What is the nature of the critical points of $f(x,y) = 4x^2 -12xy + 9y^2$.












1












$begingroup$


Discuss the nature (saddle point, extremal point) of the critical points $p_o$ of the function described by:



$f(x,y) = 4x^2 -12xy + 9y^2$.



Then,



(1) $f_1 (x,y) = 8x - 12y$



and



(2) $f_2 (x,y) -12x + 18y$.



To get the critical points of the function $f(x,y)$, I set (1) and (2) equal to $0$ and solve the simultaneous equations. So,



$8x - 12y = 0$



$-12x + 18y = 0$.



Since These two equations are scalar multiples of each other, I get that $t(frac{3}{2},1)$ is the set of critical points for the function.



Now, $f_{12} = -12$, $f_{11} = 8$, and $f_{22} = 18$. From my textbook, $Delta = (f_{12} (p_o))^2 - f_{11} (p_o)f_{22} (p_o)$, which means that $Delta = 144 -(18)(8) = 0$. My textbook says that if $Delta = 0$, then the nature of $p_o$ can't be determined by the $Delta$ formula. Are there other ways to figure out the nature of the critical points of $f$?






calculus multivariable-calculus







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    1












    $begingroup$


    Discuss the nature (saddle point, extremal point) of the critical points $p_o$ of the function described by:



    $f(x,y) = 4x^2 -12xy + 9y^2$.



    Then,



    (1) $f_1 (x,y) = 8x - 12y$



    and



    (2) $f_2 (x,y) -12x + 18y$.



    To get the critical points of the function $f(x,y)$, I set (1) and (2) equal to $0$ and solve the simultaneous equations. So,



    $8x - 12y = 0$



    $-12x + 18y = 0$.



    Since These two equations are scalar multiples of each other, I get that $t(frac{3}{2},1)$ is the set of critical points for the function.



    Now, $f_{12} = -12$, $f_{11} = 8$, and $f_{22} = 18$. From my textbook, $Delta = (f_{12} (p_o))^2 - f_{11} (p_o)f_{22} (p_o)$, which means that $Delta = 144 -(18)(8) = 0$. My textbook says that if $Delta = 0$, then the nature of $p_o$ can't be determined by the $Delta$ formula. Are there other ways to figure out the nature of the critical points of $f$?






    calculus multivariable-calculus







    share|cite|improve this question









    $endgroup$















      1












      1








      1





      $begingroup$


      Discuss the nature (saddle point, extremal point) of the critical points $p_o$ of the function described by:



      $f(x,y) = 4x^2 -12xy + 9y^2$.



      Then,



      (1) $f_1 (x,y) = 8x - 12y$



      and



      (2) $f_2 (x,y) -12x + 18y$.



      To get the critical points of the function $f(x,y)$, I set (1) and (2) equal to $0$ and solve the simultaneous equations. So,



      $8x - 12y = 0$



      $-12x + 18y = 0$.



      Since These two equations are scalar multiples of each other, I get that $t(frac{3}{2},1)$ is the set of critical points for the function.



      Now, $f_{12} = -12$, $f_{11} = 8$, and $f_{22} = 18$. From my textbook, $Delta = (f_{12} (p_o))^2 - f_{11} (p_o)f_{22} (p_o)$, which means that $Delta = 144 -(18)(8) = 0$. My textbook says that if $Delta = 0$, then the nature of $p_o$ can't be determined by the $Delta$ formula. Are there other ways to figure out the nature of the critical points of $f$?






      calculus multivariable-calculus







      share|cite|improve this question









      $endgroup$




      Discuss the nature (saddle point, extremal point) of the critical points $p_o$ of the function described by:



      $f(x,y) = 4x^2 -12xy + 9y^2$.



      Then,



      (1) $f_1 (x,y) = 8x - 12y$



      and



      (2) $f_2 (x,y) -12x + 18y$.



      To get the critical points of the function $f(x,y)$, I set (1) and (2) equal to $0$ and solve the simultaneous equations. So,



      $8x - 12y = 0$



      $-12x + 18y = 0$.



      Since These two equations are scalar multiples of each other, I get that $t(frac{3}{2},1)$ is the set of critical points for the function.



      Now, $f_{12} = -12$, $f_{11} = 8$, and $f_{22} = 18$. From my textbook, $Delta = (f_{12} (p_o))^2 - f_{11} (p_o)f_{22} (p_o)$, which means that $Delta = 144 -(18)(8) = 0$. My textbook says that if $Delta = 0$, then the nature of $p_o$ can't be determined by the $Delta$ formula. Are there other ways to figure out the nature of the critical points of $f$?






      calculus multivariable-calculus




      calculus multivariable-calculus






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      asked Dec 4 '18 at 1:08









      K.MK.M

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          $begingroup$

          $$(4x^2-12xy + 9y^2)=(2x-3y)^2ge 0$$



          Hence it attains minimum value when $2x=3y$.






          share|cite|improve this answer









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            1 Answer
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            1 Answer
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            active

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            active

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            active

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            3












            $begingroup$

            $$(4x^2-12xy + 9y^2)=(2x-3y)^2ge 0$$



            Hence it attains minimum value when $2x=3y$.






            share|cite|improve this answer









            $endgroup$


















              3












              $begingroup$

              $$(4x^2-12xy + 9y^2)=(2x-3y)^2ge 0$$



              Hence it attains minimum value when $2x=3y$.






              share|cite|improve this answer









              $endgroup$
















                3












                3








                3





                $begingroup$

                $$(4x^2-12xy + 9y^2)=(2x-3y)^2ge 0$$



                Hence it attains minimum value when $2x=3y$.






                share|cite|improve this answer









                $endgroup$



                $$(4x^2-12xy + 9y^2)=(2x-3y)^2ge 0$$



                Hence it attains minimum value when $2x=3y$.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Dec 4 '18 at 1:14









                Siong Thye GohSiong Thye Goh

                100k1466117




                100k1466117






























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