00:01
For this problem, we are asked to find all points where the given function may have a local max or min, then use the second derivative test to determine the nature of the function at those points.
00:11
So to begin, we start by taking the partial derivatives with respect to x and y.
00:17
So we get 2x minus 2y plus 4 equals 0, or rather it's going to equal 0 at the point that we're looking for, and then taking the partial derivative with respect to y, we will get negative 2x plus 6y minus 16 equals 0.
00:41
So rearranging this, we would be able to get from the partial with respect to x, y equals x plus 2, and then substituting that in to the partial with respect to y, we get negative 2x plus 6x, plus 12 minus 16.
01:02
So 12 minus 16 is going to give us a minus 4.
01:06
Then negative 2x plus 6x is going to give us a 4x.
01:14
So we have 4x minus 4 equals 0, which means that x has to equal 1.
01:18
And then that means that y is going to equal 3.
01:22
So we have our candidate point is 1 3...