We want to determine extreme values of $f(x,y)=x^3+xy^2-x^2y-y^3$. We first determine critical points by solving $\dfrac{\partial f(x,y)}{\partial x}=0$ and $\dfrac{\partial f(x,y)}{\partial y}=0$ which gives that the only critical point is $(0,0)$. Now we compute the determinant of the Hessian $$D(x,y)=(6x-2y)(2x-6y)-(2y-2x)^2$$ Hence $D(0,0)=0$ and the determinant of the Hessian test is not conclusive, so what to do next to verify existence of local and global extreme values? thank you for your help.
Asked
Active
Viewed 2.0k times
2 Answers
5
A hint:
You can factor your function and decide what's happening at $(0,0)$ by inspection.
Christian Blatter
- 226,825
-
3Can we say that $f(1/n,0)=1/n^3>0$ while $f(0,1/n)=-1/n^3<0$ hence the point $(0,0)$ can't be an extreme value? – palio Jan 19 '14 at 15:16
-
1@palio: That's it. – Christian Blatter Jan 19 '14 at 16:07
2
When your Hessian determinant is equal to zero, the second partial derivative test is indeterminant. So then you could simply look at the equation or you can develop contours around possible mins and maxs and use Gauss's Theorem to see if there are mins and maxs within them. So essentially, if the Hessian is equal to zero, you are screwed or the question is really easy and you can find the answer by inspection.
user3842604
- 21