Multi-variable extremisation problem when pure substitution fails to deliver all solutions
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I have always thought that pure, equality substitution was one of the methods in mathematics which is always guaranteed to work. Yet I appear to have found a case when it fails and if somebody here can help me understand it I would be very thankful.
Find and classify the critical points of the following functions subject to equality constraints:
$$f(x,y) = x^2 - y^2 $$
$$ x^2 + y^2 = 1 $$
Performing a Lagrangian we have:
$$mathcal{L} = x^2 - y^2 -lambda(x^2+y^2-1) $$
FOC:
$ 2x -2lambda x = 0 $
$ -2y - 2lambda y = 0 $
$x^2 + y^2 = 1 $
By pure observation solutions are:
(x,y,$lambda$) = { (1,0,1), (-1,0,1), (0,1,-1), (0,-1,-1) }
However, in this case one would be tempted to use substitution, so that the problem becomes:
$$f(x) = x^2 + (x^2-1) = 2x^2 -1 $$
giving $$f'(x) = 4x = 0 $$
Hence this substitution only leads to (x,y,$lambda$) = {(0,1,-1), (0,-1,-1) }, missing out two of the solutions! Performing the reverse substitution gives the other two.
Why is it that substitution doesn't work fully?
derivatives lagrange-multiplier maxima-minima
asked Jan 11 at 14:17
JhonnyJhonny
127110
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add a comment |
$begingroup$
I have always thought that pure, equality substitution was one of the methods in mathematics which is always guaranteed to work. Yet I appear to have found a case when it fails and if somebody here can help me understand it I would be very thankful.
Find and classify the critical points of the following functions subject to equality constraints:
$$f(x,y) = x^2 - y^2 $$
$$ x^2 + y^2 = 1 $$
Performing a Lagrangian we have:
$$mathcal{L} = x^2 - y^2 -lambda(x^2+y^2-1) $$
FOC:
$ 2x -2lambda x = 0 $
$ -2y - 2lambda y = 0 $
$x^2 + y^2 = 1 $
By pure observation solutions are:
(x,y,$lambda$) = { (1,0,1), (-1,0,1), (0,1,-1), (0,-1,-1) }
However, in this case one would be tempted to use substitution, so that the problem becomes:
$$f(x) = x^2 + (x^2-1) = 2x^2 -1 $$
giving $$f'(x) = 4x = 0 $$
Hence this substitution only leads to (x,y,$lambda$) = {(0,1,-1), (0,-1,-1) }, missing out two of the solutions! Performing the reverse substitution gives the other two.
Why is it that substitution doesn't work fully?
derivatives lagrange-multiplier maxima-minima
asked Jan 11 at 14:17
JhonnyJhonny
127110
$endgroup$
add a comment |
$begingroup$
I have always thought that pure, equality substitution was one of the methods in mathematics which is always guaranteed to work. Yet I appear to have found a case when it fails and if somebody here can help me understand it I would be very thankful.
Find and classify the critical points of the following functions subject to equality constraints:
$$f(x,y) = x^2 - y^2 $$
$$ x^2 + y^2 = 1 $$
Performing a Lagrangian we have:
$$mathcal{L} = x^2 - y^2 -lambda(x^2+y^2-1) $$
FOC:
$ 2x -2lambda x = 0 $
$ -2y - 2lambda y = 0 $
$x^2 + y^2 = 1 $
By pure observation solutions are:
(x,y,$lambda$) = { (1,0,1), (-1,0,1), (0,1,-1), (0,-1,-1) }
However, in this case one would be tempted to use substitution, so that the problem becomes:
$$f(x) = x^2 + (x^2-1) = 2x^2 -1 $$
giving $$f'(x) = 4x = 0 $$
Hence this substitution only leads to (x,y,$lambda$) = {(0,1,-1), (0,-1,-1) }, missing out two of the solutions! Performing the reverse substitution gives the other two.
Why is it that substitution doesn't work fully?
derivatives lagrange-multiplier maxima-minima
asked Jan 11 at 14:17
JhonnyJhonny
127110
$endgroup$
I have always thought that pure, equality substitution was one of the methods in mathematics which is always guaranteed to work. Yet I appear to have found a case when it fails and if somebody here can help me understand it I would be very thankful.
Find and classify the critical points of the following functions subject to equality constraints:
$$f(x,y) = x^2 - y^2 $$
$$ x^2 + y^2 = 1 $$
Performing a Lagrangian we have:
$$mathcal{L} = x^2 - y^2 -lambda(x^2+y^2-1) $$
FOC:
$ 2x -2lambda x = 0 $
$ -2y - 2lambda y = 0 $
$x^2 + y^2 = 1 $
By pure observation solutions are:
(x,y,$lambda$) = { (1,0,1), (-1,0,1), (0,1,-1), (0,-1,-1) }
However, in this case one would be tempted to use substitution, so that the problem becomes:
$$f(x) = x^2 + (x^2-1) = 2x^2 -1 $$
giving $$f'(x) = 4x = 0 $$
Hence this substitution only leads to (x,y,$lambda$) = {(0,1,-1), (0,-1,-1) }, missing out two of the solutions! Performing the reverse substitution gives the other two.
Why is it that substitution doesn't work fully?
derivatives lagrange-multiplier maxima-minima
derivatives lagrange-multiplier maxima-minima
asked Jan 11 at 14:17
JhonnyJhonny
127110
asked Jan 11 at 14:17
JhonnyJhonny
127110
asked Jan 11 at 14:17
JhonnyJhonny
127110
asked Jan 11 at 14:17
JhonnyJhonny
127110
asked Jan 11 at 14:17
JhonnyJhonny
127110
127110
add a comment |
add a comment |
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