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## Multivariable calculus

### 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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# Second partial derivative test example, part 1

A worked example of finding a classifying critical points of a two-variable function. Created by Grant Sanderson.

## Want to join the conversation?

- 5:17minor mistake you claim to have a grand total of three critical points while in fact you have 4(5 votes)
- I have the equation: x^2 + 6xy +10y^2 - 4y + 4. If I take the first derivative of x I end up with: 2x + 6y. What am I supposed to do with this? If I set it equal to zero I just keep ending up with x = -3y...?(2 votes)
- If you are looking for critical points, you will want to find the places where the tangent plane has zero slope. You will want to know where both partial df/dx and partial df/dy equal zero. In your example, you would calculate that partial df/dy is 6x +20y-4. Now you have two equations equal to zero with two variables. Just use algebra to solve the system of equations.(3 votes)

- What about inflection points? Where do they fit in in the gradient and the second partial derivative test?(1 vote)
- If you go to Desmos and you graph the two equations mentioned in this video, https://www.desmos.com/calculator/d5nrguizbn

you'll see that one of them just boils down to the horizontal line y=1 (and it's undefined at x=0) while the other one is just some hyperbola. As far as I can tell, there are only 2 intersections between the graphs, so although it made sense from the video why there should be 4 critical points, how does it square with this graphical view of the situation?(1 vote)- However, it does seem valid according to this 3d Geogebra graph for the full multivariable function f(x,y): https://www.geogebra.org/3d/fq3dutqv(1 vote)

- Can you try not to make sound of swallowing saliva when you are talking. It's disturbing.(0 votes)

## Video transcript

- [Voiceover] So one common
type of problem that you see in a number of multivariable
calculus classes will say something to the
effect of the following: find and classify all of
the critical point of, and then you'll insert some
kind of multivariable function. So first of all, this
idea of a critical point basically means anywhere where
the gradient equals zero. So you're looking for
places where the gradient of your function, at some kind of input, some specified input X and
Y that you're solving for, is equal to zero. And as I talked about in
the last couple of videos, the reason you might want
to do this is because you're hoping to maximize the function or to maybe minimize the function. And now the second requirement
of classifying those points, that's what the second
derivative test is all about. Once you find something, or
the gradient equals zero, you want to be able to determine, is it a local maximum,
is it a local minimum or is it a settle point? So let's go ahead and
work through this example. The first thing we're gonna
need to do if we're solving for when the gradient equals zero, and remember, when we say equals zero, we really mean the zero vector, but it's just a convenient
way of putting it all on one line, we take
both partial derivatives. So the partial derivative
with respect to X is, well, this first term,
when we take the derivative of three X squared times
Y with respect to X, the two hops down so we
have six times X times Y. Y cubed, well, Y looks like a constant so Y cubed looks like a
constant minus three X squared, so that two comes down. So we're subtracting off six times X. Six times X. And then again, this three Y squared term, Y looks like a constant so everything here looks like a constant
with zero derivative, as far as the X direction is concerned. And we do partial of f with respect to Y, then this first term looks
like some sort of constant, three X squared. X looks like a constant so
some kind of constant times Y. So the whole thing looks
like three X squared. The second term, minus X,
minus Y cubed, excuse me, looks like minus three
Y squared when we take the derivative. Minus three Y squared. And then this next term only
has an X so it looks like a constant as far as Y is concerned. And then this last term,
we take down the two because we're differentiating Y squared, and you'll get negative six
Y, negative six times Y. So when we are finding
the critical points, the first step is to set both
of these guys equal to zero. So this first one, when we
do set it equal to zero, we can typify a bit by
factoring out six X. So this really looks like six
X multiplied by Y minus one. And then that's what we're
setting equal to zero. And what this equation
tells us is that either it's the six X term that equals zero, in which case that would
mean X is equal to zero; or it's the case the Y
minus one equals zero, in which case that would
mean that Y equals one. So at least one of these
things has to be true. That's kind of the first
requirement that we found. Let me scroll down a little bit here. And for the second equation,
when we set it equal to zero, it's not immediately
straightforward how you would solve for this in a nice
way in terms of X and Y, but because we've already
solved one, we can kind of plug them in and say, for
example, if it was the case that X is equal to zero, and we kind of wanna see what
that turns our equation into, then we would have, well, three X squared is nothing that would be zero
and we'd just be left with negative three Y squared minus six Y is equal to zero. And that, we can factor out a
bit, so I'm gonna factor out a negative three Y. So I'll factor out negative three Y, which means that first
term just has a Y remaining and then that second term
has a two, a positive two, since I factored out negative three. So positive two and that equals zero. So what this whole situation
would imply is that either negative three Y equals
zero, which would be, which would mean Y equals zero; or it would be the case
the Y plus two equals zero, which would mean that Y
is equal to negative two. So that's the first
situation where we plug in X equals zero. Now alternatively, there's the possibility that Y equals one, so we
could say Y equals one; and what that gives us
in the entire equation, we still have that three X squared because we're kind of solving
for X now, three X squared, and then the rest of
it becomes, let's see, minus three times one squared, so minus three, we're
plugging in one for Y then we subtract off six,
plugging in that one for Y again, and that whole thing is
equal to three X squared then minus three minus six. So I'm subtracting off nine. So from here, I can
factor out a little bit, and this will be three multiplied
by X squared minus three. And what that implies then,
since this whole thing has to equal zero, what that
implies is that X squared minus three is equal to zero
so we have X is equal to plus or minus the square root of three. And maybe I should kind of specify, these are distinct things that we found. One of them was in the
circumstance where X equals zero, and then the other was where
we found in the circumstance where Y equals one. So this give us a grand
total of three different critical points because
in the first situation, where X equals zero, the
critical point that we have, well, both of them are going to have an X coordinate of zero in
them, X coordinate of zero; and the two corresponding
Y coordinates are zero or negative two. So you have zero or negative two. There's kind of two possibilities. And then there's another
two possibilities here, where if Y is equal to one, when Y is equal to one, we'll have X as positive or
negative square root of three. So we have positive square
root of three and Y equals one, and then we have negative
square root of three and Y equals one. So these, these are the critical points, critical points, which basically means
all partial derivatives are equal to zero. In then in the next video, I
will classify each of these critical points using the
second partial derivative test.