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LaGrange Multipliers

  1. Mar 21, 2005 #1
    It has been a while and trying to brush up on LaGrange points. I want to find the highest and lowest points on the ellipse of the intersection of the cone: x^2+y^2-z^2 ;subject to the single constraint: x+2y+3z=3 (plane).

    I want to find the points and I am not concerned with the minimum and maximum yet. So far, I have done the following:

    delf<x,y,z>= lambda(del)g<x,y,z>
    <2x,2y.-2z>= lambda<1,2,3> where g(x,y,z)=3

    Hence, 2x=1*lambda or (2x-1*lambda)=0
    2y=2*lambda or (2y-2*lambda)=0
    -2z=3*lambda or (-2z-3*lambda)=0

    Furthermore, lambda=2x x=lambda/2
    lambda=y or y=lambda
    lambda=(-2/3)*z z=(-3/2)*lambda

    Solving the systems I guess would be my area of problems. or putting things together: Using the constraint as a guideline, and solving for lambda, I obtain that lambda=7/3.
    And, if lambda=7/3 x=7/6

    I would say that I am at the final steps of obtaining my points for the function; however, I am confusing myself with what points are permissible. ie.: (0,0,-7/2)

    See what I am getting at? I would appreciate any guidance to putting this puzzle together
  2. jcsd
  3. Mar 21, 2005 #2


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    Science Advisor

    1) Lagrange "points" and Lagrange "multipliers" are very different things!

    2) I'm not sure what you mean by " I want to find the highest and lowest points on the ellipse of the intersection of the cone". If you mean find the highest and lowest points on the ellipse formed by the intersection of the cone x2+ y2= z2 and the intersection of the plane x+ 2y+ 3z= 3, then you are really trying to maximize and minimize z subject to the two constraints.

    For what you give, find max and min of x2+ y2- z2 subject to the constraint x+ 2y+ 3z= 3, what you've done at the beginning is exactly right: If f(x,y,z)= x2+ y2- z2, then del f= 2x i+ 2yj- 2zk. We can write the constraint as g(x,y,z)= x+ 2y+ 3z= 3 so del g= i+ 2j+ 3k is perpendicular to the plane. We must have del f= 2xi+ 2yj- 2zk= &lambda;(i+ 2j+ 3k) or 2x= &lamda;, 2y= 2&lamda;, -2z= 3&lambda;.
    I don't see how you got &lambda;= 7/3! You correctly observe that x= &lamda;/2, y= &lambda;, z= -(3/2)&lambda; so x+ 2y+ 3z= (1/2)&lambda;+ 2&lambda;- (9/2)&lambda;= (1/2+ 4/2- 9/2)&lamda;= -2&lamda;= 3: &lambda= -3/2. From that,
    x= &lamda;/2= -3/4, y= &lambda;= -3/2, and z= (-3/2)&lambda;= 9/4.

    Myself, I would solve in a slightly different way: since we don't really need to find &lambda; itself, it is simplest to eliminate &lambda; by dividing equations: The second equation divided by the first gives 2y/2x= 2&lamda;/&lambda; or y/x= 2 so y= 2x. The third equation divided by the first gives -2z/2x= 3&lamda;/&lamda; or -z/x= 3 so z= -3x. Now put those into the constraint: x+ 2y+ 3z= x+ 2(2x)+ 3(-3x)= -4x= 3. x= -3/4, y= 2x= -3/2, z= -3x= 9/4, just as before.

    Those values "go together", you don't "mix and match" with 0! The one extremal point is (-3/4, -3/2, 9/4).

    If the problem really is "find extermal values of z subject to both x2+ y2- z2= 0 and x+ 2y+ 3z= 3", do it as a "double multiplier problem: del z= k must be a linear combination of del (x2+ y2- z2) and del (x+ 2y+ 3z). That is, k= &lambda;1(2xi+ 2yj- 2zk)+ &lambda;2(i+ 2j+ 3k) or 2&lambda;1x+ 2&lambda;2= 0, 2&lambda;1y+ 2&lambda;2= 0, -2&lambda;1z+ 3&lambda;2= 1. Together with the two constraints that's 5 equations for 5 unknowns.
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