Hessian matrix, maxima and minima criteria

[Telemachus]

Homework Statement

Hi there. I've got some doubts about the maxima and minima on this function: $$f(x,y)=x \sin y$$. I've looked for critical points, and there's only one at (0,0). The thing is that when I've evaluate the second derivatives I've found that $$f_{xx}=0$$, then I have not a defined criteria on this case, cause the criteria that I have defines the concavity for $$f_{xx}>0$$ and $$f_{xx}<0$$. I don't know what happens when $$f_{xx}=0$$. I think the criteria fails, but I don't actually know what to do.

Bye there!

 

[lippyka]

Homework Equations The second derivatives of f(x,y) are f_{xx}=0 and f_{yy}=x \cos y. The Attempt at a Solution Since the second derivative f_{xx}=0, we can't use the usual criteria for finding the concavity of the function. However, we can still use the second derivative f_{yy}. We know that f_{yy}=x \cos y, so if x > 0, then f_{yy} > 0, indicating that the function is concave up. Similarly, if x < 0, then f_{yy} < 0, indicating that the function is concave down. We can also use the first derivative f'(x,y)=\sin y to determine the local extrema of the function. Since \sin y is periodic, it will have local maxima and minima at the points where its derivative is zero. Thus, the local extrema of f(x,y) occur when \sin y = 0, which means that the local maxima occur at (x,y)=(x, \pi/2 + 2n\pi) and the local minima occur at (x,y)=(x, -\pi/2 + 2n\pi).