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Lagrange Multipliers Questions Watch

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    I've done this so far:
    F(x,y) = root(xy) + lambda( x + y - 1) = 0
    Fx = 1/2(xy)^(-1/2) . y + lambda = 0
    Fy = 1/2(xy)^(-1/2) . x + lambda = 0
    Flambda = x + y - 1 = 0

    x + y - 1 = 0
    x + y = 1
    y = 1 - x

    I don't know how to continue from here :/
    Any help would be appreciated thank you
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    (Original post by Fatts13)
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    I've done this so far:
    F(x,y) = root(xy) + lambda( x + y - 1) = 0
    Fx = 1/2(xy)^(-1/2) . y + lambda = 0
    Fy = 1/2(xy)^(-1/2) . x + lambda = 0
    Flambda = x + y - 1 = 0

    x + y - 1 = 0
    x + y = 1
    y = 1 - x

    I don't know how to continue from here :/
    Any help would be appreciated thank you
    The bit in red shouldn't be present.

    I'd look to the first two equations; one minus the other gives you another relatinship between x and y.
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    (Original post by ghostwalker)
    The bit in red shouldn't be present.

    I'd look to the first two equations; one minus the other gives you another relatinship between x and y.
    This is what I got for this one:
    F(x,y, lambda) = root(xy) - lambda(x + y -1)
    Fx = 1/2(root(y/x)) - lambda
    Fy = 1/2(root(x/y) - lambda
    Flambda = 1 - x - y

    Fx = Fy
    1/2(root(y/x)) - lambda = 1/2(root(x/y) - lambda
    y/x = x/y
    y^2 = x^2
    y = x

    when x = y and Flambda = 0
    1 - x - y = 0
    1 = x + y
    Therefore x = y = 1/2
    Therefore the critical point is (1/2,1/2)

    Though my problem is I'm not exactly sure about how I got Fx and Fy :s
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    (Original post by Fatts13)
    This is what I got for this one:
    F(x,y, lambda) = root(xy) - lambda(x + y -1)
    Fx = 1/2(root(y/x)) - lambda
    Fy = 1/2(root(x/y) - lambda
    Flambda = 1 - x - y

    Fx = Fy
    1/2(root(y/x)) - lambda = 1/2(root(x/y) - lambda
    y/x = x/y
    y^2 = x^2
    y = x

    when x = y and Flambda = 0
    1 - x - y = 0
    1 = x + y
    Therefore x = y = 1/2
    Therefore the critical point is (1/2,1/2)

    Though my problem is I'm not exactly sure about how I got Fx and Fy :s
    Are you saying you differentiated, but don't know how you differentiated?
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    (Original post by ghostwalker)
    Are you saying you differentiated, but don't know how you differentiated?
    As in I differentiated but I'm not sure whether it is right as I did root(1/x) . y and made it into root(y/x)
    Is this alright to do? If so then I'm sorry ignore me
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    (Original post by Fatts13)
    As in I differentiated but I'm not sure whether it is right as I did root(1/x) . y and made it into root(y/x)
    Is this alright to do? If so then I'm sorry ignore me
    Actually, from your first post, you didn't do that.

    In the case of Fx for example, you did (correctly)

    F_x(x,y) = \frac{1}{2}\times y \times (xy)^{-1/2}+\lambda

    Which equals \frac{1}{2}\sqrt{\frac{y}{x}}+ \lambda

    Note:

    In post #3 you went from
    y^2=x^2
    to y=x
    It should have been y=x OR y=-x. But the latter has no solution when you substitute into Flambda, so you're not missing a solution this time.
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    (Original post by ghostwalker)
    Actually, from your first post, you didn't do that.

    In the case of Fx for example, you did (correctly)

    F_x(x,y) = \frac{1}{2}\times y \times (xy)^{-1/2}+\lambda

    Which equals \frac{1}{2}\sqrt{\frac{y}{x}}+ \lambda

    Note:

    In post #3 you went from
    y^2=x^2
    to y=x
    It should have been y=x OR y=-x. But the latter has no solution when you substitute into Flambda, so you're not missing a solution this time.
    Ohhh right ok, oh yeah then you rationalise to get root(y/x), understood thank you
    Ok I'll add y = -x and then state no solutions, thank you.
    I would give you a thumbs up but it's not letting me

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    for this questions, whilst its given us the constraint what would be the function?

    Sorry for being a hassle
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    (Original post by Fatts13)

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    for this questions, whilst its given us the constraint what would be the function?
    What's the distance of a point (x,y) from (0,0) ?
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    (Original post by ghostwalker)
    What's the distance of a point (x,y) from (0,0) ?
    Oh, so the distance would be x^2 + y^2 therefore that's the function?
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    (Original post by Fatts13)
    Oh, so the distance would be x^2 + y^2 therefore that's the function?
    Distance is actually the square root of that.

    BUT, when the square root is a minimum, so is x^2+y^2, so yes, go with that - it's probably easier to work with.
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    (Original post by ghostwalker)
    Distance is actually the square root of that.

    BUT, when the square root is a minimum, so is x^2+y^2, so yes, go with that - it's probably easier to work with.
    Ok so then this would be the equation:
    F(x,y,lambda) = x^2 + y^2 - lambda( y^2 - 2root3xy - x^2 - 2)
    Fx = 2x + 2root3lambday + 2lambdax
    Fy = 2y + 2root3lambdax - 2lambday
    Flambda = 2 - y^2 + 2root3xy + x^2

    But I seem to be struggling to get for example what y could be by equating Fx = Fy

    I've got to y = -x ((1 + lambda - root3lambda)/(root3lambda - 1 + lambda))
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    (Original post by Fatts13)
    Ok so then this would be the equation:
    F(x,y,lambda) = x^2 + y^2 - lambda( y^2 - 2root3xy - x^2 - 2)
    Fx = 2x + 2root3lambday + 2lambdax
    Fy = 2y + 2root3lambdax - 2lambday
    Flambda = 2 - y^2 + 2root3xy + x^2

    But I seem to be struggling to get for example what y could be by equating Fx = Fy

    I've got to y = -x ((1 + lambda - root3lambda)/(root3lambda - 1 + lambda))
    Yes, it's not as straight forward. I've not worked it through fully.

    You want ot get rid of lambda.

    Although you've been equating Fx and Fy, I don't think that's the way to go.

    You have Fx=0, so rearrange that to lambda = something.
    Similarly Fy=0.
    Now eliminate lambda between the two to get a relationship in x and y.....(1)
    You can end up with something close to your constraint.
    You should, with the constraint and (1) be able to eliminate y^2-x^2, and get xy=constant.

    Not gone further.

    Time for bed.
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    (Original post by ghostwalker)
    Yes, it's not as straight forward. I've not worked it through fully.

    You want ot get rid of lambda.

    Although you've been equating Fx and Fy, I don't think that's the way to go.

    You have Fx=0, so rearrange that to lambda = something.
    Similarly Fy=0.
    Now eliminate lambda between the two to get a relationship in x and y.....(1)
    You can end up with something close to your constraint.
    You should, with the constraint and (1) be able to eliminate y^2-x^2, and get xy=constant.

    Not gone further.

    Time for bed.
    No worries, I think I got the answer - took some time but I got there eventually after ELEMENTAL mistakes

    F(x,y,lambda) = x^2 + y^2 - lambda( y^2 - 2root3xy - x^2 - 2)
    Fx = 2x + 2root3lambday + 2lambdax
    Fy = 2y + 2root3lambdax - 2lambday
    Flambda = 2 - y^2 + 2root3xy + x^2

    Fx = 0
    2x + 2root3lambdax + 2lambdax = 0
    y = - (1 + lambda) x / root3 lambda

    Fy = 0
    2y + 2root3lambdax - 2lambday = 0
    y = root3lambda x / lambda - 1

    y =y Therefore
    - (1 + lambda) x / root3 lambda = root3lambda x / lambda - 1
    4lambda^2 - 1 = 0
    (2lamda + 1) ( 2lambda - 1) = 0
    lambda = -1/2 or 1/2

    When lambda = 1/2 we find y = -root3 x
    Substitue y = -root3 x into Flambda = 0 giving us
    x = -1/2 or 1/2

    When x = -1/2 lambda = 1/2 y =(root3)/2
    When x = 1/2 lambda = 1/2 y = - (root3)/2

    Therefore for lambda = 1/2 there are two critical points:
    (1/2, - (root3)/2) and (-1/2, (root3)/2)

    When lambda = -1/2 we find y = root3x
    Substitue y = root3 x into Flambda = 0
    2x^2 + 1 = 0 Therefore no solutions

    Therefore lambda = -1/2 is not relevant, thus leaving only two critical points:
    (1/2, - (root3)/2) and (-1/2, (root3)/2)
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    (Original post by Fatts13)
    No worries, I think I got the answer - took some time but I got there eventually after ELEMENTAL mistakes

    F(x,y,lambda) = x^2 + y^2 - lambda( y^2 - 2root3xy - x^2 - 2)
    Fx = 2x + 2root3lambday + 2lambdax
    Fy = 2y + 2root3lambdax - 2lambday
    Flambda = 2 - y^2 + 2root3xy + x^2

    Fx = 0
    2x + 2root3lambdax + 2lambdax = 0
    y = - (1 + lambda) x / root3 lambda

    Fy = 0
    2y + 2root3lambdax - 2lambday = 0
    y = root3lambda x / lambda - 1

    y =y Therefore
    - (1 + lambda) x / root3 lambda = root3lambda x / lambda - 1
    4lambda^2 - 1 = 0
    (2lamda + 1) ( 2lambda - 1) = 0
    lambda = -1/2 or 1/2

    When lambda = 1/2 we find y = -root3 x
    Substitue y = -root3 x into Flambda = 0 giving us
    x = -1/2 or 1/2

    When x = -1/2 lambda = 1/2 y =(root3)/2
    When x = 1/2 lambda = 1/2 y = - (root3)/2

    Therefore for lambda = 1/2 there are two critical points:
    (1/2, - (root3)/2) and (-1/2, (root3)/2)

    When lambda = -1/2 we find y = root3x
    Substitue y = root3 x into Flambda = 0
    2x^2 + 1 = 0 Therefore no solutions

    Therefore lambda = -1/2 is not relevant, thus leaving only two critical points:
    (1/2, - (root3)/2) and (-1/2, (root3)/2)
    Well done.

    I'm not going to check all the details.

    I notice one thing - in red -, you just cancelled the "x". You can only cancel a value out of an equation in that fashion, IF you know x=0 is not a solution. At this stage, you don't know that.

    And as a final point you really ought to check the distance for these points, (i.e. what does the function evaluate to?) to check they're the same distance, and are thus the closest points. One might have been closer than the other.
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    (Original post by ghostwalker)
    Well done.

    I'm not going to check all the details.

    I notice one thing - in red -, you just cancelled the "x". You can only cancel a value out of an equation in that fashion, IF you know x=0 is not a solution. At this stage, you don't know that.

    And as a final point you really ought to check the distance for these points, (i.e. what does the function evaluate to?) to check they're the same distance, and are thus the closest points. One might have been closer than the other.
    Ok I will take that into account. Thank You
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    Whilst this isn't a Lagrange Multiplier question, I don't know how to do this :/
    So what I've done so far:
    df/dx = (y^2 - 1) / (1 + xy) ^2 when this = 0 then y = +/- 1
    df/dy = (x^2 - 1) / (1 + xy)^2 when this = 0 then x = +/- 1
    Now I'm struggling to get the second partial derivatives, I got these but I'm not 100% sure at all:
    d^2f/dx^2 = (-2y(y^2-1)) / (yx + 1)^3
    d^2f/dy^2 =( 2(x+y) ) / (yx + 1)^3

    Sorry for the hassle
 
 
 
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