How do I find the highest and lowest points made by the union of these two functions using Lagrange Multipliers? 
Find the highest and lowest points made by the union of these two functions using Lagrange Multipliers.

$x^2+y^2+z^2 = 16$
$(x+1)^2+(y+1)^2+(z+1)^2 = 27$
I got the basics down, I used the first function as my constraint.
$2(x+1) = \lambda\times2x$
$2(y+1) = \lambda\times 2y$
$2(z+1) = \lambda\times2z$
$x^2+y^2+z^2 = 16$
and after solving I think it's all $x=y=z=\pm\dfrac{4}{\sqrt{3}}$
I just don't know how to interpret these results.
What does this mean? and how do I use this to answer 
my problem?
Thank you for your time.
 A: I think you are missing something. As @copper.hat noted you don't have nothing to optimize. I think that the two equations you uploaded are constrains, but the function you want to maximize/minimize isn't posted.
If that's not the case, add those two and you'll get that both minima and the maxima of the function is $27 + 16 = 43$.
If you want to find a solution for the system of equations, just subtract the first equation from the second and we have:
$$x + y + z = 4$$
$$x^2 + y^2 + z^2 = 16$$
Then there are infinite amount of solution. Set $x^3 + y^3 + z^3$ to any value you like and then using Newton's Identities you'll be able to obtain a cubic polynomial, where $x,y,z$ will be roots of that polynomial.
And from the thing you've posted I think you want to maximize/minimize the function
$$f(x,y,z) = (x+1)^2 + (y+1)^2 + (z+1)^2$$
under constrain:
$$g(x,y,z) = x^2 + y^2 + z^2 - 16 = 0$$
To make things even simplier expand the first function and you'll have:
$$f(x,y,z) = (x^2 + y^2 + z^2) + 2(x+y+z) + 3 = 2(x+y+z) + 19$$
Now apply Lagrange multipliers:
$$F(x,y,z,\lambda) = 2(x+y+z) + 19 - \lambda(x^2 + y^2 + z^2 - 16)$$
Now take partial derivatives and set them to 0.
$$F_x = 2 - 2x\lambda = 0$$
$$F_y = 2 - 2y\lambda = 0$$
$$F_z = 2 - 2z\lambda = 0$$
$$F_{\lambda} = x^2 + y^2 + z^2 - 16 = 0$$
Substituting $x = y = z \frac{1}{\lambda}$ from the first partial derivatives into the constarin we have:
$$\frac{3}{\lambda^2} = 16$$
$$\lambda = \pm \frac{\sqrt{3}}{4}$$
Substitute back and get that $ x = y = z = \pm \frac{4}{\sqrt{3}}$
Plug the values for $x,y,z$ into the initial function and you'll get that the maxima occurs at $f(\frac{\sqrt{3}}{4},\frac{\sqrt{3}}{4},\frac{\sqrt{3}}{4}) = 19 + 8\sqrt{3}$ and the minima at $f(\frac{- \sqrt{3}}{4}, - \frac{\sqrt{3}}{4}, - \frac{\sqrt{3}}{4}) = 19 - 8\sqrt{3}$
