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..