Finding the extrema of $f(x,y) = 2x^{4} - 3x^{2}y + y^{2}$












3












$begingroup$


$$f(x,y) = 2x^{4} - 3x^{2}y + y^{2}$$



I found the stationary points of this function using the equations -



$$frac{partial f}{partial x} = 0 qquad frac{partial f}{partial y} = 0$$



I got $(0,0)$.



Now we calculate $R$, $S$, and $T$ at $(0,0)$ where



$$R = frac{partial^{2} f}{partial x^{2} } qquad
S = frac{partial^{2} f}{partial xpartial y} qquad
T = frac{partial^{2} f}{partial y^{2} }$$



then,
$$RT - S^{2} = 0$$



This does't lead to any result. How to find nature of this stationary point?



I know what local minima, local maxima, and saddle points are, but how to find the nature of $(0,0)$?






calculus







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$endgroup$












  • $begingroup$
    Look at this post math.stackexchange.com/questions/2156583/…
    $endgroup$
    – Paul
    Dec 11 '18 at 8:28










  • $begingroup$
    You should show what you got for all of the derivatives, just in case there was a stray sign error or something.
    $endgroup$
    – Blue
    Dec 11 '18 at 8:45
















3












$begingroup$


$$f(x,y) = 2x^{4} - 3x^{2}y + y^{2}$$



I found the stationary points of this function using the equations -



$$frac{partial f}{partial x} = 0 qquad frac{partial f}{partial y} = 0$$



I got $(0,0)$.



Now we calculate $R$, $S$, and $T$ at $(0,0)$ where



$$R = frac{partial^{2} f}{partial x^{2} } qquad
S = frac{partial^{2} f}{partial xpartial y} qquad
T = frac{partial^{2} f}{partial y^{2} }$$



then,
$$RT - S^{2} = 0$$



This does't lead to any result. How to find nature of this stationary point?



I know what local minima, local maxima, and saddle points are, but how to find the nature of $(0,0)$?






calculus







share|cite|improve this question











$endgroup$












  • $begingroup$
    Look at this post math.stackexchange.com/questions/2156583/…
    $endgroup$
    – Paul
    Dec 11 '18 at 8:28










  • $begingroup$
    You should show what you got for all of the derivatives, just in case there was a stray sign error or something.
    $endgroup$
    – Blue
    Dec 11 '18 at 8:45














3












3








3





$begingroup$


$$f(x,y) = 2x^{4} - 3x^{2}y + y^{2}$$



I found the stationary points of this function using the equations -



$$frac{partial f}{partial x} = 0 qquad frac{partial f}{partial y} = 0$$



I got $(0,0)$.



Now we calculate $R$, $S$, and $T$ at $(0,0)$ where



$$R = frac{partial^{2} f}{partial x^{2} } qquad
S = frac{partial^{2} f}{partial xpartial y} qquad
T = frac{partial^{2} f}{partial y^{2} }$$



then,
$$RT - S^{2} = 0$$



This does't lead to any result. How to find nature of this stationary point?



I know what local minima, local maxima, and saddle points are, but how to find the nature of $(0,0)$?






calculus







share|cite|improve this question











$endgroup$




$$f(x,y) = 2x^{4} - 3x^{2}y + y^{2}$$



I found the stationary points of this function using the equations -



$$frac{partial f}{partial x} = 0 qquad frac{partial f}{partial y} = 0$$



I got $(0,0)$.



Now we calculate $R$, $S$, and $T$ at $(0,0)$ where



$$R = frac{partial^{2} f}{partial x^{2} } qquad
S = frac{partial^{2} f}{partial xpartial y} qquad
T = frac{partial^{2} f}{partial y^{2} }$$



then,
$$RT - S^{2} = 0$$



This does't lead to any result. How to find nature of this stationary point?



I know what local minima, local maxima, and saddle points are, but how to find the nature of $(0,0)$?






calculus




calculus






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 11 '18 at 8:44









Blue

48.4k870154




48.4k870154










asked Dec 11 '18 at 7:26









MathsaddictMathsaddict

3539




3539












  • $begingroup$
    Look at this post math.stackexchange.com/questions/2156583/…
    $endgroup$
    – Paul
    Dec 11 '18 at 8:28










  • $begingroup$
    You should show what you got for all of the derivatives, just in case there was a stray sign error or something.
    $endgroup$
    – Blue
    Dec 11 '18 at 8:45


















  • $begingroup$
    Look at this post math.stackexchange.com/questions/2156583/…
    $endgroup$
    – Paul
    Dec 11 '18 at 8:28










  • $begingroup$
    You should show what you got for all of the derivatives, just in case there was a stray sign error or something.
    $endgroup$
    – Blue
    Dec 11 '18 at 8:45
















$begingroup$
Look at this post math.stackexchange.com/questions/2156583/…
$endgroup$
– Paul
Dec 11 '18 at 8:28




$begingroup$
Look at this post math.stackexchange.com/questions/2156583/…
$endgroup$
– Paul
Dec 11 '18 at 8:28












$begingroup$
You should show what you got for all of the derivatives, just in case there was a stray sign error or something.
$endgroup$
– Blue
Dec 11 '18 at 8:45




$begingroup$
You should show what you got for all of the derivatives, just in case there was a stray sign error or something.
$endgroup$
– Blue
Dec 11 '18 at 8:45










3 Answers
3






active

oldest

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2












$begingroup$

$$f(x,y)=2x^4-3x^2y+y^2=left( y-2 {{x}^{2}}right) , left( y-{{x}^{2}}right)$$
$$fleft( varepsilon ,frac{3 {{epsilon }^{2}}}{2}right) =-frac{{{varepsilon }^{4}}}{4}<0$$
$$fleft( varepsilon ,frac{{{varepsilon }^{2}}}{2}right) =frac{3 {{varepsilon }^{4}}}{4}>0$$



Then $(0,0)$ is saddle point.






share|cite|improve this answer











$endgroup$





















    1












    $begingroup$

    $f(0,0)=0.$

    Along the line $y=0$ is $$f(x,0)=2x^4>0$$ in a neighborhood of $(0,0),$



    EDIT

    while along $y=frac 32 x^2;$ it is $$f(x,frac 32 x^2)=-frac 14 x^4<0$$ in a neighborhood of $(0,0).$

    Therefore is $(0,0)$ a saddle point.






    share|cite|improve this answer











    $endgroup$









    • 1




      $begingroup$
      $f(x,x) = 2x^4 - 3x^3 + x^2$. Also $f(x,x)>0$ is $x$ is small
      $endgroup$
      – Dylan
      Dec 11 '18 at 10:18





















    0












    $begingroup$

    From $f(x,y)=2(x^2-frac34y)^2-frac18y^2$ it's easy to see that $(0,0)$ is a saddle.






    share|cite|improve this answer









    $endgroup$













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      3 Answers
      3






      active

      oldest

      votes








      3 Answers
      3






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      2












      $begingroup$

      $$f(x,y)=2x^4-3x^2y+y^2=left( y-2 {{x}^{2}}right) , left( y-{{x}^{2}}right)$$
      $$fleft( varepsilon ,frac{3 {{epsilon }^{2}}}{2}right) =-frac{{{varepsilon }^{4}}}{4}<0$$
      $$fleft( varepsilon ,frac{{{varepsilon }^{2}}}{2}right) =frac{3 {{varepsilon }^{4}}}{4}>0$$



      Then $(0,0)$ is saddle point.






      share|cite|improve this answer











      $endgroup$


















        2












        $begingroup$

        $$f(x,y)=2x^4-3x^2y+y^2=left( y-2 {{x}^{2}}right) , left( y-{{x}^{2}}right)$$
        $$fleft( varepsilon ,frac{3 {{epsilon }^{2}}}{2}right) =-frac{{{varepsilon }^{4}}}{4}<0$$
        $$fleft( varepsilon ,frac{{{varepsilon }^{2}}}{2}right) =frac{3 {{varepsilon }^{4}}}{4}>0$$



        Then $(0,0)$ is saddle point.






        share|cite|improve this answer











        $endgroup$
















          2












          2








          2





          $begingroup$

          $$f(x,y)=2x^4-3x^2y+y^2=left( y-2 {{x}^{2}}right) , left( y-{{x}^{2}}right)$$
          $$fleft( varepsilon ,frac{3 {{epsilon }^{2}}}{2}right) =-frac{{{varepsilon }^{4}}}{4}<0$$
          $$fleft( varepsilon ,frac{{{varepsilon }^{2}}}{2}right) =frac{3 {{varepsilon }^{4}}}{4}>0$$



          Then $(0,0)$ is saddle point.






          share|cite|improve this answer











          $endgroup$



          $$f(x,y)=2x^4-3x^2y+y^2=left( y-2 {{x}^{2}}right) , left( y-{{x}^{2}}right)$$
          $$fleft( varepsilon ,frac{3 {{epsilon }^{2}}}{2}right) =-frac{{{varepsilon }^{4}}}{4}<0$$
          $$fleft( varepsilon ,frac{{{varepsilon }^{2}}}{2}right) =frac{3 {{varepsilon }^{4}}}{4}>0$$



          Then $(0,0)$ is saddle point.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Dec 11 '18 at 9:17

























          answered Dec 11 '18 at 8:32









          Aleksas DomarkasAleksas Domarkas

          1,24916




          1,24916























              1












              $begingroup$

              $f(0,0)=0.$

              Along the line $y=0$ is $$f(x,0)=2x^4>0$$ in a neighborhood of $(0,0),$



              EDIT

              while along $y=frac 32 x^2;$ it is $$f(x,frac 32 x^2)=-frac 14 x^4<0$$ in a neighborhood of $(0,0).$

              Therefore is $(0,0)$ a saddle point.






              share|cite|improve this answer











              $endgroup$









              • 1




                $begingroup$
                $f(x,x) = 2x^4 - 3x^3 + x^2$. Also $f(x,x)>0$ is $x$ is small
                $endgroup$
                – Dylan
                Dec 11 '18 at 10:18


















              1












              $begingroup$

              $f(0,0)=0.$

              Along the line $y=0$ is $$f(x,0)=2x^4>0$$ in a neighborhood of $(0,0),$



              EDIT

              while along $y=frac 32 x^2;$ it is $$f(x,frac 32 x^2)=-frac 14 x^4<0$$ in a neighborhood of $(0,0).$

              Therefore is $(0,0)$ a saddle point.






              share|cite|improve this answer











              $endgroup$









              • 1




                $begingroup$
                $f(x,x) = 2x^4 - 3x^3 + x^2$. Also $f(x,x)>0$ is $x$ is small
                $endgroup$
                – Dylan
                Dec 11 '18 at 10:18
















              1












              1








              1





              $begingroup$

              $f(0,0)=0.$

              Along the line $y=0$ is $$f(x,0)=2x^4>0$$ in a neighborhood of $(0,0),$



              EDIT

              while along $y=frac 32 x^2;$ it is $$f(x,frac 32 x^2)=-frac 14 x^4<0$$ in a neighborhood of $(0,0).$

              Therefore is $(0,0)$ a saddle point.






              share|cite|improve this answer











              $endgroup$



              $f(0,0)=0.$

              Along the line $y=0$ is $$f(x,0)=2x^4>0$$ in a neighborhood of $(0,0),$



              EDIT

              while along $y=frac 32 x^2;$ it is $$f(x,frac 32 x^2)=-frac 14 x^4<0$$ in a neighborhood of $(0,0).$

              Therefore is $(0,0)$ a saddle point.







              share|cite|improve this answer














              share|cite|improve this answer



              share|cite|improve this answer








              edited Dec 11 '18 at 11:03

























              answered Dec 11 '18 at 9:59









              user376343user376343

              3,7783827




              3,7783827








              • 1




                $begingroup$
                $f(x,x) = 2x^4 - 3x^3 + x^2$. Also $f(x,x)>0$ is $x$ is small
                $endgroup$
                – Dylan
                Dec 11 '18 at 10:18
















              • 1




                $begingroup$
                $f(x,x) = 2x^4 - 3x^3 + x^2$. Also $f(x,x)>0$ is $x$ is small
                $endgroup$
                – Dylan
                Dec 11 '18 at 10:18










              1




              1




              $begingroup$
              $f(x,x) = 2x^4 - 3x^3 + x^2$. Also $f(x,x)>0$ is $x$ is small
              $endgroup$
              – Dylan
              Dec 11 '18 at 10:18






              $begingroup$
              $f(x,x) = 2x^4 - 3x^3 + x^2$. Also $f(x,x)>0$ is $x$ is small
              $endgroup$
              – Dylan
              Dec 11 '18 at 10:18













              0












              $begingroup$

              From $f(x,y)=2(x^2-frac34y)^2-frac18y^2$ it's easy to see that $(0,0)$ is a saddle.






              share|cite|improve this answer









              $endgroup$


















                0












                $begingroup$

                From $f(x,y)=2(x^2-frac34y)^2-frac18y^2$ it's easy to see that $(0,0)$ is a saddle.






                share|cite|improve this answer









                $endgroup$
















                  0












                  0








                  0





                  $begingroup$

                  From $f(x,y)=2(x^2-frac34y)^2-frac18y^2$ it's easy to see that $(0,0)$ is a saddle.






                  share|cite|improve this answer









                  $endgroup$



                  From $f(x,y)=2(x^2-frac34y)^2-frac18y^2$ it's easy to see that $(0,0)$ is a saddle.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Dec 11 '18 at 16:53









                  Michael HoppeMichael Hoppe

                  11k31836




                  11k31836






























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