00:01
For this problem, we are asked to find the absolute maximum and minimum values for the function f of xy equals 2x squared plus y squared plus 2x minus 3y in the region where x squared plus y squared is less than or equal to 1, particularly where we are asked to use lagrange multipliers to check for extreme points on the boundary.
00:18
So to begin, we want to see if we can find any critical points of our function within the region.
00:24
So we want the gradient of f to equal 0, which means that we want 4x plus 2 to equal 0.
00:30
Which means that we must have that x equals negative 1 over 2.
00:35
And then for partial derivative with respect to y, we want 2y minus 3 to equal 0.
00:42
So we have that y must equal 3 over 2.
00:46
Evaluating our function at that point, negative 1 1⁄2, positive 3 over 2.
00:52
Let me see what we have here.
00:54
So i actually need to point something out here that these x and y points are not with in our region.
01:03
Excuse me, they are real numbers, but they're not within r as defined for us.
01:07
So we don't really need to care.
01:09
We have no critical points within our region, which then means that the absolute maximum and absolute minimum must be found somewhere along the boundary.
01:17
Along the boundary, what we do is set up the constraint equation that x squared plus y squared minus one must equal zero, and set up the lagrange multiplier equation that the gradient of f must equal lambda times the gradient of g.
01:34
So we already know the gradient of f for partial derivative with respect to x at least was 4x plus 2.
01:40
And then that must equal 2 times lambda times x, which i'll note that we can rearrange dividing both sides by 2x.
01:48
We'd have that we can write this as 2x plus 1 over x must be equal to lambda.
01:54
Then from the second equation, 2y minus 3 must be equal to 2 .2 .2 .2 .000 lambda y or two lambda y would make more sense with the emphasis there so we would then have dividing both sides by two y we have that we can write this as two y minus three over two y must be equal to lambda which in turn putting those two together we'd have that two x plus one over x must be equal to two y minus three over y then we can do here is let's see uh we can if we so choose, there are a few different ways we can go about this.
02:35
But let's multiply both sides by x and y.
02:38
So on the left hand side, we get that 2xy plus y must equal 2xy minus 3x.
02:47
Then, or excuse me, 2xy minus 3x...