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LaGrange Multiplier

  1. Mar 15, 2005 #1
    Find the points at which the function f(x,y,z)=x^8+y^8+z^8 achieves its minimum on the surface x^4+y^4+z^4=4.

    I know

    Case1: x not equal to 0, y not equal to 0, and z not equal to 0
    I get 3(4th root of 4/3 to the eigth)???

    I'm I doing this right?
  2. jcsd
  3. Mar 15, 2005 #2


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    Are you sure this doesn't belong in the homework section?

    Since grad f always points in the direction of fastest increase of f, to get to a minimum, you should go in the exact opposite direction. Of course, if you are required to stay on the surface x4+ y4+z4= 4, you can't do that. What you can do is go opposite to the component tangent to the surface. That works until you get to a point where grad f is perpendicular to the surface and has no component tangent- that's when you are at the minimum point on that surface.
    If you let g(x,y,z)= x4+ y4+ z4 then the surface is a level surface of g- and grad g is perpendicular to the surface: we must have grad f parallel to grad g which means grad f= &lamda; grad g for some number &lamda;. That's the idea of the LaGrange multiplier.

    Yes, your formulas are correct!

    If x is NOT 0, then, dividing the first equation by 8x3, we have
    x4= (1/2)λ. Similarly for y not 0 and z not 0: which means that, whatever &lamda; is, x= y= z so the x4+ y4+ z4=
    3x4= 4 so x= y= z= (4/3)1/4. The value you give is the value at that point- you were asked for the points themselves.

    However, it is not necessarily true that "either x,y,z are all 0 or they are all non-zero"! If x= 0, y= 0 but z is NOT 0, then z= 41/4 so (0, 0, 41/4) is a possible place for a minimum. If x= 0 but y and z are not 0, then as before we can show that y= z so 2y4= 4 and (0, 21/4,21/4) is possible minimum. The same is true of "y and z 0 but not x", "y 0 but not x or z", etc.
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