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Math Help - origin to surface

  1. #1
    Member billym's Avatar
    Joined
    Feb 2008
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    183

    Thumbs down origin to surface

    I'm not actually taking calc 3, I'm just taking a class where we're supposed to romp through various calc 3 topics. So, I have to use the exact methods I'm given, not alternative methods. Can anybody tell me if I'm going about this obviously simple question in at all the right way?

    On my exam I am just going to be asked how to minimize/maximize the distance from some point (most likely the origin) to the surface of some shape - that's it. I assume you just use the distance equation as f(x,y,z) and the equation of the shape as the constraint. And then just use the Lagrange multiplier method:

    Use the Lagrange multiplier method to determine the minimum distance from the origin to any point on the ellipsoid defined by:

    \frac{x^2}{4}+{y^2}{16}+z^2=1

    My work:

    Let:

    D^2=f(x,y,z)=x^2+y^2+z^2

    g(x,y,z)=\frac{x^2}{4}+{y^2}{16}+z^2-1=0

    So:

    f_x+\lambda g_x=0 \Longrightarrow \lambda=-4

    f_y+\lambda g_y=0 \Longrightarrow \lambda=-16

    f_z+\lambda g_z=0 \Longrightarrow \lambda=-1

    Assuming these are right, what am I supposed to do with three lambdas?

    How else would I do this question?
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  2. #2
    Senior Member
    Joined
    Apr 2009
    From
    Atlanta, GA
    Posts
    408

    A Matter of Interpretation

    For starters, picture the problem in 3D. You have an ellipsoid with cardinal points (\pm 2,0,0)(0,\pm\frac14,0)(0,0,\pm 1). Hopefully you can visualize that the local minima and maxima of the distance function fall exactly on these points.

    Secondly, the equations you set up are intended as a system of four equations and four unknowns:
    (2x)+\lambda(\frac12x)=0\to x=0 OR \lambda=-4
    (2y)+\lambda(32y)=0\to y=0 OR \lambda=-\frac{1}{16}
    (2z)+\lambda(2z)=0\to z=0 OR \lambda=-1
    \frac{x^2}{4}+{y^2}{16}+z^2=1

    The only logical solutions to this set of relations is (x,y,z,\lambda)=
    (\pm 2,0,0,-4)
    (0,\pm\frac14,0,-\frac{1}{16})
    (0,0,\pm 1,-1)

    The values of lambda (for your purpose) are "scrap" variables. The real purpose of the system of equations is to solve for point(s) (x,y,z). However, if you look carefully, the value |\lambda| is precisely the square of the distance at its corresponding point (x,y,z). So to answer your question, \lambda represents the output \lambda=f(x,y,z) of the function you are minimizing, for reasons beyond the scope of this question.
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