What is the local min/max of this function

Question

$$f(x,y)= x^2-3x^2y+y^3$$

There are 3 possible points: $(0,0)$, $(\tfrac{1}{3},\tfrac{1}{3})$, $(-\tfrac{1}{3},\tfrac{1}{3})$ but with the hessian of $f$ we get that only $(0,0)$ MIGHT be a local max/min and may not.

I went to desmos and I found that in a small neighborhood of $(0,0)$ it is a local minmum.

Is that right? Or we don't have local max/min in here?

Answer 1

We have $f(x,y)=x^2-3x^2y+y^3$. Then we get: $f(0,0)=0$ and

$f(0,y)=y^3$.

Therefore we have: in each neigborhood of $(0,0)$ there are points $(0,y)$ and $(0,z)$ with $f(0,y)=y^3 >0=f(0,0)$ and $f(0,z)=z^3<0=f(0,0)$.

Conclusion: in $(0,0)$ is no max. and no min. of $f$

Answer 2

there are 3 possible points (0,0) (1/3,1/3) (-1/3,1/3) but with the hessian of f we get that only (0,0) MIGHT be local .. max/min and may not.

You are right about the stationary points and the Hessian shows that the function doesn't attain local minima/maxima in $(\pm\tfrac{1}{3},\tfrac{1}{3})$.

The test with the Hessian is inconclusive about what happens in $(0,0)$, but notice that $f(0,t)=t^3$ so by moving towards the origin along the $y$-axis, it is clear that the function doesn't attain a minimum or maximum in $(0,0)$ either.

Answer 3

No it is not. We have $$f(x,y)=x^2(1-3y)+y^3$$let the point $(0,-\epsilon)$ with $\epsilon>0$ tend to $(0,0)$ therefore $$f(0,-\epsilon)=-\epsilon^3<0=f(0,0)$$which means that for small enough $\epsilon$, the point $(0,0) $ is not a local minimum.