Lagrange multipliers - missing critical points?

Question

Given $S = \left \{ (x,y,z) : z^2-xy=1\right \}$

a. Prove that there is at least one point which is closest to $(0, 0, 0)$ and find the distance.

b. In section a you should have found several "critical points". show that those which are not a global minimum are also not a local extremum.

My problem is that I only get global minimum points in section a:

Defining $f = || \ ||^2=x^2+y^2+z^2$ and using Lagrange multipliers theorem we know that for any local extremum $p$ s.t $\nabla F(p) \neq 0$, we have $\nabla F(p) = \lambda \nabla f(p)$.

So, $\lambda (2x_0, 2y_0, 2z_0) = (-y_0, -x_0, 2z_0)$, thus $\lambda = 1$, and $x_0=y_0=0$

So we get 2 solutions $(0, 0, 1), (0, 0, -1)$.

Which are both a global minimum.

Can anyone explain what am I missing?

Answer 1

$\lambda (2x, 2y, 2z) = (-y, -x, 2z)$

$2\lambda x = -y$

$2\lambda y = -x$

$\lambda z = z$

$z^2 - xy = 1$

So we also have,

$z = 0$ as a possible solution from third.

From first, $\lambda = - \frac{y}{2x}$

Plugging into second, $y^2 = x^2 \implies y = \pm x$

So given the constraint, $x = -1, y = 1, z = 0$ and $x = 1, y = -1, z = 0$ are other critical points to test.

Answer 2

For part (a), paxtibimarce and Math Lover have identified the pairs of critical points of the "distance-squared" function $ \ x^2 + y^2 + z^2 \ $ on the surface $ \ z^2 \ = \ 1 + xy \ $ as $ \ (0 \ , \ 0 \ , \ \pm 1) \ $ and $ \ (\pm 1 \ , \ \mp 1 \ , \ 0 ) \ . $ Here, we'll briefly discuss part (b) as to why there are two sets of critical points, only one of which is extremal.

The surface is an elliptic "hyperboloid of one sheet" oriented with the symmetry axis of its "throat" along the line $ \ y = x \ $ . All of the critical points are located in the narrowest part of the throat [the first pair marked in yellow, the second in light blue].

By direct calculation, we find $ \ f(0, 0, \pm 1) \ = \ 1 \ , $ while $ \ f(\pm 1, \mp 1 , 0) \ = \ 2 \ . $ The second partial derivatives all of limited help here, as the Hessian is simply a diagonal matrix with all three entries being $ \ 2 \ . $ So while this confirms that the points at distance $ \ 1 \ $ from the origin are at the global minimum distance, it doesn't tell us much about the pair at distance $ \ \sqrt{2} \ . $ (It is clear that there is no global maximum for distance from the origin.)

Instead, we need to examine the surface itself. If we add the plane $ \ y \ = \ -x \ $ , the intersection with the surface is an ellipse centered on the origin, with its minor axis along the $ \ z-$ axis and its major axis along the line $ \ y \ = \ -x \ $ . The points $ \ (\pm 1 \ , \ \mp 1 \ , \ 0 ) \ $ are the endpoints of this major axis, so their distance from the origin on this curve is maximal. However, if we intersect our surface with the plane $ \ z \ = \ 0 \ , \ $ we obtain the hyperbola $ \ xy \ = \ -1 \ , $ for which it is a familiar fact (or easily calculated) that the points under discussion lie on the curve at the minimal distance from the origin. Thus, the points $ \ (\pm 1 \ , \ \mp 1 \ , \ 0 ) \ $ serve as saddle points for the function on this surface.