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### Course: Multivariable calculus > Unit 3

Lesson 3: Optimizing multivariable functions- Multivariable maxima and minima
- Find critical points of multivariable functions
- Saddle points
- Visual zero gradient
- Warm up to the second partial derivative test
- Second partial derivative test
- Second partial derivative test intuition
- Second partial derivative test example, part 1
- Second partial derivative test example, part 2
- Classifying critical points

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# Warm up to the second partial derivative test

An example of looking for local minima in a multivariable function by finding where tangent planes are flat, along with some of the intuitions that will underly the second partial derivative test. Created by Grant Sanderson.

## Want to join the conversation?

- Can I use laplacian to find a multivariable maxima and minima? Because in the videos about the laplacian, it says that laplacian can determine whether that point is lower or higher than other points surrounding it?(17 votes)
- I think we can't. Because laplacian just evaluate the value of the area around the point which is analogous to the average value.(6 votes)

- At7:32, we conclude that neither the second partial derivative with respect to x, nor the first partial derivative with respect to x are a function of y. So, does that mean that the steepness and concavity of the graph-plane intersection do not depend on what y-value we slice at? But doesn't the curve look much steeper, for example, when we slice at y = sqrt(2) than when we slice at y = 0?(5 votes)
- Yes, neither the second partial derivative with respect to
**x**nor the first partial derivative with respect to**x**are dependent on**y**. But remember, the function of interest is dependent on**both***x* and**y**. Thus, in order to truly understand the steepness and concavity of the**entire**3d function, we must**also**examine the first and second partial derivatives with respect to**y**. I hope this cleared things up :)

If it didn't, please let me know, as this is an extremely important conceptual issue you raised and an understanding of this concept is crucial to an understanding of multi-variable calculus.(3 votes)

- So are there extreme points that would technically not count as a max, min, or a saddle-point? And if so, what do they look like?(2 votes)
- Yes, such a point can be constructed.

Here's a 1 degree-of-freedom example:

consider f(x)=sin(x)-x.

At the origin the first and second derivatives are both zero, but it is not a maximum nor a minimum.

Moving up one degree-of-freedom, consider the function:

f(x,y)=sin(x)-x+sin(y)-y.

Again, at the origin all first and second derivatives are zero, but it is not a minimum, maximum, or saddle.

I wish these comments would let me post a graph for you, as it's quite easy to see that the 1st and 2nd derivatives are zero at the origin, but that it is not a maximum nor a minimum nor a saddle. It is an inflection point.(4 votes)

- Do we strictly need the second derivative to determine if we have a minima or maxima? Why can't we just enter the solution into the original equation to find the value of the equation at those points and just compare these values to see which is the maxima, minima or saddle points?

For example the solution for $f(x,y) = x^4-4x^2+y^2$ is: ${(0, 0), (0, \sqrt{2}), (0, -\sqrt{2})}$

And...

$f(0, 0) = 0$

$f(0, \sqrt{2}) = 4-4*2+0 = -4$

$f(0, -\sqrt{2}) = 4-4*2+0 = -4$

Therefore: (0, 0) is a local maxima or saddle point

While the other points are local minima.(2 votes)- You can, but similar to the second derivative test from single-var calc, you can use the test to determine is a specific point is special, i.e. a saddle point, etc. Which you can't do by simply plugging in the numbers.(2 votes)

- wait whats the thing on the graph ?(1 vote)
- In math, we should simplify as much as possible. At5:00, if we simplify and divide both sides with 4x, we are left with xˆ2-2=0. Then, the only answer to this equation is -√2 and √2.

If I don't divide by 4x, then one of the answers can be 0.

What's wrong with me? Should I simplify it or not?(0 votes)- x should equal 0. Simplifying by dividing by 4x makes it so that x cannot equal 0, because then you would be dividing by 4*0 which would mean you would be dividing by zero. But since x does equal 0, you should not simplify it that way.(1 vote)

## Video transcript

- So in single variable calculus, if you have a function f
of x and you want to find the maximum or the
minimum of this function, what you do is you find its
derivative and you set that equal to zero. And graphically, this has
the interpretation that, you know, if you have the graph
of f, setting its derivative equal to zero means that
you're looking for places where its got a flat tangent line. So in the graph that I drew, it would be these two flat tangent lines. And then once you find these
points, you know for example, here you have one
solution that I'll call x1 and then here you have
another solution, x2, you can ask yourself the
question are these maxima, or are they minima, right? Because both of these can
have flat tangent lines. So when you do find this
and you want to understand is it a maximum or a minimum, if you're just looking at
the graph, we can tell. You can tell that this point
here is a local maximum and this point here is a local minimum. But if you weren't looking at the graph there's a nice test that
will tell you the answer. You basically look for the
second, second derivative and in this case because
the concavity is down, that second derivative is
going to be less than zero, and then over here, because
the concavity is up, that second derivative
is greater than zero. And by getting this
information about the concavity you can make a conclusion that
when the concavity is down, you're at a local maximum,
when the concavity is up, you're at a local minimum. In the case where the second,
second derivative is zero, it's undetermined. You'd have to do more
tests to figure it out. It's unknown. So in the multi-variable world, the situation is very similar. As I've talked about in
previous videos, what you'd do is you'd have some kind of function and let's say it's a
two variable function, and instead of looking for where
the derivative equals zero, you're gonna be looking for where the gradient of your function
is equal to the zero vector, which we might make bold to
emphasize that that's a vector. And that corresponds with
finding flat tangent planes. If that seems unfamiliar, go back and take a look at
the video where I introduce the idea of multi-variable
maxima and minima. But the subject of this
video is gonna be on what is analogous to this
second derivative test, where in the single variable
world, you just find the second derivative and check if it's
greater than or less than zero. How can we, in the multi-variable world, do something similar to
figure out if you have a local minimum, a local
maximum, or that new possibility of a saddle point, that I
talked about in the last video? So there is another test and it's called the second partial derivative test. I'll get to the specifics of that at the very end of this video. To set the landscape, I want
to actually talk through a specific example where we're finding when the gradient equals zero, just to see what that looks like and just to have some concrete
formulas to deal with. So, the function that
you're looking at right now is f of xy is equal to x to the fourth, minus four x squared, plus y squared. Okay, so that's the function
that we're dealing with. In order to find where
its tangent plane is flat, we're looking for where
the gradient equals zero. And remember, this is just
really a way of unpacking the requirements that
both partial derivatives, the partial derivative
of f with respect to x, at some point, and we'll
kind of write it in as we're looking for the x and
y where this is zero, and also where the partial
derivative of f with respect to y at that same point, xy is equal to zero. So the idea is that this is
gonna give us some kind of system of equations that
we can solve for x and y. So let's go ahead and actually do that. In this case, the partial
derivative with respect to x, we look up here and the only
places where x shows up, we have x to the fourth
minus four x squared, so that x to the fourth,
turns into four times x cubed, minus four x squared, that
becomes minus eight x, and then y, y just looks like a constant. So we're adding a constant
and nothing changes here. So the first requirement is that this portion is equal to zero. Now the second part, where we're looking for the partial derivative
with respect to y, the only place where y shows
up is this y squared term, so the partial derivative with
respect to y is just two y. And we're setting that equal to zero. I chose a simple example
where these partial derivative equations, you know this
one nicely only includes x and this one nicely only includes y but that's not always the case. You can imagine if you
intermingle the variables a little bit more, these
will actually intermingle Xs and Ys and it'll be a
harder thing to solve. But I just want something where we can actually start to find the solutions. So if we actually solve this
system, this equation here, the two y equals zero,
just gives us the fact that y has to equal zero. So that's nice enough, right? And then this second equation,
that four x cubed minus eight x equals zero, let's
go ahead and rewrite that where I'm going to
factor out one of the Xs and factor out a four, so
this is four x multiplied by x squared, minus
two, has to equal zero. So there's two different ways that this can equal zero, right? Either x itself is equal to zero, so that would be one
solution, x is equal to zero, or x squared minus two
is zero, which would mean x is plus or minus the square root of two. So we have x is plus or
minus the square root of two. So the solution to the
system of equations, we know that no matter
what, y has to equal zero, and then one of three
different things can happen. X equals zero, x equals
positive square root of two, or x equals negative square root of two. So this gives us three separate solutions, and I'll go ahead and write them down. Our three solutions as ordered
pairs are gonna be either zero, zero; for when x
is zero and y is zero. You have square root of two, zero. And then you have negative
square root of two, zero. These are the three different points, the three different values, for x and y that satisfy
the two requirements that both partial derivatives are zero. What that should mean on the graph then is when we look at those
three different inputs, all of those have flat tangent planes. So the first one, zero, zero,
if we kind of look above, I guess we're kind of
inside the graph here, zero, zero, is right at the origin. We can see, just looking at the graph, that that's actually a saddle point. You know, this is neither a local maximum nor a local minimum. It doesn't look like a
peak or like a valley. Then the other two, where we
kind of move along the x axis, and that guess it turns out
that this point here is directly below x equals positive
square root of two, and this other minimum is directly below x equals negative square root of two. I wouldn't have been able
to guess that just looking at the graph but we just figured it out. We can see visually that both
of those are local minima. But the question is, how could
we have figured that out, once we find these solutions,
if you didn't have the graph to look at immediately, how
could you have figured out that zero, zero corresponds
to a saddle point, and that both of these other solutions correspond to local minima? Well following the idea
of the single variable second derivative test, what you might do is take the
second partial derivatives of our function and see how
that might influence concavity. For example, if we take the
second partial derivative with respect to x, and I'll
try to squeeze it up here. Second partial derivative of the function, with respect to x, and
we're doing that twice, we're taking the second
derivative of this expression, with respect to x, so we
bring down that three, and that's gonna become 12
because three times four times x squared, 12 times
x squared minus eight, minus eight. So what this means, woah,
kind of moved that around. What this means in terms
of the graph is that if we move purely in the x direction, which means we kind of
cut it with a plane, representing a constant y value, and we look at the slice
of the graph itself, this expression will tell us the concavity at every given point. So these bottom two
points here correspond to plus and minus x equals
the square root of two. So if we go over here
and think about the case where x equals the square root of two, and we plug that in to the expression, what are we gonna get? Well, we're gonna get 12 multiplied by, if x equals square root of two, then x squared is equal to two, so that's 12 times two, minus eight. So that's 24 minus eight. We're gonna get 16. Which is a positive number,
which is why you have positive concavity at each of these points. So as far as the x direction
is concerned it feels like oh yes, both of these
have positive concavity, so they should look like local minima. Then if you plug in zero,
if instead we went over here and we said x equals zero,
then when you plug that in, you'd have 12 times zero, minus eight. And instead of 16, you would
be getting negative eight. So because we have a negative
amount that gives you this negative concavity on
the graph, which is why, as far as x is concerned, the origin looks like a local maximum. So let's actually write that down. If we kind of go down
here and we're analyzing each one of these, and we think
about what does it look like from the perspective of each variable? As far as x is concerned, that origin should look like a max, and then each of these two
points should look like minima. This is kind of what
the variable x thinks. And then the variable y,
if we do something similar, and we take the second partial
derivative with respect to y, I'll go ahead and write that over here because this'll be pretty quick, second partial derivative
with respect to y, we're taking the derivative
of this expression, with respect to y, and
that's just a constant. That's just two. And because it's positive,
it's telling you that as far as y is concerned, there's positive concavity every where. And on the graph, what that would mean, what that would mean, if you
just look at things where you kind of slicing with a constant x value to see pure movement in the y direction, there's always going to
be positive concavity. And here I've only drawn the plane where x is constantly equal to zero, but if you imagine kind of
sliding that plane around left and right, you're always
getting positive concavity. So as far as y is concerned, everything looks like a local minimum. So we kind of go down here
and you'd say everything looks like a local minimum. Minimum, minimum, and minimum. So it might be tempting here
to think that you're done, to think you found all the
information you need to. Because you say well in the x
and y direction, they disagree about whether that origin should
be a maximum or a minimum, which is why it looks like a saddle point, and then they agree, they
agree on the other two points, that both of them should
look like a minimum, which is why, which is why you could say, you think you might
say, both of these guys look like a minimum. However, that's actually not enough. There are cases, there are
examples that I could draw where doing this kind of
analysis would lead you to the wrong conclusion. You would conclude that
certain points are, you know, a local minimum when in
fact they're a saddle point. And the basic reason is that
you need to take into account information given by that other
second partial derivative. Because in the multi-variable world, you can take the partial derivative with respect to one variable, and then with respect to another. And you have to take into
account this mixed partial derivative term in order
to make full conclusions. And I'm a little bit
afraid that this video might be running long,
so I'll cut it short here and then I will give you the
second partial derivative test in it's full glory, accounting for this mixed partial derivative
term in the next video. I'll also, you know, give
intuition for where this comes in, why this comes in, why
this simple analysis that we did in this case is close
and it does give intuition but it's not quite full
and it won't give you the right conclusion always. All right, I will see you then.