Second partial derivative test is inconclusive

Question

I am trying to find the critical points of the function:

$f(x,y)=2x^4-3x^2y+y^2$

and find the Max, Min and saddle points. What I've done so far is:

$f_x=8x^3-6xy=0 , f_y=-3x^2+2y=0 , f_{xx}=24x^2-6y , f_{yy}=2 , f_{xy}=-6x$

So (0,0) is the only critical point. But using the second partial derivative test:

$\Delta(0,0)=f_{xx} . f_{yy} - f^2_{xy}=0$

Which is inconclusive. Without using MATLAB or similar software and based on calculation, how can we determine whether (0,0) is Max, Min or saddle point?

Or the general question is what to do when it's inconclusive (without using software)?

Answer 1

Plug in $x = \epsilon$, $y = 3\epsilon^2$ and you will get that the function is greater than $0$ for all $\epsilon > 0$. Plug in $x = (3/4) \epsilon$, $y = \epsilon ^2$ and you will get that the function is less than $0$ for all $\epsilon > 0$. This implies that $(0,0)$ must be a saddle point because you take take $\epsilon > 0$ arbitrarily small.

Answer 2

Hint: $$2x^4−3x^2y+y^2=(x^2-y)(2x^2-y).$$

Answer 3

Possibly easier hint (complete the square): $$ 2x^4 - 3x^2 + y^2 = (y-\frac32 x^2)^2 - \frac14x^2 $$

Answer 4

Added: Here's a way not to solve the problem (because it gives the wrong answer!)

Look at the concavity of the function at $(0,0)$ in a non-axis direction $y=ax$. $$\left.\frac{d^2}{dx^2}\right\rvert_{x=0}\left(f(x,ax)\right) = \left.\frac{d^2}{dx^2}\right\rvert_{x=0}\left(2x^4-3x^2 (a x)+(a x)^2\right)=\left. 2a^2-18ax+24x^2\right\rvert_{x=0}=2a^2.$$ This tells you that the function is concave up at $(0,0)$ in every non-axis direction, so $(0,0)$ is not a saddle point. It's not a local maximum, because $f_{yy}(0,0)>0$, so it’s a local minimum.

(The right answer is that this is a saddle point. See the answer @user2566092 gave, and the correction in the comments, if the answer hasn't been fixed.)