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Math Help - Lagrange Multipliers and Distance from point to plane

  1. #1
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    Question Lagrange Multipliers and Distance from point to plane

    I had a question on my final exam that I can't seem to get my mind off...
    Unfortunately I don't remember the equations that were given but the general question was:
    How is the maximum/minimum value (found with Lagrange multipliers) related to the distance from a point (the max/min) to a plane?
    The only reasoning I've managed to do is this:
    the plane is the constraint function, so as the distance between the point and plane goes to zero, you get closer to the max/min value of the objective function.
    To me this seems like too basic of reasoning to be on the final.

    Any other thoughts?
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  2. #2
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    Why would distnce magically decrease? What's a Tangent and what's a Normal?
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  3. #3
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    Quote Originally Posted by TKHunny View Post
    Why would distnce magically decrease?
    If you are trying to find the extreme point of a surface subject to a constraint, the limit of the distance between the extreme point and the constraint surface would approach zero as the constraint approaches the point.

    I`m not sure how finding the normal to the surface would provide anything... just a vector, not a distance, you don`t have any points on the plane.
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  4. #4
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    Hello, kateburns!

    Without a specific problem, it's difficult to discuss the approach and solution.


    Could it be something like this?

    Find the minimum distance from the point (3, 1, -2)
    . . to the plane 2x + 3y - 6z + 7 .= .0.


    Let P(x,y,z) be a point on the plane.


    Its distance from Q(3,1,-2) is given by:
    . . . . . . . . . ____________________________
    . . . d . = . √(x - 3)^2 + (y - 1)^2 + (z + 2)^2

    or: .D . = . (x - 3)^2 + (y - 1)^2 + (z + 2)^2


    The constraint is: .2x + 3y - 6z + 7 .= .0


    The function to minimize is:

    . . f(x,y,L) . = . (x - 3)^2 + (y - 1)^2 + (z + 2)^2 + L(2x + 3y - 6z + 7)

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