Why does the method of Lagrange multiplier fail?
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Let $mathbf{x}$ and $mathbf{y}$ be two vectors in $mathbb{R}^2$, and $f$ be a function defined by $f(mathbf{x},mathbf{y}) := mathbf{x} cdot mathbf{y}$. Can this function be minimised subject to the conditions $parallel mathbf{x} parallel = parallel mathbf{y} parallel = 1$, with the method of Lagrange multiplier?
Attempt
I failed to used Lagrange multiplier to solve this question. The primary reason is that as I try to express $f$ as a function of four variables $x_1, x_2, x_3, x_4$, the gradient of $g_1$ and the gradient of $g_2$ has the last two entries being $0$. While the gradient of $f$ does not have zero elements. Thus, it seems there can be no $lambda_1$ and $lambda_2$ that equates the three gradient vectors. Am I correct? Or am I grossly simplifying the question when I express $f$ as a function of four variables?
multivariable-calculus optimization lagrange-multiplier
asked Nov 23 at 18:11
hephaes
1658
|
show 2 more comments
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0
down vote
favorite
Question
Let $mathbf{x}$ and $mathbf{y}$ be two vectors in $mathbb{R}^2$, and $f$ be a function defined by $f(mathbf{x},mathbf{y}) := mathbf{x} cdot mathbf{y}$. Can this function be minimised subject to the conditions $parallel mathbf{x} parallel = parallel mathbf{y} parallel = 1$, with the method of Lagrange multiplier?
Attempt
I failed to used Lagrange multiplier to solve this question. The primary reason is that as I try to express $f$ as a function of four variables $x_1, x_2, x_3, x_4$, the gradient of $g_1$ and the gradient of $g_2$ has the last two entries being $0$. While the gradient of $f$ does not have zero elements. Thus, it seems there can be no $lambda_1$ and $lambda_2$ that equates the three gradient vectors. Am I correct? Or am I grossly simplifying the question when I express $f$ as a function of four variables?
multivariable-calculus optimization lagrange-multiplier
asked Nov 23 at 18:11
hephaes
1658
4
When you have 2 constraints the associated system of equations is $nabla f = lambda nabla g_1 + mu nabla g_2$ rather than $nabla f = lambda nabla g_1$ AND $nabla f = mu g_2$
– user25959
Nov 23 at 18:22
2
See example here: mathonline.wikidot.com/…
– user25959
Nov 23 at 18:23
1
Also another way : Notice that $$- |x| ~ |y| leq langle x ,y rangle$$
– Red shoes
Nov 23 at 18:27
1
@Redshoes Probably it was his intent to prove that inequality with Lagrange multipliers
– Federico
Nov 23 at 18:29
@Federico If that would be orginal question, I would fix $y$ and just minimize $x.y$ over the constraint $|x|^2 = 1$. Then we have a mode with less variable and one constraint (only one Lagrange multiplier ). so much easier to solve.
– Red shoes
Nov 23 at 18:54
|
show 2 more comments
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Question
Let $mathbf{x}$ and $mathbf{y}$ be two vectors in $mathbb{R}^2$, and $f$ be a function defined by $f(mathbf{x},mathbf{y}) := mathbf{x} cdot mathbf{y}$. Can this function be minimised subject to the conditions $parallel mathbf{x} parallel = parallel mathbf{y} parallel = 1$, with the method of Lagrange multiplier?
Attempt
I failed to used Lagrange multiplier to solve this question. The primary reason is that as I try to express $f$ as a function of four variables $x_1, x_2, x_3, x_4$, the gradient of $g_1$ and the gradient of $g_2$ has the last two entries being $0$. While the gradient of $f$ does not have zero elements. Thus, it seems there can be no $lambda_1$ and $lambda_2$ that equates the three gradient vectors. Am I correct? Or am I grossly simplifying the question when I express $f$ as a function of four variables?
multivariable-calculus optimization lagrange-multiplier
asked Nov 23 at 18:11
hephaes
1658
Question
Let $mathbf{x}$ and $mathbf{y}$ be two vectors in $mathbb{R}^2$, and $f$ be a function defined by $f(mathbf{x},mathbf{y}) := mathbf{x} cdot mathbf{y}$. Can this function be minimised subject to the conditions $parallel mathbf{x} parallel = parallel mathbf{y} parallel = 1$, with the method of Lagrange multiplier?
Attempt
I failed to used Lagrange multiplier to solve this question. The primary reason is that as I try to express $f$ as a function of four variables $x_1, x_2, x_3, x_4$, the gradient of $g_1$ and the gradient of $g_2$ has the last two entries being $0$. While the gradient of $f$ does not have zero elements. Thus, it seems there can be no $lambda_1$ and $lambda_2$ that equates the three gradient vectors. Am I correct? Or am I grossly simplifying the question when I express $f$ as a function of four variables?
multivariable-calculus optimization lagrange-multiplier
multivariable-calculus optimization lagrange-multiplier
asked Nov 23 at 18:11
hephaes
1658
asked Nov 23 at 18:11
hephaes
1658
asked Nov 23 at 18:11
hephaes
1658
asked Nov 23 at 18:11
hephaes
1658
asked Nov 23 at 18:11
hephaes
1658
1658
4
When you have 2 constraints the associated system of equations is $nabla f = lambda nabla g_1 + mu nabla g_2$ rather than $nabla f = lambda nabla g_1$ AND $nabla f = mu g_2$
– user25959
Nov 23 at 18:22
2
See example here: mathonline.wikidot.com/…
– user25959
Nov 23 at 18:23
1
Also another way : Notice that $$- |x| ~ |y| leq langle x ,y rangle$$
– Red shoes
Nov 23 at 18:27
1
@Redshoes Probably it was his intent to prove that inequality with Lagrange multipliers
– Federico
Nov 23 at 18:29
@Federico If that would be orginal question, I would fix $y$ and just minimize $x.y$ over the constraint $|x|^2 = 1$. Then we have a mode with less variable and one constraint (only one Lagrange multiplier ). so much easier to solve.
– Red shoes
Nov 23 at 18:54
|
show 2 more comments
4
When you have 2 constraints the associated system of equations is $nabla f = lambda nabla g_1 + mu nabla g_2$ rather than $nabla f = lambda nabla g_1$ AND $nabla f = mu g_2$
– user25959
Nov 23 at 18:22
2
See example here: mathonline.wikidot.com/…
– user25959
Nov 23 at 18:23
1
Also another way : Notice that $$- |x| ~ |y| leq langle x ,y rangle$$
– Red shoes
Nov 23 at 18:27
1
@Redshoes Probably it was his intent to prove that inequality with Lagrange multipliers
– Federico
Nov 23 at 18:29
@Federico If that would be orginal question, I would fix $y$ and just minimize $x.y$ over the constraint $|x|^2 = 1$. Then we have a mode with less variable and one constraint (only one Lagrange multiplier ). so much easier to solve.
– Red shoes
Nov 23 at 18:54
4
4
When you have 2 constraints the associated system of equations is $nabla f = lambda nabla g_1 + mu nabla g_2$ rather than $nabla f = lambda nabla g_1$ AND $nabla f = mu g_2$
– user25959
Nov 23 at 18:22
When you have 2 constraints the associated system of equations is $nabla f = lambda nabla g_1 + mu nabla g_2$ rather than $nabla f = lambda nabla g_1$ AND $nabla f = mu g_2$
– user25959
Nov 23 at 18:22
2
2
See example here: mathonline.wikidot.com/…
– user25959
Nov 23 at 18:23
See example here: mathonline.wikidot.com/…
– user25959
Nov 23 at 18:23
1
1
Also another way : Notice that $$- |x| ~ |y| leq langle x ,y rangle$$
– Red shoes
Nov 23 at 18:27
Also another way : Notice that $$- |x| ~ |y| leq langle x ,y rangle$$
– Red shoes
Nov 23 at 18:27
1
1
@Redshoes Probably it was his intent to prove that inequality with Lagrange multipliers
– Federico
Nov 23 at 18:29
@Redshoes Probably it was his intent to prove that inequality with Lagrange multipliers
– Federico
Nov 23 at 18:29
@Federico If that would be orginal question, I would fix $y$ and just minimize $x.y$ over the constraint $|x|^2 = 1$. Then we have a mode with less variable and one constraint (only one Lagrange multiplier ). so much easier to solve.
– Red shoes
Nov 23 at 18:54
@Federico If that would be orginal question, I would fix $y$ and just minimize $x.y$ over the constraint $|x|^2 = 1$. Then we have a mode with less variable and one constraint (only one Lagrange multiplier ). so much easier to solve.
– Red shoes
Nov 23 at 18:54
|
show 2 more comments
1 Answer
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Let $f,g_1,g_2:mathbb R^{2d}tomathbb R$ be given by
$$
f(x,y)=xcdot y qquad g_1(x,y)=|x|^2 qquad g_2(x,y)=|y|^2.
$$
You want to minimize $f$ subject to $g_1=g_2=1$. Then
$$
nabla f(x,y) = (y,x) qquad nabla g_1(x,y) = (2x, 0) qquad nabla g_2(x,y) = (0, 2y).
$$
From Lagrange you get
$$
(y,x) = lambda_1(2x,0) + lambda_2(0,2y) = 2(lambda_1 x, lambda_2 y).
$$
From this you get that $x$ and $y$ bust be linearly dependent, hence $y=pm x$ because of the constraint. $y=-x$ is a global minimizer, $y=x$ is a global maximizer.
answered Nov 23 at 18:28
Federico
4,184512
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
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active
oldest
votes
up vote
1
down vote
accepted
Let $f,g_1,g_2:mathbb R^{2d}tomathbb R$ be given by
$$
f(x,y)=xcdot y qquad g_1(x,y)=|x|^2 qquad g_2(x,y)=|y|^2.
$$
You want to minimize $f$ subject to $g_1=g_2=1$. Then
$$
nabla f(x,y) = (y,x) qquad nabla g_1(x,y) = (2x, 0) qquad nabla g_2(x,y) = (0, 2y).
$$
From Lagrange you get
$$
(y,x) = lambda_1(2x,0) + lambda_2(0,2y) = 2(lambda_1 x, lambda_2 y).
$$
From this you get that $x$ and $y$ bust be linearly dependent, hence $y=pm x$ because of the constraint. $y=-x$ is a global minimizer, $y=x$ is a global maximizer.
answered Nov 23 at 18:28
Federico
4,184512
add a comment |
up vote
1
down vote
accepted
Let $f,g_1,g_2:mathbb R^{2d}tomathbb R$ be given by
$$
f(x,y)=xcdot y qquad g_1(x,y)=|x|^2 qquad g_2(x,y)=|y|^2.
$$
You want to minimize $f$ subject to $g_1=g_2=1$. Then
$$
nabla f(x,y) = (y,x) qquad nabla g_1(x,y) = (2x, 0) qquad nabla g_2(x,y) = (0, 2y).
$$
From Lagrange you get
$$
(y,x) = lambda_1(2x,0) + lambda_2(0,2y) = 2(lambda_1 x, lambda_2 y).
$$
From this you get that $x$ and $y$ bust be linearly dependent, hence $y=pm x$ because of the constraint. $y=-x$ is a global minimizer, $y=x$ is a global maximizer.
answered Nov 23 at 18:28
Federico
4,184512
add a comment |
up vote
1
down vote
accepted
up vote
1
down vote
accepted
Let $f,g_1,g_2:mathbb R^{2d}tomathbb R$ be given by
$$
f(x,y)=xcdot y qquad g_1(x,y)=|x|^2 qquad g_2(x,y)=|y|^2.
$$
You want to minimize $f$ subject to $g_1=g_2=1$. Then
$$
nabla f(x,y) = (y,x) qquad nabla g_1(x,y) = (2x, 0) qquad nabla g_2(x,y) = (0, 2y).
$$
From Lagrange you get
$$
(y,x) = lambda_1(2x,0) + lambda_2(0,2y) = 2(lambda_1 x, lambda_2 y).
$$
From this you get that $x$ and $y$ bust be linearly dependent, hence $y=pm x$ because of the constraint. $y=-x$ is a global minimizer, $y=x$ is a global maximizer.
answered Nov 23 at 18:28
Federico
4,184512
Let $f,g_1,g_2:mathbb R^{2d}tomathbb R$ be given by
$$
f(x,y)=xcdot y qquad g_1(x,y)=|x|^2 qquad g_2(x,y)=|y|^2.
$$
You want to minimize $f$ subject to $g_1=g_2=1$. Then
$$
nabla f(x,y) = (y,x) qquad nabla g_1(x,y) = (2x, 0) qquad nabla g_2(x,y) = (0, 2y).
$$
From Lagrange you get
$$
(y,x) = lambda_1(2x,0) + lambda_2(0,2y) = 2(lambda_1 x, lambda_2 y).
$$
From this you get that $x$ and $y$ bust be linearly dependent, hence $y=pm x$ because of the constraint. $y=-x$ is a global minimizer, $y=x$ is a global maximizer.
answered Nov 23 at 18:28
Federico
4,184512
answered Nov 23 at 18:28
Federico
4,184512
answered Nov 23 at 18:28
Federico
4,184512
answered Nov 23 at 18:28
Federico
4,184512
4,184512
add a comment |
add a comment |
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4
When you have 2 constraints the associated system of equations is $nabla f = lambda nabla g_1 + mu nabla g_2$ rather than $nabla f = lambda nabla g_1$ AND $nabla f = mu g_2$
– user25959
Nov 23 at 18:22
2
See example here: mathonline.wikidot.com/…
– user25959
Nov 23 at 18:23
1
Also another way : Notice that $$- |x| ~ |y| leq langle x ,y rangle$$
– Red shoes
Nov 23 at 18:27
1
@Redshoes Probably it was his intent to prove that inequality with Lagrange multipliers
– Federico
Nov 23 at 18:29
@Federico If that would be orginal question, I would fix $y$ and just minimize $x.y$ over the constraint $|x|^2 = 1$. Then we have a mode with less variable and one constraint (only one Lagrange multiplier ). so much easier to solve.
– Red shoes
Nov 23 at 18:54