Question: Use the bordered Hessian test to show that $f(x,y,z)=x^2+y^2+z^2$ under $g(x,y,z)=z-xy-2=0$ has two minimum points and no maximum (find the two points), and explain how this is possible.
I'm having trouble with the "explain how this is possible part".
What I did:
a. Found the minimum points $(-1,1,1) , (1,-1,1)$.
b. Found one critical point, $(0,0,2)$, which fails the bordered Hessian test (as $|\tilde{H_2}|=0$).
c. Found the complex point $(\sqrt{3}i,\sqrt{3}i,-1)$, but we don't deal with complex numbers in this course.
$f(-1,1,1)=f(1,-1,1)=3$, and $f(0,0,2)=4$, so $(0,0,2)$ might be a maximum, even though the Hessian test fails, but I don't know if this answer is sufficient.
Is there another explanation I'm missing?
Best Answer
Note that the function $f$ is the distance function squared. So a (local) maximum of $f$ that lies on the surface $g(x,y,z)=0$ would be a point that (locally) lies the farthest from the origin. Make a plot of $g(x,y,z)=z-xy-2=0$ and you will see that for every point on the surface, you can take another point on the surface that lies further away. One reason is that $g(x,y,z)=0$ is unbounded. Also around the point (0,0,2), there are points on $g(x,y,z)=0$ that are further away from (0,0,0) (take any point with $z$ little bit bigger than 2).
Edit: You 'll see that the surface $g(x,y,z)=0$ looks like a saddle surface. Think of the saddle as being "squeezed" together in the horizontal direction (this is a bit of an exageration, but it might help). Therefore two points on both sides of the saddle $(-1,1,1)$ and $(1,-1,1)$ are being the closest to the origin. A 2nd and more algebraic explanation goes like this: by interchanging $x$ and $y$ you get another point on $g(x,y,z)=0$ that lies on the same distance from the origin.