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Lagrange multiplier problem

  1. Jun 21, 2005 #1
    This problem was given in my calc class during the semester,
    "Find the lowest point on the intersection of the sphere x^2+y^2 +z^2 = 30 and the cone 2*x^2 +y^2 = c^2". I dont know how to solve this problem with lagrange multipliers.
    How is it done?
    Thanks!

    Callisto
     
  2. jcsd
  3. Jun 21, 2005 #2
    first of all you need to determin of which function you are gonna search the extrema (minima or maxima, each of these two can be searched seperately as you know)...Can you find this function ? If that is done just apply the algoritm of Lagrangian multiplyers

    regards
    marlon
     
  4. Jun 21, 2005 #3

    HallsofIvy

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    I assume the "c2" was a typo and the second equation is actually 2x2+ y2= z2. You want to find the "lowest" point on the curve of intersection which means that you want to minimize z itself with the two constraints:
    x2+ y2+ z2= 30 and 2x2+ y2= z2.

    Then if these as level surfaces of the functions g(x,y,z= x2+ y2+ z2= 30 and h(x,y,z)= 2x2+ y2- z2= 0. The gradient vector of the first is 2xi+ 2yj+ 2zk and the gradient vector of the second is 4xi+ 2yj- 2zk. The gradient of the object function f(x,y,z)= z is just k.
    Using "Lagrange multipliers", we try to make that gradient a linear combination of the other two: k= λ(2xi+ 2yj+ 2zk)+ ρ(4xi+ 2yj- 2zk). That has 5 "unknowns", x, y, z, λ, and ρ (although you only need to find x, y, and z) and we have 5 equations:
    the i component: 0= 2λx+ 4ρx
    the j component: 0= 2λy+ 2ρy
    the k component:1= 2λx- 2ρz
    the equation of the sphere: x2+ y2+ z2d= 30
    the equation of the cone: 2x2+ y2= z2.

    Solve those equations for x, y, and z.
     
    Last edited: Jun 21, 2005
  5. Jun 21, 2005 #4
    thanks for your replies'

    from 0 = 2*lambda*x+ 4*rho*x
    = 2*x(lambda +2*rho) = 0 when x = 0
    if x = 0 then y^2 = z^2 so z = y = +/-sqrt(15)
    so the min value is -sqrt(15),
    that must mean the max value is sqrt(15)

    is this right?
     
  6. Jun 22, 2005 #5
    ok this makes my post look quite stupid :)

    i was under the impression that we were not allowed to give complete answers, or have the PF-rules changed ?

    marlon
     
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