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The critical points and saddle points of the function $$f\left(x,y\right)=x^{3}+y^{2}-3xy$$ are $A=\left(0,0\right)$ and $B=\left(\frac{3}{2},\frac{9}{4}\right)$. The hessian is $$ H=\begin{bmatrix}\begin{array}{rr} 6x & -3\\ -3 & 2 \end{array}\end{bmatrix}. $$

$A$: Principal minors are $0$ and $-9$, so $A$ is a saddle point.

$B$: Principal minors are $9$ and $9$, so $B$ is a local minimum.

I would like to check the result from Wolfram Alpha, $B$ was okay but it gave $A$ as a min (or max) as in the pictures.

I also add the plot of the function which shows a saddle point as well.

Is my solution wrong?

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mert
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  • Your question is good. I recommend changing your title "An optimization problem question" into "The critical points and saddle points of the function $f(x,y)=x^3+y^2−3xy$ are $A=(0,0)$ and $B=\left(\frac{3}{2},\frac{9}{4}\right)$.The hessian is...". Many people might have the same question as yours and it is easier to search it on internet....Did you get that question from a book you read? – fitzmerl duron Feb 13 '18 at 15:59
  • Thanks for your advice. I do not know the resource of the question, it is asked me to solve by someone. – mert Feb 13 '18 at 16:05
  • I see........... – fitzmerl duron Feb 13 '18 at 16:11

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It is beautifully seen at the picture that $A(0,0)$ is a saddle point. Something like mountain pass. Can you see this?

szw1710
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