Mixing
An exercise using Lagrange's Multipliers
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I am having some trouble finding the max and min of the below problem using Lagrange multipliers:
$$ f(x,y) = x^2 -2xy + 3y^2 \ text{subject to} x^2 + 2y^2 + x + y = 0$$
Is there a trick or something i am missing? that was a question in one of my friends test and no one was able to solve it.
--------------------------------------------------------//--------------------------------------------------------
i get:
$$ 2x-2y = lambda(2x+1) \ -2x+6y = lambda(4y+1) \ x^2 + 2y^2 + x + y = 0$$
i can't solve it, is there a shortcut?
multivariable-calculus
asked Jan 7 at 23:56
Robert William HanksRobert William Hanks
44959
$endgroup$
|
show 10 more comments
$begingroup$
I am having some trouble finding the max and min of the below problem using Lagrange multipliers:
$$ f(x,y) = x^2 -2xy + 3y^2 \ text{subject to} x^2 + 2y^2 + x + y = 0$$
Is there a trick or something i am missing? that was a question in one of my friends test and no one was able to solve it.
--------------------------------------------------------//--------------------------------------------------------
i get:
$$ 2x-2y = lambda(2x+1) \ -2x+6y = lambda(4y+1) \ x^2 + 2y^2 + x + y = 0$$
i can't solve it, is there a shortcut?
multivariable-calculus
asked Jan 7 at 23:56
Robert William HanksRobert William Hanks
44959
$endgroup$
2
$begingroup$
Where did you get stuck? Did you compute the partial derivatives? Could you solve the system of equations obtained?
$endgroup$
– A. Pongrácz
Jan 8 at 0:08
$begingroup$
i am stuck at the system of equations
$endgroup$
– Robert William Hanks
Jan 8 at 0:46
$begingroup$
Yes, you should revise the rules of derivation.
$endgroup$
– A. Pongrácz
Jan 8 at 0:49
$begingroup$
@A. Pongrácz i just did
$endgroup$
– Robert William Hanks
Jan 8 at 0:49
1
$begingroup$
@RobertWilliamHanks: I added a solution below. It is rather tedious, but I see no obvious simplification that would yield an easier path.
$endgroup$
– copper.hat
Jan 8 at 2:06
|
show 10 more comments
$begingroup$
I am having some trouble finding the max and min of the below problem using Lagrange multipliers:
$$ f(x,y) = x^2 -2xy + 3y^2 \ text{subject to} x^2 + 2y^2 + x + y = 0$$
Is there a trick or something i am missing? that was a question in one of my friends test and no one was able to solve it.
--------------------------------------------------------//--------------------------------------------------------
i get:
$$ 2x-2y = lambda(2x+1) \ -2x+6y = lambda(4y+1) \ x^2 + 2y^2 + x + y = 0$$
i can't solve it, is there a shortcut?
multivariable-calculus
asked Jan 7 at 23:56
Robert William HanksRobert William Hanks
44959
$endgroup$
I am having some trouble finding the max and min of the below problem using Lagrange multipliers:
$$ f(x,y) = x^2 -2xy + 3y^2 \ text{subject to} x^2 + 2y^2 + x + y = 0$$
Is there a trick or something i am missing? that was a question in one of my friends test and no one was able to solve it.
--------------------------------------------------------//--------------------------------------------------------
i get:
$$ 2x-2y = lambda(2x+1) \ -2x+6y = lambda(4y+1) \ x^2 + 2y^2 + x + y = 0$$
i can't solve it, is there a shortcut?
multivariable-calculus
multivariable-calculus
asked Jan 7 at 23:56
Robert William HanksRobert William Hanks
44959
asked Jan 7 at 23:56
Robert William HanksRobert William Hanks
44959
edited Jan 8 at 0:54
Robert William Hanks
asked Jan 7 at 23:56
Robert William HanksRobert William Hanks
44959
asked Jan 7 at 23:56
Robert William HanksRobert William Hanks
44959
asked Jan 7 at 23:56
Robert William HanksRobert William Hanks
44959
44959
2
$begingroup$
Where did you get stuck? Did you compute the partial derivatives? Could you solve the system of equations obtained?
$endgroup$
– A. Pongrácz
Jan 8 at 0:08
$begingroup$
i am stuck at the system of equations
$endgroup$
– Robert William Hanks
Jan 8 at 0:46
$begingroup$
Yes, you should revise the rules of derivation.
$endgroup$
– A. Pongrácz
Jan 8 at 0:49
$begingroup$
@A. Pongrácz i just did
$endgroup$
– Robert William Hanks
Jan 8 at 0:49
1
$begingroup$
@RobertWilliamHanks: I added a solution below. It is rather tedious, but I see no obvious simplification that would yield an easier path.
$endgroup$
– copper.hat
Jan 8 at 2:06
|
show 10 more comments
2
$begingroup$
Where did you get stuck? Did you compute the partial derivatives? Could you solve the system of equations obtained?
$endgroup$
– A. Pongrácz
Jan 8 at 0:08
$begingroup$
i am stuck at the system of equations
$endgroup$
– Robert William Hanks
Jan 8 at 0:46
$begingroup$
Yes, you should revise the rules of derivation.
$endgroup$
– A. Pongrácz
Jan 8 at 0:49
$begingroup$
@A. Pongrácz i just did
$endgroup$
– Robert William Hanks
Jan 8 at 0:49
1
$begingroup$
@RobertWilliamHanks: I added a solution below. It is rather tedious, but I see no obvious simplification that would yield an easier path.
$endgroup$
– copper.hat
Jan 8 at 2:06
2
2
$begingroup$
Where did you get stuck? Did you compute the partial derivatives? Could you solve the system of equations obtained?
$endgroup$
– A. Pongrácz
Jan 8 at 0:08
$begingroup$
Where did you get stuck? Did you compute the partial derivatives? Could you solve the system of equations obtained?
$endgroup$
– A. Pongrácz
Jan 8 at 0:08
$begingroup$
i am stuck at the system of equations
$endgroup$
– Robert William Hanks
Jan 8 at 0:46
$begingroup$
i am stuck at the system of equations
$endgroup$
– Robert William Hanks
Jan 8 at 0:46
$begingroup$
Yes, you should revise the rules of derivation.
$endgroup$
– A. Pongrácz
Jan 8 at 0:49
$begingroup$
Yes, you should revise the rules of derivation.
$endgroup$
– A. Pongrácz
Jan 8 at 0:49
$begingroup$
@A. Pongrácz i just did
$endgroup$
– Robert William Hanks
Jan 8 at 0:49
$begingroup$
@A. Pongrácz i just did
$endgroup$
– Robert William Hanks
Jan 8 at 0:49
1
1
$begingroup$
@RobertWilliamHanks: I added a solution below. It is rather tedious, but I see no obvious simplification that would yield an easier path.
$endgroup$
– copper.hat
Jan 8 at 2:06
$begingroup$
@RobertWilliamHanks: I added a solution below. It is rather tedious, but I see no obvious simplification that would yield an easier path.
$endgroup$
– copper.hat
Jan 8 at 2:06
|
show 10 more comments
3 Answers
3
active
oldest
votes
$begingroup$
I assume you are looking for real solutions.
The Lagrangian function is:
$$
L(x,y,lambda)= x^2-2xy + 3y^2-lambda(x^2+2y^2+x+y)
$$
Taking derivatives with respect to $x, y, lambda$ respectively yields:
$$
2x-2y -2lambda x -lambda=0$$
$$-2x+6y-4 lambda y -lambda = 0$$
$$x^2+2y^2+x+y=0$$
To solve the system (three equations, three unknowns), note that:
$lambda = frac{2x-2y}{2x+1} = frac{-2x+6y}{4y+1}$. Therefore:
$$
(2x-2y)(1+4y) = (2x+1) (6y-2x)
$$
or
$$
4x^2-8y^2-4xy+4x-8y=0 ,(I)
$$
Multiplying the constraint by 4 gives:
$$4x^2 + 8y^2 + 4x + 4y=0. , (II)$$
Deducting (I) from (II) yields:
$$16y^2 + 4xy +12y=0$$
Therefore, either $y=0 Rightarrow x=0$ or $x=-1$, or $16y+4x+12=0$. Combine the last equation with the constraint to find additional solutions, and you will get none (they are complex).
So there are two solutions for $(x,y)$, namely $(0,0)$ and $(-1,0)$
You should check also the second order conditions at the optimum, e.g. bordered Hessian, or KKT conditions.
Plugging these values into the objective function gives: $f(0,0)=0$, $f(-1,0)=3$. The point (0,0) is a minimum; (-1,0) is a maximum.
answered Jan 8 at 0:48
pendermathpendermath
54410
$endgroup$
$begingroup$
I think the conclusion is a bit hasty. Just like in case of univariate functions, just because the derivative is zero at $x$, it doesn't mean that $x$ is any kind of extremal point.
$endgroup$
– A. Pongrácz
Jan 8 at 0:51
$begingroup$
@copper.hat Actually, the author of the post is very unclear about where his problems arose. He made mistakes in finding the equations as well, so this answer is clearly helpful. I see that pendermath has extended the answer; but the problem I pointed out is still valid. It is not clear at all if the two points you have found are extremal or not. And the fact that $f(0,0)=0$ and $f(-1,0)=3$ does not imply that the former is a minimum and the latter is a maximum. It could be the other way around, or it could be that some of these are not even local extrema.
$endgroup$
– A. Pongrácz
Jan 8 at 0:55
$begingroup$
@A.Pongrácz: The author has added an elaboration.
$endgroup$
– copper.hat
Jan 8 at 0:56
$begingroup$
@copper.hat But it is still not addressing my problem.
$endgroup$
– A. Pongrácz
Jan 8 at 0:58
$begingroup$
Sorry, but the "Therefore" part in the end is still absurd.
$endgroup$
– A. Pongrácz
Jan 8 at 1:01
|
show 3 more comments
$begingroup$
The constraint defines a compact set so we know a solution exists.
The Lagrange equations are
$E_1:2x-2y+ lambda (1+2x) = 0$,
$E_2: 6y-2x + lambda (1+4y) = 0$.
Simplifying the equation $(1+4y)E_1-(1+2x)E_2 = 0$ gives
$(1+x+y)(2y-x) = 0$.
If we let $y=-(1+x)$ in the constraint we get $(x+1)(3x+1) = 0$ which yields
the candidates $(-1,0), (-{1 over 3} , -{2 over 3})$. A tedious computation shows that
$f$ has value $1$ at both points.
If we let $x=2y$ in the constraint we get $y(2y+1) = 0$ which yields the candidates
$(0,0),(-1, -{1 over 2})$. Another computation shows that $f$ has value $0, {3 over 4}$ respectively from which we see that
the $min$ is $0$ at $(0,0)$ and the $max$ is $1$ at either $(-1,0), (-{1 over 3} , -{2 over 3})$.
Notes:
Note that the quadratic form $f(x,y)$ is positive definite and that $(0,0)$
satisfies the constraint. Hence $(0,0)$ is a (in fact, the) minimiser since $f(0,0) = 0$.
answered Jan 8 at 1:39
copper.hatcopper.hat
126k559160
$endgroup$
add a comment |
$begingroup$
Starting from pendermath's answer, considering the equations
$$2x-2y -2lambda x -lambda=0tag1$$
$$-2x+6y-4 lambda y -lambda = 0tag2$$
$$x^2+2y^2+x+y=0tag3$$ solve $(1)$ and $(2)$ for $x$ and $y$. This gives
$$x=-frac{lambda }{2 lambda -1}qquad text{and} qquad y=-frac{lambda }{2 (2 lambda -1)}=frac x 2$$ assuming $lambda neq frac 12$.
Plug in $(3)$ to get
$$frac{3 (lambda -1) lambda }{2 (2 lambda-1 )^2}=0$$ so the candidates $lambda=0$ for which $x=y=0$ and $lambda=1$ for which $x=-1$ and $y=frac 12$
answered Jan 8 at 4:48
Claude LeiboviciClaude Leibovici
120k1157132
$endgroup$
$begingroup$
I think you are missing a minus sign on the last y.
$endgroup$
– copper.hat
Jan 9 at 15:56
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
I assume you are looking for real solutions.
The Lagrangian function is:
$$
L(x,y,lambda)= x^2-2xy + 3y^2-lambda(x^2+2y^2+x+y)
$$
Taking derivatives with respect to $x, y, lambda$ respectively yields:
$$
2x-2y -2lambda x -lambda=0$$
$$-2x+6y-4 lambda y -lambda = 0$$
$$x^2+2y^2+x+y=0$$
To solve the system (three equations, three unknowns), note that:
$lambda = frac{2x-2y}{2x+1} = frac{-2x+6y}{4y+1}$. Therefore:
$$
(2x-2y)(1+4y) = (2x+1) (6y-2x)
$$
or
$$
4x^2-8y^2-4xy+4x-8y=0 ,(I)
$$
Multiplying the constraint by 4 gives:
$$4x^2 + 8y^2 + 4x + 4y=0. , (II)$$
Deducting (I) from (II) yields:
$$16y^2 + 4xy +12y=0$$
Therefore, either $y=0 Rightarrow x=0$ or $x=-1$, or $16y+4x+12=0$. Combine the last equation with the constraint to find additional solutions, and you will get none (they are complex).
So there are two solutions for $(x,y)$, namely $(0,0)$ and $(-1,0)$
You should check also the second order conditions at the optimum, e.g. bordered Hessian, or KKT conditions.
Plugging these values into the objective function gives: $f(0,0)=0$, $f(-1,0)=3$. The point (0,0) is a minimum; (-1,0) is a maximum.
answered Jan 8 at 0:48
pendermathpendermath
54410
$endgroup$
$begingroup$
I think the conclusion is a bit hasty. Just like in case of univariate functions, just because the derivative is zero at $x$, it doesn't mean that $x$ is any kind of extremal point.
$endgroup$
– A. Pongrácz
Jan 8 at 0:51
$begingroup$
@copper.hat Actually, the author of the post is very unclear about where his problems arose. He made mistakes in finding the equations as well, so this answer is clearly helpful. I see that pendermath has extended the answer; but the problem I pointed out is still valid. It is not clear at all if the two points you have found are extremal or not. And the fact that $f(0,0)=0$ and $f(-1,0)=3$ does not imply that the former is a minimum and the latter is a maximum. It could be the other way around, or it could be that some of these are not even local extrema.
$endgroup$
– A. Pongrácz
Jan 8 at 0:55
$begingroup$
@A.Pongrácz: The author has added an elaboration.
$endgroup$
– copper.hat
Jan 8 at 0:56
$begingroup$
@copper.hat But it is still not addressing my problem.
$endgroup$
– A. Pongrácz
Jan 8 at 0:58
$begingroup$
Sorry, but the "Therefore" part in the end is still absurd.
$endgroup$
– A. Pongrácz
Jan 8 at 1:01
|
show 3 more comments
$begingroup$
I assume you are looking for real solutions.
The Lagrangian function is:
$$
L(x,y,lambda)= x^2-2xy + 3y^2-lambda(x^2+2y^2+x+y)
$$
Taking derivatives with respect to $x, y, lambda$ respectively yields:
$$
2x-2y -2lambda x -lambda=0$$
$$-2x+6y-4 lambda y -lambda = 0$$
$$x^2+2y^2+x+y=0$$
To solve the system (three equations, three unknowns), note that:
$lambda = frac{2x-2y}{2x+1} = frac{-2x+6y}{4y+1}$. Therefore:
$$
(2x-2y)(1+4y) = (2x+1) (6y-2x)
$$
or
$$
4x^2-8y^2-4xy+4x-8y=0 ,(I)
$$
Multiplying the constraint by 4 gives:
$$4x^2 + 8y^2 + 4x + 4y=0. , (II)$$
Deducting (I) from (II) yields:
$$16y^2 + 4xy +12y=0$$
Therefore, either $y=0 Rightarrow x=0$ or $x=-1$, or $16y+4x+12=0$. Combine the last equation with the constraint to find additional solutions, and you will get none (they are complex).
So there are two solutions for $(x,y)$, namely $(0,0)$ and $(-1,0)$
You should check also the second order conditions at the optimum, e.g. bordered Hessian, or KKT conditions.
Plugging these values into the objective function gives: $f(0,0)=0$, $f(-1,0)=3$. The point (0,0) is a minimum; (-1,0) is a maximum.
answered Jan 8 at 0:48
pendermathpendermath
54410
$endgroup$
$begingroup$
I think the conclusion is a bit hasty. Just like in case of univariate functions, just because the derivative is zero at $x$, it doesn't mean that $x$ is any kind of extremal point.
$endgroup$
– A. Pongrácz
Jan 8 at 0:51
$begingroup$
@copper.hat Actually, the author of the post is very unclear about where his problems arose. He made mistakes in finding the equations as well, so this answer is clearly helpful. I see that pendermath has extended the answer; but the problem I pointed out is still valid. It is not clear at all if the two points you have found are extremal or not. And the fact that $f(0,0)=0$ and $f(-1,0)=3$ does not imply that the former is a minimum and the latter is a maximum. It could be the other way around, or it could be that some of these are not even local extrema.
$endgroup$
– A. Pongrácz
Jan 8 at 0:55
$begingroup$
@A.Pongrácz: The author has added an elaboration.
$endgroup$
– copper.hat
Jan 8 at 0:56
$begingroup$
@copper.hat But it is still not addressing my problem.
$endgroup$
– A. Pongrácz
Jan 8 at 0:58
$begingroup$
Sorry, but the "Therefore" part in the end is still absurd.
$endgroup$
– A. Pongrácz
Jan 8 at 1:01
|
show 3 more comments
$begingroup$
I assume you are looking for real solutions.
The Lagrangian function is:
$$
L(x,y,lambda)= x^2-2xy + 3y^2-lambda(x^2+2y^2+x+y)
$$
Taking derivatives with respect to $x, y, lambda$ respectively yields:
$$
2x-2y -2lambda x -lambda=0$$
$$-2x+6y-4 lambda y -lambda = 0$$
$$x^2+2y^2+x+y=0$$
To solve the system (three equations, three unknowns), note that:
$lambda = frac{2x-2y}{2x+1} = frac{-2x+6y}{4y+1}$. Therefore:
$$
(2x-2y)(1+4y) = (2x+1) (6y-2x)
$$
or
$$
4x^2-8y^2-4xy+4x-8y=0 ,(I)
$$
Multiplying the constraint by 4 gives:
$$4x^2 + 8y^2 + 4x + 4y=0. , (II)$$
Deducting (I) from (II) yields:
$$16y^2 + 4xy +12y=0$$
Therefore, either $y=0 Rightarrow x=0$ or $x=-1$, or $16y+4x+12=0$. Combine the last equation with the constraint to find additional solutions, and you will get none (they are complex).
So there are two solutions for $(x,y)$, namely $(0,0)$ and $(-1,0)$
You should check also the second order conditions at the optimum, e.g. bordered Hessian, or KKT conditions.
Plugging these values into the objective function gives: $f(0,0)=0$, $f(-1,0)=3$. The point (0,0) is a minimum; (-1,0) is a maximum.
answered Jan 8 at 0:48
pendermathpendermath
54410
$endgroup$
I assume you are looking for real solutions.
The Lagrangian function is:
$$
L(x,y,lambda)= x^2-2xy + 3y^2-lambda(x^2+2y^2+x+y)
$$
Taking derivatives with respect to $x, y, lambda$ respectively yields:
$$
2x-2y -2lambda x -lambda=0$$
$$-2x+6y-4 lambda y -lambda = 0$$
$$x^2+2y^2+x+y=0$$
To solve the system (three equations, three unknowns), note that:
$lambda = frac{2x-2y}{2x+1} = frac{-2x+6y}{4y+1}$. Therefore:
$$
(2x-2y)(1+4y) = (2x+1) (6y-2x)
$$
or
$$
4x^2-8y^2-4xy+4x-8y=0 ,(I)
$$
Multiplying the constraint by 4 gives:
$$4x^2 + 8y^2 + 4x + 4y=0. , (II)$$
Deducting (I) from (II) yields:
$$16y^2 + 4xy +12y=0$$
Therefore, either $y=0 Rightarrow x=0$ or $x=-1$, or $16y+4x+12=0$. Combine the last equation with the constraint to find additional solutions, and you will get none (they are complex).
So there are two solutions for $(x,y)$, namely $(0,0)$ and $(-1,0)$
You should check also the second order conditions at the optimum, e.g. bordered Hessian, or KKT conditions.
Plugging these values into the objective function gives: $f(0,0)=0$, $f(-1,0)=3$. The point (0,0) is a minimum; (-1,0) is a maximum.
answered Jan 8 at 0:48
pendermathpendermath
54410
edited Jan 8 at 1:17
answered Jan 8 at 0:48
pendermathpendermath
54410
answered Jan 8 at 0:48
pendermathpendermath
54410
answered Jan 8 at 0:48
pendermathpendermath
54410
54410
$begingroup$
I think the conclusion is a bit hasty. Just like in case of univariate functions, just because the derivative is zero at $x$, it doesn't mean that $x$ is any kind of extremal point.
$endgroup$
– A. Pongrácz
Jan 8 at 0:51
$begingroup$
@copper.hat Actually, the author of the post is very unclear about where his problems arose. He made mistakes in finding the equations as well, so this answer is clearly helpful. I see that pendermath has extended the answer; but the problem I pointed out is still valid. It is not clear at all if the two points you have found are extremal or not. And the fact that $f(0,0)=0$ and $f(-1,0)=3$ does not imply that the former is a minimum and the latter is a maximum. It could be the other way around, or it could be that some of these are not even local extrema.
$endgroup$
– A. Pongrácz
Jan 8 at 0:55
$begingroup$
@A.Pongrácz: The author has added an elaboration.
$endgroup$
– copper.hat
Jan 8 at 0:56
$begingroup$
@copper.hat But it is still not addressing my problem.
$endgroup$
– A. Pongrácz
Jan 8 at 0:58
$begingroup$
Sorry, but the "Therefore" part in the end is still absurd.
$endgroup$
– A. Pongrácz
Jan 8 at 1:01
|
show 3 more comments
$begingroup$
I think the conclusion is a bit hasty. Just like in case of univariate functions, just because the derivative is zero at $x$, it doesn't mean that $x$ is any kind of extremal point.
$endgroup$
– A. Pongrácz
Jan 8 at 0:51
$begingroup$
@copper.hat Actually, the author of the post is very unclear about where his problems arose. He made mistakes in finding the equations as well, so this answer is clearly helpful. I see that pendermath has extended the answer; but the problem I pointed out is still valid. It is not clear at all if the two points you have found are extremal or not. And the fact that $f(0,0)=0$ and $f(-1,0)=3$ does not imply that the former is a minimum and the latter is a maximum. It could be the other way around, or it could be that some of these are not even local extrema.
$endgroup$
– A. Pongrácz
Jan 8 at 0:55
$begingroup$
@A.Pongrácz: The author has added an elaboration.
$endgroup$
– copper.hat
Jan 8 at 0:56
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@copper.hat But it is still not addressing my problem.
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– A. Pongrácz
Jan 8 at 0:58
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Sorry, but the "Therefore" part in the end is still absurd.
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– A. Pongrácz
Jan 8 at 1:01
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I think the conclusion is a bit hasty. Just like in case of univariate functions, just because the derivative is zero at $x$, it doesn't mean that $x$ is any kind of extremal point.
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– A. Pongrácz
Jan 8 at 0:51
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I think the conclusion is a bit hasty. Just like in case of univariate functions, just because the derivative is zero at $x$, it doesn't mean that $x$ is any kind of extremal point.
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– A. Pongrácz
Jan 8 at 0:51
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@copper.hat Actually, the author of the post is very unclear about where his problems arose. He made mistakes in finding the equations as well, so this answer is clearly helpful. I see that pendermath has extended the answer; but the problem I pointed out is still valid. It is not clear at all if the two points you have found are extremal or not. And the fact that $f(0,0)=0$ and $f(-1,0)=3$ does not imply that the former is a minimum and the latter is a maximum. It could be the other way around, or it could be that some of these are not even local extrema.
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– A. Pongrácz
Jan 8 at 0:55
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@copper.hat Actually, the author of the post is very unclear about where his problems arose. He made mistakes in finding the equations as well, so this answer is clearly helpful. I see that pendermath has extended the answer; but the problem I pointed out is still valid. It is not clear at all if the two points you have found are extremal or not. And the fact that $f(0,0)=0$ and $f(-1,0)=3$ does not imply that the former is a minimum and the latter is a maximum. It could be the other way around, or it could be that some of these are not even local extrema.
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– A. Pongrácz
Jan 8 at 0:55
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@A.Pongrácz: The author has added an elaboration.
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– copper.hat
Jan 8 at 0:56
$begingroup$
@A.Pongrácz: The author has added an elaboration.
$endgroup$
– copper.hat
Jan 8 at 0:56
$begingroup$
@copper.hat But it is still not addressing my problem.
$endgroup$
– A. Pongrácz
Jan 8 at 0:58
$begingroup$
@copper.hat But it is still not addressing my problem.
$endgroup$
– A. Pongrácz
Jan 8 at 0:58
$begingroup$
Sorry, but the "Therefore" part in the end is still absurd.
$endgroup$
– A. Pongrácz
Jan 8 at 1:01
$begingroup$
Sorry, but the "Therefore" part in the end is still absurd.
$endgroup$
– A. Pongrácz
Jan 8 at 1:01
|
show 3 more comments
$begingroup$
The constraint defines a compact set so we know a solution exists.
The Lagrange equations are
$E_1:2x-2y+ lambda (1+2x) = 0$,
$E_2: 6y-2x + lambda (1+4y) = 0$.
Simplifying the equation $(1+4y)E_1-(1+2x)E_2 = 0$ gives
$(1+x+y)(2y-x) = 0$.
If we let $y=-(1+x)$ in the constraint we get $(x+1)(3x+1) = 0$ which yields
the candidates $(-1,0), (-{1 over 3} , -{2 over 3})$. A tedious computation shows that
$f$ has value $1$ at both points.
If we let $x=2y$ in the constraint we get $y(2y+1) = 0$ which yields the candidates
$(0,0),(-1, -{1 over 2})$. Another computation shows that $f$ has value $0, {3 over 4}$ respectively from which we see that
the $min$ is $0$ at $(0,0)$ and the $max$ is $1$ at either $(-1,0), (-{1 over 3} , -{2 over 3})$.
Notes:
Note that the quadratic form $f(x,y)$ is positive definite and that $(0,0)$
satisfies the constraint. Hence $(0,0)$ is a (in fact, the) minimiser since $f(0,0) = 0$.
answered Jan 8 at 1:39
copper.hatcopper.hat
126k559160
$endgroup$
add a comment |
$begingroup$
The constraint defines a compact set so we know a solution exists.
The Lagrange equations are
$E_1:2x-2y+ lambda (1+2x) = 0$,
$E_2: 6y-2x + lambda (1+4y) = 0$.
Simplifying the equation $(1+4y)E_1-(1+2x)E_2 = 0$ gives
$(1+x+y)(2y-x) = 0$.
If we let $y=-(1+x)$ in the constraint we get $(x+1)(3x+1) = 0$ which yields
the candidates $(-1,0), (-{1 over 3} , -{2 over 3})$. A tedious computation shows that
$f$ has value $1$ at both points.
If we let $x=2y$ in the constraint we get $y(2y+1) = 0$ which yields the candidates
$(0,0),(-1, -{1 over 2})$. Another computation shows that $f$ has value $0, {3 over 4}$ respectively from which we see that
the $min$ is $0$ at $(0,0)$ and the $max$ is $1$ at either $(-1,0), (-{1 over 3} , -{2 over 3})$.
Notes:
Note that the quadratic form $f(x,y)$ is positive definite and that $(0,0)$
satisfies the constraint. Hence $(0,0)$ is a (in fact, the) minimiser since $f(0,0) = 0$.
answered Jan 8 at 1:39
copper.hatcopper.hat
126k559160
$endgroup$
add a comment |
$begingroup$
The constraint defines a compact set so we know a solution exists.
The Lagrange equations are
$E_1:2x-2y+ lambda (1+2x) = 0$,
$E_2: 6y-2x + lambda (1+4y) = 0$.
Simplifying the equation $(1+4y)E_1-(1+2x)E_2 = 0$ gives
$(1+x+y)(2y-x) = 0$.
If we let $y=-(1+x)$ in the constraint we get $(x+1)(3x+1) = 0$ which yields
the candidates $(-1,0), (-{1 over 3} , -{2 over 3})$. A tedious computation shows that
$f$ has value $1$ at both points.
If we let $x=2y$ in the constraint we get $y(2y+1) = 0$ which yields the candidates
$(0,0),(-1, -{1 over 2})$. Another computation shows that $f$ has value $0, {3 over 4}$ respectively from which we see that
the $min$ is $0$ at $(0,0)$ and the $max$ is $1$ at either $(-1,0), (-{1 over 3} , -{2 over 3})$.
Notes:
Note that the quadratic form $f(x,y)$ is positive definite and that $(0,0)$
satisfies the constraint. Hence $(0,0)$ is a (in fact, the) minimiser since $f(0,0) = 0$.
answered Jan 8 at 1:39
copper.hatcopper.hat
126k559160
$endgroup$
The constraint defines a compact set so we know a solution exists.
The Lagrange equations are
$E_1:2x-2y+ lambda (1+2x) = 0$,
$E_2: 6y-2x + lambda (1+4y) = 0$.
Simplifying the equation $(1+4y)E_1-(1+2x)E_2 = 0$ gives
$(1+x+y)(2y-x) = 0$.
If we let $y=-(1+x)$ in the constraint we get $(x+1)(3x+1) = 0$ which yields
the candidates $(-1,0), (-{1 over 3} , -{2 over 3})$. A tedious computation shows that
$f$ has value $1$ at both points.
If we let $x=2y$ in the constraint we get $y(2y+1) = 0$ which yields the candidates
$(0,0),(-1, -{1 over 2})$. Another computation shows that $f$ has value $0, {3 over 4}$ respectively from which we see that
the $min$ is $0$ at $(0,0)$ and the $max$ is $1$ at either $(-1,0), (-{1 over 3} , -{2 over 3})$.
Notes:
Note that the quadratic form $f(x,y)$ is positive definite and that $(0,0)$
satisfies the constraint. Hence $(0,0)$ is a (in fact, the) minimiser since $f(0,0) = 0$.
answered Jan 8 at 1:39
copper.hatcopper.hat
126k559160
edited Jan 8 at 2:13
answered Jan 8 at 1:39
copper.hatcopper.hat
126k559160
answered Jan 8 at 1:39
copper.hatcopper.hat
126k559160
answered Jan 8 at 1:39
copper.hatcopper.hat
126k559160
126k559160
add a comment |
add a comment |
$begingroup$
Starting from pendermath's answer, considering the equations
$$2x-2y -2lambda x -lambda=0tag1$$
$$-2x+6y-4 lambda y -lambda = 0tag2$$
$$x^2+2y^2+x+y=0tag3$$ solve $(1)$ and $(2)$ for $x$ and $y$. This gives
$$x=-frac{lambda }{2 lambda -1}qquad text{and} qquad y=-frac{lambda }{2 (2 lambda -1)}=frac x 2$$ assuming $lambda neq frac 12$.
Plug in $(3)$ to get
$$frac{3 (lambda -1) lambda }{2 (2 lambda-1 )^2}=0$$ so the candidates $lambda=0$ for which $x=y=0$ and $lambda=1$ for which $x=-1$ and $y=frac 12$
answered Jan 8 at 4:48
Claude LeiboviciClaude Leibovici
120k1157132
$endgroup$
$begingroup$
I think you are missing a minus sign on the last y.
$endgroup$
– copper.hat
Jan 9 at 15:56
add a comment |
$begingroup$
Starting from pendermath's answer, considering the equations
$$2x-2y -2lambda x -lambda=0tag1$$
$$-2x+6y-4 lambda y -lambda = 0tag2$$
$$x^2+2y^2+x+y=0tag3$$ solve $(1)$ and $(2)$ for $x$ and $y$. This gives
$$x=-frac{lambda }{2 lambda -1}qquad text{and} qquad y=-frac{lambda }{2 (2 lambda -1)}=frac x 2$$ assuming $lambda neq frac 12$.
Plug in $(3)$ to get
$$frac{3 (lambda -1) lambda }{2 (2 lambda-1 )^2}=0$$ so the candidates $lambda=0$ for which $x=y=0$ and $lambda=1$ for which $x=-1$ and $y=frac 12$
answered Jan 8 at 4:48
Claude LeiboviciClaude Leibovici
120k1157132
$endgroup$
$begingroup$
I think you are missing a minus sign on the last y.
$endgroup$
– copper.hat
Jan 9 at 15:56
add a comment |
$begingroup$
Starting from pendermath's answer, considering the equations
$$2x-2y -2lambda x -lambda=0tag1$$
$$-2x+6y-4 lambda y -lambda = 0tag2$$
$$x^2+2y^2+x+y=0tag3$$ solve $(1)$ and $(2)$ for $x$ and $y$. This gives
$$x=-frac{lambda }{2 lambda -1}qquad text{and} qquad y=-frac{lambda }{2 (2 lambda -1)}=frac x 2$$ assuming $lambda neq frac 12$.
Plug in $(3)$ to get
$$frac{3 (lambda -1) lambda }{2 (2 lambda-1 )^2}=0$$ so the candidates $lambda=0$ for which $x=y=0$ and $lambda=1$ for which $x=-1$ and $y=frac 12$
answered Jan 8 at 4:48
Claude LeiboviciClaude Leibovici
120k1157132
$endgroup$
Starting from pendermath's answer, considering the equations
$$2x-2y -2lambda x -lambda=0tag1$$
$$-2x+6y-4 lambda y -lambda = 0tag2$$
$$x^2+2y^2+x+y=0tag3$$ solve $(1)$ and $(2)$ for $x$ and $y$. This gives
$$x=-frac{lambda }{2 lambda -1}qquad text{and} qquad y=-frac{lambda }{2 (2 lambda -1)}=frac x 2$$ assuming $lambda neq frac 12$.
Plug in $(3)$ to get
$$frac{3 (lambda -1) lambda }{2 (2 lambda-1 )^2}=0$$ so the candidates $lambda=0$ for which $x=y=0$ and $lambda=1$ for which $x=-1$ and $y=frac 12$
answered Jan 8 at 4:48
Claude LeiboviciClaude Leibovici
120k1157132
answered Jan 8 at 4:48
Claude LeiboviciClaude Leibovici
120k1157132
answered Jan 8 at 4:48
Claude LeiboviciClaude Leibovici
120k1157132
answered Jan 8 at 4:48
Claude LeiboviciClaude Leibovici
120k1157132
120k1157132
$begingroup$
I think you are missing a minus sign on the last y.
$endgroup$
– copper.hat
Jan 9 at 15:56
add a comment |
$begingroup$
I think you are missing a minus sign on the last y.
$endgroup$
– copper.hat
Jan 9 at 15:56
$begingroup$
I think you are missing a minus sign on the last y.
$endgroup$
– copper.hat
Jan 9 at 15:56
$begingroup$
I think you are missing a minus sign on the last y.
$endgroup$
– copper.hat
Jan 9 at 15:56
add a comment |
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$begingroup$
Where did you get stuck? Did you compute the partial derivatives? Could you solve the system of equations obtained?
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– A. Pongrácz
Jan 8 at 0:08
$begingroup$
i am stuck at the system of equations
$endgroup$
– Robert William Hanks
Jan 8 at 0:46
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Yes, you should revise the rules of derivation.
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– A. Pongrácz
Jan 8 at 0:49
$begingroup$
@A. Pongrácz i just did
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– Robert William Hanks
Jan 8 at 0:49
1
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@RobertWilliamHanks: I added a solution below. It is rather tedious, but I see no obvious simplification that would yield an easier path.
$endgroup$
– copper.hat
Jan 8 at 2:06