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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?

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  • $\begingroup$ Jam ka question he ? $\endgroup$ – Daman deep Feb 20 at 9:53
  • $\begingroup$ @Damandeep log in krke dekhle response aagye $\endgroup$ – Abhay Feb 20 at 9:58
  • $\begingroup$ Bhai aa gya pr inhone mere answers marks galat diye he.recent post dekh mera. add kr facebook.com/damanisdeep $\endgroup$ – Daman deep Feb 20 at 10:01
  • $\begingroup$ @Damandeep done $\endgroup$ – Abhay Feb 20 at 10:09
  • $\begingroup$ message dala dekh baat kr mujse answers nikalte he mene mathematical stats diya tha $\endgroup$ – Daman deep Feb 20 at 10:41
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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 $\left(-\tfrac{2}{3},-\tfrac{2}{3}\right)$:

enter image description here

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