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# Constrained optimization introduction

Video transcript

- [Instructor] Hey everyone, so in the next couple of videos, I'm going to be talking about a different sort of optimization problem, something called a Constrained
Optimization problem, and an example of this is
something where you might see, you might be asked to maximize some kind of multi-variable function, and let's just say it was the function f of x,y is equal to x squared, times y. But that's not all you're asked to do, you're subject to a certain constraint where you're only allowed values of x and y on a certain set, and I'm just going to say, the set of all values of x and y, such that, x squared plus
y squared, equals one. And this is something you might recognize as the unit circle, this particular constraint
that I've put on here, this is the unit circle. So one way that you might think
about a problem like this, you know, you're maximizing a
certain two-variable function, is to first think of the
graph of that function. That's what I have pictured here, is the graph of f of x,y,
equals x squared, times y. And now this constraint,
x squared plus y squared, is basically just a
subset of the x,y plane. So if we look at it head on here, and we look at the x,y plane, this circle represents all of the points x,y, such that, this holds. And what I've actually drawn here isn't the circle on the x,y plane, but I've projected it up onto the graph. So this is showing you
basically the values where this constraint holds, and also what they look like when graphed. So the way you can think
about a problem like this, is that you're looking on this circle, this kind of projected
circle onto the graph, and looking for the highest points. And you might notice kind of here, there's sort of a peak
on that wiggly circle, and over here there's another one, and then the low points would be, you know, around that
point and around over here. Now this is good, and I think this a nice way to sort of wrap your head around
what this problem is asking, but there's actually a
better way to visualize it, in terms of finding the actual solution, and that's to look only in the x,y plane, rather than trying to graph things, and just limit our perspective
to the input space. So what I have here are the contour lines for f of x,y equals x
squared plus y squared. And if you're unfamiliar with contour lines or a contour map, I have a video on that, you can go back and take a look, it's going to be pretty crucial for the next couple videos
to have a feel for that. But basically, each one
of these lines represents a certain constant value for f. So for example, one of them, one of them might represent
all of the values of x and y, where f of x,y is equal to, you know, two, right, so if you looked
at all of the values of x and y where this is true, you'd find yourself on one of these lines, and each line represents
a different possible value for what this constant here actually is. So what I'm going to do here, is I'm actually going to just zoom in on one particular contour line, right. So this here is something
that I'm going to vary, where I'm going to be able to change what the constant we're
setting f equal to is, and look at how the contour
line changes as a result. So for example, if I
put it around here-ish, what you're looking at
is the contour line, the contour line for f of x,y equals 0.1. So all of the values on
these two blue lines here tell you what values
of x and y satisfy 0.1. But on the other hand, I
could also shift this guy up, and maybe I'll shift it up, I'm going to set to where that constant is actually equal to one, so that would be kind of an alternative, we'll just kind of separate over here. That would be the line where f of x,y is set equal to one, itself. And the main thing I
want to highlight here, is that at some values, like 0.1, this contour line
intersects with the circle, it intersects with our constraint. And let's just thing
about what that means, if there's a point, x and y, on that intersection there, that basically gives us a
pair of numbers, x and y, such that, this is true, that fact that f of x,y equals 0.1, and also that x squared
plus y squared equals one. So it means this is something that actually exists and is possible. And in fact, we can see
that there is four different pairs of numbers where that's true, where they intersect here, where they intersect over here, and then the other two, kind of symmetrically on that side. But on the other hand, if
we look as this other world, where we shift up to the
line f of x,y equals one, this never intersects with the constraint. So what that means is x,y, the pairs of numbers
that satisfy this guy, are off the constraint, they're off of that circle, x squared plus y squared equals one. So what that tells us, as we try to maximize this function, subject to this constraint, is that we can never get as high as one. 0.1 would be achievable, and in fact, if we kind of go back to
that, and we look at 0.1, if I upped that value, and you know, changed
it to the line where, instead what you're looking at is 0.2, that's also possible, because it intersects with the circle. And in fact, you could
play around with it, and increase it a little bit more, and if I go to 0.3, instead, and I go over here and I say,
0.3, that's also possible. And what we're basically trying to do is find the maximum value
that we can put here, the maximum value so that if we look at the line that represents f
of x,y equals that value, it still intersects with the circle. And the key here, the key observation, is that that maximum value happens when these guys are tangent. And in the next video, I'll start going into the details of how we can use that observation, this notion of tangency,
to solve the problem, to find the actual value of x and y that maximizes this,
subject to the constraint. But in the interim, I kind
of want you to mull on that, and think a little bit about
how you might use that. What does tangency mean here? How can you take advantage
of certain other notions that we've learned about
in multi-variable calculus, like, hint, hint, the gradient, to actually solve something like this. So with that, I will see you next video.