paolo Guiotto: Mean point… Then, what are the possible candidates with attended with this armor?
25:26:510Paolo Guiotto: The possible candidates are…
25:29:460Paolo Guiotto: Since if the point is in the interior, it must be a stationary point, one candidate is 0.11, because if it is in the interior, it must be 11. So it means that 11 must be considered as a potential minimum, okay? We don't know yet. Then.
25:49:880Paolo Guiotto: moss.
25:53:780Paolo Guiotto: B… One… off.
25:57:860Paolo Guiotto: 0.11, which is the stationary one. Then.
26:01:860Paolo Guiotto: We don't know where is it, but if it is on S… One.
26:07:700Paolo Guiotto: The minimum is this one, is 0.00.
26:13:590Paolo Guiotto: If it is on S2, it is again 00. If it is on S3, it is this one. So what are the candidates that we gain are 00,
26:25:800Paolo Guiotto: 0, 0, 2 times, so again, 0, 0, and 3 halves, 3 halves.
26:32:730Paolo Guiotto: Now, how can I decide? Well, first of all, I want you to notice that the search has restricted to 3 points.
26:40:700Paolo Guiotto: Okay? So now I know that minimum is one of these three guys.
26:45:790Paolo Guiotto: So at this point, since they are only 3, I just need to compute F.
26:50:330Paolo Guiotto: F at 1, 1 is, remind that it is X plus Y minus X times Y. So 1 plus 1, 2,
26:58:740Paolo Guiotto: Minus 1 times 1, so it is equal to 2 minus 1, 1.
27:04:180Paolo Guiotto: At 00F, it is equal to 0.
27:07:460Paolo Guiotto: At 3 half, 3 half, it is equal to 3 half plus 3 half.
27:11:930Paolo Guiotto: Minus 3 halves times 3 halves.
27:16:60Paolo Guiotto: So, it is, 3F plus 3F is 3.
27:21:380Paolo Guiotto: Minus 3 minus 9 fourths, whatever it is,
27:26:880Paolo Guiotto: It is, 9 of, no.
27:29:920Paolo Guiotto: 12, right?
27:32:840Paolo Guiotto: 12 minus 9, 3… forth, right?
27:36:940Paolo Guiotto: So, the minimum… the values taken by F at these three points are these ones.
27:44:40Paolo Guiotto: Now you can see which one is a minimum.
27:46:770Paolo Guiotto: Because 1, 0, 3, 4th, this one is the minimum point.
27:52:130Paolo Guiotto: Okay?
27:55:160Paolo Guiotto: And for the maximum, we do a similar argument.
27:59:110Paolo Guiotto: If XY Ease.
28:04:940Paolo Guiotto: Max Boyd.
28:08:330Paolo Guiotto: Ben…
28:12:540Paolo Guiotto: Moss.
28:14:260Paolo Guiotto: B… one off.
28:18:260Paolo Guiotto: well, one can be two, three, I don't know, it's not necessarily unique. One off.
28:24:450Paolo Guiotto: Now, what are the candidates? We say, if it is a minimum, and the same for the maximum, we have these possibilities.
28:32:870Paolo Guiotto: If it is in the interior, it is 1-1. So 1-1 is now a possibility for the maximum.
28:42:910Paolo Guiotto: Then, if it is not in the interior, it is on the boundary. On the boundary, it can be in the first part, we call this one, and then we have 0.30.
28:54:290Paolo Guiotto: can be in the second part, then we have 0.03, or it can be in the third part, and the third part, we have again these two points, 03 and 3 zero. So, it means that the alternatives are 1103, and 3-0.
29:10:850Paolo Guiotto: So, you now see what is… even if this, this, this technique of searching for minimum-maximum is quite different from
29:19:350Paolo Guiotto: solving the problem in one variable, no? Where you have a nice way to do, you study sine, you see, well, function is decreasing, increasing, etc. Here, you don't do that, but you restrict the search to a few number of points.
29:33:00Paolo Guiotto: And hopefully, if there are not so many, you can evaluate and do a direct
29:38:330Paolo Guiotto: comparison. So we already know that this has value 1. At 03, we have X plus Y minus X times Y is 3, and also this is 3. As you can see, these two
29:51:00Paolo Guiotto: are maximum values.
29:53:380Paolo Guiotto: So this means that these two guys are maximum points.
29:57:200Paolo Guiotto: So these are… Max.
30:00:420Paolo Guiotto: points.
30:03:270Paolo Guiotto: And you send the exercise.
30:06:40Paolo Guiotto: So, as you can see, none of the points is in the interior in this problem.
30:11:120Paolo Guiotto: So we have data. If you look at the figure, you'll remind the main was this kind of triangles.
30:19:100Paolo Guiotto: with the insider.
30:21:300Paolo Guiotto: We have seen that the,
30:24:390Paolo Guiotto: The max points are this one and this one. These are… the max… points.
30:32:610Paolo Guiotto: While the minimum point is 0, so it is this one.
30:42:270Paolo Guiotto: So actually, in this problem, the condition gradient F equals zero
30:47:870Paolo Guiotto: Does not produce any minimum-maximum bond.
30:52:60Paolo Guiotto: It can happen.
30:55:90Paolo Guiotto: Okay?
30:56:800Paolo Guiotto: So now, we have to do practice on this kind of,
31:02:500Paolo Guiotto: on these kind of problems, let me…
31:06:640Paolo Guiotto: leave you with some exercise. Maybe, yeah, since we have a few minutes, we can… do something.
31:15:100Paolo Guiotto: So the exercises to do would be…
31:22:920Paolo Guiotto: Do.
31:25:960Paolo Guiotto: Exercise, well, maybe we start with the… with this 298.
31:33:120Paolo Guiotto: 9, 10…
31:38:380Paolo Guiotto: Well, the 11, these are optimization on R2 we will see, maybe on Monday, so…
31:46:100Paolo Guiotto: Actually, 2910 is made of several functions, so let's start the 298. Maybe we start, you continue at home along the weekend, and then you come back with the solution, and let's see.
32:01:690Paolo Guiotto: If you… what you… what you have done, huh? So we have here FXY equal X squared times 1 minus Y,
32:10:850Paolo Guiotto: on D, which is this set, set of points XY in R2, such that X squared plus modulus of Y is less or equal than 4.
32:28:910Paolo Guiotto: Well, there is a question, one, determine the sine of F, number two.
32:37:880Paolo Guiotto: stationary points… of F… On the… And number 3, Mean Max.
32:51:280Paolo Guiotto: of F only.
32:54:560Paolo Guiotto: Maybe, let's put number 0, draw the…
33:00:830Paolo Guiotto: So we start with the solution, with the first question, Drohi.
33:07:650Paolo Guiotto: Which is not difficult in this case. So we have that point XY belongs to domain D. You see that there is a unique condition. It is X squared plus absolute value of Y less or equal than 4.
33:22:880Paolo Guiotto: how can I, plot this?
33:26:100Paolo Guiotto: For example, I could say modulus of Y less or equal than 4 minus X squared.
33:33:550Paolo Guiotto: Which means, Y… Below 4 minus X squared.
33:39:820Paolo Guiotto: and above, minus this, no? It is better if you write in this form, because maybe
33:46:670Paolo Guiotto: instead of writing X squared minus 4, which is the same.
33:52:60Paolo Guiotto: Because this means what? I am in plane XY.
33:59:980Paolo Guiotto: Let's look at this. Why less or equal than 4 minus X squared? Why equal 4 minus X squared means a parabola.
34:08:180Paolo Guiotto: minus X squared is, pointing downward, plus 4. It means that for X equals 0, it crosses the y-axis at,
34:16:740Paolo Guiotto: 0.4.
34:18:800Paolo Guiotto: It is equal to 0 for X equal plus minus 2, so let's say minus 2 plus 2.
34:24:570Paolo Guiotto: And it is a parabola pointing downward like that.
34:32:510Paolo Guiotto: Okay? So this is where Y is equal. I want Y less or equal, so it is below, I have to discard everything which is above the parabola. So the black region here is discard.
34:47:330Paolo Guiotto: Now, I have the second inequality, which is why above
34:52:140Paolo Guiotto: That's why I prefer to write for this part with the minus like that, because it is the opposite, no? So it is specular, it is like this, so this is minus 4. You have another parabola, which is just that one, reflected with respect to the
35:09:70Paolo Guiotto: x-axis, and I have to take points which are on the parabola or above the parabola, so I discard these points out here.
35:18:800Paolo Guiotto: So, as you can see, We have that region between these two parabolas, so that is the domain.
35:25:690Paolo Guiotto: Including the two parabolas, the two pieces of parabolas.
35:33:520Paolo Guiotto: So the domain is all this.
35:38:710Paolo Guiotto: Okay, so that's the… Okay, question one was determine the sign of the function.
35:50:20Paolo Guiotto: It's just a curiosity here, maybe.
35:54:10Paolo Guiotto: But it is also to make practice in solving equations, inequalities with several variables. When I have to write FXY greater or equal than zero, this means what?
36:07:380Paolo Guiotto: the function is X squared times 1 minus y greater or equal than 0. Since x squared is always greater or equal than zero.
36:17:150Paolo Guiotto: This means 1 minus Y greater or equal than 0.
36:20:750Paolo Guiotto: Which is why less or equal than 1.
36:25:240Paolo Guiotto: Y equal 1, what does it mean? We are in plane, so this is still a condition for X and Y, no?
36:32:60Paolo Guiotto: So, now let's, draw only the domain.
36:38:140Paolo Guiotto: So we have the… With two parabolas, like that.
36:51:720Paolo Guiotto: So, this is, 4 minus 4, this is 2 minus 2. Domain is,
37:00:00Paolo Guiotto: Maybe this is too strong.
37:02:260Paolo Guiotto: Let's take this one.
37:04:220Paolo Guiotto: Domain is what is inside these two parabolas.
37:08:510Paolo Guiotto: And Y less or equal 1 is what? Y equal 1 is an horizontal line.
37:14:670Paolo Guiotto: is this one. Y equal 1.
37:18:600Paolo Guiotto: Why less or equal means below, so I have this part here.
37:29:990Paolo Guiotto: This part in blue is… Y less or equal 1, including this segment here.
37:39:560Paolo Guiotto: So…
37:40:740Paolo Guiotto: Sine of F. F is positive if and only if I am in the blue region. So, F positive here.
37:50:670Paolo Guiotto: And, of course, by difference, this will be F-negative.
37:55:650Paolo Guiotto: Okay?
37:58:480Paolo Guiotto: Okay, well, let's do just the stationary points, so I leave you with this. Gradient F equals 0.
38:06:130Paolo Guiotto: Because it is, A little bit faster.
38:09:770Paolo Guiotto: If and only if…
38:11:330Paolo Guiotto: F is X squared times 1 minus Y. We have here. Derivative with respect to X is 2X.
38:19:300Paolo Guiotto: times 1 minus Y.
38:21:870Paolo Guiotto: derivative respect to Y is… X squared times minus 1, so minus X squared.
38:31:550Paolo Guiotto: Now, this is equal to 0, factor 0, if and only if we have… the second line is minus X squared equals 0, the first line is 2x times 1 minus y equals 0.
38:45:480Paolo Guiotto: As you can see, the second becomes just X equals 0.
38:50:550Paolo Guiotto: that when you plug in, the first line gives 0 equals zero, something always true, we can discard. And so X equals 0 means…
39:01:220Paolo Guiotto: 0.0Y.
39:03:550Paolo Guiotto: Since we are looking for stationary points in DE, So, these points are where
39:11:660Paolo Guiotto: 0Y they are on the Y axis, so they are here.
39:17:600Paolo Guiotto: These are .0Y.
39:19:960Paolo Guiotto: If you want those who are in D, you have to take Y between minus 4 and 4, okay?
39:27:410Paolo Guiotto: So 0Y with Y between minus 4 and 4.
39:33:940Paolo Guiotto: to have… the .0Y in the… Okay?
39:42:650Paolo Guiotto: Now, we have done the most difficult part of the exercise.
39:48:260Paolo Guiotto: You have to do the last part, okay? Now, determine the minimum, maximum points of F on D. The argument is analogous to that one we have seen. Of course, you have to adapt to this case, so it's not a straightforward, repetition of the previous argument. We will complete the solution Monday morning.
40:08:60Paolo Guiotto: Okay, have a nice weekend.