Main content

### 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

© 2024 Khan AcademyTerms of usePrivacy PolicyCookie Notice

# Second partial derivative test example, part 2

Continuing the worked example from the previous video, now classifying each critical point. Created by Grant Sanderson.

## Want to join the conversation?

- at5:30, when Grant compute the mixed partial derivative, he doesn't multiply sqrt(3) by 6: (Df/Dyx)=6x = 6*sqrt(3) = sqrt(108),
**so FxFy-Fxy^2 = -108**

I understand that in this case, the magnitude of the mixed partial Derivative didn't matter since the other term =0, but is there another reason he didn't compute or might've he forgotten?(52 votes)- No. Chiarandini's right. All results are -108 except for one that's 36>0. I think Grant's very stressed doing these videos! That's too bad because he's doing an awesome job otherwise.(27 votes)

- Wouldn't (0,0) also be the global maximum because it is the only maximum?(6 votes)
- No, because it is a local maximum only. Take the graph of y=x^3-x and compute the local maximum and draw the graph (with wolframalpha or so). Then you can easily see that only one local maximum exists, but that this only local maximum clearly is not the global maximum. The same principle applies to multivariable functions. If the function goes towards infinity at some point the gradient never gets to 0 because the function-value keeps getting higher and higher (towards infinity) and we only "see" it when the gradient is 0 (because that's what we calculate). I hope you got my explanation, my english isn't that well :X(15 votes)

- at5:43should it be -6*6*3=-108?, since the second derivative xy=6x. why 6 was missed?(9 votes)
- because the x value for that critical point (0,-2) was 0, so we sub in 0 into the mixed derivative, 6x = 6(0) = 0. The video is correct.(0 votes)

- Hi! Thanks for sharing this new awesome MultiVariable Calculus course. Are you going to upload some video when the Second Partial Derivative Test is equal to zero? Last exam I had something like: find a,b,c Real Numbers, such that ax-by-2+2xy+cx^2-y^2. How should I proceed when I see this? Thanks!(5 votes)
- In this example what if fxx were positive and fyy is negative, would that be a min or max(3 votes)
- It would be neither, it would be a saddle point. You can't have a max or a min if fxx and fyy have different signs, and that's true for any function, not just this one.(4 votes)

- At1:56, Grant forgot to write the mixed partial derivative as d^2F in the numerator, right?(3 votes)
- I initially thought that the 2nd-derivative test was simply the determinant of the Hessian; when the top right and bottom left terms of the Hessian are the same, then it is the same as the final term in the 2nd-derivative test. But in this case (if I've done my math correctly), the Hessian's top right and bottom left terms are not equal (6x and 6y). In this case, the determinant of the Hessian is NOT the same as the 2nd-derivative test.

Am I mistaken? Are there any connections between the determinant of the Hessian and the topography of the graph?(2 votes)- It is the determinant of the Hessian, both mixed partial derivatives are 6x, idk where you got 6y from but we're only considering cases where the second derivatives are continuous around the critical point, in which case the mixed partial derivatives are equivalent according to Schwarz's theorem.(2 votes)

- How is it possible to have multiple saddle points and only one maximum?(2 votes)
- Why did you take fxx not fyy sign to determin the max or min?(1 vote)
- Since H is positive, the point is a relative maximum or minimum, meaning the signs of the pure second partial derivatives agree. Because we already know the signs are the same, there is no need to determine the sign of both when one is enough.(2 votes)

## Video transcript

- [Voiceover] In the
last video, we were given a multi variable function, and asked to find and classify
all of its critical points. So critical points,
just means finding where the gradient is equal to zero, and we found four
different points for that. I have them down here. They were zero zero, zero negative two, square root of three and one, and negative square root of three and one. So then the next step
is to classify those. And that requires the second
partial derivative test. So what I'm gonna go ahead and do, is copy down the partial derivatives. Since we already computed those. Copy, and then just kind
of paste them down here. Where we can start to use them for the second partial derivatives. So I'm gonna clean things up a little bit, and we don't need this
simplification of it. So we've got our partial derivatives. Now since we know we want to apply the second partial derivative test. We've got to first just
compute all of the different second partial derivatives
of our function. That's just kind of the first thing to do. So let's go ahead and do it. The second partial
derivative of the function with respect to x twice in a row. Will take the partial
derivative with respect to x, and then do it with respect to x again. So this first term looks
like six times a variable times a constant, so it'll just be six times that constant. And then the second term. The derivative of negative
six x, is just negative six. Moving right along. When we do the second partial derivative with respect to y twice in a row. We take the partial
derivative with respect to y, and then do it again. So this x squared term looks like nothing. It looks like a constant
as far as y is concerned so we ignore it. The derivative of negative three y squared is negative six times y, and then the derivative of negative six y is just negative six. And then we can't forget
that last crucially important mixed partial derivative term. Which is the partial derivative of f. Where first we do it with respect to x, and then with respect to y. The order doesn't really
matter in this case since it's a perfectly
ordinary polynomial function. So we could do it either way, but I'm just gonna choose
to take a look at this guy and differentiate it with respect to y. So the derivative of the
first term with respect to y is six x, six x. And then that second term
looks like a constant with respect to y, so that's all we have. So now what we're gonna do is plug in each of the critical points to the special second partial derivative test expression. And to remind you of what that is. That expression is we take
the second partial derivative with respect to x twice. And I'll just write it with a kind of shorter notation using subscripts. Then we multiply that by the
second partial derivative with respect to x, and
then we subtract off. Subtract off the mixed partial
derivative term squared. So let's go ahead and do
that for each of our points. So when we do this at the point zero zero. Zero zero, what we end up getting. Plugging that into the partial derivative with respect to x twice. Six times zero is zero, so
that's just negative six. So that gives us negative
six multiplied by. When we plug it into
this partial derivative with respect to y squared. Again, that y goes to zero. So we're left with just negative six. And then we subtract off the
mixed partial derivative term. Which in this case is zero. Cause when we plug in x
equals zero, we get zero. So we're subtracting off zero squared. And that entire thing equals negative six times negative six is 36, 36. And we'll get to analyzing what it means that that's positive in just a moment, but let's just kind of get
all of them on the board so we can kind of start
doing this with all of them. If we do this with zero and negative two. Zero and negative two. Then once we plug in y equals negative two to this expression. This time I'll write it out. Six times negative two minus six, so that's negative 12 minus six. We'll get negative 18, negative 18. Then when we plug it into
the partial derivative of f with respect to y squared. Again, I'll kinda write it out. We have negative six times y is equal to negative two minus six. So now we have negative
six times negative two, so that's positive 12 minus six. So that will be a positive
six that we plug in here. And then for the mixed partial derivative. Again, x is equal to zero. So the mixed partial
derivative is just gonna look like zero when we do this. So we're subtracting off zero squared and we get negative 18 times six. And geez what's 18 times six. So that's gonna be 36 times three. So that's the same as 90. Plus 18, so I think that's 108. Negative 108, and the specific
magnitude won't matter. It's gonna be the sign that's important. And this is definitely negative. So now kind of moving right along. These examples can take quite a while. If we plug in square root of three one. Square root of three one, what we get. Now instead of plugging
in y equals negative two. We're plugging in y equals one. So that'll be six times one minus six. So the whole thing is just zero. And then for the partial derivative with respect to y squared. Instead of plugging in negative two. Now we're plugging in y equals one. So we have negative
six times one minus six so the whole thing is negative 12. So negative 12, and now for the mixed partial derivative term. Which is six x. X is equal to the square root of three. So now we're subtracting
off the square root of three squared. So what that equals is, this first part is just entirely zero, and we're subtracting off three. So that's negative three. And then we have square root of three. No, no we don't, that's what we just did. Now we have negative
square root of three one. And this will be very similar
'cause this first term just had a y and we plugged in a y. So it's also gonna be zero. For totally the same reasons,
and same deal over here. The value of y didn't change. So that's also gonna be negative 12. Doesn't really matter cause we're multiplying it by zero, right? And then over here, now we're plugging in negative square root of three, and that's gonna have the same square. So again we're just subtracting off three. So what does the second partial
derivative test tell us? Once we express this term. If it's greater than zero. We have a max or a min. That's what the test tells us. And then if it's less than zero. If it's less than zero
we have a saddle point. So in this case, the only
term that's greater than zero is this first one, is this first one. And to analyse whether it's
a maximum or a minimum. Notice that the partial derivative with respect to x twice in a row, or with respect to y twice
in a row was negative. Which indicates a sort
of negative concavity. Meaning this corresponds to a maximum. So this guy corresponds
to a local maximum. Now all of the other three
gave us negative numbers. So all of these other three
give us saddle points. Saddle points. So the answer to the question. The original find and
classify such and such points is that we found four
different critical points. Four different critical points. Zero zero, zero negative two,
square root of three one, and negative square root of three one. And all of them are saddle
points except for zero zero. Which is a local maximum. And all of that is
something that we can tell without even looking at
the graph of the function. And with that I will see you next video.