00:01
For this problem, we are asked to find the local extreme values of f of xy equals x squared y on the line x plus y equals 3.
00:08
So effectively, we have a constraint equation, g of xy, equals x plus y minus 3 equals 0.
00:16
So we set up the lagrange multiplier equation, lambda, or excuse me, we have gradient of f must be equal to lambda times the gradient of g, which then means that we have 2xy must be equal to lambda, we then also have that x squared must be equal to lambda.
00:35
So in that case, if 2xy equals x squared, then we can divide both sides, or actually i'll correct myself here, let's factor or subtract x squared from both sides, then factor out an x.
00:51
So we can write this as x times 2y minus x, must be equal to 0, which means that either we have x equals 0, or we have 2y equals x in which case we have y equals x over 2 i'll note that if we have x equals 0 then that would also have to give us let's see here that would give us that y must be equal to 3 in which case we have f of 0 3 gives a value of 0 that's not necessarily an extreme value though then we have the other possibility, y equals x over 2.
01:32
Oh, actually, no, let me correct myself...