Do absolute maxima and minima exist in this question?

Question

I was solving following question related to lagrange multipliers

Find absolute extreme values of $f(x,y,z)=z$ subject to the constraints $x^2+y^2=z^2$ and $x+y+z=24$

By the method of lagrange multipliers I got two values of $f$, $ z=\frac{48}{2+\sqrt2},\frac{48}{2-\sqrt2}$,where one would be absolute minima and other absolute maxima.

But our constraints form a non-bounded region so we can't guarantee the existence of absolute maxima and minima.So my question is Do we have any of the two(abs maxima and minima) in this question?

Answer 1

I think you are right. Solving the constraints we get: $2xy-48(x+y)+24^2=0$ , which is a hyperbola on the plane $x+y+z=24$. From the first equation we also get $x=\frac{48y-24^2}{2y-48}$, from this we see that for $y\to 24$, $x\to \infty$ . From your first constrain this maximises/ minimises $z$, arbitrary large. So $f=z$ , has no maximum or minimum on the hyperbola.