00:01
When we are trying to find local maximum and local minima using multiple variables, we want to be able to find where the both derivatives equal zero at the same time.
00:12
We know that when we're doing one dimensional calculus, we only need f of x equal to zero, f sub x equals zero, the diverse derivative, but we also need an and fy is equal to zero.
00:23
We need both to happen for there to be a maximum or minima.
00:27
So in order to do this, we have to first take the derivatives with respect to x and with respect to y.
00:34
So we direct derivative with respect to x here, which is going to be just 3x squared minus 6x minus 9.
00:51
As well, we then have for f sub y, which is quite interesting, we have 3y squared.
01:02
Minus 6y.
01:04
And we need to see where both of these hold.
01:07
In the first one, which is quite interesting, we set this equal to zero, we set this equal to zero.
01:12
As you can see, there's no x in the f y, and there's no y in the fx.
01:16
So what's interesting is this just individually holds.
01:19
So we take and see what happens at both of these, and then we combine all of the possibilities.
01:27
I'll show you what happens when we go for it.
01:29
So this first equation is not too bad.
01:31
We can factor throughout the three, we get x squared minus three x, minus two x, minus three is equal to zero.
01:41
We can foil this pretty quickly.
01:43
We can, we can get this into a nice quadratic form by noting that this is going to be just negative three and a positive one would get that nicely.
01:55
And so therefore, we get x is equal to three and x is equal to negative one.
02:00
So at x equal to three and x equal to negative one, we have these.
02:03
Now, we would initially what we would usually do is we would plug these guys into the fy equation individually and then find the y that fit.
02:12
But since we don't have any values, whatever y we find will work with either one of the xs we find.
02:18
So this starts to become very interesting.
02:20
So we're going to have probably four different points.
02:23
If we solve this out, we can factor out a 3y of this one as well.
02:28
And we would get a y minus 2.
02:32
So we get y is equal to zero.
02:33
So we get y is equal to zero and y is equal to two.
02:38
And so therefore, our values will be, our points are going to be just the combinations of all of these.
02:48
So we have 3 -0, 3 -2, negative -1 -0, negative -1 -2.
02:55
And the reason why is that regardless of whatever value of x i pick, the ys will work in the fy case.
03:02
So we have to conclude all of those values.
03:05
And so then lastly, well, we must.
03:08
Look at are the, this determinant that we have to do to identify it.
03:16
So we have to look at the second derivative with respect to x times the second derivative with respect to y, multiply by the mixed derivative squared.
03:24
If this is less than zero, it's a saddle point.
03:29
If it's greater than zero, we then have to look individually at fxx and f y y.
03:36
If they're both positive, this is a minimum...