Constraint Qualification in Lagrange
$begingroup$
$f(x,y)=x^2-y^2$ subject to single constraint $g(x,y)=1-x-y=0$
For this question, I understand that the Constraint Qualification holds, since, rank of $D(g(x,y))=1$ everywhere. Solving Lagrange would suggest that no critical points exist.
For this example, my book states, "Since the constraint qualification holds everywhere, it must be the case that global maxima and global minima fail to exist in problem".
I cannot seem to conceptually grasp the meaning for this statement. In context of this, how are we supposed to find global maxima or minima? Please help me understand the significance of constraint qualification for global maxima or minima.
optimization self-learning lagrange-multiplier maxima-minima
asked Dec 2 '18 at 13:21
Shinjini RanaShinjini Rana
8916
$endgroup$
add a comment |
$begingroup$
$f(x,y)=x^2-y^2$ subject to single constraint $g(x,y)=1-x-y=0$
For this question, I understand that the Constraint Qualification holds, since, rank of $D(g(x,y))=1$ everywhere. Solving Lagrange would suggest that no critical points exist.
For this example, my book states, "Since the constraint qualification holds everywhere, it must be the case that global maxima and global minima fail to exist in problem".
I cannot seem to conceptually grasp the meaning for this statement. In context of this, how are we supposed to find global maxima or minima? Please help me understand the significance of constraint qualification for global maxima or minima.
optimization self-learning lagrange-multiplier maxima-minima
asked Dec 2 '18 at 13:21
Shinjini RanaShinjini Rana
8916
$endgroup$
$begingroup$
The quote of your book is literal or a translation?
$endgroup$
– Martín-Blas Pérez Pinilla
Dec 2 '18 at 14:53
add a comment |
$begingroup$
$f(x,y)=x^2-y^2$ subject to single constraint $g(x,y)=1-x-y=0$
For this question, I understand that the Constraint Qualification holds, since, rank of $D(g(x,y))=1$ everywhere. Solving Lagrange would suggest that no critical points exist.
For this example, my book states, "Since the constraint qualification holds everywhere, it must be the case that global maxima and global minima fail to exist in problem".
I cannot seem to conceptually grasp the meaning for this statement. In context of this, how are we supposed to find global maxima or minima? Please help me understand the significance of constraint qualification for global maxima or minima.
optimization self-learning lagrange-multiplier maxima-minima
asked Dec 2 '18 at 13:21
Shinjini RanaShinjini Rana
8916
$endgroup$
$f(x,y)=x^2-y^2$ subject to single constraint $g(x,y)=1-x-y=0$
For this question, I understand that the Constraint Qualification holds, since, rank of $D(g(x,y))=1$ everywhere. Solving Lagrange would suggest that no critical points exist.
For this example, my book states, "Since the constraint qualification holds everywhere, it must be the case that global maxima and global minima fail to exist in problem".
I cannot seem to conceptually grasp the meaning for this statement. In context of this, how are we supposed to find global maxima or minima? Please help me understand the significance of constraint qualification for global maxima or minima.
optimization self-learning lagrange-multiplier maxima-minima
optimization self-learning lagrange-multiplier maxima-minima
asked Dec 2 '18 at 13:21
Shinjini RanaShinjini Rana
8916
asked Dec 2 '18 at 13:21
Shinjini RanaShinjini Rana
8916
asked Dec 2 '18 at 13:21
Shinjini RanaShinjini Rana
8916
asked Dec 2 '18 at 13:21
Shinjini RanaShinjini Rana
8916
asked Dec 2 '18 at 13:21
Shinjini RanaShinjini Rana
8916
8916
$begingroup$
The quote of your book is literal or a translation?
$endgroup$
– Martín-Blas Pérez Pinilla
Dec 2 '18 at 14:53
add a comment |
$begingroup$
The quote of your book is literal or a translation?
$endgroup$
– Martín-Blas Pérez Pinilla
Dec 2 '18 at 14:53
$begingroup$
The quote of your book is literal or a translation?
$endgroup$
– Martín-Blas Pérez Pinilla
Dec 2 '18 at 14:53
$begingroup$
The quote of your book is literal or a translation?
$endgroup$
– Martín-Blas Pérez Pinilla
Dec 2 '18 at 14:53
add a comment |
1 Answer
1
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$begingroup$
In your example, the set defined by the constraint is unbounded, so is possible that no global extremum exists. Taking $y = 1 - x$, then
$$f(x,y) = x^2 - (1 - x)^2 = 2x - 1$$
and obviously this is the case.
answered Dec 2 '18 at 14:50
Martín-Blas Pérez PinillaMartín-Blas Pérez Pinilla
34.2k42871
$endgroup$
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
In your example, the set defined by the constraint is unbounded, so is possible that no global extremum exists. Taking $y = 1 - x$, then
$$f(x,y) = x^2 - (1 - x)^2 = 2x - 1$$
and obviously this is the case.
answered Dec 2 '18 at 14:50
Martín-Blas Pérez PinillaMartín-Blas Pérez Pinilla
34.2k42871
$endgroup$
add a comment |
$begingroup$
In your example, the set defined by the constraint is unbounded, so is possible that no global extremum exists. Taking $y = 1 - x$, then
$$f(x,y) = x^2 - (1 - x)^2 = 2x - 1$$
and obviously this is the case.
answered Dec 2 '18 at 14:50
Martín-Blas Pérez PinillaMartín-Blas Pérez Pinilla
34.2k42871
$endgroup$
add a comment |
$begingroup$
In your example, the set defined by the constraint is unbounded, so is possible that no global extremum exists. Taking $y = 1 - x$, then
$$f(x,y) = x^2 - (1 - x)^2 = 2x - 1$$
and obviously this is the case.
answered Dec 2 '18 at 14:50
Martín-Blas Pérez PinillaMartín-Blas Pérez Pinilla
34.2k42871
$endgroup$
In your example, the set defined by the constraint is unbounded, so is possible that no global extremum exists. Taking $y = 1 - x$, then
$$f(x,y) = x^2 - (1 - x)^2 = 2x - 1$$
and obviously this is the case.
answered Dec 2 '18 at 14:50
Martín-Blas Pérez PinillaMartín-Blas Pérez Pinilla
34.2k42871
answered Dec 2 '18 at 14:50
Martín-Blas Pérez PinillaMartín-Blas Pérez Pinilla
34.2k42871
answered Dec 2 '18 at 14:50
Martín-Blas Pérez PinillaMartín-Blas Pérez Pinilla
34.2k42871
answered Dec 2 '18 at 14:50
Martín-Blas Pérez PinillaMartín-Blas Pérez Pinilla
34.2k42871
34.2k42871
add a comment |
add a comment |
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$begingroup$
The quote of your book is literal or a translation?
$endgroup$
– Martín-Blas Pérez Pinilla
Dec 2 '18 at 14:53