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Extremizing multivariable functions

  1. Nov 14, 2011 #1
    1. The problem statement, all variables and given/known data
    Find extreme values for
    [tex]f(x, y, z)=x+y+z[/tex]
    subject to:

    Therefore, since c>0, exclude the origin.

    Let [tex]L(x, y, z, \lambda{)}=x+y+z+\lambda{(}x^4+y^4+z^4-c)[/tex]
    and thus
    [tex]L_1(x, y, z, \lambda{)}=1+4x^3\lambda[/tex]
    [tex]L_2(x, y, z, \lambda{)}=1+4y^3\lambda[/tex]
    [tex]L_3(x, y, z, \lambda{)}=1+4z^3\lambda[/tex]
    [tex]L_4(x, y, z, \lambda{)}=x^4+y^4+z^4-c[/tex]
    with solutions
    and since c>0, the question has no solutions.
  2. jcsd
  3. Nov 14, 2011 #2


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  4. Nov 14, 2011 #3

    Ray Vickson

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    The equation [itex] 3((-1/4\lambda{)}^{1/3})^4 = c[/itex] can certainly be solved for [itex] \lambda [/itex] as a function of [itex]c. [/itex] Then plug that solution into your formulas for x, y and z.

  5. Nov 15, 2011 #4


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    I prefer a little different way of thinking about it. To find maximum or minimum of f(x,y,z) subject to the constraint [itex]g(x,y,z)= constant[/itex], find (x,y,z) where the gradients are parallel: [itex]\nabla f= \lambda\nabla g[/itex]. For this problem that gives
    [itex]1= 4\lambda x^3[/itex], [itex]1= 4\lambda y^3[/itex], [itex]1= 4\lambda z^3[/itex]
    (These are equivalent to your [itex]L_1= 1+ 4\lambda x^2= 0[/itex], etc. although you did not write the "= 0".)

    Now, eliminate [itex]\lambda[/itex] by dividing one equation by another. That gives [itex]1= (x/y)^3[/itex] and [itex]1= (x/z)^3[/itex] which give x= y= z. Put that into [itex]x^4+ y^4+ z^4= c[/itex].
  6. Nov 15, 2011 #5
    Thanks for the responses guys.
    [tex](x, y, z)=(\frac{c^{1/4}}{3^{1/4}}, \frac{c^{1/4}}{3^{1/4}}, \frac{c^{1/4}}{3^{1/4}})[/tex]
  7. Nov 15, 2011 #6
    I'd just like to say that when I do it following the original path, I get
    [tex](x, y, z)=(-c^{1/4}/3^{1/4}, -c^{1/4}/3^{1/4}, -c^{1/4}/3^{1/4})[/tex]
  8. Nov 16, 2011 #7
    We know that
    however, this does not really follow the method HallsofIvy suggested..
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