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So, today we are going to
continue looking at critical
points,
and we'll learn how to actually
decide whether a typical point
is a minimum,
maximum, or a saddle point.
So, that's the main topic for
today.
So, remember yesterday,
we looked at critical points of
functions of several variables.
And, so a critical point
functions, we have two values,
x and y.
That's a point where the
partial derivatives are both
zero.
And, we've seen that there's
various kinds of critical
points.
There's local minima.
So, maybe I should show the
function on this contour
plot,there is local maxima,
which are like that.
And, there's saddle points
which are neither minima nor
maxima.
And, of course,
if you have a real function,
then it would be more
complicated.
It will have several critical
points.
So, this example here,
well, you see on the plot that
there is two maxima.
And, there is in the middle,
between them,
a saddle point.
And, actually,
you can see them on the contour
plot.
On the contour plot,
you see the maxima because the
level curves become circles that
now down and shrink to the
maximum.
And, you can see the saddle
point because here you have this
level curve that makes a figure
eight.
It crosses itself.
And, if you move up or down
here, so along the y direction,
the values of the function will
decrease.
Along the x direction,
the values will increase.
So, you can see usually quite
easily where are the critical
points just by looking either at
the graph or at the contour
plots.
So, the only thing with the
contour plots is you need to
read the values to tell a
minimum from a maximum because
the contour plots look the same.
Just, of course,
in one case,
the values increase,
and in another one they
decrease.
So, the question -- -- is,
how do we decide -- -- between
the various possibilities?
So, local minimum,
local maximum,
or saddle point.
So, and, in fact,
why do we care?
Well, the other question is how
do we find the global
minimum/maximum of a function?
So, here what I should point
out, well,
first of all,
to decide where the function is
the largest,
in general you'll have actually
to compare the values.
For example,
here, if you want to know,
what is the maximum of this
function?
Well, we have two obvious
candidates.
We have this local maximum and
that local maximum.
And, the question is,
which one is the higher of the
two?
Well, in this case,
actually, there is actually a
tie for maximum.
But, in general,
you would have to compute the
function at both points,
and compare the values if you
know that it's three at one of
them and four at the other.
Well, four wins.
The other thing that you see
here is if you are looking for
the minimum of this function,
well, the minimum is not going
to be at any of the critical
points.
So, where's the minimum?
Well, it looks like the minimum
is actually out there on the
boundary or at infinity.
So, that's another feature.
The global minimum or maximum
doesn't have to be at a critical
point.
It could also be,
somehow, on the side in some
limiting situation where one
variable stops being in the
allowed rang of values or goes
to infinity.
So, we have to actually check
the boundary and the infinity
behavior of our function to know
where, actually,
the minimum and maximum will
be.
So, in general,
I should point out,
these should occur either at
the critical point or on the
boundary or at infinity.
So, by that,
I mean on the boundary of a
domain of definition that we are
considering.
And so, we have to try both.
OK, but so we'll get back to
that.
For now, let's try to focus on
the question of,
you know, what's the type of
the critical point?
So, we'll use something that's
known as the second derivative
test.
And, in principle,
well, the idea is kind of
similar to what you do with the
function of one variable,
namely, the function of one
variable.
If the derivative is zero,
then you know that you should
look at the second derivative.
And, that will tell you whether
it's curving up or down whether
you have a local max and the
local min.
And, the main problem here is,
of course, we have more
possible situations,
and we have several
derivatives.
So, we have to think a bit
harder about how we'll decide.
But, it will again involve the
second derivative.
OK, so let's start with just an
easy example that will be useful
to us because actually it will
provide the basis for the
general method.
OK, so we are first going to
consider a case where we have a
function that's actually just
quadratic.
So, let's say I have a
function, W of (x,y) that's of
the form ax^2 bxy cy^2.
OK, so this guy has a critical
point at the origin because if
you take the derivative with
respect to x,
well, and if you plug x equals
y equals zero,
you'll get zero,
and same with respect to y.
You can also see,
if you try to do a linear
approximation of this,
well, all these guys are much
smaller than x and y when x and
y are small.
So, the linear approximation,
the tangent plane to the graph
is really just w=0.
OK, so, how do we do it?
Well, yesterday we actually did
an example.
It was a bit more complicated
than that, but let me do it,
so remember,
we were looking at something
that started with x^2 2xy 3y^2.
And, there were other terms.
But, let's forget them now.
And, what we did is we said,
well, we can rewrite this as (x
y)^2 2y^2.
And now, this is a sum of two
squares.
So, each of these guys has to
be nonnegative.
And so, the origin will be a
minimum.
Well, it turns out we can do
something similar in general no
matter what the values of a,
b, and c are.
We'll just try to first
complete things to a square.
OK, so let's do that.
So, in general,
well, let me be slightly less
general, and let me assume that
a is not zero because otherwise
I can't do what I'm going to do.
So, I'm going to write this as
a times x^2 plus b over axy.
And then I have my cy^2.
And now this looks like the
beginning of the square of
something, OK,
just like what we did over
there.
So, what is it the square of?
Well, you'd start with x plus I
claim if I put b over 2a times y
and I square it,
then see the cross term two
times x times b over 2a y will
become b over axy.
Of course, now I also get some
y squares out of this.
How many y squares do I get?
Well, I get b^2 over 4a^2 times
a.
So, I get b2 over 4a y^2.
So, and I want,
in fact, c times y^2.
So, the number of y^2 that I
should add is c minus b^2 over
4a.
OK, let's see that again.
If I expand this thing,
I will get ax^2 plus a times b
over 2a times 2xy.
That's going to be my bxy.
But, I also get b^2 over 4a^2
y^2 times a.
That's b^2 over 4ay^2.
And, that cancels out with this
guy here.
And then, I will be left with
cy^2.
OK, do you see it kind of?
OK, if not, well,
try expanding this square
again.
OK, maybe I'll do it just to
convince you.
But, so if I expand this,
I will get A times,
let me put that in a different
color because you shouldn't
write that down.
It's just to convince you again.
So, if you don't see it yet,
let's expend this thing.
We'll get a times x^2 plus a
times 2xb over 2ay.
Well, the two A's cancel out.
We get bxy plus a times the
square of that's going to be b^2
over 4a^2 y^2 plus cy^2 minus
b^2 over 4ay^2.
Here, the a and the a
simplifies, and now these two
terms simplify and give me just
cy^2 in the end.
OK, and that's kind of
unreadable after I've canceled
everything,
but if you follow it,
you see that basically I've
just rewritten my initial
function.
OK, is that kind of OK?
I mean, otherwise there's just
no substitute.
You'll have to do it yourself,
I'm afraid.
OK, so, let me continue to play
with this.
So, I'm just going to put this
in a slightly different form
just to clear the denominators.
OK, so, I will instead write
this as one over 4a times the
big thing.
So, I'm going to just put 4a^2
times x plus b over 2ay squared.
OK, so far I have the same
thing as here.
I just introduced the 4a that
cancels out, plus for the other
one, I'm just clearing the
denominator.
I end up with (4ac-b^2)y^2.
OK, so that's a lot of terms.
But, what does it look like?
Well, it looks like,
so we have some constant
factors, and here we have a
square, and here we have a
square.
So, basically,
we've written this as a sum of
two squares, well,
a sum or a difference of two
squares.
And, maybe that's what we need
to figure out to know what kind
of point it is because,
see, if you take a sum of two
squares,
that you will know that each
square takes nonnegative values.
And you will have,
the function will always take
nonnegative values.
So, the origin will be a
minimum.
Well, if you have a difference
of two squares that typically
you'll have a saddle point
because depending on whether one
or the other is larger,
you will have a positive or a
negative quantity.
OK, so I claim there's various
cases to look at.
So, let's see.
So, in fact,
I claim there will be three
cases.
And, that's good news for us
because after all,
we want to distinguish between
three possibilities.
So, let's first do away with
the most complicated one.
What if 4ac minus b^2 is
negative?
Well, if it's negative,
then it means what I have
between the brackets is,
so the first guy is obviously a
positive quantity,
while the second one will be
something negative times y2.
So, it will be a negative
quantity.
OK, so one term is positive.
The other is negative.
That tells us we actually have
a saddle point.
We have, in fact,
written our function as a
difference of two squares.
OK, is that convincing?
So, if you want,
what I could do is actually I
could change my coordinates,
have new coordinates called u
equals x b over 2ay,
and v, actually,
well, I could keep y,
and that it would look like the
difference of squares directly.
OK, so that's the first case.
The second case is where
4ac-b^2 = 0.
Well, what happens if that's
zero?
Then it means that this term
over there goes away.
So, what we have is just one
square.
OK, so what that means is
actually that our function
depends only on one direction of
things.
In the other direction,
it's going to actually be
degenerate.
So, for example,
forget all the clutter in
there.
Say I give you just the
function of two variables,
w equals just x^2.
So, that means it doesn't
depend on y at all.
And, if I try to plot the
graph, it will look like,
well, x is here.
So, it will depend on x in that
way, but it doesn't depend on y
at all.
So, what the graph looks like
is something like that.
OK, basically it's a valley
whose bottom is completely flat.
So, that means,
actually, we have a degenerate
critical point.
It's called degenerate because
there is a direction in which
nothing happens.
And, in fact,
you have critical points
everywhere along the y axis.
Now, whether the square that we
have is x or something else,
namely, x plus b over 2a y,
it doesn't matter.
I mean, it will still get this
degenerate behavior.
But there's a direction in
which nothing happens because we
just have the square of one
quantity.
I'm sure that 300 students
means 300 different ring tones,
but I'm not eager to hear all
of them, thanks.
[LAUGHTER]
OK, so, this is what's called a
degenerate critical point,
and [LAUGHTER].
OK, so basically we'll leave it
here.
We won't actually try to figure
out further what happens,
and the reason for that is that
when you have an actual
function,
a general function,
not just one that's quadratic
like this,
then there will actually be
other terms maybe involving
higher powers,
maybe x^3 or y^3 or things like
that.
And then, they will mess up
what happens in this valley.
And, it's a situation where we
won't be able,
actually, to tell automatically
just by looking at second
derivatives what happens.
See, for example,
in a function of one variable,
if you have just a function of
one variable,
say, f of x equals x to the
five,
well, if you try to decide what
type of point the origin is,
you're going to take the second
derivative.
It will be zero,
and then you can conclude.
Those things depend on higher
order derivatives.
So, we just won't like that
case.
We just won't try to figure out
what's going on here.
Now, the last situation is if
4ac-b^2 is positive.
So, then, that means that
actually we've written things.
The big bracket up there is a
sum of two squares.
So, that means that we've
written w as one over 4a times
plus something squared plus
something else squared,
OK?
So, these guys have the same
sign, and that means that this
term here will always be greater
than or equal to zero.
And that means that we should
either have a maximum or
minimum.
How we find out which one it is?
Well, we look at the sign of a,
exactly.
OK?
So, there's two sub-cases.
One is if a is positive,
then, this quantity overall
will always be nonnegative.
And that means we have a
minimum, OK?
And, if a is negative on the
other hand,
so that means that we multiply
this positive quantity by a
negative number,
we get something that's always
negative.
So, zero is actually the
maximum.
OK, is that clear for everyone?
Yes?
Sorry, yeah,
so I said in the example w
equals x^2, it doesn't depend on
y.
So, the more general situation
is w equals some constant.
Well, I guess it's a times (x b
over 2a times y)^2.
So, it does depend on x and y,
but it only depends on this
combination.
OK, so if I choose to move in
some other perpendicular
direction,
in the direction where this
remains constant,
so maybe if I set x equals
minus b over 2a y,
then this remains zero all the
time.
So, there's a degenerate
direction in which I stay at the
minimum or maximum,
or whatever it is that I have.
OK, so that's why it's called
degenerate.
There is a direction in which
nothing happens.
OK, yes?
Yes, yeah, so that's a very
good question.
So, there's going to be the
second derivative test.
Why do not have derivatives yet?
Well, that's because I've been
looking at this special example
where we have a function like
this.
And, so I don't actually need
to take derivatives yet.
But, secretly,
that's because a,
b, and c will be the second
derivatives of the function,
actually, 2a,
b, and 2c.
So now, we are going to go to
general function.
And there, instead of having
these coefficients a,
b, and c given to us,
we'll have to compute them as
second derivatives.
OK, so here,
I'm basically setting the stage
for what will be the actual
criterion we'll use using second
derivatives.
Yes?
So, yeah, so what you have a
degenerate critical point,
it could be a degenerate
minimum, or a degenerate maximum
depending on the sign of a.
But, in general,
once you start having
functions, you don't really know
what will happen anymore.
It could also be a degenerate
saddle, and so on.
So, we won't really be able to
tell.
Yes?
It is possible to have a
degenerate saddle point.
For example,
if I gave you x^3 y^3,
you can convince yourself that
if you take x and y to be
negative, it will be negative.
If x and y are positive,
it's positive.
And, it has a very degenerate
critical point at the origin.
So, that's a degenerate saddle
point.
We don't see it here because
that doesn't happen if you have
only quadratic terms like that.
You need to have higher-order
terms to see it happen.
OK.
OK, so let's continue.
Before we continue,
but see, I wanted to point out
one small thing.
So, here, we have the magic
quantity, 4ac minus b^2.
You've probably seen that
before in your life.
Yet, it looks like the
quadratic formula,
except that one involves
b^2-4ac.
But that's really the same
thing.
OK, so let's see,
where does the quadratic
formula come in here?
Well, let me write things
differently.
OK, so we've manipulated
things, and got into a
conclusion.
But, let me just do a different
manipulation,
and write this now instead as
y^2 times a times x over y
squared plus b(x over y) plus c.
OK, see, that's the same thing
that I had before.
Well, so now this quantity here
is always nonnegative.
What about this one?
Well, of course,
this one depends on x over y.
It means it depends on which
direction you're going to move
away from the origin,
which ratio between x and y you
will consider.
But, I claim there's two
situations.
One is, so, let's try to
reformulate things.
So, if a discriminate here is
positive, then it means that
these have roots and these have
solutions.
And, that means that this
quantity can be both positive
and negative.
This quantity takes positive
and negative values.
One way to convince yourself is
just to, you know,
plot at^2 bt c.
You know that there's two roots.
So, it might look like this,
or might look like that
depending on the sign of a.
But, in either case,
it will take values of both
signs.
So, that means that your
function will take values of
both signs.
The value takes both positive
and negative values.
And, so that means we have a
saddle point,
while the other situation,
when b^2-4ac is negative -- --
means that this equation is
quadratic never takes the value,
zero.
So, it's always positive or
it's always negative,
depending on the sign of a.
So, the other case is if
b^2-4ac is negative,
then the quadratic doesn't have
a solution.
And it could look like this or
like that depending on whether a
is positive or a is negative.
So, in particular,
that means that ax over y2 plus
bx over y plus c is always
positive or always negative
depending on the sign of a.
And then, that tells us that
our function,
w, will be always positive or
always negative.
And then we'll get a minimum or
maximum.
OK, we'll have a min or a max
depending on which situation we
are in.
OK, so that's another way to
derive the same answer.
And now, you see here why the
discriminate plays a role.
That's because it exactly tells
you whether this quadratic
quantity has always the same
sign,
or whether it can actually
cross the value,
zero, when you have the root of
a quadratic.
OK, so hopefully at this stage
you are happy with one of the
two explanations,
at least.
And now, you are willing to
believe, I hope,
that we have basically a way of
deciding what type of critical
point we have in the special
case of a quadratic function.
OK, so, now what do we do with
the general function?
Well, so in general,
we want to look at second
derivatives.
OK, so now we are getting to
the real stuff.
So, how many second derivatives
do we have?
That's maybe the first thing we
should figure out.
Well, we can take the
derivative first with respect to
x, and then again with respect
to x.
OK, that gives us something we
denote by partial square f over
partial x squared or fxx.
Then, there's another one which
is fxy, which means you take the
derivative with respect to x,
and then with respect to y.
Another thing you can do,
is do first derivative respect
to y, and then with respect to
x.
That would be fyx.
Well, good news.
These are actually always equal
to each other.
OK, so it's the fact that we
will admit, it's actually not
very hard to check.
So these are always the same.
We don't need to worry about
which one we do.
That's one computation that we
won't need to do.
We can save a bit of effort.
And then, we have the last one,
namely, the second partial with
respect to y and y fyy.
OK, so we have three of them.
So, what does the second
derivative test say?
It says, say that you have a
critical point (x0,
y0) of a function of two
variables, f,
and then let's compute the
partial derivatives.
So, let's call capital A the
second derivative with respect
to x.
Let's call capital B the second
derivative with respect to x and
y.
And C equals fyy at this point,
OK?
So, these are just numbers
because we first compute the
second derivative,
and then we plug in the values
of x and y at the critical
point.
So, these will just be numbers.
And now, what we do is we look
at the quantity AC-B^2.
I am not forgetting the four.
You will see why there isn't
one.
So, if AC-B^2 is positive,
then there's two sub-cases.
If A is positive,
then it's local minimum.
The second case,
so, still, if AC-B^2 is
positive, but A is negative,
then it's going to be a local
maximum.
And, if AC-B^2 is negative,
then it's a saddle point,
and finally,
if AC-B^2 is zero,
then we actually cannot
compute.
We don't know whether it's
going to be a minimum,
a maximum, or a saddle.
We know it's degenerate in some
way, but we don't know what type
of point it is.
OK, so that's actually what you
need to remember.
If you are formula oriented,
that's all you need to remember
about today.
But, let's try to understand
why, how this comes out of what
we had there.
OK, so, I think maybe I
actually want to keep,
so maybe I want to keep this
middle board because it actually
has,
you know, the recipe that we
found before the quadratic
function.
So, let me move directly over
there and try to relate our old
recipe with the new.
OK, you are easily amused.
OK, so first,
let's check that these two
things say the same thing in the
special case that we are looking
at.
OK, so let's verify in the
special case where the function
was ax^2 bxy cy^2.
So -- Well, what is the second
derivative with respect to x and
x?
If I take the second derivative
with respect to x and x,
so first I want to take maybe
the derivative with respect to
x.
But first, let's take the first
partial, Wx.
That will be 2ax by, right?
So, Wxx will be,
well, let's take a partial with
respect to x again.
That's 2a.
Wxy, I take the partial respect
to y, and we'll get b.
OK, now we need,
also, the partial with respect
to y.
So, Wy is bx 2cy.
In case you don't believe what
I told you about the mixed
partials, Wyx,
well, you can check.
And it's, again, b.
So, they are,
indeed, the same thing.
And, Wyy will be 2c.
So, if we now look at these
quantities, that tells us,
well, big A is two little a,
big B is little b,
big C is two little c.
So, AC-B^2 is what we used to
call four little ac minus b2.
OK, ooh.
[LAUGHTER]
So, now you can compare the
cases.
They are not listed in the same
order just to make it harder.
So, we said first,
so the saddle case is when
AC-B^2 in big letters is
negative, that's the same as
4ac-b2 in lower case is
negative.
The case where capital AC-B2 is
positive, local min and local
max corresponds to this one.
And, the case where we can't
conclude was what used to be the
degenerate one.
OK, so at least we don't seem
to have messed up when copying
the formula.
Now, why does that work more
generally than that?
Well, the answer that is,
again, Taylor approximation.
Aww.
OK, so let me just do here
quadratic approximation.
So, quadratic approximation
tells me the following thing.
It tells me,
if I have a function,
f of xy, and I want to
understand the change in f when
I change x and y a little bit.
Well, there's the first-order
terms.
There is the linear terms that
by now you should know and be
comfortable with.
That's fx times the change in x.
And then, there's fy times the
change in y.
OK, that's the starting point.
But now, of course,
if x and y, sorry,
if we are at the critical
point, then that's going to be
zero at the critical point.
So, that term actually goes
away, and that's also zero at
the critical point.
So, that term also goes away.
OK, so linear approximation is
really no good.
We need more terms.
So, what are the next terms?
Well, the next terms are
quadratic terms,
and so I mean,
if you remember the Taylor
formula for a function of a
single variable,
there was the derivative times
x minus x0 plus one half of a
second derivative times x-x0^2.
And see, this side here is
really Taylor approximation in
one variable looking only at x.
But of course,
we also have terms involving y,
and terms involving
simultaneously x and y.
And, these terms are fxy times
change in x times change in y
plus one half of fyy(y-y0)^2.
There's no one half in the
middle because,
in fact, you would have two
terms, one for xy,
one for yx, but they are the
same.
And then, if you want to
continue, there is actually
cubic terms involving the third
derivatives, and so on,
but we are not actually looking
at them.
And so, now,
when we do this approximation,
well, the type of critical
point remains the same when we
replace the function by this
approximation.
And so, we can apply the
argument that we used to deduce
things in the quadratic case.
In fact, it still works in the
general case using this
approximation formula.
So -- The general case reduces
to the quadratic case.
And now, you see actually why,
well, here you see,
again, how this coefficient
which we used to call little a
is also one half of capital A.
And same here:
this coefficient is what we
call capital B or little b,
and this coefficient here is
what we called little c or one
half of capital C.
And then, when you replace
these into the various cases
that we had here,
you end up with the second
derivative test.
So, what about the degenerate
case?
Why can't we just say,
well, it's going to be a
degenerate critical point?
So, the reason is that this
approximation formula is
reasonable only if the higher
order terms are negligible.
OK, so in fact,
secretly, there's more terms.
This is only an approximation.
There would be terms involving
third derivatives,
and maybe even beyond that.
And, so it is not to generate
case,
they don't actually matter
because the shape of the
function,
the shape of the graph,
is actually determined by the
quadratic terms.
But, in the degenerate case,
see, if I start with this and I
add something even very,
very small along the y axis,
then that can be enough to bend
this very slightly up or
slightly down,
and turn my degenerate point in
to either a minimum or a saddle
point.
And, I won't be able to tell
until I go further in the list
of derivatives.
So, in the degenerate case,
what actually happens depends
on the higher order derivatives.
So, we will need to analyze
things more carefully.
Well, we're not going to bother
with that in this class.
So, we'll just say,
well, we cannot compute,
OK?
I mean, you have to realize
that in real life,
you have to be extremely
unlucky for this quantity to end
up being exactly 0.
[LAUGHTER]
Well, if that happens,
then what you should do is
maybe try by inspection.
See if there's a good reason
why the function should always
be positive or always be
negative, or something.
Or, you know,
plot it on a computer and see
what happens.
But, otherwise we can't compute.
OK, so let's do an example.
So, probably I should leave
this on so that we still have
the test with us.
And, instead,
OK, so I'll do my example here.
OK, so just an example.
Let's look at f of (x,
y) = x y 1/xy,
where x and y are positive.
So, I'm looking only at the
first quadrant.
OK, I mean, I'm doing this
because I don't want the
denominator to become zero.
So, I'm just looking at the
situation.
So, let's look first for,
so, the question will be,
what are the minimum and
maximum of this function?
So, the first thing we should
do to answer this question is
look for critical points,
OK?
So, for that,
we have to compute the first
derivatives.
OK, so fx is one minus one over
x^2y, OK?
Take the derivative of one over
x, that's negative one over x^2.
And, we'll want to set that
equal to zero.
And fy is one minus one over
xy^2.
And, we want to set that equal
to zero.
So, what are the equations we
have to solve?
Well, I guess x^2y equals one,
I mean, if I move this guy over
here I get one over x^2y equals
one.
That's x^2y equals one,
and xy^2 equals one.
What do you get by comparing
these two?
Well, x and y should both be,
OK, so yeah,
I agree with you that one and
one is a solution.
Why is it the only one?
So, first, if I divide this one
by that one, I get x over y
equals one.
So, it tells me x equals y.
And then, if x equals y,
then if I put that into here,
it will give me y^3 equals one,
which tells me y equals one,
and therefore,
x equals one as well.
OK, so, there's only one
solution.
There's only one critical
point, which is going to be
(1,1).
OK, so, now here's where you do
a bit of work.
What do you think of that
critical point?
OK, I see some valid votes.
I see some, OK,
I see a lot of people answering
four.
[LAUGHTER]
that seems to suggest that
maybe you haven't completed the
second derivative yet.
Yes, I see someone giving the
correct answer.
I see some people not giving
quite the correct answer.
I see more and more correct
answers.
OK, so let's see.
To figure out what type of
point is, we should compute the
second partial derivatives.
So, fxx is, what do we get what
we take the derivative of this
with respect to x?
Two over x^3y, OK?
So, at our point, a will be 2.
Fxy will be one over x^2y^2.
So, B will be one.
And, Fyy is going to be two
over xy^3.
So, C will be two.
And so that tells us,
well, AC-B^2 is four minus one.
Sorry, I should probably use a
different blackboard for that.
AC-B2 is two times two minus
1^2 is three.
It's positive.
That tells us we are either a
local minimum or local maximum.
And, A is positive.
So, it's a local minimum.
And, in fact,
you can check it's the global
minimum.
What about the maximum?
Well, if a maximum is not
actually at a critical point,
it's on the boundary,
or at infinity.
See, so we have actually to
check what happens when x and y
go to zero or to infinity.
Well, if that happens,
if x or y goes to infinity,
then the function goes to
infinity.
Also, if x or y goes to zero,
then one over xy goes to
infinity.
So, the maximum,
well, the function goes to
infinity when x goes to infinity
or y goes to infinity,
or x and y go to zero.
So, it's not at a critical
point.
OK, so, in general,
we have to check both the
critical points and the
boundaries to decide what
happens.
OK, the end.
Have a nice weekend.
