How to solve this system of equations systematically?
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
edited Dec 10 '18 at 20:33
gt6989b
33k22452
asked Dec 10 '18 at 20:21
Hello_World
3,90321630
add a comment |
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
edited Dec 10 '18 at 20:33
gt6989b
33k22452
asked Dec 10 '18 at 20:21
Hello_World
3,90321630
6
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.
– Paul
Dec 10 '18 at 20:29
add a comment |
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
edited Dec 10 '18 at 20:33
gt6989b
33k22452
asked Dec 10 '18 at 20:21
Hello_World
3,90321630
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
edited Dec 10 '18 at 20:33
gt6989b
33k22452
asked Dec 10 '18 at 20:21
Hello_World
3,90321630
edited Dec 10 '18 at 20:33
gt6989b
33k22452
asked Dec 10 '18 at 20:21
Hello_World
3,90321630
edited Dec 10 '18 at 20:33
gt6989b
33k22452
edited Dec 10 '18 at 20:33
gt6989b
33k22452
edited Dec 10 '18 at 20:33
gt6989b
33k22452
33k22452
asked Dec 10 '18 at 20:21
Hello_World
3,90321630
asked Dec 10 '18 at 20:21
Hello_World
3,90321630
asked Dec 10 '18 at 20:21
Hello_World
3,90321630
3,90321630
6
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.
– Paul
Dec 10 '18 at 20:29
add a comment |
6
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.
– Paul
Dec 10 '18 at 20:29
6
6
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.
– Paul
Dec 10 '18 at 20:29
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.
– Paul
Dec 10 '18 at 20:29
add a comment |
2 Answers
2
active
oldest
votes
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
$$S_1 cap S_2 = {x = 0, y = 0, (2+3x+4y = 0)cap(1+x+3y = 0)} = {x = 0, y = 0}cup {x=-frac 25, y = -frac 15}$$
answered Dec 10 '18 at 21:02
Cesareo
8,3013516
add a comment |
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 $(x=0)lor(y=0)lor(x=-2/5land y=-1/5)$
answered Dec 10 '18 at 20:52
Shubham Johri
3,961717
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
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
$$S_1 cap S_2 = {x = 0, y = 0, (2+3x+4y = 0)cap(1+x+3y = 0)} = {x = 0, y = 0}cup {x=-frac 25, y = -frac 15}$$
answered Dec 10 '18 at 21:02
Cesareo
8,3013516
add a comment |
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
$$S_1 cap S_2 = {x = 0, y = 0, (2+3x+4y = 0)cap(1+x+3y = 0)} = {x = 0, y = 0}cup {x=-frac 25, y = -frac 15}$$
answered Dec 10 '18 at 21:02
Cesareo
8,3013516
add a comment |
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
$$S_1 cap S_2 = {x = 0, y = 0, (2+3x+4y = 0)cap(1+x+3y = 0)} = {x = 0, y = 0}cup {x=-frac 25, y = -frac 15}$$
answered Dec 10 '18 at 21:02
Cesareo
8,3013516
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
$$S_1 cap S_2 = {x = 0, y = 0, (2+3x+4y = 0)cap(1+x+3y = 0)} = {x = 0, y = 0}cup {x=-frac 25, y = -frac 15}$$
answered Dec 10 '18 at 21:02
Cesareo
8,3013516
answered Dec 10 '18 at 21:02
Cesareo
8,3013516
answered Dec 10 '18 at 21:02
Cesareo
8,3013516
answered Dec 10 '18 at 21:02
Cesareo
8,3013516
8,3013516
add a comment |
add a comment |
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 $(x=0)lor(y=0)lor(x=-2/5land y=-1/5)$
answered Dec 10 '18 at 20:52
Shubham Johri
3,961717
add a comment |
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 $(x=0)lor(y=0)lor(x=-2/5land y=-1/5)$
answered Dec 10 '18 at 20:52
Shubham Johri
3,961717
add a comment |
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 $(x=0)lor(y=0)lor(x=-2/5land y=-1/5)$
answered Dec 10 '18 at 20:52
Shubham Johri
3,961717
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 $(x=0)lor(y=0)lor(x=-2/5land y=-1/5)$
answered Dec 10 '18 at 20:52
Shubham Johri
3,961717
answered Dec 10 '18 at 20:52
Shubham Johri
3,961717
answered Dec 10 '18 at 20:52
Shubham Johri
3,961717
answered Dec 10 '18 at 20:52
Shubham Johri
3,961717
3,961717
add a comment |
add a comment |
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6
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.
– Paul
Dec 10 '18 at 20:29