What are the saddle points of the function $f(x,y)=x^3+2xy+y^3$

Question

What are the saddle points of the function $f(x,y)=x^3+2xy+y^3$?

First I'll try to find the set of critical points of this:

we have :

$f_x=3x^2+2y=0 \rightarrow y=\frac{-3x^2}{2}$

$f_y=2x+3y^2=0 \rightarrow2x+ \frac{27x^4}{4} = 0 \rightarrow x=0$ or $\frac{-2}{3}$

for $x=0 \rightarrow y=0$ and for $ x =\frac{-2}{3}\rightarrow y=\frac{-2}{3}$

So the only critical points are $(0,0)$ and $(\frac{-2}{3},\frac{-2}{3})$

$f_{xx}= 6x$

$f_{yy}=6y$

$f_{xy}=2$

$D=36xy – 4$

Clearly only $(0,0)$ satisfies $D<0$ so it is the only saddle point. Is this correct?

Answer 1

Clearly only $(0,0)$ satisfies $D<0$ so it is the only saddle point. Is this correct?

Correct, well done!

On the contour plot (WolframAlpha), you can nicely see the saddle point behavior at $(0,0)$ and the maximum at $\left(-\tfrac{2}{3},-\tfrac{2}{3}\right)$: