Why does the method of Lagrange multiplier fail?











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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







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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















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







share|cite|improve this question


















  • 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













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







share|cite|improve this question













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






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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














  • 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










1 Answer
1






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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.






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    1 Answer
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    up vote
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    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.






    share|cite|improve this answer

























      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.






      share|cite|improve this answer























        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.






        share|cite|improve this answer












        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.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Nov 23 at 18:28









        Federico

        4,184512




        4,184512






























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