Lagrange Multipliers Questions

Announcements
    • Thread Starter
    Offline

    2
    ReputationRep:
    Name:  Screen Shot 2017-03-19 at 18.35.38.png
Views: 8
Size:  18.7 KB

    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
    Offline

    3
    (Original post by Fatts13)
    Name:  Screen Shot 2017-03-19 at 18.35.38.png
Views: 8
Size:  18.7 KB

    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.
    • Thread Starter
    Offline

    2
    ReputationRep:
    (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
    Offline

    3
    (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?
    • Thread Starter
    Offline

    2
    ReputationRep:
    (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
    Offline

    3
    (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.
    • Thread Starter
    Offline

    2
    ReputationRep:
    (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

    Also Name:  Screen Shot 2017-03-20 at 22.06.50.png
Views: 3
Size:  14.5 KB
    for this questions, whilst its given us the constraint what would be the function?

    Sorry for being a hassle
    Offline

    3
    (Original post by Fatts13)

    Also Name:  Screen Shot 2017-03-20 at 22.06.50.png
Views: 3
Size:  14.5 KB
    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) ?
    • Thread Starter
    Offline

    2
    ReputationRep:
    (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?
    Offline

    3
    (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.
    • Thread Starter
    Offline

    2
    ReputationRep:
    (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))
    Offline

    3
    (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.
    • Thread Starter
    Offline

    2
    ReputationRep:
    (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)
    Offline

    3
    (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.
    • Thread Starter
    Offline

    2
    ReputationRep:
    (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
    Name:  Screen Shot 2017-03-21 at 17.29.36.png
Views: 2
Size:  16.8 KB
    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
 
 
 
Write a reply… Reply
Submit reply

Register

Thanks for posting! You just need to create an account in order to submit the post
  1. this can't be left blank
    that username has been taken, please choose another Forgotten your password?
  2. this can't be left blank
    this email is already registered. Forgotten your password?
  3. this can't be left blank

    6 characters or longer with both numbers and letters is safer

  4. this can't be left empty
    your full birthday is required
  1. Oops, you need to agree to our Ts&Cs to register
  2. Slide to join now Processing…

Updated: March 21, 2017
TSR Support Team
Poll
Spring onions - Yay or nay?
Useful resources

Make your revision easier

Maths

Maths Forum posting guidelines

Not sure where to post? Read the updated guidelines here

Equations

How to use LaTex

Writing equations the easy way

Student revising

Study habits of A* students

Top tips from students who have already aced their exams

Study Planner

Create your own Study Planner

Never miss a deadline again

Polling station sign

Thinking about a maths degree?

Chat with other maths applicants

Can you help? Study help unanswered threads

Groups associated with this forum:

View associated groups
Study resources

The Student Room, Get Revising and Marked by Teachers are trading names of The Student Room Group Ltd.

Register Number: 04666380 (England and Wales), VAT No. 806 8067 22 Registered Office: International House, Queens Road, Brighton, BN1 3XE

Quick reply
Reputation gems: You get these gems as you gain rep from other members for making good contributions and giving helpful advice.