Stationary points in local optimization

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SUMMARY

The discussion focuses on identifying stationary points of the function \(f(x,y)=3x^{2}y+y^{3}-3x^{2}-3y^{2}+2\). The user initially found three stationary points by solving the partial derivatives \(f_{x}=6xy-6x\) and \(f_{y}=3x^{2}+3y^{2}-6y\) set to zero. However, upon further analysis, it was determined that there are actually four stationary points, as one value of \(y\) yields two corresponding values of \(x\). The correct stationary points are derived from the equations \(x(y-1)=0\) and \(x^2+(y-1)^2=1\).

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Yankel
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Hello again,

I have a small problem. I am looking for local minimum and maximum points of the function:

\[f(x,y)=3x^{2}y+y^{3}-3x^{2}-3y^{2}+2\]

The first question was how many stationary points are there. I have found the derivatives by x and y:

\[f_{x}=6xy-6x\]

\[f_{y}=3x^{2}+3y^{2}-6y\]

and compared them to 0. I found 3 points: y=0,1,2.

According to the attached answers, there should be 4. There is either a mistake in the answer attached, or I am missing a point. Can you help me solve the system of two equations please to find all the points ? Thank you !

- - - Updated - - -

Ok, I couldn't find a DELETE button, so I will answer my own question.

When I put the 3 (!) values of y back in the equations, for one of them, x got 2 values, bringing the sum of points to 4. My mistake.
 
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We don't allow users to delete threads (which is what would happen to a thread if the first post in it is deleted) because that could potentially cause valuable content to be destroyed if a user decided to delete their thread after getting help because they are trying to keep their professor from finding the thread. :D

Your partials look correct, and from them we obtain:

$$x(y-1)=0$$

$$x^2+(y-1)^2=1$$

And as you found, when from the first equation we take $x=0$, we find $$y=1\pm1$$, and for $$y=1$$ we find $$x=\pm1$$ for a total of 4 points. :D
 

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