- [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.