00:02
All right, our goal is to find the local extreme values of our function x squared plus y squared, but subject to the condition that g of xy equals zero in our function you can see here for g of xy.
00:17
So we're going to do it by the method of lagrange multipliers, and our goal is to basically take the gradient of f equal to lambda times a gradient of g, and we'll use this to be able to get our solution.
00:28
So first thing we want to do is go ahead and find a gradient with respect to f.
00:35
So we're going to take the partial derivative of f and remember this is f, partial derivative with respect to x, so that will give me 2x i .5, and then plus the partial derivative of f with respect to y, so that will give me 2 .y, j half.
00:50
Now we'll do a similar thing, but we're going to do it for g, and we have a lambda multiplier.
00:54
So we're going to take the derivative of g with respect to x and we'll get 2x minus 2, but we have a lambda multiplier.
01:03
And that's i -hat.
01:05
And then we have a lambda multiplier and the partial derivative of g with respect to y is 2 -y minus 4.
01:15
And that's j -hat.
01:16
Okay, so our method says that we now match up the i -hat terms.
01:21
The i -hat terms have to be equal.
01:25
So we're going to set 2x equal to lambda times 2x minus 2.
01:30
We can clean this one up a bit and divide through by 2.
01:33
So x is equal to lambda times x minus 1.
01:38
So that's one expression.
01:39
Let's now do the second one.
01:41
Here's the second one.
01:42
We're going to match up 2y and lambda 2y minus 4.
01:46
So we get 2y equals lambda 2y minus 4.
01:51
We can do the similar trick to last time and divide through by two.
01:56
So y then is lambda y minus two.
02:00
Okay, so we're going to go ahead and solve both for lambda and then we'll set those equal to each other.
02:10
So let's do that.
02:12
Okay, so let's take a look at this here.
02:14
If i solve for lambda, i get x minus 1 over x.
02:19
And on the second one below, if i solve for a lambda, i will get y over y minus two.
02:28
And i think i just did the other one upside down.
02:31
Sorry, this is upside down.
02:33
The other one, i can divide both sides by x minus one.
02:36
So actually have the reciprocal.
02:39
Okay, so i have lambda equals x over x minus one and lambda equals y over y minus two.
02:45
So since lambdas are equal to lambdas, therefore these other expressions, are also equal to each other.
02:51
So x over x minus one has to equal y over y minus two.
02:56
So i'm going to cross multiply x y minus two x equals x y minus y.
03:04
Cool thing is i can subtract the x y from both sides.
03:08
Very handy...