Maximum distance from the origin to the surface I am having trouble getting the maximum distance from the origin to the surface 
$$
    \frac{x^4}{16} +\frac{y^4}{81} + z^4 = 1
$$
Knowing I have to maximize $x^2 +y^2+ z^2$ and that the constrain is the above surface.
My Lagrangian is: 
$$
    L(x,y,z,\lambda) = x^2+y^2+z^2 +\lambda(1- \frac{x^4}{16} - \frac{y^4}{81} - z^4).
$$
And doing the partial derivatives i get:
$$
  \frac{\partial L}{\partial x} = 2x-\frac{x^3\lambda}{4}; \\
  \frac{\partial L}{\partial y} = 2y-\frac{4y^3\lambda}{81};\\
   \frac{\partial L}{\partial z} = 2z-4z^3\lambda;
$$
After getting that, I don't know how to proceed in order to get the maximum distance.
Thank you so much.

From there you have a system of four equations, so you can use substitution to eliminate variables and solve it.
$2x-\frac{x^3\lambda}{4}=0 \\ 2y-\frac{y^3\lambda}{81}=0 \\ 2z-4z^3\lambda=0 \\ \frac{x^4}{16}+\frac{y^4}{81}+z^4=1$
Using the third one:
$2z-4z^3\lambda = 0 \\ 1-2z^2\lambda=0 \\ \lambda=\frac{1}{2z^2}$
Plugging that into the first equation:
$2x-\frac{x^3}{8z^2}=0 \\ 16z^2=x^2 \\ 4|z|=|x|$
Using the value for $\lambda$ in the second equation:
$2y-\frac{y^3}{162z^2}=0 \\ 324z^2 = y^2 \\ 18|z|=|y| $
Then putting all the values in terms of z in the last equation:
$ \frac{(4z)^4}{16}+\frac{(18z)^4}{81}+z^4=1 \\ 16z^4+1296z^4+z^4=1 $
$ 1313z^4 = 1$
$z=\pm\frac{1}{{}^4\sqrt{1313}}, y = \pm\frac{18}{{}^4\sqrt{1313}}, x=\pm\frac{4}{{}^4\sqrt{1313}} $
Please double check my math, it looks like it's correct, but I'm not positive.
