**Multi-variable Optimization & the Second Derivative Test — Transcript & Summary | SozAI**
Source: https://sozai.app/transcript/multi-variable-optimization-second-derivative-test/

Learn how to find maxima, minima, and saddle points in multivariable functions using the first and second derivative tests.

## Key Takeaways

- Critical points occur where both partial derivatives are zero.
- The second derivative test uses the Hessian determinant to classify critical points.
- A positive Hessian determinant with positive f\_xx indicates a local minimum.
- A positive Hessian determinant with negative f\_xx indicates a local maximum.
- A negative Hessian determinant indicates a saddle point.

## What the video covers

- Introduction to optimization for functions of two variables and how to find maxima, minima, and saddle points.
- Comparison of single variable critical points (max, min, neither) with multivariable cases.
- Explanation of the new phenomenon in multivariable calculus called a saddle point.
- Use of partial derivatives to find candidate points where maxima or minima may occur.
- First derivative test states both partial derivatives must be zero at local maxima or minima.
- Example of finding critical points for the function f(x,y) = x^2 + y^2.
- Introduction to the second derivative test for multivariable functions and its conditions.
- Role of second partial derivatives and the discriminant (D = f\_xx \* f\_yy - (f\_xy)^2) in classifying critical points.
- Explanation of concavity in each direction and how it relates to maxima, minima, and saddle points.
- Discussion of cases where the second derivative test is inconclusive and the importance of further analysis.

## Chapters

1. 00:00 Introduction to Optimization in Multivariable Functions
2. 00:54 Single Variable Critical Points Recap
3. 01:47 Saddle Points in Multivariable Calculus
4. 02:39 Using Partial Derivatives to Find Critical Points
5. 03:31 First Derivative Test for Multivariable Functions
6. 04:18 Example: Finding Critical Points of x² + y²
7. 05:10 Introduction to the Second Derivative Test
8. 06:06 Concavity and Second Partial Derivatives
9. 07:05 Role of the Hessian and Mixed Partial Derivatives
10. 08:47 Summary and Further Resources

Answers

## Questions about this video

What is a saddle point in multivariable calculus?

A saddle point is a critical point where the function behaves like a maximum in one direction and a minimum in another, which cannot occur in single variable calculus.

How do you find candidate points for maxima or minima in multivariable functions?

Candidate points are found by setting both partial derivatives of the function with respect to each variable equal to zero.

What does the second derivative test tell us about critical points?

The second derivative test uses the Hessian determinant and second partial derivatives to classify critical points as maxima, minima, saddle points, or inconclusive.

## Full Transcript — Download SRT & Markdown

00:03

Speaker A

In this video, we're gonna study optimization. If I have some particular function that depends on an x and y, how do I find where this function is a max? So how do I find where it has a minimum?

00:14

Speaker A

And sometimes, how do I find where it has this interesting new phenomenon called a saddle? In the single variable case, there were actually three different major possibilities. The first of them was you have a minimum. So some point, for example,

00:28

Speaker A

a function x squared, and it would go up. At the value of x equal to 0, you would have a minimum location. The derivative was 0, you had a horizontal tangent line, but the function appeared to sort of bow out

00:40

Speaker A

and flatten off at that point. Likewise, we had maximums and also a third case where it appears the derivative was 0, you had a horizontal tangent line at the value of x equal to 0, but in this third case, an example of

00:54

Speaker A

which is someone like y, it would be x cubed, but it was neither a maximum nor a minimum. It went up and I leveled off and then went the other way. No, the first thing to say is all three of these scenarios

01:05

Speaker A

are peep themselves in the multivariable calculus case. You can have a minimum, you can have a maximum, and that neither situation can also occur. You can be going, not flattening off, and then going up again, but there's a new

01:18

Speaker A

situation that can happen in the multivariable calculus case that couldn't happen before, and that's the idea of a saddle. When I have a saddle, when I look at one of those directions, for example, I've plugged away. If I go along the x

01:33

Speaker A

axis, I get this yellow curve. Then it appears to be a minimum according to the yellow curve, but if you go parallel now to the y axis at the point 0, 0, you get this blue curve which is going downwards

01:47

Speaker A

and then it appears to be a maximum. The idealist saddle is that in one direction it appears to be a maximum, the other appears to be a minimum. When you have both those properties, you get a saddle, and single variable isn't relevant because you

02:00

Speaker A

only ever had one direction that you could go. Now, these two curves, the yellow and blue ones, the yellow one is what happens when I just plug in the value of y. It would is around if I have a multivariable function f of

02:14

Speaker A

x and y, and I plug in y producer, I know these single variable functions, and you can see its graph. Likewise, for the blue one, if I plug in now this time x equal to zero, I also get a single variable

02:25

Speaker A

function f of zero because there are single variable functions. I can use all the analysis from first or calculus, but where to find that something was a minimum, you took the derivative, set it equal to 0, and those gave you candidates

02:39

Speaker A

to be minimums, maximums, or neither. So in both of these cases, I can pick the derivative with respect to x of the yellow curve or the derivative with respect to y of the blue curve, and both of those are

02:51

Speaker A

derivatives of single variable functions. Now, one of the big features of single variable calculus was just because the derivative was 0 didn't tell you it was a max or a minimum or neither. You had to do more and else beyond that, but setting

03:03

Speaker A

the derivatives equal to 0 gives you candidates to be maximums and candidates to be minimums. In the same exact story, it may be true here is just that when I take the derivatives of these yellow and blue curves, the derivatives say of f of

03:17

Speaker A

x comma 0, that is going to turn into a partial derivative in the multivariable case. So this is my statement of the so-called first derivative test. It says if you have a local max or min at a point, then the following must be true. It

03:31

Speaker A

must be true that both of the partials are 0 at that particular point. So this is the exact analog of the previous theorem in single variable calculus saying that if the function had a maximum at one point, the

03:43

Speaker A

derivative had to be 0 at that particular point. Now, let's see an example of this. So I need to give the parabola x squared plus y squared, so I want to go and figure out, well, where is the parcel throat is equal to 0? Where is

03:56

Speaker A

my candidates to be a maximum or a minimum? Well, you take the first part, so privilege is to exit that serum like you do the next is equal to 0 taking the second partial derivative. This is going to do to wife take it

04:07

Speaker A

respect to y. It said that it would is really good wide loser just because I tell you that the partials are both so doesn't tell you it was a maximum or a minimum or neither. We have to investigate a little bit

04:18

Speaker A

further, but in this particular case, if I just look at what the function is, well, the function is always positive or equal to zero of this one spot zero zero. So that one spot zero zero discovered has to be a minimum because the thumb just

04:32

Speaker A

always bigger around it. But notice how I'm doing a bit of an ad arguing here. I had to go and investigate the certain specifics of the function. You changed function and this analysis might be harder or smaller. It's easy to find

04:44

Speaker A

candidates. It's easy to find candidates to be maxims or minimums that the partial is equal to zero, but how do I really know that they're going to be maximums or minimums in a general case where I can't just do this easy little

04:57

Speaker A

bit of analysis? But for that, we have the analog of this second derivative test, and Europeans it's a lot too much, so we have to unpack it. Basically, what it says in the following: first of all, assume that your functions are nice, so what an

05:10

Speaker A

Imagi first and second partial derivatives all are going to exist, and there it would exist in some open disk around your point. So you've got some nicely functions. Then it gives you conditions to tell you whether you've got a max, a min, a saddle, or when the

05:25

Speaker A

test fails and is inconclusive and you cannot use the test. So the test is not complete, tells you sometimes you can use this test and sometimes it's inconclusive, but a good number of times we can use the test, so it's worthwhile

05:38

Speaker A

knowing. Okay, the conditions here are a little bit weird. There's something about the second derivative at x x being negative, being positive, but then there's also this weird expression in pink, the second partial respect to x times the second

05:52

Speaker A

partial respect to y minus the square of the mixed partials f xy and whether this is greater than zero or less. What's going on? So let's try to break it down piece by piece. First, I'm gonna study that's an example

06:06

Speaker A

of a minimum here. Now, what I've highlighted in yellow here is the curve when I plug in y equal to zero, so I have f of x zero. Then the second derivative of that curve tells me the concavity of

06:23

Speaker A

it. If the second derivative is positive, it tells me that this yellow curve is facing up, it's concave. If the yellow curve here is concave down, the second derivative will be negative. Indeed, we use this analysis of concavity

06:38

Speaker A

in single variable calculus to tell whether or not our particular point that was a candidate remax or a minimum was a maxima or minima, and so indeed it makes a lot of sense that we demand that the second partial is going to be positive

06:51

Speaker A

at this point down at the bottom is going to be a minimum. And likewise, if it's going to be a maximum, you'd expect the exact opposite, that the second partial has to be negative, which means it's concave down. However, studying just

07:05

Speaker A

the person with the x is not enough. You might think, okay, well, why don't you go and also look at the second partial factor y? So if you go the other direction, if you fix your x value or y

07:15

Speaker A

is changing, look at the second derivative. That is a concave up, concave down. So that is actually done a little bit in the middle of the pink expression. The pink expression is where f y y first appears, but it's a little bit complex, so

07:29

Speaker A

let's switch to the third exam area that could happen to see why we might have this. Let's look now at the stat. Well, in the case of the saddle, what we have is that my second partial respective x

07:41

Speaker A

is positive, but my second partials but 2y is negative. So this means it's sort of concave up in one way, concave down the other, consistent with my notion of a saddle point. Then if I go and look at

07:54

Speaker A

what the product of these two things are, the partial of f respective x twice, the partial of y with about two.

08:09

Speaker A

opposite sign so for instance in the case when they have opposite sign like one is concave up in the one direction and concave down in the other so positive second derivative the one Russian negative second derivative the other then the product FX

08:23

Speaker A

and at YY will be negative since we're then subtracting something square is always positive in the case of a saddle what you have is that if these mixed partials or opposite directions for sure it's going to be a saddle so we think

08:38

Speaker A

it's so far we justify why we have the FX X being less than zero and greater than zero with differential between maxims minimums and we've also seen the sort of first part of my pig expression the F xx times FY why why the sign of

08:51

Speaker A

that might be important for telling us whether or not we have a saddle but what about the other part of this expression consider the following weird example ok so this is a new example it is a weird type of saddle it's actually the

09:05

Speaker A

function x squared plus y squared minus 3x y know if I go along here and I specify that I am on the x axis I plug in Y for the 0 then what I just get is the curve x squared that's what I've

09:18

Speaker A

highlighted in yellow and x squared its second partial derivative is just 2 which is positive it's concave up as you go along the x-axis alright but what happens when you go along the y-axis well I lose the exact

09:32

Speaker A

same thing if you're going along the y axis of this now it's also concave up indeed the function is y squared along the y axis and its second derivative expected Y is 2 so whether you all on the X whether you go on the Y is concave

09:47

Speaker A

up in both directions which might make you think that it should be a minimum and yet if I choose this somewhat clever third what if I go along the path y equal to X map so I'm here I'm

10:02

Speaker A

restricting to y equal to X and if I plug it back in then what I get out of this is x squared plus x squared minus 3x squared I get minus x squared minus x squared is concave down

10:14

Speaker A

so basically when I want to be x axis with concave up going on the y axis is concave up but going along y equal to X it was concave down now watch what happens to the mixed partial the f sub x

10:27

Speaker A

y well first of all I'm going to say that that's taking the first version but the Y of I pick or soap derivatives for 2x 2x minus 3y I take that parcel and I get the value of minus 3 when you swear that you're gonna

10:42

Speaker A

get nine which is just much bigger than the two times two you have for FXX times FY y so in this case considering that FX y the so-called mixed partial squared even allowed us to see that this was a

10:57

Speaker A

saddle even though the analysis along x and y respectively could not see that with a saddle indeed it can be going down in some direction that's not the X and the y axis so the way to think about

11:08

Speaker A

the mixed partial is that as I change my access I in for example parts of respect to X partial derivative with respect to Y now as I studied that it changes as I move off of the x axis and indeed you

11:23

Speaker A

might recall that F X Y was equal to my ex so likewise you can study the partial respect to Y and then see how that changes if you move the X which is now moving off of the y axis so the mixed

11:35

Speaker A

partials are a way of figuring out what happens when we're not just constrained to the axes that's what FXX I have to why why does the mixed partials tell us well how big are these changes if I go

11:47

Speaker A

off a little on X and then I globe along Y that captures changes for example in y equal to X which is not along either of the axes and so if this you change X and then as you change Y which is picking

12:00

Speaker A

you off the axes if that change is very large then the FX Y squared is gonna be very large and indeed then you can have a saddle even way along both axes it appears to be a maximum or appear to be

12:12

Speaker A

a minimum now exactly proving this that the grotesque feel on the intuition that I've given you is beyond the scope of this particular video I can tell you that it comes from the generalization of Taylor's theorem we've seen in single

12:26

Speaker A

variable calculus to the multivariable calculus case and indeed we're gonna have this expression is related to something called the hestian and that will be our way to formally demonstrate that these conditions are valid ways to decide maximum or minimum or a sample from the

12:45

Speaker A

computational perspective our method is now relatively straightforward you find all the candidates by the first derivative test that is you go along and you figure out where is the partial present at X equal to 0 partial reified to y 0 and when both of those occur then

12:59

Speaker A

the second derivative test gives us the conditions to classify those when is it the case is gonna be a maximum a minimum a saddle or unfortunately sometimes the test fails and we have to go further we have to do more analysis that we can get

13:15

Speaker A

from just this test if you have a question about this video leave it down in the comments below real mathematicians here we appreciate algorithms so let's just hope the YouTube l know about by giving this video a like and finally if you want to

13:27

Speaker A

watch more multivariable calculus videos this video is part of a larger playlist a multivariable calculus so you can check out those videos here and we'll do some more math in the next video

Topics: multivariable optimization second derivative test partial derivatives saddle point critical points Hessian matrix concavity multivariable calculus local maxima local minima

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