The Question: Find the minimum distance between the origin and the surface $x^2y -z^2 +9 = 0$.
I've been able to find the critical points when $x =0$ and when x is not equal to zero but lamda is equal to negative 1. How do I go about finding the critical points when x is not equal to zero and lamda is not equal to -1?
We are attempting to minimize the distance function. In order to remove square roots, it is usually much easier to minimize the square of the distance function: $$f(\mathbf x) = \rho^2(\mathbf{x},0) = x^2 + y^2 + z^2.$$ Our other constraint is the level curve $$ g(\mathbf x) = x^2y - z^2 + 9 = 0.$$ We create a system of equations with four unknowns by considering the system $$ \begin{align} \lambda \nabla f &= \nabla g \\ g(\mathbf x) &= 0 \end{align}$$ which expands and simplifies to $$ \begin{align} \lambda x &= xy \\ \lambda 2y &= x^2 \\ \lambda z &= -z \\ x^2y - z^2 + 9 &= 0 \end{align}$$
Now we need to begin to break it up into cases.
Suppose $x \neq 0$:
In that case, the first equation gives that $\lambda = y$. Substitution into the second gives that $2y^2 = x^2$, or that $x = \sqrt 2 y = \sqrt 2 \lambda$.
If $z = 0$, then the level curve constraint gives us that $2y^3 = 9$. This lets us solve for $y$, and therefore $x$ (and we are supposing that $z = 0$).
If $z \neq 0$, then the third equation gives us that $\lambda = -1$, which means that $y = -1$ and $x = -\sqrt 2$. Substitution into the level curve constraint lets us solve for $z$, and we have all the cases covered for when $x \neq 0$. $\blacktriangle$
Suppose $x = 0$:
Then the level curve constraing gives us that $9 = z^2$, so $z = \pm 3$. As $z \neq 0$, the third equation gives $\lambda = -1$. The second equation now reads $-2y = 0$, which means that $y = 0$. So the two points that arise here are $(0,0,\pm 3)$. $\blacktriangle$
We've covered all the cases, and all the solutions are now evident. To determine the actual minimum distance, the easiest way is to now plug in the five critical points and see which is the minimum. $\diamondsuit$