00:01
So in this question we have a function of two variables.
00:04
F of x and y is x minus y multiplied by 4 minus x y.
00:13
And we want to find the local maxima, minima and saddle points of the function.
00:20
So first of all, we're going to take its derivative because at a local maximum minimum or saddle point, the derivatives in both directions are zero.
00:33
So first of all, i'm going to take the x derivative, which i'm going to denote this way, f x of x and y.
00:41
And i'm going to take the derivative with respect to x now, and i'm going to do the product rule.
00:47
So this is going to give me, so firstly taking the derivative of this and multiplying it by this, i get 4 minus xy.
00:54
Now i take the derivative of this and multiply by this, and this gives us minus y, x minus y, which is 4, minus 2, xy plus y squared.
01:11
And i want to find the zeros of this.
01:19
So this is going to be equal to zero at any saddle points.
01:24
And i'm going to use the quadratic formula.
01:28
So the quadratic formula tells us that when we have something like a t squared plus bt plus c equals zero, then this is solved by t is equal to minus b plus or minus root b squared minus 4 a c divided by 2a so i'm going to apply that to this equation up here so a is going to be 1 b is going to be 2x and c is going to be 4 minus 2x in fact for b so this tells us that y critical is going to be so minus b is 2x plus or minus root so b squared is 4x squared.
02:16
And then we need to do 4, so minus 4 times a.
02:21
So a is 1.
02:21
So minus 4 times c is minus 16.
02:28
And then we divide by 2a, so we have to divide by 2.
02:31
Now i can put that, just put that two through.
02:34
So that gets rid of that, gets rid of that, and that's going to become a 4.
02:41
So yep, this is this is telling us that the y critical values.
02:48
So why is equal to 0 when x is equal to you know why why is equal to the derivative the x derivative of the function is equal to 0 when y is related to x in this way so now this tells us when the when the x derivative is zero so this is equation 1 so now i'm going to take the y derivative of the function f subscript y of x and y uh so this is this is going to give me minus 1 times 4 minus xy, and then x minus y times minus x.
03:46
So this is minus 4 plus xy, minus x squared plus xy is equal to 0.
03:56
Now i'm going to put a minus sign through this to get x squared minus 2 xy plus 4 is equal to 0.
04:06
Now notice that this is exactly the same equation as before.
04:15
And we could have seen this basically because you can replace x with y in this equation up here and all that happens is that the function goes to minus the function.
04:26
But if the function goes to minus the function, then its derivative goes to minus its derivative.
04:31
So if in all the places the derivative is zero, it's still going to be zero.
04:36
So we can swap x and y in this equation.
04:39
And it's all still going to work.
04:43
So x crit is going to be y plus or minus root, y squared minus 4.
04:55
So this is equation 2.
04:57
And so whenever equation 2 holds, the y derivative of the function is 0.
05:02
So at all its critical points, equations 1 and 2 are both going to hold.
05:08
And that's going to allow us to solve for the critical points.
05:13
So if 1 and 2 holds, i'm going to put the y value into this equation.
05:22
So we're going to have x, so critical points, when.
05:35
So first of all, i'm going to put equation 1 into equation 2.
05:41
So x equals x plus or minus root, x squared minus 4, plus or minus root, and now i have to square this and take away four.
05:58
But remember that y squared minus four, if it obeys this equation, y squared plus four is 2xy.
06:08
So y squared minus four is going to be 2xy minus eight.
06:15
But actually, i don't think that's going to simplify the situation much.
06:20
So let's just square the thing that we've got here.
06:24
X plus or minus root, x squared minus four.
06:32
Squared minus 4.
06:35
And so this sign i'm going to give the subscript 1 and this sign i'm going to give the subscript 2 because they're independent of each other.
06:46
But this sign is the same as this sign.
06:52
So now let's do a bit of work on this.
06:56
So x plus or minus 1 root x squared minus 4 plus or minus 2 root and now i can square this out x squared and then plus or minus 2x root x squared minus 4 this is plus or minus 1 and then we're going to get plus x squared minus 4 but then this minus 4 is going to turn into a minus 8 so x plus or minus 1 root root and x squared minus 4 plus or minus 2 root x squared.
08:01
So 2x squared plus minus 1, 2x root x squared, minus 4 plus x squared.
08:16
Sorry, i've got the x squared already.
08:19
So it's just going to be minus 8.
08:25
So x is going to be equal to all this.
08:32
I've got x equals this stuff, and that means that i can take this x off to get zero.
08:38
And that also means that this plus or minus one is going to be eliminated.
08:45
So zero equals, and so i'm eliminating this plus or minus, but doing that will change the sign of this, but this is already independent of the other sign, which means that i can just, it doesn't matter.
09:05
So this is going to be plus minus 1, 2.
09:13
I'll just say that this is the product of them.
09:18
So root x squared minus 4 plus minus 1 2, root 2x squared, plus or minus 2x root x squared minus 4, minus 8.
09:34
Now, this, in order for, so these things are both strictly positive, or well, not strictly positive, but so let's, let's, so this is going to have to be a minus sign unless both of these things are zero.
09:56
So let's see whether they can both be zero.
09:59
So this is going to be equal to zero if x is plus or minus two.
10:05
So let's put plus 2 into here.
10:09
So 2 times 2 squared plus 2 times 2.
10:18
Oh, well, that second bit is going to be 0...