A Matter of Interpretation

For starters, picture the problem in 3D. You have an ellipsoid with cardinal points $\displaystyle (\pm 2,0,0)(0,\pm\frac14,0)(0,0,\pm 1)$. Hopefully you can visualize that the local minima and maxima of the distance function fall exactly on these points.

Secondly, the equations you set up are intended as a system of four equations and four unknowns:

$\displaystyle (2x)+\lambda(\frac12x)=0\to x=0$ OR $\displaystyle \lambda=-4$

$\displaystyle (2y)+\lambda(32y)=0\to y=0$ OR $\displaystyle \lambda=-\frac{1}{16}$

$\displaystyle (2z)+\lambda(2z)=0\to z=0$ OR $\displaystyle \lambda=-1$

$\displaystyle \frac{x^2}{4}+{y^2}{16}+z^2=1$

The only logical solutions to this set of relations is $\displaystyle (x,y,z,\lambda)=$

$\displaystyle (\pm 2,0,0,-4)$

$\displaystyle (0,\pm\frac14,0,-\frac{1}{16})$

$\displaystyle (0,0,\pm 1,-1)$

The values of lambda (for your purpose) are "scrap" variables. The real purpose of the system of equations is to solve for point(s) $\displaystyle (x,y,z)$. However, if you look carefully, the value $\displaystyle |\lambda|$ is precisely the square of the distance at its corresponding point $\displaystyle (x,y,z)$. So to answer your question, $\displaystyle \lambda$ represents the output $\displaystyle \lambda=f(x,y,z)$ of the function you are minimizing, for reasons beyond the scope of this question.