Extremizing multivariable functions

[sandy.bridge]

Homework Statement

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

Therefore, since c>0, exclude the origin.

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

 

[Dick]

(-1)^(4/3)=(+1).

 

[Ray Vickson]

Homework Statement

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

Therefore, since c>0, exclude the origin.

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

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

RGV

 

[HallsofIvy]

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

Now, eliminate \lambda by dividing one equation by another. That gives 1= (x/y)^3 and 1= (x/z)^3 which give x= y= z. Put that into x^4+ y^4+ z^4= c.

 

[sandy.bridge]

Thanks for the responses guys.
(x, y, z)=(\frac{c^{1/4}}{3^{1/4}}, \frac{c^{1/4}}{3^{1/4}}, \frac{c^{1/4}}{3^{1/4}})

 

[sandy.bridge]

I'd just like to say that when I do it following the original path, I get
(x, y, z)=(-c^{1/4}/3^{1/4}, -c^{1/4}/3^{1/4}, -c^{1/4}/3^{1/4})

 

[sandy.bridge]

Since
x=y=z
and
0=1+4x^3\lambda
then
3(-1/(4\lambda{)})^{4/3}=c
hence
\lambda{=}1/4(3/c)^{3/4}
We know that
0=1+4x^3\lambda
so
0=1+4x^3(1/4(3/c)^{3/4})
and
x=-c^{1/4}/3^{1/4}
however, this does not really follow the method HallsofIvy suggested..