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Math Help - Help With Lagrange Multipliers 2 Questions

  1. #1
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    Angry Help With Lagrange Multipliers 2 Questions

    I need help with these two questions

    1.
    Use Lagrange multipliers to show that, of all the triangles inscribed in a circle of radius
    R, the equilateral triangle has the largest perimeter.



    For this I know that the equation for the unit circle is
    x^2+y^2 = 1
    But i'm not sure what the constraint is.
    Is it that for a triangle to be on a circle, each side must be <= to the diameter?



    2.
    In dimensions 2 and 3, the Lagrange condition can be replaced by a cross product condition. If a is an extremum for f(x) subject to a constraint g(x) = 0, then Gr[f] Gr[g] = 0.

    the x is a cross product sign
    Gr is the gradient
    fx is derivative of f w.r.t. x

    If f and g are functions of 2 variables, Gr[f] Gr[g] =
    det
    [e1 e2 e3]
    [fx fy 0]
    [gx gy 0]
    = (fx gy − fy gx) e3.

    Use the cross product condition to find the extrema of f(x, y) = x y on the unit circle.
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  2. #2
    Grand Panjandrum
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    Re: Help With Lagrange Multipliers 2 Questions

    Quote Originally Posted by Sneaky View Post
    I need help with these two questions

    1.
    Use Lagrange multipliers to show that, of all the triangles inscribed in a circle of radius
    R, the equilateral triangle has the largest perimeter.



    For this I know that the equation for the unit circle is
    x^2+y^2 = 1
    But i'm not sure what the constraint is.
    Is it that for a triangle to be on a circle, each side must be <= to the diameter?
    You only need consider triangles inscribed in the unit circle, so:

    Take the vertices of the triangle be to be (0,0) , (\cos(\theta_1) , \sin(\theta_1)) , and (\cos(\theta_2) , \sin(\theta_2)) with \theta_i >0, \theta_i<2 \pi,\ \ i=1,2 and \theta_1<\theta_2

    There is a neater way of expressing this but the above will do.

    CB
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  3. #3
    MHF Contributor

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    Re: Help With Lagrange Multipliers 2 Questions

    Quote Originally Posted by Sneaky View Post
    2.
    In dimensions 2 and 3, the Lagrange condition can be replaced by a cross product condition. If a is an extremum for f(x) subject to a constraint g(x) = 0, then Gr[f] Gr[g] = 0.

    the x is a cross product sign
    Gr is the gradient
    fx is derivative of f w.r.t. x

    If f and g are functions of 2 variables, Gr[f] Gr[g] =
    det
    [e1 e2 e3]
    [fx fy 0]
    [gx gy 0]
    = (fx gy − fy gx) e3.

    Use the cross product condition to find the extrema of f(x, y) = x y on the unit circle.
    Okay, ON the unit circle g(x,y)= x^2+ y^2- 1= 0 and you are given that f(x,y)= xy.
    What does f_xg_y- f_yg_x= 0 give you?
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  4. #4
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    Re: Help With Lagrange Multipliers 2 Questions

    Quote Originally Posted by HallsofIvy View Post
    Okay, ON the unit circle g(x,y)= x^2+ y^2- 1= 0 and you are given that f(x,y)= xy.
    What does f_xg_y- f_yg_x= 0 give you?
    I got

    (2y^2 - 2x^2)e_3 = 0

    so
    f_xg_y- f_yg_x= 0
    is
    2y^2 - 2x^2= 0

    does that mean the lagrange equation is

    h(x,y,\lambda) = xy - \lambda(2y^2 - 2x^2)
    ?
    Last edited by Sneaky; November 17th 2011 at 08:51 AM.
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  5. #5
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    Re: Help With Lagrange Multipliers 2 Questions

    Quote Originally Posted by CaptainBlack View Post
    You only need consider triangles inscribed in the unit circle, so:

    Take the vertices of the triangle be to be (0,0) , (\cos(\theta_1) , \sin(\theta_1)) , and (\cos(\theta_2) , \sin(\theta_2)) with \theta_i >0, \theta_i<2 \pi,\ \ i=1,2 and \theta_1<\theta_2

    There is a neater way of expressing this but the above will do.

    CB

    So this means that I am maximizing the distance between the vertexes so that the distance is <= diameter of the circle?

    Circle with radius R

    Triangle with sides A, B, C

    Point_1: (0,0)
    Point_2: (cos(\theta_1),sin(\theta_1))
    Point_3: (cos(\theta_2),sin(\theta_2))

    A = \sqrt{     (0-cos(\theta_1))^2 + (0-sin(\theta_1))^2      }
    B = \sqrt{(cos(\theta_1)-(cos(\theta_2))^2 + (sin(\theta_1)-sin(\theta_2))^2}
    C = \sqrt{(cos(\theta_2)-0)^2 + (sin(\theta_2)-0)^2}

    Constraints are
    A <= 2R
    B <= 2R
    C <= 2R


    So then does the Lagrange equation become this?

    h(\theta_1,\theta_2,\lambda_1,\lambda_2,\lambda_3) = A+B+C - \lambda_1(A-2R) - \lambda_2(B-2R) - \lambda_3(C-2R)
    Last edited by Sneaky; November 17th 2011 at 09:37 AM.
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