How to solve this system of equations systematically?












4












$begingroup$


This might seem a trivial problem, but I have some trouble in arranging the data. So suppose you are given $f(x,y)=x^2y^2(1+x+2y)$ and you want to find it's critical points. Thus we find
$$frac{partial f}{partial x}(x,y)=xy^2(2+3x+4y)textrm{ and }frac{partial f}{partial y}(x,y)=2x^2y(1+x+3y).$$



Now we set $f_x= 0$ and $f_y=0.$ Thus we get a system
$$
begin{split}
xy^2(2+3x+4y) &=0\
2x^2y(1+x+3y) &=0
end{split}
$$



Now we have a bunch of cases. The way I think about this is as follows:
$$((x=0)lor(y=0)lor(3x+4y=-2))land((x=0)lor(y=0)lor(x+3y=-1)).$$
Then I consider each possibility separately, but this seems to be slow and sometimes I forget some solutions. Thus I was wondering if there are any other methods which one can use to solve such type of problems.






multivariable-calculus systems-of-equations maxima-minima







share|cite|improve this question











$endgroup$








  • 6




    $begingroup$
    If $x=0$ then y can be anything. Similarly if $y=0$ then x can be anything. If neither x nor y is 0 then you can divide through by those variables leaving two linear simultaneous equations.
    $endgroup$
    – Paul
    Dec 10 '18 at 20:29
















4












$begingroup$


This might seem a trivial problem, but I have some trouble in arranging the data. So suppose you are given $f(x,y)=x^2y^2(1+x+2y)$ and you want to find it's critical points. Thus we find
$$frac{partial f}{partial x}(x,y)=xy^2(2+3x+4y)textrm{ and }frac{partial f}{partial y}(x,y)=2x^2y(1+x+3y).$$



Now we set $f_x= 0$ and $f_y=0.$ Thus we get a system
$$
begin{split}
xy^2(2+3x+4y) &=0\
2x^2y(1+x+3y) &=0
end{split}
$$



Now we have a bunch of cases. The way I think about this is as follows:
$$((x=0)lor(y=0)lor(3x+4y=-2))land((x=0)lor(y=0)lor(x+3y=-1)).$$
Then I consider each possibility separately, but this seems to be slow and sometimes I forget some solutions. Thus I was wondering if there are any other methods which one can use to solve such type of problems.






multivariable-calculus systems-of-equations maxima-minima







share|cite|improve this question











$endgroup$








  • 6




    $begingroup$
    If $x=0$ then y can be anything. Similarly if $y=0$ then x can be anything. If neither x nor y is 0 then you can divide through by those variables leaving two linear simultaneous equations.
    $endgroup$
    – Paul
    Dec 10 '18 at 20:29














4












4








4





$begingroup$


This might seem a trivial problem, but I have some trouble in arranging the data. So suppose you are given $f(x,y)=x^2y^2(1+x+2y)$ and you want to find it's critical points. Thus we find
$$frac{partial f}{partial x}(x,y)=xy^2(2+3x+4y)textrm{ and }frac{partial f}{partial y}(x,y)=2x^2y(1+x+3y).$$



Now we set $f_x= 0$ and $f_y=0.$ Thus we get a system
$$
begin{split}
xy^2(2+3x+4y) &=0\
2x^2y(1+x+3y) &=0
end{split}
$$



Now we have a bunch of cases. The way I think about this is as follows:
$$((x=0)lor(y=0)lor(3x+4y=-2))land((x=0)lor(y=0)lor(x+3y=-1)).$$
Then I consider each possibility separately, but this seems to be slow and sometimes I forget some solutions. Thus I was wondering if there are any other methods which one can use to solve such type of problems.






multivariable-calculus systems-of-equations maxima-minima







share|cite|improve this question











$endgroup$




This might seem a trivial problem, but I have some trouble in arranging the data. So suppose you are given $f(x,y)=x^2y^2(1+x+2y)$ and you want to find it's critical points. Thus we find
$$frac{partial f}{partial x}(x,y)=xy^2(2+3x+4y)textrm{ and }frac{partial f}{partial y}(x,y)=2x^2y(1+x+3y).$$



Now we set $f_x= 0$ and $f_y=0.$ Thus we get a system
$$
begin{split}
xy^2(2+3x+4y) &=0\
2x^2y(1+x+3y) &=0
end{split}
$$



Now we have a bunch of cases. The way I think about this is as follows:
$$((x=0)lor(y=0)lor(3x+4y=-2))land((x=0)lor(y=0)lor(x+3y=-1)).$$
Then I consider each possibility separately, but this seems to be slow and sometimes I forget some solutions. Thus I was wondering if there are any other methods which one can use to solve such type of problems.






multivariable-calculus systems-of-equations maxima-minima




multivariable-calculus systems-of-equations maxima-minima






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 10 '18 at 20:33









gt6989b

34.2k22455




34.2k22455










asked Dec 10 '18 at 20:21









Hello_WorldHello_World

4,13121831




4,13121831








  • 6




    $begingroup$
    If $x=0$ then y can be anything. Similarly if $y=0$ then x can be anything. If neither x nor y is 0 then you can divide through by those variables leaving two linear simultaneous equations.
    $endgroup$
    – Paul
    Dec 10 '18 at 20:29














  • 6




    $begingroup$
    If $x=0$ then y can be anything. Similarly if $y=0$ then x can be anything. If neither x nor y is 0 then you can divide through by those variables leaving two linear simultaneous equations.
    $endgroup$
    – Paul
    Dec 10 '18 at 20:29








6




6




$begingroup$
If $x=0$ then y can be anything. Similarly if $y=0$ then x can be anything. If neither x nor y is 0 then you can divide through by those variables leaving two linear simultaneous equations.
$endgroup$
– Paul
Dec 10 '18 at 20:29




$begingroup$
If $x=0$ then y can be anything. Similarly if $y=0$ then x can be anything. If neither x nor y is 0 then you can divide through by those variables leaving two linear simultaneous equations.
$endgroup$
– Paul
Dec 10 '18 at 20:29










2 Answers
2






active

oldest

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1












$begingroup$

For $xy^2(2+3x+4y) =0$ we have the set of solutions



$$S_1 = {x = 0, y = 0, 2+3x+4y = 0}$$



For $2x^2y(1+x+3y) =0$ we have the set of solutions



$$S_2 = {x = 0, y = 0, 1+x+3y = 0}$$



So for the system of equations we have



$$S_1 cap S_2 = {x = 0, y = 0, (2+3x+4y = 0)cap(1+x+3y = 0)} = {x = 0, y = 0}cup {x=-frac 25, y = -frac 15}$$






share|cite|improve this answer









$endgroup$





















    0












    $begingroup$

    Use the converse of the distributive property:



    $((x=0)lor(y=0)lor(3x+4y=-2))land((x=0)lor(y=0)lor(x+3y=-1))\equiv(x=0)lor(y=0)lor[(3x+4y=-2)land(x+3y=-1)]$



    $3x+4y+2=0=x+3y+1$ is just a pair of straight lines (linear equations) intersecting at $(-2/5,-1/5)$. Therefore, you have $(x=0)lor(y=0)lor(x=-2/5land y=-1/5)$






    share|cite|improve this answer









    $endgroup$













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      2 Answers
      2






      active

      oldest

      votes








      2 Answers
      2






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      1












      $begingroup$

      For $xy^2(2+3x+4y) =0$ we have the set of solutions



      $$S_1 = {x = 0, y = 0, 2+3x+4y = 0}$$



      For $2x^2y(1+x+3y) =0$ we have the set of solutions



      $$S_2 = {x = 0, y = 0, 1+x+3y = 0}$$



      So for the system of equations we have



      $$S_1 cap S_2 = {x = 0, y = 0, (2+3x+4y = 0)cap(1+x+3y = 0)} = {x = 0, y = 0}cup {x=-frac 25, y = -frac 15}$$






      share|cite|improve this answer









      $endgroup$


















        1












        $begingroup$

        For $xy^2(2+3x+4y) =0$ we have the set of solutions



        $$S_1 = {x = 0, y = 0, 2+3x+4y = 0}$$



        For $2x^2y(1+x+3y) =0$ we have the set of solutions



        $$S_2 = {x = 0, y = 0, 1+x+3y = 0}$$



        So for the system of equations we have



        $$S_1 cap S_2 = {x = 0, y = 0, (2+3x+4y = 0)cap(1+x+3y = 0)} = {x = 0, y = 0}cup {x=-frac 25, y = -frac 15}$$






        share|cite|improve this answer









        $endgroup$
















          1












          1








          1





          $begingroup$

          For $xy^2(2+3x+4y) =0$ we have the set of solutions



          $$S_1 = {x = 0, y = 0, 2+3x+4y = 0}$$



          For $2x^2y(1+x+3y) =0$ we have the set of solutions



          $$S_2 = {x = 0, y = 0, 1+x+3y = 0}$$



          So for the system of equations we have



          $$S_1 cap S_2 = {x = 0, y = 0, (2+3x+4y = 0)cap(1+x+3y = 0)} = {x = 0, y = 0}cup {x=-frac 25, y = -frac 15}$$






          share|cite|improve this answer









          $endgroup$



          For $xy^2(2+3x+4y) =0$ we have the set of solutions



          $$S_1 = {x = 0, y = 0, 2+3x+4y = 0}$$



          For $2x^2y(1+x+3y) =0$ we have the set of solutions



          $$S_2 = {x = 0, y = 0, 1+x+3y = 0}$$



          So for the system of equations we have



          $$S_1 cap S_2 = {x = 0, y = 0, (2+3x+4y = 0)cap(1+x+3y = 0)} = {x = 0, y = 0}cup {x=-frac 25, y = -frac 15}$$







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 10 '18 at 21:02









          CesareoCesareo

          8,9093516




          8,9093516























              0












              $begingroup$

              Use the converse of the distributive property:



              $((x=0)lor(y=0)lor(3x+4y=-2))land((x=0)lor(y=0)lor(x+3y=-1))\equiv(x=0)lor(y=0)lor[(3x+4y=-2)land(x+3y=-1)]$



              $3x+4y+2=0=x+3y+1$ is just a pair of straight lines (linear equations) intersecting at $(-2/5,-1/5)$. Therefore, you have $(x=0)lor(y=0)lor(x=-2/5land y=-1/5)$






              share|cite|improve this answer









              $endgroup$


















                0












                $begingroup$

                Use the converse of the distributive property:



                $((x=0)lor(y=0)lor(3x+4y=-2))land((x=0)lor(y=0)lor(x+3y=-1))\equiv(x=0)lor(y=0)lor[(3x+4y=-2)land(x+3y=-1)]$



                $3x+4y+2=0=x+3y+1$ is just a pair of straight lines (linear equations) intersecting at $(-2/5,-1/5)$. Therefore, you have $(x=0)lor(y=0)lor(x=-2/5land y=-1/5)$






                share|cite|improve this answer









                $endgroup$
















                  0












                  0








                  0





                  $begingroup$

                  Use the converse of the distributive property:



                  $((x=0)lor(y=0)lor(3x+4y=-2))land((x=0)lor(y=0)lor(x+3y=-1))\equiv(x=0)lor(y=0)lor[(3x+4y=-2)land(x+3y=-1)]$



                  $3x+4y+2=0=x+3y+1$ is just a pair of straight lines (linear equations) intersecting at $(-2/5,-1/5)$. Therefore, you have $(x=0)lor(y=0)lor(x=-2/5land y=-1/5)$






                  share|cite|improve this answer









                  $endgroup$



                  Use the converse of the distributive property:



                  $((x=0)lor(y=0)lor(3x+4y=-2))land((x=0)lor(y=0)lor(x+3y=-1))\equiv(x=0)lor(y=0)lor[(3x+4y=-2)land(x+3y=-1)]$



                  $3x+4y+2=0=x+3y+1$ is just a pair of straight lines (linear equations) intersecting at $(-2/5,-1/5)$. Therefore, you have $(x=0)lor(y=0)lor(x=-2/5land y=-1/5)$







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Dec 10 '18 at 20:52









                  Shubham JohriShubham Johri

                  5,192717




                  5,192717






























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