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.
[Math] Finding extreme values when the determinant of the Hessian at a critical point is zero.
multivariable-calculus
Related Solutions
The point $(0,0)$ is a saddle point for $n>0$.
To see this, look at the function along the lines $y=x$ and $y=-x$: $$y=\ \ x\quad \ \to \quad f(t) = 2 x^{2 n}\quad\quad\quad\quad\quad\ \ \ $$ $$y=-x\quad \ \to \quad f(t) = (-x)^{2 n} - 4 n x^2 + x^{2 n}$$
Now take the second derivative on the line $y=-x$ to get $-8n<0$, and take the $2n$-th derivative on the line $y=x$ to get $2(2n!)>0$, thus showing this point is a saddle point and not a local maximum or minimum.
To find the minima, note that $f_x =0, f_y=0$ leads to: $$2 n x^{2 n-1} = -2 n y^{2 n-1} \quad \to \quad y = -x$$ Using the fact that $2n-1$ is odd. Now plug this into $f_x=0$ to get:
$$x\to \pm2^{-1/(2-2 n)}, \ y=-x$$ You can verify these are indeed minima.
In the graphs below:
$f(x,y)$ along the lines $y=x$ and $y=-x$ for $n=3$.
There isn't a global maximum or minimum, at least not when considered on $\mathbb{R}^{2}$. Set $y=0$ and you can easily see that your function neither has an upper nor lower bound.
Best Answer
A hint:
You can factor your function and decide what's happening at $(0,0)$ by inspection.