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Need help understanding Lagrange multipliers at a more fundamental level.

  1. Oct 8, 2012 #1
    I understand that for Lagrange multipliers,

    [itex] ∇f = λ∇g [/itex]

    And that you can use this to solve for extreme values.

    I have a set of questions because I don't understand these on a basic level.

    1. How do you determine whether it is a max, min, or saddle point, especially when you only get one extreme value/critical point.

    2. Why does this work? Could someone help paint a picture or better description of why you can find these critical points using Lagrange multipliers?

    3. Is there a more significant purpose for Lagrange multipliers?

    You may use any problem where you have either [itex] f(x,y) [/itex] with the constraint [itex] g(x,y) = k [/itex] or with [itex] f(x,y,z) [/itex] with the constraint [itex] g(x,y,z) = k [/itex]

    Both would be preferred; The former preferred for a basic understanding, the latter for a more complex example.

    Any help would be appreciated, I have a quiz and test over it this week.
  2. jcsd
  3. Oct 8, 2012 #2


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    By comparing values of points close by the critical point.

    Suppose you want to maximize some function f(x,y,z) without any constraint. The "gradient" of f, [itex]\nabla f[/itex] is a vector pointing in the direction of fastest increase. So, starting at any given point, calculate [itex]\nabla f[/itex] at that point, and move in its direction. Keep doing that until you can no longer do it. The only reason you could "no longer do it" is because the gradient vector does not have a 'direction'- and that would be when [itex]\nabla f= 0[/itex].

    Now, suppose we have the constraint g(x,y,z)= constant, so we are constrained to stay on some surface satisfying that equation. We can still caculate [itex]\nabla f[/itex], but if it does not happen to be tangent to the surface, we can not "move in its direction". What we can do is find its projectin tangent to the surface and move in that direction. We can continue doing that until there is no projection of [itex]\nabla f[/itex] tangent to the surface. That will happen if and only if [itex]\nabla f[/itex] is itself normal to the surface. But, of course, for any surface g(x,y,z)= constant, [itex]\nabla g[/itex] is perpendicular to the surface so we are saying that [itex]\nabla f[/itex] and [itex]\nabla g[/itex] are both perpendicular to the surface- they are parallel and so one is a multiple of the other: [itex]\nabla f= \lambda\nabla g[/itex] for some constant, [itex]\lambda[/itex].

    Not sure what you mean by this but in the theory of partial differential equations, "Green's functions" can be derived in a manner similar to lagrange multipliers.

  4. Oct 9, 2012 #3
    So, is the max and min found a max and min based on the constraint, and not a regular max/min of [itex] f(x,y) [/itex] or [itex] f(x,y,z) [/itex]?
  5. Oct 9, 2012 #4
    Nevermind. I understand now. I spent 2 seconds on the wikipedia page for it, and I finally had that "Oh my God. I get it." moment.
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