1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: 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


    User Avatar
    Science Advisor
    Homework Helper

  4. Nov 14, 2011 #3

    Ray Vickson

    User Avatar
    Science Advisor
    Homework Helper

    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


    User Avatar
    Science Advisor

    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..
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook