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Math for Economists - Lecture 13

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Math for Economists - Lecture 13
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Transkript: Englisch(automatisch erzeugt)
Okay, so final is a week from today. I feel obligated to announce that since we haven't seen each other for a little while. So today we'll do new material, Friday new material, and then Monday will be review. I should get everything done by Friday. In fact, I might even end a little early.
So what are we doing right now? We're in section 12.3, and we're doing—it's worth it to just review a little bit. So the scenario in Math 2A is this. If I have f of x equals, let's say, 3 minus x squared, so we can actually graph that.
It goes like this. Okay, and then what if I restrict the interval?
So on a closed interval—okay, so what should I make the interval? Let's say from 1 to 3. So it's a closed interval. When I say closed, that means there's a particular meaning in Math for the word closed.
In this case, what it means is that the endpoints are included. So if I said an open interval, I'd have the parentheses there, and if I say closed interval, it has the brackets. And if I say half closed, you know, it's got one bracket and one open side.
Okay, so that's just the word closed. But you can see what's going on here. If I do 1 here, when I plug in 1, I get the value 2, and then here's 3. The function continues on down there. That's at negative 6.
So you can see that the extreme values of this, if we want to find the max and the min on this interval, there's the max and there's the min. But the method is what we really want to expose. And so the method is we're going to use the extreme value theorem.
And this says if you have f of x on a closed interval, and the function needs to be continuous, but, you know, in Math classes like this, they usually are, so it doesn't come up. But if you have f of x on a closed interval, the max and the min exist
and occur at either a critical point or the boundary.
So the reason it's called the extreme value theorem is because you were going to get an extreme value. If I make this an open interval, you might not get an extreme value.
For instance, you wouldn't in this case. If I make this an open interval, then there's a hole here, you see. So you never actually arrive at the extreme value. So it's crucial that it's a closed interval, that you actually have boundary points. So for this problem here, what you do is you just go ahead. There's two places where it could occur, either the critical point or the boundary.
And so we can go get the derivative. The derivative is negative 2x, and then set that equal to zero. That says that x equals zero. So we did get a critical point. That's right here. But you see, the critical point doesn't occur in the interval that I'm interested in, so I reject it.
Okay? This is not in that interval. So even though it is a critical point in this particular context, it's not relevant to what I'm trying to do. So now my only potential points for the max or the min are going to be at the extremes, or at the boundaries of this interval.
And you can see that in the picture, too. That's where the max is and that's where the min is. But we just test these points. So if you didn't have a picture, what you would do is you just plug in the boundary points. f of 1 is equal to 3 minus 1 is 2. And f of 3 is equal to 3 minus 9, which is negative 6.
So this is the max and this is the min. Which we already knew from the picture, but it's good to just practice the method. So sometimes when you go get your critical point, it's just not in the interval that you're interested in, and so you have to reject it and then focus on the boundaries.
So for math 4, let's write down the extreme value theorem for math 4. It's the same idea. You have f of x, but here in math 4 we allow ourselves to have more variables.
So you might have f of xy, which is x, here's y. So there's a function here. And then what's the equivalent of the interval now in this context?
What we're going to have is a, so now instead of a closed interval, it's a closed rectangle. So we have f of xy, we're just going to basically repeat this, but we'll put it in this context.
So we have f of xy on, I can't say a closed interval now, I have to say on a closed rectangle. And being closed means that it has its edges.
Because you could have an open rectangle where one of the sides is dotted and it's not included, but in this case we're going to have a closed rectangle and it's continuous. I might as well just add more of this continuous. And it has to be differentiable because I have to be able to take the derivatives, but all these things are going to be given in our problem.
I just want to focus on the closed part. So we have a closed rectangle, the max or the min exist and occur at either a critical point,
here I'll list them out. Now, so the second part here, it's either a critical point or the boundary. Here the boundary is just two points because it's just the edges of the interval. Here when I go to the boundary it's much more complicated
because the boundary includes these edges and these corners. So let's put the third one here. This third one here represents the edges, like in this case, that's where you're just plugging in the real extreme points which are the corners. So the corners represent the edges of the interval. And then the extra thing that we have in here is we may get critical points along the edges.
And if you want to think of the edges, each edge here is like one of those problems because each edge is a little interval. One of the variables is constant on that interval. And so you're really just reverting back to this problem. That's why it's good to just go do this problem and see what the deal is.
Okay, so the extra thing that I need that I didn't have here is I might have a critical point, the edges.
Okay, we can make them plural. Okay, so these are the three possibilities. We didn't have this possibility in the, or if you want to think of it as the other possibility just had two and three. And we have this extra one here. Okay, so what could happen here as we plug in points from inside here, what we get is, you know, we might get a little something, a little flap up there, but it's not flat.
It's a flap that has some shape to it, some three-dimensional shape. So you might get a little maximum here on the edge. You might get a minimum over here on the edge.
So those are what we're trying to discover. Okay, so that's the recipe for how to do this. So let's do, I'm going to do three examples and we just kind of have all the little possibilities. And by the time you watch me do these, you should be pretty good at it
because it really is the same thing over and over again. All right, so let's have this example here. f of xy is equal to 20 plus 4x plus 6y minus x squared minus y squared.
So right away, you don't really know what that looks like, so we have to rely on the calculus and the algebra to get the job done. And then I want to restrict the x's and y's to a rectangle. So let's just say we're restricting it between 0 and 1. So if it helps you, I mean, it's hard to keep track of all this in your head,
go and write down what the interval, not the interval, the rectangle is. Okay, so here's x and here's y. There's 1. So this is my, this is my rectangle r. And then I'm not really going to be interested in the interior so much.
I'm going to go try to find critical points. If some critical points land inside there, then I'll be interested in them. But really, the reason why I draw this is because I want to label the edges. Let's call that L1, L2, L3, L4. You don't have to call this L1. That's just the way I do it. I just start and go counterclockwise there.
You have to pick something to start with. So I don't care whether you call them L1, L2, or E1 for edge. I use L for line segment. Okay, so now that's the setup. What we're going to have to do is just go through those three things. We're going to try to find the critical points in the traditional way.
So that's taking the gradient and then setting it equal to 0. So let's get that here. The gradient of f, that's f sub x, f sub y. When I do the gradient here, I get 4 minus 2x. And then the derivative with respect to y, that's 0, 0.
I get 6. And then minus 2y. And then to find the critical points, we set that equal to 0. Okay, and then this guy equals 0. That tells me that 4 minus 2x equals 0.
I'm just going to say what it is because I ran out of space there. You can see that x is equal to 2 there. I'll just write it down below. So I have 4 minus 2x equals 0. That implies that 2x equals 4, which means x equals 2. And then I have 6 minus 2y equals 0.
That's 2y equals 6, so y equals 3. So I did get a critical point using my traditional way. However, 2, 3, if I were to graph it, 2, 3 is off the...
It's out of my region. Yes? Sorry, how did you get the picture? It's given in the problem. Will it always be given like that? Yeah, the command, I didn't include this here. It's find the max and the min on r.
So I give you the rectangle and I say go find the max and the min. Yeah, this is given in the problem. Okay, so since it's given in the problem, when I go and get this critical point, it has to be in there, because I'm only interested in things that are happening inside that rectangle.
So 2, 3 is not in r, so we don't include this in our list of potential points where the max and the min could occur. It turns out, if you want to check that, you can go take the second derivative.
You'll find out that that's a maximum for this function. But you see, I'm not interested in the global maximum. I'm only interested in what the max and the min is on that little rectangle. For practice, you can just go see that that is a maximum on your own. Okay, so now what am I left with? I did this, but I didn't get anything for this.
Yeah, any questions? Okay, so now I move on to this. Now let's find the critical points on the edges. And each one of those edges, it's like doing four separate problems, and they're all just like the Math 2A example. So let's go through that. It's very repetitive. Okay, so on L1, this is what I was describing to you last time, is that when you decide that you're on a particular line segment,
one of the variables is constant on that line segment. So immediately decide what the case is. For L1, you see the y value is moving up and down, but the x value is constant at 1. So here, the defining feature of L1 is that x equals 1.
And we can just go through all of them and do that. For L2, the defining feature is that y equals 1. And for L3, the defining feature over there is that x equals 0. And for L4, the defining feature, you see at the bottom segment there, x is moving, but y is stuck at 0.
So if you like, you can just go through each one of those and put down the defining feature. What happens is as soon as you set one of the variables equal to a constant, then your function just becomes a function of one variable, and you go right back to the Math 2A example. So when x equals 1, I plug in 1 for x,
and my function becomes 20 plus 4 plus 6y minus 1 minus y squared. Let me put that over a little bit. I probably didn't leave myself enough space. So f is equal to 20 plus 4 plus 6y minus 1 minus y squared,
so we can simplify all that to 23 plus 6y minus y squared.
Okay, so now I've got one variable, so let's just take the derivative to see if there's any critical points along that segment from 0 to 1. When I take the derivative here, I get 6 minus 2y, and set it equal to 0, I get 2y equals 6, I get y equals 3. Okay, but y equals 3 is not part of that rectangle,
so this is out, okay? This is not a member of R, so I'm not going to include that. Even though I did get a critical point, it's not part of my, just like over here, right? I solved it and I got x equals 0. It wasn't in the interval, so I just disregarded, or discarded.
Okay, for L2, now y is equal to 1, so f is equal to 20 plus 4x plus 6 minus x squared plus 1, so that's equal to, let's see, minus 1.
So 25 plus 4x minus x squared. Take the derivative, I get negative 2x plus 4, so x equals 2. Set it equal to 0, x equals 2. Okay, but again, we're only interested in x's between 0 and 1, so x equals 2 is not a member of 0, 1,
which is not a member of R. So we have to discard it. So far, we have no potential points for our maxes and our mins. So if you just kind of look ahead, this is what's going to happen in each case. The only one we're going to get is the corners, in the same way that the only ones we got here were the edges.
We're just looking ahead, that's what's going to happen. Let's fill in the details here, just for practice. So when x equals 0, this is a little easier, we get 20 minus, or plus 6y minus y squared, and then f prime is the same as we've had every time,
minus 2y, set it equal to 0, y equals 3, this is not a member of 0, 1. So I don't get any critical points that are valid for me here. Okay, same thing here, when y is equal to 0, I get 20 minus, or plus 4x minus x squared,
the derivative is 4 minus 2x, set it equal to 0, x equals 2, which is not in the interval. You see, it's much more complicated in this one, because we have to go through all this process, but you see the repetition. Once you, you know, when you do a few of these, you start to just be able to look ahead
and see what's going to happen. Okay, so once again, so so far we have no points, and the only thing left to check are the corners. So now let's check the corners, and here I don't have to do any calculus, I'm just literally plugging in the values,
so I have 0, 0, I have 0, 1, I have 1, 0, and I have 1, 1. So let's check f, at 0, 0 I get 20, and I'm plugging in back into that function there. 0, 1, so x is 0 but y is 1,
so it looks like I get 25 for that, check, check me on that. Looks good. And then x is 1, but y is 0, so here I'm going to get f is 23, or is it 20, yeah it's 23.
And then at 1, 1 it's going to be 24, 30 minus 2 is 28. Okay, so now we reveal what the max and the min are. Here's the min, and here's the max. And this is where they occur,
they occur at 0, 0 and at 1, 1.
Any questions about that? Okay, let's kind of go through a couple more of these, just for practice. If you're able to do all the ones that I do in lecture, you'll be pretty set for the final exam. So that's the place to start.
If you want to challenge yourself, you can do the ones, the suggested exercises in the book, but if you're limited on time and you're just getting ready for the exam, I think you'll be okay if you just do the ones from lecture.
Okay, so there's my rectangle, we might as well draw it. Okay, X1, let's label that as X1. And there's X2.
X1 is between 0 and 3, and X2 from 0 to 2. That's our rectangle. We might as well label L1, L2, L3, L4.
We're going to have to go through each of those. Okay, so first things first, let's go see if there's any critical points that happen to occur in the middle of the rectangle. Otherwise, then we'll be relegated to the edges. So first things first, the command for this is find the max and the min on R.
Okay, so let's get the gradient. The gradient is, my partial derivatives are F1, F2.
So F1 will be 2X1 and then minus 2X2 and then plus 0. And then the derivative with respect to the second variable, that's 0.
I get negative 2X1 and then plus 2. Okay, so what can we get from this? We're going to set this all equal to 0.
This one down here can be solved immediately. Let's put it, well, let's just do it in order. So I have 2X1 minus 2X2 equals 0. That tells me that X1 equals X2. So I have a relationship between X1 and X2 from this if I just bring this over to the other side and divide by 2.
And then the second one, 2X1 plus 2 equals 0. That implies that X1 equals 1. So X1 equals 1, but X1 equals X2. So that implies X2 equals 1. So I have a critical point and that 1, 1 is in fact in the rectangle.
So I do have a valid critical point. If you want, you can kind of check F as you go. As soon as you find a point that's potentially valid, just plug in F and see what it is. If I do F of 1, 1, F of 1, 1 is 1, minus 2 is negative 1, plus 2 is 1.
So that's the first part of it, just finding the critical points. Now we're going to move on to the edges.
So for L1, the defining feature for L1 over here is that X2 the whole time,
I'm sorry, X1 the whole time is equal to 3. X2 varies, but X1 is stuck at 3. So let's plug that into the function. F is going to be 9 minus 6X2 plus 2X2.
So that's 9 minus 4X2. Plugging in X1 equals 3 along that line. So let's take our derivative. F prime is equal to negative 4. So what do we do with the derivative?
We set it equal to 0, it's never going to be equal to 0. The derivative is negative 4 there. So what you want to think of is along that, we've revealed something there, along this edge the thing is decreasing. So you can kind of think of the maximum as, well, it's decreasing along this edge here. So as X1 goes from 0, or I'm sorry, as X2 goes from 0 up to 2, it's decreasing.
Okay, so, but the bottom line is for this we get no critical point. L2, so this one's slightly different because in the last one we got critical points every time,
but they didn't occur in the interval so we had to reject them. Here we just don't even get one. Okay, on L2 I think it's going to work a little differently. So on L2 the X1 variable is moving, but X2 is equal to 2 there the whole time.
So let's plug that X2 into the function. F becomes X1 squared minus 4X1 plus 4. That's the new F. Now when I take the derivative, F prime is 2X1 minus 4,
and that gives us, we set that equal to 0, I get X1 is equal to 2. So you put these two points together at the point 2 for X1 and 1 for X2, we have a critical point. We can put that in here.
What just happened here? Oh wait, yeah, that's right. I wrote it. X2 is 2, X2 is 2. That's not a 2.
I mean that's not a 1. Okay, so here's our other critical point right here. I knew it had to be on the edge there. And if you like, you can check F at this point. F of 2, 2, that's equal to 4 minus 8 is negative 4, plus 4 is 0.
So between the two that I have so far, that's the max and that's the min so far, but we might get others. L3. Okay, L3 is this little segment there. The defining feature for that is that X1 is equal to 0 the whole time.
So now when I plug in X1 equals 0, I just get F is really simple, it's just 2X2. It's 0 plus 0 plus 2X2. And when I take the derivative here, I just get 2, which is never equal to 0, so I get no critical point.
Okay, just like L1, same way. L4, the last one. Okay, the defining feature down here, see X1 is moving, but X2 is stuck at 0.
When I plug in X2 equals 0, I get F is equal to X1 squared. And then pick the derivative, set it equal to 0, that gives me X1 equals 0. So I have a critical point at 0, 0. That's my critical point. And then we might as well check F at that point, so we got this critical point right here.
And F of 0, 0 is 0.
Okay, so we've done step 2. We got the original critical point, we checked that, that gave us a value of 1. We got two other critical points along the edges, those gave us values of 0 for the function. And then the last step is to check the corners. What are the corners? We have 0, 0.
We already checked 0, 0, just as there's a little overlap. It just so happened that one of our critical points was a corner. And let's check F at 0 at that point, we already knew that. Then we have the point 3, 0, this corner over the lower right.
3, 0, I get 9 minus 0 plus 0, so F equals 9 there. And then I have the point 3, 2, F of 3, 2.
F of 3, 2, I get 9, minus 12 is negative 3, plus 4 is 1. Check that. Okay, I get 9, minus 12 is negative 3, plus 4 is 1.
And then the last one is 0, 2, that's the last corner. And what's F there? The first two terms are 0, then I get 4 there.
Okay, so at this point I've done everything, I got all the critical points, I just checked which is the biggest and which is the smallest. Looks like this is going to be the max right here. Okay, F equals 9 is the max. And then I don't have any negative values, I have two places where we got 0.
So this is the min here, and this is also the min. So let me just write down the formal, I don't have any space there on those boards, so let me just write down what the final answer is. So the minimum value is 0, and then it occurs at two locations, 0, 0 and 2, 0.
And then the maximum, that occurs at 3, 0.
3, 2? What do I have here? Yeah, 2, 2, thanks. And then this one occurs at 3, 0.
Okay, that's everything all written out in all its grandeur. Do I have an error here? No? Yeah. Because I just couldn't, I set the derivative equal to 0 and it couldn't equal 0. I didn't get a variable here, so I couldn't solve anything.
This just says that it has a slope of 2 along that whole edge, so it's never going to hit a maximum. But notice the difference, here I had an x to solve for. So when I solved for the x, I got a number and then I could say I had a critical point here. Here there's no x to solve for, so it could never equal 0. For the same reason as L2.
When I took the derivative, I got something to solve for. And then I found out x1, 0, and then I pair that up with that one, so I get 0, 0. The good news is when you do this, you really feel a sense of satisfaction because you're working for a while
and you're working out all these little details and you come to the conclusion at the end, you feel like, yeah, I really, really did some math there.
Any other questions? All right, let's just do one more. We can go a little faster through this one since we've seen the process a bunch of times now.
We'll have a simpler function here.
We're going to do the saddle. Oops, that's not the saddle. That's the saddle.
Okay, so let me give you the interval. Let's say that x1 is between negative 1 and 1 and so is x2. So let's immediately write down the picture here. Here's negative 1, here's positive 1, positive 1, negative 1.
We're talking about this rectangle. Call this, I'm not very creative, I just keep calling the same ones L1, L2, L3, L4.
Okay, so let's go find the critical point. We can do this, these derivatives will be quick, so it'll work out a little more quickly overall. Okay, the gradient is f1, f2. That's going to be 2x1, negative 2x2, and then we set that equal to 0.
So x1 equals 0, 2x2 equals 0. That implies x2 equals 0. So I have the critical point 0, 0, and it does occur inside the rectangle, so I need to use it as a candidate.
And we might as well check f at that point. Okay, so let's check L1.
For L1, the defining feature here is that x1 is stuck at 1 there the whole time on L1.
So then f is equal to, if x1 is 1, then I put 1 in there, I get 1 minus x2 squared. Then the derivative is equal to negative 2x2, set it equal to 0, x2 equals 0.
Okay, so that is between 0 and 1, so I've got the point 1, 0. And that is on there. 1, 0. Oh, this is going to be a pain. I'm going to put all the critical points in now so I don't have to keep pulling this thing down.
So those are going to be all our critical points. And there's the first one, 1, 0. They really do just work out exactly the same way. I mean, if plugging in one constant does one thing, then when you go over to the other side,
you're essentially plugging the same variable in as a constant, so it's going to give you the same results. You start to notice some patterns in this. Okay, L2, the defining feature up there is that x2 is equal to 1. So then f is equal to x1 squared minus 1. The derivative is 2x1, set it equal to 0, that gives me x1 equals 0.
So now I got the point 0, 1. And there you go. Okay, then L3.
L3 is, the defining feature for L3 is that x1 is equal to negative 1. So then f becomes 1 minus x2 squared. So f prime is negative 2x2, set it equal to 0, x2 equals 0, and we get the point negative 1, 0.
So that's that x there. And then for L4, the last one, that's where x1 is moving, but x2 is stuck at negative 1.
So f becomes x1 squared minus 1, f prime, and then x1 equals 0. Okay, so we're done with that, we've got 0, negative 1. We can check f at all these points here.
f of 1, 0 is equal to 1. f of 0, 1 is equal to 1. f of negative 1, 0, that's equal to 1. And f of 0, negative 1, that's all equal to 1. So we have a nice easy function to plug all those things in pretty simply.
I'll take that back. This one's negative 1, and this one's negative 1 as well. So be careful, there is a minus in there.
Okay, so so far, if this was my complete list, I would have maxes occurring here and minimums occurring here because 0 is in between, right? So more extreme is 1, that's higher, and negative 1 is lower, so that one won't be counted. Okay, but I need to at least check the corners just to make sure that I didn't get some other values that beat out those.
So the corners, okay, those are going to occur at, there's one at 1, negative 1. There's one at 1, 1. So let's plug those in. f of 1, negative 1, that's equal to 0, and you plug that in.
And f of 1, 1, that's equal to 0 as well, so we haven't beat any of those. And then I've got negative 1, 1, f of negative 1, 1, that's also equal to 0.
And f of negative 1, let's be consistent here, that's the other corner, negative 1, negative 1, f of that is also equal to 0. So this one here is going to be the max, or sorry, the min, so it's negative 1. This is a max, this is a min, and this is a max.
So the maximum is 1, and it occurs at two locations, and the minimum is negative 1, and it occurs at two locations as well.
Once you get this down, I can't really mess you up, because what I could do to make it harder is just make the algebra harder. That's the only thing, I can't make the process any harder.
You find the critical point, if it's in the region you keep it, you find the critical points on the edges, then you get the corner points, you collect all those things and check the function. So I can't really change the method, all I can do is just make the functions a lot harder to deal with, but then that just adds unnecessary algebra. So this is really the level that you need to demonstrate that you studied
this, you learned the method, and you can expect functions that are of this caliber. As soon as I start putting cubes and things, it just becomes impossible. Okay, so any questions about that? Now we need to move on to the last topic, section 13.1.
So this is the culmination of it all. In a way, this was too. The next section is one step up from this, and it doesn't really have an analogous thing from Math 2A.
This truly is just unique to multivariable calculus, Lagrange multipliers.
Lagrange was a mathematician, so this is a person's name.
And then we'll find out what these multipliers are in a minute. But what I want to do is I want to take our standard example that I always start with, our paraboloid, and I want to work out a really detailed example of all this. Before I even tell you what the formulas are, I just want to set up the problem, and then we'll kind of discover what the method is, and then we'll apply it to more general functions.
Okay, so here's our setting. We'll take, we'll start with the paraboloid because we already know that one. It's something familiar, we can write out all the pictures, and we can see exactly what's going on. If I start with one where we don't know the pictures, then I can't show you what's going on.
I can only do the algebra, which is not as enlightening. So let me remind you of the picture here.
So there's our parabola. Let's label the axes here. X, Y, and this is F or Z.
It's often Z, but here we have F equals that. Okay, so as it stands right now, there's nothing new. So I have to give you a new command for you to be able to have a new problem. So here's the new command. Find the max and min, if it exists, just try to find them.
The max and the min subject to the constraint X plus Y equals two.
Okay, so what's different about this is compare since we have another example of what we were doing. Over here, I didn't have the word constraint, but I was constrained.
I was put into a little region, but we don't use that language. You just say on that region. Now, I want to find the maximum and minimum, but it's not a region that I'm interested in. I'm interested in this line. I'm only picking points along this line and then seeing what the maximum min is on this paraboloid.
So let's see what that is. If I do, this is really Y equals negative X plus two. Let's graph that. Negative X plus two is when X is zero, Y is two.
And when Y is zero, X is two. So that's the, let's put the, label the axis here.
There's X and there's Y. Now let's, see this is good practice for your brain. See if you can put this in the XY plane over there. Can you? So the X has to match up with the X. So I'm going to put it down in the floor there.
I'm going to match up the X and the Y. Okay, so what's going on here? I have two in the X direction. I have two in the Y direction. And then I connect those values. So now you see this line here is playing the role of the rectangle.
Up here, the rectangle was like this. Here's your rectangle. And then I have some function and I only plug in values from there so I get a little rectangle on the function. And I want to know what the max is up there on that little rectangle.
So let's do the same thing for this. I'm only allowed to plug in points from this line. So try to imagine, what's it going to look like on here? Can I draw it on here? Can you see it in your mind? When I plug in points, so let's say I plug in this point here
and then I plug in this point and so on. What I end up with is a little parabola on the paraboloid. So it's like this. Take this and just bend it up. Paste it. Take this and paste it on there and then it's like a string
and I want to paste the string on the vase. Think of this as a vase sitting on your table and you've got a string here and you're going to take the string and you're going to paste it on there and there's going to be a little minimum here. We want to discover that minimum. You can also see from this picture that there is no maximum. This goes forever out this way so as I keep plugging in points
I just get points higher and higher up there. So there is no maximum but there is a minimum. So there's the min and there's no max. So what we want to do is we want to discover the method of how to find this thing.
Now where is this min occurring? It's, you see, it's further up on the graph. In fact, if I plug in 1, 1 here, that's going to be the, I think that's going to be where it is, the 1, 1. That's where it's going to occur.
So this is going to be at 2 here. So you have to imagine that on this we have all these levels. Each level, as I go up, it intersects this graph and this is the minimum one. But you see, these levels below here, they never intersect.
So what we want to do is let's look at, this is why I say this topic puts everything together here. So what I want to do is I'm going to put the level sets here over on this graph and let's see what happens. In fact, I think I'm just going to reproduce it again and we'll have a bigger version of it so we can really see what's going on.
Any questions so far, just about the setup? Okay.
Okay, I'm going to reproduce this a little bigger here. X, there's Y. Let's go 1 here and 2, 1 and 2.
So this guy right here is X plus Y equals 2. And where have we ever had, can you imagine a situation where we had a bunch of XY stuff
and we set it equal to a constant? When do we do that? When do we take a function and set it equal to a constant and then graph it? What's the concept that we're doing in that case? Very good, level sets.
So what I can do is I can pretend, so this is like having a function equal to a constant. So there's a bigger function going on here and this is just one level set of that function. So I'm going to call this G of XY equals X plus Y and this is the level set of G, G equals 2.
But now I'm making it more confusing by doing that but you see that gives us a little advantage. If I view this as just a level set of a larger function
then I can do things like take the gradient of that guy. I can take the gradient of this. And we'll do that in a minute. But what I want to do is I want to start to introduce the level sets of the other function. So let's go slowly here. If I have F equals 0, what is the level set for that?
That's just this little, let's do this in white here. So that's the level set for F equals 0, right? Now what happens if I go up one?
So if I go to, so for F equals 0, let me just write it out. So that's X plus Y squared, we've done this before, that's why I don't want to go through all the level sets again but this gave you a single point. And then the next level set, F equals 1, that's X squared plus Y squared equals 1. So that's the unit circle, so I can draw that.
So this is for F equals 1, that little, that unit circle. But you see I haven't intersected yet. I haven't intersected any of this. So what that is is it's the first few levels there and I haven't gotten up to that minimum point yet. Now the question is, when do we hit this for the very first time?
That's going to be the minimum value, right? So this is F equals 1. If I go out, say to F equals 2, then what's that? F equals, or sorry, F equals 4. If I go out to F equals 4, then that's a circle of radius 2.
So let's draw that one in. So now I have a circle of radius 2. Now I've intersected it, okay, I've intersected it, but at this level of F equals 4, I've already intersected it.
My goal is to find the very first one that touches that line. So it's not 1 because this doesn't ever intersect it. Once I get to 4, yeah, that intersects it, but see if you can agree with me that the very first one,
this is the very first level that touches this line right there. That's the very first level. See, this is lower, this is 0, here's 1, here's 4, so somewhere in between 1 and 4, it hits it for the very first time. And this point right here where they're tangent is what we're looking for.
That's what we want to find. We want to find the very first level. So that's what I'm drawing here, compare that to the levels there. You see those little concentric circles? As I, this is the bottom, and then this is one unit up, so then I go up one, and then here's two units up, but I've already passed over it, and then here's the magic one.
I want to find out when it exactly kisses this, just tangent. Okay, so how do we find this? This magic point here, this, we're going to use the fact that
the gradient is always perpendicular to the level sets. And I mentioned that when we first talked about level sets, but we can see it again. So let's take a look at the gradient of G here. That's equal to 1, 1. So that 1, 1, this is the gradient of G.
It's perpendicular to that line. The gradient is always perpendicular, let me write that down here. The gradient is always perpendicular to the level set.
So that's the gradient of G, and what's going to happen with the gradient of F? See, right there at that point where they're just tangent like that, the gradients line up. That's the magic point, the gradients, because if they're tangent,
then the gradient of F, let's draw it with another color here, it'll also be perpendicular to its level set. So right at that point, these two guys have the same angle because they're tangent there, the gradient of F will be pointing in the same direction.
That's the magic point. Okay, so I'm just showing you what's going on. Now I need to show you how we find this. What's the algebra? And that's really what you're going to be interested in because you're not always going to be able to draw the pictures, you're not always going to be able to know every little thing that's going on. What you need is the method for finding this magic point where the level set matches up with the constraint function,
just at that kissing point, that tangent point. Okay, so what does it mean when two vectors are lined up? Let's write that down.
What does it mean when two vectors line up? Try to answer that in your mind. If I told you I had two vectors and they were pointing in the same direction, what could you say about one relative to the other? So here's a vector, and I tell you that I have another vector w
that's in the same direction as this. They're lined up. Well, if w is in the same direction, then it's just a multiple. It's a scalar multiple of it because it keeps the same direction, it just maybe has a different length. So let me write that down a little more precisely.
Okay, so the gradient of f, this is your original function,
and the gradient of g, this, we have to keep things straight because now we have two functions floating around. This is what's called your constraint. And you know that from the original description of the problem. Find the max and min subject to the constraint.
So the constraint is this other g function, and then f is your original function. So these are...
they're pointing in the same direction when the gradient of f is equal to a scalar multiple of the other one and lambda is a constant. Now it's a little beyond the scope of this class but
the fact that we use lambda is not a coincidence we also use lambda when we're doing eigenvalues so this is a form of an eigenvalue in fact it looks kind of similar when we did eigenvalues before we did AX equals lambda X. So it's a little bit like that. You can think of AX as F and this
X as G. But anyway, so I just want to point out that it's not an accident that we're using lambda in this case. But it's not that important that you know that. What we need to do is be able to solve this equation. So when the vectors are pointing together we have this property that
one vector is a constant multiple or scalar multiple of another vector. By the way this lambda, this is called the Lagrange multiplier. That's where the name comes from. That lambda is the Lagrange multiplier.
You never really solve for lambda. I mean you can and we might even do it in one of our examples but the bottom line is to find that minimum value. Remember that's our goal over there is to find this minimum value. Okay now here's the punchline. So solving
this equation will find the locations of the max of the min of F subject. So that's the bottom line. Okay so that
what you want to do is once you've learned this you'll want to go back and
look at this example and go through all the details. When you're first seeing it for the first time it kind of just you know might just go over your head a little bit. So now what we need to do is we want to see that this solving that equation. First of all I want to learn how to solve it. And then I'm telling you that when you solve this
this is where these vectors are all lined up right at that tangent point and that'll give you your maxes and min. It'll give you the locations of the maxes and min. Okay so now what we need to do is we need to do that. We need to learn how to deal with that equation.
So let's go back and start over with the problem and just work through it as if we didn't have all this discussion here. We're just going to use this fact here. Okay so we have
f of xy is x squared plus y squared and we're doing this subject to the constraint x plus y equals two. So immediately take this. This is going to become your g function, the constraint.
So immediately you just take all the stuff that's not part of the constant and set it equal to g. So g of xy equals x plus y. So that's why I say the constraint g equals a constant. See I have the function g equals a constant.
Let's find the minimum value because we know in advance here there is no maximum. So let's just find the minimum value. Okay now I'm telling you that if you solve that equation it will find the location of this minimum value because
of this picture I drew. It's right where those gradients line up. So we just want to find where the gradients line up and we'll be there at that location. Okay so let's do that. If I'm going to solve that equation I have to know what the different gradients are. The gradient of f, that's 2x2y.
That's not a big mystery. And then the gradient of g, I'm going to take the gradient over here, I get 1, 1. One for x and one for y. Okay so now I've got the two things, the two vectors and now I want to find out the locations of x and y where these things
are lined up. So gradient of f equals lambda gradient of g that means I have 2x2y equals lambda times 1, 1 and that gives me, you can think of that as lambda times lambda. So that gives me two equations.
I've got 2x equals lambda and I've got 2y equals lambda. Okay so let's just pause for a second.
I want to find x and y but look what I have here. I have a system of equations but I've got three variables. I've got two equations and three variables. Do I have any hope of finding the answers to those three variables if I only have two equations?
It's under-determined, right? That's an under-determined system. So if I really want to get a value for x and y I need a third equation. Where's that third equation going to come from? It's going to come from g but this is g. This is the third equation.
This is my third equation. See I haven't used this information yet. What I did was I ignored the two for a second and I said okay I have this function and I took the gradient but I haven't used this information. So that's the last piece of information that's going to go with this system of equations that's going to allow me to solve for x and y.
Three variables, three equations and in fact one of the variables isn't even in this equation so this becomes irrelevant. Okay so let's see how we do that. Let's go do the algebra for that. Let me reproduce that. I have 2x equals lambda. I have 2y equals
lambda and I have x plus y equals 2. And from that I need to find the value of x and y. So what I recommend you doing is, see we're interested in x and y so lambda is irrelevant. So just solve for lambda and plug it into the
other equation. So here's what I mean. So since lambda equals 2x, why don't I just take that 2x and just put it in for lambda there. So lambda equals 2x and 2y equals lambda that implies that 2y equals 2x and then that implies that y equals x.
So from these two equations you can get a relationship between x and y then you take that relationship and you plug it into here and that'll tell you what x and y are. So x plus y equals 2. Let's put y equals x. Let's put that in there. We get 2x equals 2.
x equals 1. Now if y equals x then we have y equals 1. So this is the location where those gradients line up. The location, the max or the min, I don't know what it is yet. It's just the location.
This is just the location of that special point. I still have to do a little bit more work to figure out whether it was a max or a min or not. It's at f of 1, 1
is equal to 2. If you just plug in to x squared plus y squared you get 2. Now I just need to plug in any other point on that constraint to see whether that's a max or a min. Plug in any other
point x plus y equals 2. So how about 0, 2? I just want to compare. If this is a max or a min let's say it's a max then when I plug that in this will be bigger. If it's a min
then when I plug this in this one will be bigger. That'll be smaller. So f of 0, 2 when you plug that in you get 4 so that means that this must have been a min. So the location of the min is at 1, 1.
That's the picture right here. This is 1, 1 right here. So we just found that location and then when we did that we weren't sure if it was a max or a min yet but I knew the location was either a max or a min. So I just test another point and I find out
that the original one was smaller so it had to be a minimum. Okay so it's a little bit tough the first time through because I gotta explain everything a little bit of the theory and we're also doing an example so now let's take- yes? How do you catch why you picked 0, 2? Okay so I just want to pick any other point
on this line here so anything that works I could have had 2, 0. I don't want to pick 1, 1 because I already had it so I just pick one of them to be 0. I could do 4, negative 2 anything just to compare.
Okay let's
let's do a new example. Any questions? Where's my- oh there it is. Okay that's what I want.
Let's extract out the method that we just used here
and this will have a recipe that we can follow. Okay so what do we do here? Here's the recipe for Lagrange. So first let's have a
a generic picture. So we're talking about just some function of two variables and
I'm going to find the max and the min of f of x, y subject to the constraint g of x, y equals k.
I'm doing this all in generality here so I don't have to get caught up in one example. Okay so what was this guy in that last example? That was the line there, right? So the line is very specific. If I want to have it in general
let's take that line and grab it, grab it by the end there and just go like that. So rather than having a line there I could just have some general squiggle. So that's g of x, y
equals k. That's my little constraint there. So what happens when I go and I'm only allowed to pick points from here and plug them in up here so what I get is a little squiggle up there and I want to find potentially the min and the max. That's my goal to find the min and the max there up there and then how do I do it?
So let's look at what we did here. We took the level sets of the original function and when the level sets just got right up to that constraint and just kissed it and was just tangent to it that's where the maximum and minimum occurred. So how do I draw the level sets? The level sets of that function
let's put them in red here. So the level sets, I don't know what they look like, this is just generic but let's say they're, you know, like this. These level sets. Well at some point you get right up and you just kiss one of them and then maybe you have another one over here that just kisses it right there and those two points where the level sets just come up and are tangent
see that's a good calculus where it is tangent. Everything is always tangent, right? That's where the magic occurs at the tangent spot. Those two locations give you the maxes and the mins and the way you find them
so let's do this. Let's write out the recipe here. To find the max and min for F subject to G we have a three-step process.
So what was the first thing we did? We solved where the gradient of F is lined up with the gradient of G and that's the way we say it mathematically. I can say it in English that they're lined up but mathematically it means that this vector is a
scalar multiple of that vector. When we solved this here if you look back, this is this case right here. We had two X equals lambda two Y equals lambda and from that all we got was a relationship between X and Y.
All we did was just get X and equals Y. I didn't get a value yet and then the next step was to take that and plug it into the constraint. So that's what I'm going to do next. So plug solution to one into
the constraint and solve. And then the third step was plug solution to part two into
F to see max. So because in that one example we just did we had a little
it was part of the third step it was this little extra goody that we had to check an extra point here. So I'll put in check additional
on the constraint if necessary.
As we'll see in some of these other examples it doesn't always happen so that's why it's a little extra step. You can make it step four if you want but it's really part of step three because you've already got your answer in step three. You plug it into the function
and in that last case we didn't have a true answer so we had to check an additional point. Okay so there's our little method here. Now let's have a nice example and we'll just go through this recipe and you'll see that it's really not that bad. It's only bad when you have no idea what I'm talking about which is at the beginning. Now you have a little bit of an idea what I'm
talking about. Okay so let's have a new example. Here's f of x is 2x plus y. Can you visualize what that is? What if I tell you it's linear?
You see it's linear? It's a word that we're supposed to be familiar with in this class. It's linear. I don't have any powers so this is the two variable version of a line which is a plane. So this is just a plane. I don't know what it is exactly. We can graph it a little bit and I will in a second. I'll show you what was going on here.
But first there's the function that in order for it to be this setting I need to have a constraint. So I'm gonna find the max and the min subject to the constraint
x squared plus y squared equals one. Can you visualize that? So we've seen this a bunch of times now. I might even still have it on the board over there. X squared plus y squared equals one. That's a circle, right? This is the unit circle. So my constraint is the unit circle.
We've got that function. So let's start to go, let's go to work here. First things first, let's write this down. This is my g, right? This is the g equals a constant. So let's write that down. g of x y is x squared plus y squared. And the first step in my recipe here
is to take the gradient of f and set it equal to lambda times the gradient of g. And just see what we get from that. Okay, I can't do that until I get the gradients. So let's, let me go to a clean board here.
So gradient of f equals lambda times the gradient of g. Let's go get these things. The gradient of f, that's f sub x
f sub y. When I do the derivative with respect to x, I just get two and then one. And the gradient of g is equal to 2x 2y.
gx gy. So let's get a little picture of what's going on here just so you can see it. Okay, so here's the function, the original function is linear. It's a plane. So just picture it like this. Okay, there's a plane. That's f.
And then g of x y equals one. That x squared plus y squared equals one. That's the unit circle. So that's sitting down there in the floor, the unit circle. That's x squared plus y squared equals one.
And so then I just take points along here, plug them in. What do I get here? I get something like that. Every time I plug in a point up here, it just gives me a little circle up there on that plane. And I'm interested in finding
this is the min and that's the max. So what I do is I take the level sets of this guy. So here's all the level sets of this and I want the one that's tangent here and tangent back there. That's what I'm looking for.
And that occurs when I solve this equation. That's what's happening here. Okay, so let's solve it.
Alright, so gradient of f equals lambda gradient of g. That means I have two one equals lambda times two x two y. So that gives me two equations. I have two equals two x.
That's just from the top line. Just multiply through by lambda. So I have two equals two lambda x. And I have one equals two lambda y. What I recommend doing is this is going to work for almost I think all the examples I might give you. All the ones in lecture and
any one that you get on a test is just solve for lambda in one of them and plug it into the other equation. Because then what happens is you don't have lambda anymore and you just have x and y then you can get a relationship between x and y. That's step two here. Once I've solved this, then in step two I'll take that relationship and I plug it back into the constraint and that gives me the actual numbers.
Okay, so for this one here I can say that lambda see if you agree with me, lambda is equal to one over x. If I solve for lambda there. Now I'm going to take that see this is lambda equals one over x and I'm going to put it in right there.
Okay, so then I have one equals two lambda is one over x times y. So that means that if I solve this is one equals two y over x, so that means that x equals two y.
See now I have a relationship between x and y. Now I'm going to, so this is step one here. Now in step two I take that relationship and I plug it into x squared plus y squared equals one. That's the constraint. So let's take
that two y and put it in for x there. So I have two y squared plus y squared equals one. So five y squared equals one.
That implies that y is equal to plus or minus one over root five. Okay, so I have y equals one over root five and I have y equals negative one over root five. Those are my two values and x is equal to two y.
So this implies over here that x equals two over root five and over here it implies that x equals negative two over root five. So what I've found here is two points. I've got the point
two over root five, one over root five, and the point negative two over root five, negative one over root five. Those are my two magic points. So let's look back at the recipe. I just plugged in the solution to one into the constraint and solved it and I got answers.
Those are my final answers. Those are the points where the gradients are lined up. The last step says plug solution two into the function to see the maximum in. So let's do that. Step three here is f of two
over root five, one over root five. What's that equal to? I've got two x plus y. So that's equal to, let's put it here, so that's two times two over root five plus one over root five. That's equal to
five over root five. See if you can appreciate this. That's equal to root five. If I multiply top and bottom by root five I get this. And that's the first point. And then the second point is f of negative two over root five, negative one over root five.
That's equal to two times negative over root five plus or minus one over root five. So that's negative five over root five, which is negative root five.
So you see now it's revealed to us which is the max and which is the min. The magic occurs at these points and then the function values root five at this point and it's negative root five at that point. So that means this is the max and this is the min. And there's no need to test another point because it's revealed to us which one is the max and the min.
So the picture that goes with this, I already have the three-dimensional picture, but let's do the two-dimensional one.
So there's, that's g of x, that's just x squared plus y squared equals one. That's the constraint.
And then the level sets, let's look at f of x, y, that's two x plus y. What are the level sets of this? This is good practice anyway. We've got a final in a week. So the level sets, let's say I have f equals zero.
That means I have y equals negative two x. So that's this graph right here. Goes through here. This is one, this is negative one. It's got a slope of negative two. This is f equals zero. And as I do these level sets for f,
let's just do f equals two, that's two x plus y. So y is equal to negative two x plus two. So that's now this graph here. It's just a bunch, the level sets of this function here are just a bunch of parallel lines.
And the magic ones that we found were the ones that just went right up to the constraint. And right here. Oops, I'm not parallel there. And then, so we got this point here,
and we got this point over here too. So right there where those are tangent,
this is, this is where f equals root five, and this is where f equals negative root five. That's the level set for those points, and they're just tangent to that circle at those points. And it can put in the values here. This is negative two over root five, negative one over root five,
that's this point right here, and this one there is two over root five, one over root five. That's what we just found. And the gradients are all lined up there. Because the tangents, they're tangents so the perpendiculars have to be in the same line. That's what we just did there.
Okay, I got a few more examples of this to do, but I'm going to save them for next time. That's enough for today. I'll give you a lot today.