What is the local min/max of this function












0














$$f(x,y)= x^2-3x^2y+y^3$$



There are 3 possible points: $(0,0)$, $(tfrac{1}{3},tfrac{1}{3})$, $(-tfrac{1}{3},tfrac{1}{3})$ but with the hessian of $f$ we get that only $(0,0)$ MIGHT be a local max/min and may not.



I went to desmos and I found that in a small neighborhood of $(0,0)$ it is a local minmum.



Is that right? Or we don't have local max/min in here?






multivariable-calculus







share|cite|improve this question
























  • Do you mean $f(x,y)=x^2-3xy+y^3$?
    – Mostafa Ayaz
    Nov 29 at 12:46










  • nope 3x^2 * y in the middle
    – Razi Awad
    Nov 29 at 12:47
















0














$$f(x,y)= x^2-3x^2y+y^3$$



There are 3 possible points: $(0,0)$, $(tfrac{1}{3},tfrac{1}{3})$, $(-tfrac{1}{3},tfrac{1}{3})$ but with the hessian of $f$ we get that only $(0,0)$ MIGHT be a local max/min and may not.



I went to desmos and I found that in a small neighborhood of $(0,0)$ it is a local minmum.



Is that right? Or we don't have local max/min in here?






multivariable-calculus







share|cite|improve this question
























  • Do you mean $f(x,y)=x^2-3xy+y^3$?
    – Mostafa Ayaz
    Nov 29 at 12:46










  • nope 3x^2 * y in the middle
    – Razi Awad
    Nov 29 at 12:47














0












0








0







$$f(x,y)= x^2-3x^2y+y^3$$



There are 3 possible points: $(0,0)$, $(tfrac{1}{3},tfrac{1}{3})$, $(-tfrac{1}{3},tfrac{1}{3})$ but with the hessian of $f$ we get that only $(0,0)$ MIGHT be a local max/min and may not.



I went to desmos and I found that in a small neighborhood of $(0,0)$ it is a local minmum.



Is that right? Or we don't have local max/min in here?






multivariable-calculus







share|cite|improve this question















$$f(x,y)= x^2-3x^2y+y^3$$



There are 3 possible points: $(0,0)$, $(tfrac{1}{3},tfrac{1}{3})$, $(-tfrac{1}{3},tfrac{1}{3})$ but with the hessian of $f$ we get that only $(0,0)$ MIGHT be a local max/min and may not.



I went to desmos and I found that in a small neighborhood of $(0,0)$ it is a local minmum.



Is that right? Or we don't have local max/min in here?






multivariable-calculus




multivariable-calculus






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Nov 29 at 12:59









StackTD

22k1947




22k1947










asked Nov 29 at 12:42









Razi Awad

124




124












  • Do you mean $f(x,y)=x^2-3xy+y^3$?
    – Mostafa Ayaz
    Nov 29 at 12:46










  • nope 3x^2 * y in the middle
    – Razi Awad
    Nov 29 at 12:47


















  • Do you mean $f(x,y)=x^2-3xy+y^3$?
    – Mostafa Ayaz
    Nov 29 at 12:46










  • nope 3x^2 * y in the middle
    – Razi Awad
    Nov 29 at 12:47
















Do you mean $f(x,y)=x^2-3xy+y^3$?
– Mostafa Ayaz
Nov 29 at 12:46




Do you mean $f(x,y)=x^2-3xy+y^3$?
– Mostafa Ayaz
Nov 29 at 12:46












nope 3x^2 * y in the middle
– Razi Awad
Nov 29 at 12:47




nope 3x^2 * y in the middle
– Razi Awad
Nov 29 at 12:47










3 Answers
3






active

oldest

votes


















0














We have $f(x,y)=x^2-3x^2y+y^3$. Then we get: $f(0,0)=0$ and



$f(0,y)=y^3$.



Therefore we have: in each neigborhood of $(0,0)$ there are points $(0,y)$ and $(0,z)$ with $f(0,y)=y^3 >0=f(0,0)$ and $f(0,z)=z^3<0=f(0,0)$.



Conclusion: in $(0,0)$ is no max. and no min. of $f$






share|cite|improve this answer





























    0















    there are 3 possible points (0,0) (1/3,1/3) (-1/3,1/3)
    but with the hessian of f we get that only (0,0) MIGHT be local .. max/min and may not.




    You are right about the stationary points and the Hessian shows that the function doesn't attain local minima/maxima in $(pmtfrac{1}{3},tfrac{1}{3})$.



    The test with the Hessian is inconclusive about what happens in $(0,0)$, but notice that $f(0,t)=t^3$ so by moving towards the origin along the $y$-axis, it is clear that the function doesn't attain a minimum or maximum in $(0,0)$ either.






    share|cite|improve this answer





















    • thank you too :)
      – Razi Awad
      Nov 29 at 13:01



















    0














    No it is not. We have $$f(x,y)=x^2(1-3y)+y^3$$let the point $(0,-epsilon)$ with $epsilon>0$ tend to $(0,0)$ therefore $$f(0,-epsilon)=-epsilon^3<0=f(0,0)$$which means that for small enough $epsilon$, the point $(0,0) $ is not a local minimum.






    share|cite|improve this answer





















    • thank you as well :)
      – Razi Awad
      Nov 29 at 13:03










    • you're welcome........
      – Mostafa Ayaz
      Nov 29 at 13:06











    Your Answer





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    3 Answers
    3






    active

    oldest

    votes








    3 Answers
    3






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    0














    We have $f(x,y)=x^2-3x^2y+y^3$. Then we get: $f(0,0)=0$ and



    $f(0,y)=y^3$.



    Therefore we have: in each neigborhood of $(0,0)$ there are points $(0,y)$ and $(0,z)$ with $f(0,y)=y^3 >0=f(0,0)$ and $f(0,z)=z^3<0=f(0,0)$.



    Conclusion: in $(0,0)$ is no max. and no min. of $f$






    share|cite|improve this answer


























      0














      We have $f(x,y)=x^2-3x^2y+y^3$. Then we get: $f(0,0)=0$ and



      $f(0,y)=y^3$.



      Therefore we have: in each neigborhood of $(0,0)$ there are points $(0,y)$ and $(0,z)$ with $f(0,y)=y^3 >0=f(0,0)$ and $f(0,z)=z^3<0=f(0,0)$.



      Conclusion: in $(0,0)$ is no max. and no min. of $f$






      share|cite|improve this answer
























        0












        0








        0






        We have $f(x,y)=x^2-3x^2y+y^3$. Then we get: $f(0,0)=0$ and



        $f(0,y)=y^3$.



        Therefore we have: in each neigborhood of $(0,0)$ there are points $(0,y)$ and $(0,z)$ with $f(0,y)=y^3 >0=f(0,0)$ and $f(0,z)=z^3<0=f(0,0)$.



        Conclusion: in $(0,0)$ is no max. and no min. of $f$






        share|cite|improve this answer












        We have $f(x,y)=x^2-3x^2y+y^3$. Then we get: $f(0,0)=0$ and



        $f(0,y)=y^3$.



        Therefore we have: in each neigborhood of $(0,0)$ there are points $(0,y)$ and $(0,z)$ with $f(0,y)=y^3 >0=f(0,0)$ and $f(0,z)=z^3<0=f(0,0)$.



        Conclusion: in $(0,0)$ is no max. and no min. of $f$







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Nov 29 at 12:54









        Fred

        44.1k1644




        44.1k1644























            0















            there are 3 possible points (0,0) (1/3,1/3) (-1/3,1/3)
            but with the hessian of f we get that only (0,0) MIGHT be local .. max/min and may not.




            You are right about the stationary points and the Hessian shows that the function doesn't attain local minima/maxima in $(pmtfrac{1}{3},tfrac{1}{3})$.



            The test with the Hessian is inconclusive about what happens in $(0,0)$, but notice that $f(0,t)=t^3$ so by moving towards the origin along the $y$-axis, it is clear that the function doesn't attain a minimum or maximum in $(0,0)$ either.






            share|cite|improve this answer





















            • thank you too :)
              – Razi Awad
              Nov 29 at 13:01
















            0















            there are 3 possible points (0,0) (1/3,1/3) (-1/3,1/3)
            but with the hessian of f we get that only (0,0) MIGHT be local .. max/min and may not.




            You are right about the stationary points and the Hessian shows that the function doesn't attain local minima/maxima in $(pmtfrac{1}{3},tfrac{1}{3})$.



            The test with the Hessian is inconclusive about what happens in $(0,0)$, but notice that $f(0,t)=t^3$ so by moving towards the origin along the $y$-axis, it is clear that the function doesn't attain a minimum or maximum in $(0,0)$ either.






            share|cite|improve this answer





















            • thank you too :)
              – Razi Awad
              Nov 29 at 13:01














            0












            0








            0







            there are 3 possible points (0,0) (1/3,1/3) (-1/3,1/3)
            but with the hessian of f we get that only (0,0) MIGHT be local .. max/min and may not.




            You are right about the stationary points and the Hessian shows that the function doesn't attain local minima/maxima in $(pmtfrac{1}{3},tfrac{1}{3})$.



            The test with the Hessian is inconclusive about what happens in $(0,0)$, but notice that $f(0,t)=t^3$ so by moving towards the origin along the $y$-axis, it is clear that the function doesn't attain a minimum or maximum in $(0,0)$ either.






            share|cite|improve this answer













            there are 3 possible points (0,0) (1/3,1/3) (-1/3,1/3)
            but with the hessian of f we get that only (0,0) MIGHT be local .. max/min and may not.




            You are right about the stationary points and the Hessian shows that the function doesn't attain local minima/maxima in $(pmtfrac{1}{3},tfrac{1}{3})$.



            The test with the Hessian is inconclusive about what happens in $(0,0)$, but notice that $f(0,t)=t^3$ so by moving towards the origin along the $y$-axis, it is clear that the function doesn't attain a minimum or maximum in $(0,0)$ either.







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Nov 29 at 12:55









            StackTD

            22k1947




            22k1947












            • thank you too :)
              – Razi Awad
              Nov 29 at 13:01


















            • thank you too :)
              – Razi Awad
              Nov 29 at 13:01
















            thank you too :)
            – Razi Awad
            Nov 29 at 13:01




            thank you too :)
            – Razi Awad
            Nov 29 at 13:01











            0














            No it is not. We have $$f(x,y)=x^2(1-3y)+y^3$$let the point $(0,-epsilon)$ with $epsilon>0$ tend to $(0,0)$ therefore $$f(0,-epsilon)=-epsilon^3<0=f(0,0)$$which means that for small enough $epsilon$, the point $(0,0) $ is not a local minimum.






            share|cite|improve this answer





















            • thank you as well :)
              – Razi Awad
              Nov 29 at 13:03










            • you're welcome........
              – Mostafa Ayaz
              Nov 29 at 13:06
















            0














            No it is not. We have $$f(x,y)=x^2(1-3y)+y^3$$let the point $(0,-epsilon)$ with $epsilon>0$ tend to $(0,0)$ therefore $$f(0,-epsilon)=-epsilon^3<0=f(0,0)$$which means that for small enough $epsilon$, the point $(0,0) $ is not a local minimum.






            share|cite|improve this answer





















            • thank you as well :)
              – Razi Awad
              Nov 29 at 13:03










            • you're welcome........
              – Mostafa Ayaz
              Nov 29 at 13:06














            0












            0








            0






            No it is not. We have $$f(x,y)=x^2(1-3y)+y^3$$let the point $(0,-epsilon)$ with $epsilon>0$ tend to $(0,0)$ therefore $$f(0,-epsilon)=-epsilon^3<0=f(0,0)$$which means that for small enough $epsilon$, the point $(0,0) $ is not a local minimum.






            share|cite|improve this answer












            No it is not. We have $$f(x,y)=x^2(1-3y)+y^3$$let the point $(0,-epsilon)$ with $epsilon>0$ tend to $(0,0)$ therefore $$f(0,-epsilon)=-epsilon^3<0=f(0,0)$$which means that for small enough $epsilon$, the point $(0,0) $ is not a local minimum.







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Nov 29 at 13:01









            Mostafa Ayaz

            13.6k3836




            13.6k3836












            • thank you as well :)
              – Razi Awad
              Nov 29 at 13:03










            • you're welcome........
              – Mostafa Ayaz
              Nov 29 at 13:06


















            • thank you as well :)
              – Razi Awad
              Nov 29 at 13:03










            • you're welcome........
              – Mostafa Ayaz
              Nov 29 at 13:06
















            thank you as well :)
            – Razi Awad
            Nov 29 at 13:03




            thank you as well :)
            – Razi Awad
            Nov 29 at 13:03












            you're welcome........
            – Mostafa Ayaz
            Nov 29 at 13:06




            you're welcome........
            – Mostafa Ayaz
            Nov 29 at 13:06


















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