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Extremizing multivariable functions

  • #1
798
1

Homework Statement


Find extreme values for
[tex]f(x, y, z)=x+y+z[/tex]
subject to:
[tex]x^4+y^4+z^4=c>0[/tex]

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
[tex]x=y=z=(-1/4\lambda{)}^{1/3}[/tex]
and
[tex]0=((-1/4\lambda{)}^{1/3})^4+((-1/4\lambda{)}^{1/3})^4+((-1/4\lambda{)}^{1/3})^4-c=3((-1/4\lambda{)}^{1/3})^4-c[/tex]
[tex]c=-1/(4c^{3/4})[/tex]
and since c>0, the question has no solutions.
 

Answers and Replies

  • #2
Dick
Science Advisor
Homework Helper
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(-1)^(4/3)=(+1).
 
  • #3
Ray Vickson
Science Advisor
Homework Helper
Dearly Missed
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1,728

Homework Statement


Find extreme values for
[tex]f(x, y, z)=x+y+z[/tex]
subject to:
[tex]x^4+y^4+z^4=c>0[/tex]

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
[tex]x=y=z=(-1/4\lambda{)}^{1/3}[/tex]
and
[tex]0=((-1/4\lambda{)}^{1/3})^4+((-1/4\lambda{)}^{1/3})^4+((-1/4\lambda{)}^{1/3})^4-c=3((-1/4\lambda{)}^{1/3})^4-c[/tex]
[tex]c=-1/(4c^{3/4})[/tex]
and since c>0, the question has no solutions.
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.

RGV
 
  • #4
HallsofIvy
Science Advisor
Homework Helper
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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].
 
  • #5
798
1
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]
 
  • #6
798
1
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]
 
  • #7
798
1
Since
[tex]x=y=z[/tex]
and
[tex]0=1+4x^3\lambda[/tex]
then
[tex]3(-1/(4\lambda{)})^{4/3}=c[/tex]
hence
[tex]\lambda{=}1/4(3/c)^{3/4}[/tex]
We know that
[tex]0=1+4x^3\lambda[/tex]
so
[tex]0=1+4x^3(1/4(3/c)^{3/4})[/tex]
and
[tex]x=-c^{1/4}/3^{1/4}[/tex]
however, this does not really follow the method HallsofIvy suggested..
 

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