1 00:00:01 --> 00:00:03 The following content is provided under a Creative 2 00:00:03 --> 00:00:05 Commons license. Your support will help MIT 3 00:00:05 --> 00:00:08 OpenCourseWare continue to offer high quality educational 4 00:00:08 --> 00:00:13 resources for free. To make a donation or to view 5 00:00:13 --> 00:00:18 additional materials from hundreds of MIT courses, 6 00:00:18 --> 00:00:23 visit MIT OpenCourseWare at ocw.mit.edu. 7 00:00:23 --> 00:00:28 So, today we are going to continue looking at critical 8 00:00:28 --> 00:00:31 points, and we'll learn how to actually 9 00:00:31 --> 00:00:33 decide whether a typical point is a minimum, 10 00:00:33 --> 00:00:37 maximum, or a saddle point. So, that's the main topic for 11 00:00:37 --> 00:00:41 today. So, remember yesterday, 12 00:00:41 --> 00:00:50 we looked at critical points of functions of several variables. 13 00:00:50 --> 00:00:58 And, so a critical point functions, we have two values, 14 00:00:58 --> 00:01:05 x and y. That's a point where the 15 00:01:05 --> 00:01:11 partial derivatives are both zero. 16 00:01:11 --> 00:01:15 And, we've seen that there's various kinds of critical 17 00:01:15 --> 00:01:20 points. There's local minima. 18 00:01:20 --> 00:01:24 So, maybe I should show the function on this contour 19 00:01:24 --> 00:01:28 plot,there is local maxima, which are like that. 20 00:01:28 --> 00:01:35 And, there's saddle points which are neither minima nor 21 00:01:35 --> 00:01:37 maxima. And, of course, 22 00:01:37 --> 00:01:41 if you have a real function, then it would be more 23 00:01:41 --> 00:01:45 complicated. It will have several critical 24 00:01:45 --> 00:01:48 points. So, this example here, 25 00:01:48 --> 00:01:54 well, you see on the plot that there is two maxima. 26 00:01:54 --> 00:01:58 And, there is in the middle, between them, 27 00:01:58 --> 00:02:00 a saddle point. And, actually, 28 00:02:00 --> 00:02:02 you can see them on the contour plot. 29 00:02:02 --> 00:02:07 On the contour plot, you see the maxima because the 30 00:02:07 --> 00:02:12 level curves become circles that now down and shrink to the 31 00:02:12 --> 00:02:15 maximum. And, you can see the saddle 32 00:02:15 --> 00:02:18 point because here you have this level curve that makes a figure 33 00:02:18 --> 00:02:20 eight. It crosses itself. 34 00:02:20 --> 00:02:25 And, if you move up or down here, so along the y direction, 35 00:02:25 --> 00:02:28 the values of the function will decrease. 36 00:02:28 --> 00:02:32 Along the x direction, the values will increase. 37 00:02:32 --> 00:02:37 So, you can see usually quite easily where are the critical 38 00:02:37 --> 00:02:42 points just by looking either at the graph or at the contour 39 00:02:42 --> 00:02:44 plots. So, the only thing with the 40 00:02:44 --> 00:02:47 contour plots is you need to read the values to tell a 41 00:02:47 --> 00:02:51 minimum from a maximum because the contour plots look the same. 42 00:02:51 --> 00:02:53 Just, of course, in one case, 43 00:02:53 --> 00:02:56 the values increase, and in another one they 44 00:02:56 --> 00:03:03 decrease. So, the question -- -- is, 45 00:03:03 --> 00:03:17 how do we decide -- -- between the various possibilities? 46 00:03:17 --> 00:03:23 So, local minimum, local maximum, 47 00:03:23 --> 00:03:26 or saddle point. 48 00:03:26 --> 00:03:38 49 00:03:38 --> 00:03:44 So, and, in fact, why do we care? 50 00:03:44 --> 00:03:55 Well, the other question is how do we find the global 51 00:03:55 --> 00:04:05 minimum/maximum of a function? So, here what I should point 52 00:04:05 --> 00:04:07 out, well, first of all, 53 00:04:07 --> 00:04:09 to decide where the function is the largest, 54 00:04:09 --> 00:04:12 in general you'll have actually to compare the values. 55 00:04:12 --> 00:04:14 For example, here, if you want to know, 56 00:04:14 --> 00:04:16 what is the maximum of this function? 57 00:04:16 --> 00:04:19 Well, we have two obvious candidates. 58 00:04:19 --> 00:04:22 We have this local maximum and that local maximum. 59 00:04:22 --> 00:04:24 And, the question is, which one is the higher of the 60 00:04:24 --> 00:04:26 two? Well, in this case, 61 00:04:26 --> 00:04:30 actually, there is actually a tie for maximum. 62 00:04:30 --> 00:04:32 But, in general, you would have to compute the 63 00:04:32 --> 00:04:34 function at both points, and compare the values if you 64 00:04:34 --> 00:04:36 know that it's three at one of them and four at the other. 65 00:04:36 --> 00:04:40 Well, four wins. The other thing that you see 66 00:04:40 --> 00:04:43 here is if you are looking for the minimum of this function, 67 00:04:43 --> 00:04:47 well, the minimum is not going to be at any of the critical 68 00:04:47 --> 00:04:49 points. So, where's the minimum? 69 00:04:49 --> 00:04:53 Well, it looks like the minimum is actually out there on the 70 00:04:53 --> 00:04:56 boundary or at infinity. So, that's another feature. 71 00:04:56 --> 00:04:59 The global minimum or maximum doesn't have to be at a critical 72 00:04:59 --> 00:05:01 point. It could also be, 73 00:05:01 --> 00:05:05 somehow, on the side in some limiting situation where one 74 00:05:05 --> 00:05:09 variable stops being in the allowed rang of values or goes 75 00:05:09 --> 00:05:13 to infinity. So, we have to actually check 76 00:05:13 --> 00:05:19 the boundary and the infinity behavior of our function to know 77 00:05:19 --> 00:05:23 where, actually, the minimum and maximum will 78 00:05:23 --> 00:05:27 be. So, in general, 79 00:05:27 --> 00:05:37 I should point out, these should occur either at 80 00:05:37 --> 00:05:48 the critical point or on the boundary or at infinity. 81 00:05:48 --> 00:05:52 So, by that, I mean on the boundary of a 82 00:05:52 --> 00:05:55 domain of definition that we are considering. 83 00:05:55 --> 00:06:00 And so, we have to try both. OK, but so we'll get back to 84 00:06:00 --> 00:06:04 that. For now, let's try to focus on 85 00:06:04 --> 00:06:09 the question of, you know, what's the type of 86 00:06:09 --> 00:06:16 the critical point? So, we'll use something that's 87 00:06:16 --> 00:06:21 known as the second derivative test. 88 00:06:21 --> 00:06:25 And, in principle, well, the idea is kind of 89 00:06:25 --> 00:06:29 similar to what you do with the function of one variable, 90 00:06:29 --> 00:06:32 namely, the function of one variable. 91 00:06:32 --> 00:06:34 If the derivative is zero, then you know that you should 92 00:06:34 --> 00:06:38 look at the second derivative. And, that will tell you whether 93 00:06:38 --> 00:06:41 it's curving up or down whether you have a local max and the 94 00:06:41 --> 00:06:44 local min. And, the main problem here is, 95 00:06:44 --> 00:06:46 of course, we have more possible situations, 96 00:06:46 --> 00:06:48 and we have several derivatives. 97 00:06:48 --> 00:06:52 So, we have to think a bit harder about how we'll decide. 98 00:06:52 --> 00:06:56 But, it will again involve the second derivative. 99 00:06:56 --> 00:07:01 OK, so let's start with just an easy example that will be useful 100 00:07:01 --> 00:07:06 to us because actually it will provide the basis for the 101 00:07:06 --> 00:07:10 general method. OK, so we are first going to 102 00:07:10 --> 00:07:15 consider a case where we have a function that's actually just 103 00:07:15 --> 00:07:20 quadratic. So, let's say I have a 104 00:07:20 --> 00:07:28 function, W of (x,y) that's of the form ax^2 bxy cy^2. 105 00:07:28 --> 00:07:32 OK, so this guy has a critical point at the origin because if 106 00:07:32 --> 00:07:36 you take the derivative with respect to x, 107 00:07:36 --> 00:07:38 well, and if you plug x equals y equals zero, 108 00:07:38 --> 00:07:42 you'll get zero, and same with respect to y. 109 00:07:42 --> 00:07:44 You can also see, if you try to do a linear 110 00:07:44 --> 00:07:47 approximation of this, well, all these guys are much 111 00:07:47 --> 00:07:50 smaller than x and y when x and y are small. 112 00:07:50 --> 00:07:55 So, the linear approximation, the tangent plane to the graph 113 00:07:55 --> 00:07:59 is really just w=0. OK, so, how do we do it? 114 00:07:59 --> 00:08:03 Well, yesterday we actually did an example. 115 00:08:03 --> 00:08:09 It was a bit more complicated than that, but let me do it, 116 00:08:09 --> 00:08:13 so remember, we were looking at something 117 00:08:13 --> 00:08:19 that started with x^2 2xy 3y^2. And, there were other terms. 118 00:08:19 --> 00:08:23 But, let's forget them now. And, what we did is we said, 119 00:08:23 --> 00:08:28 well, we can rewrite this as (x y)^2 2y^2. 120 00:08:28 --> 00:08:31 And now, this is a sum of two squares. 121 00:08:31 --> 00:08:35 So, each of these guys has to be nonnegative. 122 00:08:35 --> 00:08:40 And so, the origin will be a minimum. 123 00:08:40 --> 00:08:44 Well, it turns out we can do something similar in general no 124 00:08:44 --> 00:08:47 matter what the values of a, b, and c are. 125 00:08:47 --> 00:08:50 We'll just try to first complete things to a square. 126 00:08:50 --> 00:08:55 OK, so let's do that. So, in general, 127 00:08:55 --> 00:09:01 well, let me be slightly less general, and let me assume that 128 00:09:01 --> 00:09:08 a is not zero because otherwise I can't do what I'm going to do. 129 00:09:08 --> 00:09:20 So, I'm going to write this as a times x^2 plus b over axy. 130 00:09:20 --> 00:09:25 And then I have my cy^2. And now this looks like the 131 00:09:25 --> 00:09:28 beginning of the square of something, OK, 132 00:09:28 --> 00:09:31 just like what we did over there. 133 00:09:31 --> 00:09:39 So, what is it the square of? Well, you'd start with x plus I 134 00:09:39 --> 00:09:45 claim if I put b over 2a times y and I square it, 135 00:09:45 --> 00:09:52 then see the cross term two times x times b over 2a y will 136 00:09:52 --> 00:09:57 become b over axy. Of course, now I also get some 137 00:09:57 --> 00:10:01 y squares out of this. How many y squares do I get? 138 00:10:01 --> 00:10:05 Well, I get b^2 over 4a^2 times a. 139 00:10:05 --> 00:10:11 So, I get b2 over 4a y^2. So, and I want, 140 00:10:11 --> 00:10:17 in fact, c times y^2. So, the number of y^2 that I 141 00:10:17 --> 00:10:22 should add is c minus b^2 over 4a. 142 00:10:22 --> 00:10:27 OK, let's see that again. If I expand this thing, 143 00:10:27 --> 00:10:33 I will get ax^2 plus a times b over 2a times 2xy. 144 00:10:33 --> 00:10:39 That's going to be my bxy. But, I also get b^2 over 4a^2 145 00:10:39 --> 00:10:44 y^2 times a. That's b^2 over 4ay^2. 146 00:10:44 --> 00:10:47 And, that cancels out with this guy here. 147 00:10:47 --> 00:10:52 And then, I will be left with cy^2. 148 00:10:52 --> 00:10:58 OK, do you see it kind of? OK, if not, well, 149 00:10:58 --> 00:11:04 try expanding this square again. 150 00:11:04 --> 00:11:06 OK, maybe I'll do it just to convince you. 151 00:11:06 --> 00:11:11 But, so if I expand this, I will get A times, 152 00:11:11 --> 00:11:16 let me put that in a different color because you shouldn't 153 00:11:16 --> 00:11:19 write that down. It's just to convince you again. 154 00:11:19 --> 00:11:25 So, if you don't see it yet, let's expend this thing. 155 00:11:25 --> 00:11:35 We'll get a times x^2 plus a times 2xb over 2ay. 156 00:11:35 --> 00:11:42 Well, the two A's cancel out. We get bxy plus a times the 157 00:11:42 --> 00:11:53 square of that's going to be b^2 over 4a^2 y^2 plus cy^2 minus 158 00:11:53 --> 00:11:59 b^2 over 4ay^2. Here, the a and the a 159 00:11:59 --> 00:12:06 simplifies, and now these two terms simplify and give me just 160 00:12:06 --> 00:12:09 cy^2 in the end. OK, and that's kind of 161 00:12:09 --> 00:12:12 unreadable after I've canceled everything, 162 00:12:12 --> 00:12:19 but if you follow it, you see that basically I've 163 00:12:19 --> 00:12:24 just rewritten my initial function. 164 00:12:24 --> 00:12:29 OK, is that kind of OK? I mean, otherwise there's just 165 00:12:29 --> 00:12:32 no substitute. You'll have to do it yourself, 166 00:12:32 --> 00:12:38 I'm afraid. OK, so, let me continue to play 167 00:12:38 --> 00:12:43 with this. So, I'm just going to put this 168 00:12:43 --> 00:12:48 in a slightly different form just to clear the denominators. 169 00:12:48 --> 00:12:56 OK, so, I will instead write this as one over 4a times the 170 00:12:56 --> 00:13:03 big thing. So, I'm going to just put 4a^2 171 00:13:03 --> 00:13:10 times x plus b over 2ay squared. OK, so far I have the same 172 00:13:10 --> 00:13:13 thing as here. I just introduced the 4a that 173 00:13:13 --> 00:13:19 cancels out, plus for the other one, I'm just clearing the 174 00:13:19 --> 00:13:28 denominator. I end up with (4ac-b^2)y^2. 175 00:13:28 --> 00:13:32 OK, so that's a lot of terms. But, what does it look like? 176 00:13:32 --> 00:13:35 Well, it looks like, so we have some constant 177 00:13:35 --> 00:13:38 factors, and here we have a square, and here we have a 178 00:13:38 --> 00:13:39 square. So, basically, 179 00:13:39 --> 00:13:44 we've written this as a sum of two squares, well, 180 00:13:44 --> 00:13:47 a sum or a difference of two squares. 181 00:13:47 --> 00:13:51 And, maybe that's what we need to figure out to know what kind 182 00:13:51 --> 00:13:55 of point it is because, see, if you take a sum of two 183 00:13:55 --> 00:13:57 squares, that you will know that each 184 00:13:57 --> 00:14:01 square takes nonnegative values. And you will have, 185 00:14:01 --> 00:14:04 the function will always take nonnegative values. 186 00:14:04 --> 00:14:07 So, the origin will be a minimum. 187 00:14:07 --> 00:14:10 Well, if you have a difference of two squares that typically 188 00:14:10 --> 00:14:13 you'll have a saddle point because depending on whether one 189 00:14:13 --> 00:14:18 or the other is larger, you will have a positive or a 190 00:14:18 --> 00:14:24 negative quantity. OK, so I claim there's various 191 00:14:24 --> 00:14:32 cases to look at. So, let's see. 192 00:14:32 --> 00:14:34 So, in fact, I claim there will be three 193 00:14:34 --> 00:14:37 cases. And, that's good news for us 194 00:14:37 --> 00:14:40 because after all, we want to distinguish between 195 00:14:40 --> 00:14:45 three possibilities. So, let's first do away with 196 00:14:45 --> 00:14:52 the most complicated one. What if 4ac minus b^2 is 197 00:14:52 --> 00:14:56 negative? Well, if it's negative, 198 00:14:56 --> 00:15:00 then it means what I have between the brackets is, 199 00:15:00 --> 00:15:06 so the first guy is obviously a positive quantity, 200 00:15:06 --> 00:15:10 while the second one will be something negative times y2. 201 00:15:10 --> 00:15:13 So, it will be a negative quantity. 202 00:15:13 --> 00:15:23 OK, so one term is positive. The other is negative. 203 00:15:23 --> 00:15:31 That tells us we actually have a saddle point. 204 00:15:31 --> 00:15:35 We have, in fact, written our function as a 205 00:15:35 --> 00:15:40 difference of two squares. OK, is that convincing? 206 00:15:40 --> 00:15:42 So, if you want, what I could do is actually I 207 00:15:42 --> 00:15:47 could change my coordinates, have new coordinates called u 208 00:15:47 --> 00:15:50 equals x b over 2ay, and v, actually, 209 00:15:50 --> 00:15:55 well, I could keep y, and that it would look like the 210 00:15:55 --> 00:16:02 difference of squares directly. OK, so that's the first case. 211 00:16:02 --> 00:16:12 The second case is where 4ac-b^2 = 0. 212 00:16:12 --> 00:16:18 Well, what happens if that's zero? 213 00:16:18 --> 00:16:21 Then it means that this term over there goes away. 214 00:16:21 --> 00:16:25 So, what we have is just one square. 215 00:16:25 --> 00:16:29 OK, so what that means is actually that our function 216 00:16:29 --> 00:16:32 depends only on one direction of things. 217 00:16:32 --> 00:16:36 In the other direction, it's going to actually be 218 00:16:36 --> 00:16:38 degenerate. So, for example, 219 00:16:38 --> 00:16:40 forget all the clutter in there. 220 00:16:40 --> 00:16:45 Say I give you just the function of two variables, 221 00:16:45 --> 00:16:49 w equals just x^2. So, that means it doesn't 222 00:16:49 --> 00:16:53 depend on y at all. And, if I try to plot the 223 00:16:53 --> 00:16:58 graph, it will look like, well, x is here. 224 00:16:58 --> 00:17:04 So, it will depend on x in that way, but it doesn't depend on y 225 00:17:04 --> 00:17:10 at all. So, what the graph looks like 226 00:17:10 --> 00:17:18 is something like that. OK, basically it's a valley 227 00:17:18 --> 00:17:22 whose bottom is completely flat. So, that means, 228 00:17:22 --> 00:17:24 actually, we have a degenerate critical point. 229 00:17:24 --> 00:17:28 It's called degenerate because there is a direction in which 230 00:17:28 --> 00:17:30 nothing happens. And, in fact, 231 00:17:30 --> 00:17:38 you have critical points everywhere along the y axis. 232 00:17:38 --> 00:17:42 Now, whether the square that we have is x or something else, 233 00:17:42 --> 00:17:46 namely, x plus b over 2a y, it doesn't matter. 234 00:17:46 --> 00:17:48 I mean, it will still get this degenerate behavior. 235 00:17:48 --> 00:17:56 But there's a direction in which nothing happens because we 236 00:17:56 --> 00:18:02 just have the square of one quantity. 237 00:18:02 --> 00:18:06 I'm sure that 300 students means 300 different ring tones, 238 00:18:06 --> 00:18:09 but I'm not eager to hear all of them, thanks. 239 00:18:09 --> 00:18:18 [LAUGHTER] OK, so, this is what's called a 240 00:18:18 --> 00:18:28 degenerate critical point, and [LAUGHTER]. 241 00:18:28 --> 00:18:33 OK, so basically we'll leave it here. 242 00:18:33 --> 00:18:38 We won't actually try to figure out further what happens, 243 00:18:38 --> 00:18:42 and the reason for that is that when you have an actual 244 00:18:42 --> 00:18:44 function, a general function, 245 00:18:44 --> 00:18:46 not just one that's quadratic like this, 246 00:18:46 --> 00:18:50 then there will actually be other terms maybe involving 247 00:18:50 --> 00:18:54 higher powers, maybe x^3 or y^3 or things like 248 00:18:54 --> 00:18:56 that. And then, they will mess up 249 00:18:56 --> 00:19:00 what happens in this valley. And, it's a situation where we 250 00:19:00 --> 00:19:03 won't be able, actually, to tell automatically 251 00:19:03 --> 00:19:06 just by looking at second derivatives what happens. 252 00:19:06 --> 00:19:09 See, for example, in a function of one variable, 253 00:19:09 --> 00:19:12 if you have just a function of one variable, 254 00:19:12 --> 00:19:14 say, f of x equals x to the five, 255 00:19:14 --> 00:19:18 well, if you try to decide what type of point the origin is, 256 00:19:18 --> 00:19:20 you're going to take the second derivative. 257 00:19:20 --> 00:19:23 It will be zero, and then you can conclude. 258 00:19:23 --> 00:19:26 Those things depend on higher order derivatives. 259 00:19:26 --> 00:19:29 So, we just won't like that case. 260 00:19:29 --> 00:19:34 We just won't try to figure out what's going on here. 261 00:19:34 --> 00:19:40 Now, the last situation is if 4ac-b^2 is positive. 262 00:19:40 --> 00:19:45 So, then, that means that actually we've written things. 263 00:19:45 --> 00:19:52 The big bracket up there is a sum of two squares. 264 00:19:52 --> 00:20:00 So, that means that we've written w as one over 4a times 265 00:20:00 --> 00:20:08 plus something squared plus something else squared, 266 00:20:08 --> 00:20:12 OK? So, these guys have the same 267 00:20:12 --> 00:20:18 sign, and that means that this term here will always be greater 268 00:20:18 --> 00:20:22 than or equal to zero. And that means that we should 269 00:20:22 --> 00:20:24 either have a maximum or minimum. 270 00:20:24 --> 00:20:29 How we find out which one it is? Well, we look at the sign of a, 271 00:20:29 --> 00:20:30 exactly. OK? 272 00:20:30 --> 00:20:35 So, there's two sub-cases. One is if a is positive, 273 00:20:35 --> 00:20:40 then, this quantity overall will always be nonnegative. 274 00:20:40 --> 00:20:54 And that means we have a minimum, OK? 275 00:20:54 --> 00:20:58 And, if a is negative on the other hand, 276 00:20:58 --> 00:21:01 so that means that we multiply this positive quantity by a 277 00:21:01 --> 00:21:04 negative number, we get something that's always 278 00:21:04 --> 00:21:10 negative. So, zero is actually the 279 00:21:10 --> 00:21:18 maximum. OK, is that clear for everyone? 280 00:21:18 --> 00:21:29 Yes? Sorry, yeah, 281 00:21:29 --> 00:21:34 so I said in the example w equals x^2, it doesn't depend on 282 00:21:34 --> 00:21:37 y. So, the more general situation 283 00:21:37 --> 00:21:44 is w equals some constant. Well, I guess it's a times (x b 284 00:21:44 --> 00:21:48 over 2a times y)^2. So, it does depend on x and y, 285 00:21:48 --> 00:21:51 but it only depends on this combination. 286 00:21:51 --> 00:21:54 OK, so if I choose to move in some other perpendicular 287 00:21:54 --> 00:21:58 direction, in the direction where this 288 00:21:58 --> 00:22:02 remains constant, so maybe if I set x equals 289 00:22:02 --> 00:22:06 minus b over 2a y, then this remains zero all the 290 00:22:06 --> 00:22:08 time. So, there's a degenerate 291 00:22:08 --> 00:22:11 direction in which I stay at the minimum or maximum, 292 00:22:11 --> 00:22:15 or whatever it is that I have. OK, so that's why it's called 293 00:22:15 --> 00:22:17 degenerate. There is a direction in which 294 00:22:17 --> 00:22:29 nothing happens. OK, yes? 295 00:22:29 --> 00:22:31 Yes, yeah, so that's a very good question. 296 00:22:31 --> 00:22:33 So, there's going to be the second derivative test. 297 00:22:33 --> 00:22:36 Why do not have derivatives yet? Well, that's because I've been 298 00:22:36 --> 00:22:39 looking at this special example where we have a function like 299 00:22:39 --> 00:22:41 this. And, so I don't actually need 300 00:22:41 --> 00:22:43 to take derivatives yet. But, secretly, 301 00:22:43 --> 00:22:46 that's because a, b, and c will be the second 302 00:22:46 --> 00:22:49 derivatives of the function, actually, 2a, 303 00:22:49 --> 00:22:52 b, and 2c. So now, we are going to go to 304 00:22:52 --> 00:22:54 general function. And there, instead of having 305 00:22:54 --> 00:22:57 these coefficients a, b, and c given to us, 306 00:22:57 --> 00:23:00 we'll have to compute them as second derivatives. 307 00:23:00 --> 00:23:03 OK, so here, I'm basically setting the stage 308 00:23:03 --> 00:23:07 for what will be the actual criterion we'll use using second 309 00:23:07 --> 00:23:13 derivatives. Yes? 310 00:23:13 --> 00:23:16 So, yeah, so what you have a degenerate critical point, 311 00:23:16 --> 00:23:20 it could be a degenerate minimum, or a degenerate maximum 312 00:23:20 --> 00:23:23 depending on the sign of a. But, in general, 313 00:23:23 --> 00:23:26 once you start having functions, you don't really know 314 00:23:26 --> 00:23:30 what will happen anymore. It could also be a degenerate 315 00:23:30 --> 00:23:36 saddle, and so on. So, we won't really be able to 316 00:23:36 --> 00:23:40 tell. Yes? 317 00:23:40 --> 00:23:43 It is possible to have a degenerate saddle point. 318 00:23:43 --> 00:23:46 For example, if I gave you x^3 y^3, 319 00:23:46 --> 00:23:49 you can convince yourself that if you take x and y to be 320 00:23:49 --> 00:23:53 negative, it will be negative. If x and y are positive, 321 00:23:53 --> 00:23:55 it's positive. And, it has a very degenerate 322 00:23:55 --> 00:23:59 critical point at the origin. So, that's a degenerate saddle 323 00:23:59 --> 00:24:01 point. We don't see it here because 324 00:24:01 --> 00:24:04 that doesn't happen if you have only quadratic terms like that. 325 00:24:04 --> 00:24:12 You need to have higher-order terms to see it happen. 326 00:24:12 --> 00:24:23 OK. OK, so let's continue. 327 00:24:23 --> 00:24:27 Before we continue, but see, I wanted to point out 328 00:24:27 --> 00:24:30 one small thing. So, here, we have the magic 329 00:24:30 --> 00:24:34 quantity, 4ac minus b^2. You've probably seen that 330 00:24:34 --> 00:24:37 before in your life. Yet, it looks like the 331 00:24:37 --> 00:24:40 quadratic formula, except that one involves 332 00:24:40 --> 00:24:43 b^2-4ac. But that's really the same 333 00:24:43 --> 00:24:47 thing. OK, so let's see, 334 00:24:47 --> 00:24:57 where does the quadratic formula come in here? 335 00:24:57 --> 00:25:00 Well, let me write things differently. 336 00:25:00 --> 00:25:03 OK, so we've manipulated things, and got into a 337 00:25:03 --> 00:25:08 conclusion. But, let me just do a different 338 00:25:08 --> 00:25:14 manipulation, and write this now instead as 339 00:25:14 --> 00:25:23 y^2 times a times x over y squared plus b(x over y) plus c. 340 00:25:23 --> 00:25:28 OK, see, that's the same thing that I had before. 341 00:25:28 --> 00:25:35 Well, so now this quantity here is always nonnegative. 342 00:25:35 --> 00:25:39 What about this one? Well, of course, 343 00:25:39 --> 00:25:43 this one depends on x over y. It means it depends on which 344 00:25:43 --> 00:25:45 direction you're going to move away from the origin, 345 00:25:45 --> 00:25:48 which ratio between x and y you will consider. 346 00:25:48 --> 00:25:51 But, I claim there's two situations. 347 00:25:51 --> 00:25:57 One is, so, let's try to reformulate things. 348 00:25:57 --> 00:26:04 So, if a discriminate here is positive, then it means that 349 00:26:04 --> 00:26:10 these have roots and these have solutions. 350 00:26:10 --> 00:26:19 And, that means that this quantity can be both positive 351 00:26:19 --> 00:26:24 and negative. This quantity takes positive 352 00:26:24 --> 00:26:31 and negative values. One way to convince yourself is 353 00:26:31 --> 00:26:37 just to, you know, plot at^2 bt c. 354 00:26:37 --> 00:26:43 You know that there's two roots. So, it might look like this, 355 00:26:43 --> 00:26:48 or might look like that depending on the sign of a. 356 00:26:48 --> 00:26:52 But, in either case, it will take values of both 357 00:26:52 --> 00:26:54 signs. So, that means that your 358 00:26:54 --> 00:26:56 function will take values of both signs. 359 00:26:56 --> 00:27:04 360 00:27:04 --> 00:27:13 The value takes both positive and negative values. 361 00:27:13 --> 00:27:21 And, so that means we have a saddle point, 362 00:27:21 --> 00:27:28 while the other situation, when b^2-4ac is negative -- -- 363 00:27:28 --> 00:27:36 means that this equation is quadratic never takes the value, 364 00:27:36 --> 00:27:39 zero. So, it's always positive or 365 00:27:39 --> 00:27:42 it's always negative, depending on the sign of a. 366 00:27:42 --> 00:27:48 So, the other case is if b^2-4ac is negative, 367 00:27:48 --> 00:27:53 then the quadratic doesn't have a solution. 368 00:27:53 --> 00:27:58 And it could look like this or like that depending on whether a 369 00:27:58 --> 00:28:03 is positive or a is negative. So, in particular, 370 00:28:03 --> 00:28:12 that means that ax over y2 plus bx over y plus c is always 371 00:28:12 --> 00:28:21 positive or always negative depending on the sign of a. 372 00:28:21 --> 00:28:23 And then, that tells us that our function, 373 00:28:23 --> 00:28:25 w, will be always positive or always negative. 374 00:28:25 --> 00:28:28 And then we'll get a minimum or maximum. 375 00:28:28 --> 00:28:40 376 00:28:40 --> 00:28:44 OK, we'll have a min or a max depending on which situation we 377 00:28:44 --> 00:28:47 are in. OK, so that's another way to 378 00:28:47 --> 00:28:51 derive the same answer. And now, you see here why the 379 00:28:51 --> 00:28:55 discriminate plays a role. That's because it exactly tells 380 00:28:55 --> 00:28:59 you whether this quadratic quantity has always the same 381 00:28:59 --> 00:29:04 sign, or whether it can actually 382 00:29:04 --> 00:29:12 cross the value, zero, when you have the root of 383 00:29:12 --> 00:29:16 a quadratic. OK, so hopefully at this stage 384 00:29:16 --> 00:29:20 you are happy with one of the two explanations, 385 00:29:20 --> 00:29:23 at least. And now, you are willing to 386 00:29:23 --> 00:29:26 believe, I hope, that we have basically a way of 387 00:29:26 --> 00:29:30 deciding what type of critical point we have in the special 388 00:29:30 --> 00:29:32 case of a quadratic function. 389 00:29:32 --> 00:29:58 390 00:29:58 --> 00:30:05 OK, so, now what do we do with the general function? 391 00:30:05 --> 00:30:19 Well, so in general, we want to look at second 392 00:30:19 --> 00:30:24 derivatives. OK, so now we are getting to 393 00:30:24 --> 00:30:27 the real stuff. So, how many second derivatives 394 00:30:27 --> 00:30:29 do we have? That's maybe the first thing we 395 00:30:29 --> 00:30:32 should figure out. Well, we can take the 396 00:30:32 --> 00:30:39 derivative first with respect to x, and then again with respect 397 00:30:39 --> 00:30:44 to x. OK, that gives us something we 398 00:30:44 --> 00:30:54 denote by partial square f over partial x squared or fxx. 399 00:30:54 --> 00:31:00 Then, there's another one which is fxy, which means you take the 400 00:31:00 --> 00:31:05 derivative with respect to x, and then with respect to y. 401 00:31:05 --> 00:31:09 Another thing you can do, is do first derivative respect 402 00:31:09 --> 00:31:12 to y, and then with respect to x. 403 00:31:12 --> 00:31:17 That would be fyx. Well, good news. 404 00:31:17 --> 00:31:22 These are actually always equal to each other. 405 00:31:22 --> 00:31:26 OK, so it's the fact that we will admit, it's actually not 406 00:31:26 --> 00:31:30 very hard to check. So these are always the same. 407 00:31:30 --> 00:31:33 We don't need to worry about which one we do. 408 00:31:33 --> 00:31:36 That's one computation that we won't need to do. 409 00:31:36 --> 00:31:43 We can save a bit of effort. And then, we have the last one, 410 00:31:43 --> 00:31:51 namely, the second partial with respect to y and y fyy. 411 00:31:51 --> 00:32:00 OK, so we have three of them. So, what does the second 412 00:32:00 --> 00:32:02 derivative test say? 413 00:32:02 --> 00:32:16 414 00:32:16 --> 00:32:22 It says, say that you have a critical point (x0, 415 00:32:22 --> 00:32:27 y0) of a function of two variables, f, 416 00:32:27 --> 00:32:34 and then let's compute the partial derivatives. 417 00:32:34 --> 00:32:41 So, let's call capital A the second derivative with respect 418 00:32:41 --> 00:32:45 to x. Let's call capital B the second 419 00:32:45 --> 00:32:49 derivative with respect to x and y. 420 00:32:49 --> 00:32:55 And C equals fyy at this point, OK? 421 00:32:55 --> 00:32:59 So, these are just numbers because we first compute the 422 00:32:59 --> 00:33:02 second derivative, and then we plug in the values 423 00:33:02 --> 00:33:04 of x and y at the critical point. 424 00:33:04 --> 00:33:14 So, these will just be numbers. And now, what we do is we look 425 00:33:14 --> 00:33:21 at the quantity AC-B^2. I am not forgetting the four. 426 00:33:21 --> 00:33:26 You will see why there isn't one. 427 00:33:26 --> 00:33:31 So, if AC-B^2 is positive, then there's two sub-cases. 428 00:33:31 --> 00:33:39 If A is positive, then it's local minimum. 429 00:33:39 --> 00:33:50 430 00:33:50 --> 00:33:56 The second case, so, still, if AC-B^2 is 431 00:33:56 --> 00:34:04 positive, but A is negative, then it's going to be a local 432 00:34:04 --> 00:34:11 maximum. And, if AC-B^2 is negative, 433 00:34:11 --> 00:34:17 then it's a saddle point, and finally, 434 00:34:17 --> 00:34:24 if AC-B^2 is zero, then we actually cannot 435 00:34:24 --> 00:34:28 compute. We don't know whether it's 436 00:34:28 --> 00:34:33 going to be a minimum, a maximum, or a saddle. 437 00:34:33 --> 00:34:37 We know it's degenerate in some way, but we don't know what type 438 00:34:37 --> 00:34:40 of point it is. OK, so that's actually what you 439 00:34:40 --> 00:34:43 need to remember. If you are formula oriented, 440 00:34:43 --> 00:34:45 that's all you need to remember about today. 441 00:34:45 --> 00:34:53 But, let's try to understand why, how this comes out of what 442 00:34:53 --> 00:34:59 we had there. OK, so, I think maybe I 443 00:34:59 --> 00:35:05 actually want to keep, so maybe I want to keep this 444 00:35:05 --> 00:35:06 middle board because it actually has, 445 00:35:06 --> 00:35:09 you know, the recipe that we found before the quadratic 446 00:35:09 --> 00:35:12 function. So, let me move directly over 447 00:35:12 --> 00:35:16 there and try to relate our old recipe with the new. 448 00:35:16 --> 00:35:43 449 00:35:43 --> 00:35:50 OK, you are easily amused. OK, so first, 450 00:35:50 --> 00:35:57 let's check that these two things say the same thing in the 451 00:35:57 --> 00:36:01 special case that we are looking at. 452 00:36:01 --> 00:36:12 OK, so let's verify in the special case where the function 453 00:36:12 --> 00:36:22 was ax^2 bxy cy^2. So -- Well, what is the second 454 00:36:22 --> 00:36:28 derivative with respect to x and x? 455 00:36:28 --> 00:36:31 If I take the second derivative with respect to x and x, 456 00:36:31 --> 00:36:34 so first I want to take maybe the derivative with respect to 457 00:36:34 --> 00:36:37 x. But first, let's take the first 458 00:36:37 --> 00:36:46 partial, Wx. That will be 2ax by, right? 459 00:36:46 --> 00:36:50 So, Wxx will be, well, let's take a partial with 460 00:36:50 --> 00:36:54 respect to x again. That's 2a. 461 00:36:54 --> 00:37:02 Wxy, I take the partial respect to y, and we'll get b. 462 00:37:02 --> 00:37:06 OK, now we need, also, the partial with respect 463 00:37:06 --> 00:37:13 to y. So, Wy is bx 2cy. 464 00:37:13 --> 00:37:17 In case you don't believe what I told you about the mixed 465 00:37:17 --> 00:37:21 partials, Wyx, well, you can check. 466 00:37:21 --> 00:37:24 And it's, again, b. So, they are, 467 00:37:24 --> 00:37:30 indeed, the same thing. And, Wyy will be 2c. 468 00:37:30 --> 00:37:39 So, if we now look at these quantities, that tells us, 469 00:37:39 --> 00:37:46 well, big A is two little a, big B is little b, 470 00:37:46 --> 00:37:55 big C is two little c. So, AC-B^2 is what we used to 471 00:37:55 --> 00:38:04 call four little ac minus b2. OK, ooh. 472 00:38:04 --> 00:38:07 [LAUGHTER] So, now you can compare the 473 00:38:07 --> 00:38:10 cases. They are not listed in the same 474 00:38:10 --> 00:38:14 order just to make it harder. So, we said first, 475 00:38:14 --> 00:38:20 so the saddle case is when AC-B^2 in big letters is 476 00:38:20 --> 00:38:26 negative, that's the same as 4ac-b2 in lower case is 477 00:38:26 --> 00:38:30 negative. The case where capital AC-B2 is 478 00:38:30 --> 00:38:35 positive, local min and local max corresponds to this one. 479 00:38:35 --> 00:38:40 And, the case where we can't conclude was what used to be the 480 00:38:40 --> 00:38:44 degenerate one. OK, so at least we don't seem 481 00:38:44 --> 00:38:48 to have messed up when copying the formula. 482 00:38:48 --> 00:38:56 Now, why does that work more generally than that? 483 00:38:56 --> 00:39:03 Well, the answer that is, again, Taylor approximation. 484 00:39:03 --> 00:39:16 Aww. OK, so let me just do here 485 00:39:16 --> 00:39:22 quadratic approximation. So, quadratic approximation 486 00:39:22 --> 00:39:25 tells me the following thing. It tells me, 487 00:39:25 --> 00:39:30 if I have a function, f of xy, and I want to 488 00:39:30 --> 00:39:37 understand the change in f when I change x and y a little bit. 489 00:39:37 --> 00:39:40 Well, there's the first-order terms. 490 00:39:40 --> 00:39:43 There is the linear terms that by now you should know and be 491 00:39:43 --> 00:39:51 comfortable with. That's fx times the change in x. 492 00:39:51 --> 00:39:56 And then, there's fy times the change in y. 493 00:39:56 --> 00:40:00 OK, that's the starting point. But now, of course, 494 00:40:00 --> 00:40:03 if x and y, sorry, if we are at the critical 495 00:40:03 --> 00:40:09 point, then that's going to be zero at the critical point. 496 00:40:09 --> 00:40:16 So, that term actually goes away, and that's also zero at 497 00:40:16 --> 00:40:22 the critical point. So, that term also goes away. 498 00:40:22 --> 00:40:24 OK, so linear approximation is really no good. 499 00:40:24 --> 00:40:27 We need more terms. So, what are the next terms? 500 00:40:27 --> 00:40:35 Well, the next terms are quadratic terms, 501 00:40:35 --> 00:40:38 and so I mean, if you remember the Taylor 502 00:40:38 --> 00:40:42 formula for a function of a single variable, 503 00:40:42 --> 00:40:46 there was the derivative times x minus x0 plus one half of a 504 00:40:46 --> 00:40:51 second derivative times x-x0^2. And see, this side here is 505 00:40:51 --> 00:40:55 really Taylor approximation in one variable looking only at x. 506 00:40:55 --> 00:40:57 But of course, we also have terms involving y, 507 00:40:57 --> 00:41:00 and terms involving simultaneously x and y. 508 00:41:00 --> 00:41:10 And, these terms are fxy times change in x times change in y 509 00:41:10 --> 00:41:17 plus one half of fyy(y-y0)^2. There's no one half in the 510 00:41:17 --> 00:41:20 middle because, in fact, you would have two 511 00:41:20 --> 00:41:24 terms, one for xy, one for yx, but they are the 512 00:41:24 --> 00:41:26 same. And then, if you want to 513 00:41:26 --> 00:41:29 continue, there is actually cubic terms involving the third 514 00:41:29 --> 00:41:32 derivatives, and so on, but we are not actually looking 515 00:41:32 --> 00:41:34 at them. And so, now, 516 00:41:34 --> 00:41:39 when we do this approximation, well, the type of critical 517 00:41:39 --> 00:41:45 point remains the same when we replace the function by this 518 00:41:45 --> 00:41:48 approximation. And so, we can apply the 519 00:41:48 --> 00:41:53 argument that we used to deduce things in the quadratic case. 520 00:41:53 --> 00:41:55 In fact, it still works in the general case using this 521 00:41:55 --> 00:41:57 approximation formula. 522 00:41:57 --> 00:42:12 523 00:42:12 --> 00:42:26 So -- The general case reduces to the quadratic case. 524 00:42:26 --> 00:42:31 And now, you see actually why, well, here you see, 525 00:42:31 --> 00:42:36 again, how this coefficient which we used to call little a 526 00:42:36 --> 00:42:41 is also one half of capital A. And same here: 527 00:42:41 --> 00:42:47 this coefficient is what we call capital B or little b, 528 00:42:47 --> 00:42:52 and this coefficient here is what we called little c or one 529 00:42:52 --> 00:42:57 half of capital C. And then, when you replace 530 00:42:57 --> 00:43:02 these into the various cases that we had here, 531 00:43:02 --> 00:43:06 you end up with the second derivative test. 532 00:43:06 --> 00:43:08 So, what about the degenerate case? 533 00:43:08 --> 00:43:11 Why can't we just say, well, it's going to be a 534 00:43:11 --> 00:43:16 degenerate critical point? So, the reason is that this 535 00:43:16 --> 00:43:20 approximation formula is reasonable only if the higher 536 00:43:20 --> 00:43:24 order terms are negligible. OK, so in fact, 537 00:43:24 --> 00:43:27 secretly, there's more terms. This is only an approximation. 538 00:43:27 --> 00:43:30 There would be terms involving third derivatives, 539 00:43:30 --> 00:43:34 and maybe even beyond that. And, so it is not to generate 540 00:43:34 --> 00:43:37 case, they don't actually matter 541 00:43:37 --> 00:43:39 because the shape of the function, 542 00:43:39 --> 00:43:42 the shape of the graph, is actually determined by the 543 00:43:42 --> 00:43:45 quadratic terms. But, in the degenerate case, 544 00:43:45 --> 00:43:49 see, if I start with this and I add something even very, 545 00:43:49 --> 00:43:53 very small along the y axis, then that can be enough to bend 546 00:43:53 --> 00:43:56 this very slightly up or slightly down, 547 00:43:56 --> 00:44:00 and turn my degenerate point in to either a minimum or a saddle 548 00:44:00 --> 00:44:03 point. And, I won't be able to tell 549 00:44:03 --> 00:44:06 until I go further in the list of derivatives. 550 00:44:06 --> 00:44:14 So, in the degenerate case, what actually happens depends 551 00:44:14 --> 00:44:20 on the higher order derivatives. 552 00:44:20 --> 00:44:38 553 00:44:38 --> 00:44:42 So, we will need to analyze things more carefully. 554 00:44:42 --> 00:44:45 Well, we're not going to bother with that in this class. 555 00:44:45 --> 00:44:52 So, we'll just say, well, we cannot compute, 556 00:44:52 --> 00:44:54 OK? I mean, you have to realize 557 00:44:54 --> 00:44:57 that in real life, you have to be extremely 558 00:44:57 --> 00:45:02 unlucky for this quantity to end up being exactly 0. 559 00:45:02 --> 00:45:03 [LAUGHTER] Well, if that happens, 560 00:45:03 --> 00:45:05 then what you should do is maybe try by inspection. 561 00:45:05 --> 00:45:08 See if there's a good reason why the function should always 562 00:45:08 --> 00:45:11 be positive or always be negative, or something. 563 00:45:11 --> 00:45:16 Or, you know, plot it on a computer and see 564 00:45:16 --> 00:45:23 what happens. But, otherwise we can't compute. 565 00:45:23 --> 00:45:33 OK, so let's do an example. So, probably I should leave 566 00:45:33 --> 00:45:39 this on so that we still have the test with us. 567 00:45:39 --> 00:45:42 And, instead, OK, so I'll do my example here. 568 00:45:42 --> 00:46:20 569 00:46:20 --> 00:46:30 OK, so just an example. Let's look at f of (x, 570 00:46:30 --> 00:46:37 y) = x y 1/xy, where x and y are positive. 571 00:46:37 --> 00:46:39 So, I'm looking only at the first quadrant. 572 00:46:39 --> 00:46:42 OK, I mean, I'm doing this because I don't want the 573 00:46:42 --> 00:46:46 denominator to become zero. So, I'm just looking at the 574 00:46:46 --> 00:46:50 situation. So, let's look first for, 575 00:46:50 --> 00:46:55 so, the question will be, what are the minimum and 576 00:46:55 --> 00:47:03 maximum of this function? So, the first thing we should 577 00:47:03 --> 00:47:12 do to answer this question is look for critical points, 578 00:47:12 --> 00:47:15 OK? So, for that, 579 00:47:15 --> 00:47:19 we have to compute the first derivatives. 580 00:47:19 --> 00:47:34 OK, so fx is one minus one over x^2y, OK? 581 00:47:34 --> 00:47:39 Take the derivative of one over x, that's negative one over x^2. 582 00:47:39 --> 00:47:44 And, we'll want to set that equal to zero. 583 00:47:44 --> 00:47:50 And fy is one minus one over xy^2. 584 00:47:50 --> 00:47:54 And, we want to set that equal to zero. 585 00:47:54 --> 00:47:59 So, what are the equations we have to solve? 586 00:47:59 --> 00:48:05 Well, I guess x^2y equals one, I mean, if I move this guy over 587 00:48:05 --> 00:48:09 here I get one over x^2y equals one. 588 00:48:09 --> 00:48:14 That's x^2y equals one, and xy^2 equals one. 589 00:48:14 --> 00:48:18 What do you get by comparing these two? 590 00:48:18 --> 00:48:21 Well, x and y should both be, OK, so yeah, 591 00:48:21 --> 00:48:24 I agree with you that one and one is a solution. 592 00:48:24 --> 00:48:27 Why is it the only one? So, first, if I divide this one 593 00:48:27 --> 00:48:29 by that one, I get x over y equals one. 594 00:48:29 --> 00:48:34 So, it tells me x equals y. And then, if x equals y, 595 00:48:34 --> 00:48:40 then if I put that into here, it will give me y^3 equals one, 596 00:48:40 --> 00:48:44 which tells me y equals one, and therefore, 597 00:48:44 --> 00:48:50 x equals one as well. OK, so, there's only one 598 00:48:50 --> 00:48:54 solution. There's only one critical 599 00:48:54 --> 00:48:58 point, which is going to be (1,1). 600 00:48:58 --> 00:49:09 OK, so, now here's where you do a bit of work. 601 00:49:09 --> 00:49:18 What do you think of that critical point? 602 00:49:18 --> 00:49:25 OK, I see some valid votes. I see some, OK, 603 00:49:25 --> 00:49:28 I see a lot of people answering four. 604 00:49:28 --> 00:49:30 [LAUGHTER] that seems to suggest that 605 00:49:30 --> 00:49:34 maybe you haven't completed the second derivative yet. 606 00:49:34 --> 00:49:37 Yes, I see someone giving the correct answer. 607 00:49:37 --> 00:49:41 I see some people not giving quite the correct answer. 608 00:49:41 --> 00:49:43 I see more and more correct answers. 609 00:49:43 --> 00:49:49 OK, so let's see. To figure out what type of 610 00:49:49 --> 00:49:52 point is, we should compute the second partial derivatives. 611 00:49:52 --> 00:50:02 So, fxx is, what do we get what we take the derivative of this 612 00:50:02 --> 00:50:11 with respect to x? Two over x^3y, OK? 613 00:50:11 --> 00:50:25 So, at our point, a will be 2. Fxy will be one over x^2y^2. 614 00:50:25 --> 00:50:37 So, B will be one. And, Fyy is going to be two 615 00:50:37 --> 00:50:42 over xy^3. So, C will be two. 616 00:50:42 --> 00:50:51 And so that tells us, well, AC-B^2 is four minus one. 617 00:50:51 --> 00:51:02 Sorry, I should probably use a different blackboard for that. 618 00:51:02 --> 00:51:06 AC-B2 is two times two minus 1^2 is three. 619 00:51:06 --> 00:51:10 It's positive. That tells us we are either a 620 00:51:10 --> 00:51:17 local minimum or local maximum. And, A is positive. 621 00:51:17 --> 00:51:21 So, it's a local minimum. And, in fact, 622 00:51:21 --> 00:51:23 you can check it's the global minimum. 623 00:51:23 --> 00:51:29 What about the maximum? Well, if a maximum is not 624 00:51:29 --> 00:51:32 actually at a critical point, it's on the boundary, 625 00:51:32 --> 00:51:35 or at infinity. See, so we have actually to 626 00:51:35 --> 00:51:39 check what happens when x and y go to zero or to infinity. 627 00:51:39 --> 00:51:42 Well, if that happens, if x or y goes to infinity, 628 00:51:42 --> 00:51:44 then the function goes to infinity. 629 00:51:44 --> 00:51:48 Also, if x or y goes to zero, then one over xy goes to 630 00:51:48 --> 00:51:51 infinity. So, the maximum, 631 00:51:51 --> 00:51:59 well, the function goes to infinity when x goes to infinity 632 00:51:59 --> 00:52:05 or y goes to infinity, or x and y go to zero. 633 00:52:05 --> 00:52:07 So, it's not at a critical point. 634 00:52:07 --> 00:52:10 OK, so, in general, we have to check both the 635 00:52:10 --> 00:52:13 critical points and the boundaries to decide what 636 00:52:13 --> 00:52:15 happens. OK, the end. 637 00:52:15 --> 00:52:18 Have a nice weekend. 638 00:52:18 --> 00:52:23