Second derivative test. Worked example conclusion
Computing the second partial derivatives can help us determine what type of point it is. What is the derivative of this function with respect to x? If we set A equal to 2 over x cubed y, B will be 1 and C will be 2. This tells us that AC minus B squared is equal to 4 minus 1. AC minus B squared is 2 times 2 minus squared, which equals 3. This tells us that we are either at a local minimum or a local maximum. We can see that A is positive and thus, it is a local minimum. In fact, it is the is global minimum. f_xx=2/x^3y f_xy=1/x^2y^2 f_yy=2/xy^3 A=2 B=1 C=2 AC-B^2=2*2-1^2=3>0 local minimum A>0 Max: f-> infinity when x-> infinity or y-> infinity or x,y->0 The maximum is not at a critical point. It’s on the boundary or at infinity, so we need to check what happens when x and y go to 0 as both approaches infinity. If x or y goes to infinity, then the function goes to infinity; also if x or y goes to 0, then one over xy goes to infinity. The function goes to infinity when x goes to infinity or y goes to infinity or x and y go to 0. So it’s not at a critical point. In general, we have to check for both critical points and boundaries to decide what happens
