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

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?

• Jam ka question he ? – Daman deep Feb 20 at 9:53
• @Damandeep log in krke dekhle response aagye – Abhay Feb 20 at 9:58
• Bhai aa gya pr inhone mere answers marks galat diye he.recent post dekh mera. add kr facebook.com/damanisdeep – Daman deep Feb 20 at 10:01
• @Damandeep done – Abhay Feb 20 at 10:09
• message dala dekh baat kr mujse answers nikalte he mene mathematical stats diya tha – Daman deep Feb 20 at 10:41

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 $$\left(-\tfrac{2}{3},-\tfrac{2}{3}\right)$$: