MHB Stationary points in local optimization

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The discussion revolves around finding stationary points for the function f(x,y)=3x²y+y³-3x²-3y²+2. Initially, three stationary points were identified based on the derivatives f_x and f_y, but the expected total was four. Upon further analysis, it was discovered that one value of y leads to two corresponding values of x, confirming the total of four stationary points. The equations derived from the partial derivatives are x(y-1)=0 and x²+(y-1)²=1. The clarification emphasizes the importance of thorough verification in optimization problems.
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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