# Find local max min or saddle points for $f(x,y)=3y-y^3-3(x^2)y$

Find local max min or saddle points for $f(x,y)=3y-y^3-3(x^2)y$ I know how to solve this problem, just having trouble finding critical pts when i set $f_x = 0$ and $f_y =0$. We have: $$f_x=-6xy = 0$$ $$f_y=3-3y^2-3x^2 = 0$$

So, we have $f(x,y)=3y-y^3-3yx^2$. Hence, for critical points $(x,y)$, we have,

$$\begin{cases}\dfrac{\partial f}{\partial x}=-6xy=0\\ \dfrac{\partial f}{\partial y}=3-3y^2-3x^2=0\end{cases}$$

First equation implies that either $x=0$ or $y=0$ or both $x,y=0$.

Now, if $x=y=0$, then the second equation is not satisfied, so $(0,0)$ is not a critical point. Then,

$$x=0\implies 3=3y^2\implies y^2=1\implies y=\pm 1\\ y=0\implies 3=3x^2\implies x^2=1\implies x=\pm 1$$

So, we have our required critical points as $(0,1),(0,-1),(1,0),(-1,0)$.

• For (1,0) D>0 but fxx=0 so would that make it not a critcal point? – Charlene Apr 27 '15 at 13:44
• @Charlene, I didn't understand what you meant! Can you please elaborate? – Prasun Biswas Apr 27 '15 at 13:50
• To find out if 1,0 is a local min max or saddle point you find out what D equals and if its negative its a saddle and if positive then depends on whether fxx is pos or neg. but if fxx =0 then idk if its max min saddle of none – Charlene Apr 27 '15 at 13:52
• @Charlene, I'm getting $D(x,y)=36y^2-36x^2$ which gives me $D(1,0)=-36\lt 0$ which implies that $(1,0)$ is a saddle point of $f$. – Prasun Biswas Apr 27 '15 at 14:08