00:01
For this problem, we are asked to find and classify using the second partial derivative test, the critical points of the function f of xy equals 6xy plus 4y minus lon of xy squared.
00:12
To do this, we first want to find the critical points by solving for when the gradient of f will equal 0.
00:18
So the first element of the gradient is going to be 6y minus 1 over x.
00:23
The second element is going to be 6x plus 4 minus 2 over y.
00:29
Now, if we want to solve for when this equals zero, we have one solution, that being when x equals 2 over 3, and y equals 1 over 4.
00:39
Having our critical point now, we proceed with applying our second partial derivative test.
00:45
So, to do that, we need the second partial x, which is going to be 1 over x squared...