find extrema of a multivariable function Lagrange

Question

I need to evaluate the extrema of this function :

$f(x,y,z)=\frac{1}{x^2+y^2+z^2}$

restricted on: $R=\{x^2-y^2-z^2+16<=0\}$ \ $\{0,0,0\}$

boundary : $\theta R=\{x^2-y^2-z^2+16=0\}$ \ $\{0,0,0\}$

using the Lagrange multiplier method : $$ \left\{ \begin{aligned} \frac{-2x}{x^2+y^2+z^2}=2x\lambda\\ \frac{-2y}{x^2+y^2+z^2}=-2y\lambda\\ \frac{-2z}{x^2+y^2+z^2}=-2z\lambda\\ x^2-y^2-z^2+16=0 \end{aligned} \right. $$

I found out --> $(0,0,\pm 4),(0,\pm 4,0),(\pm 4,0,0)$

Using the second derivate test I got :

$(0,0,\pm 4)$ -> Max

$(\pm 4,0,0),(0,\pm 4,0)$ -> Saddle points

eventually : $f(\pm 4,0,0)=f(0,\pm 4,0)=f(0,0,\pm 4)=\frac{1}{16}$

The image of this function is : $f(R)=[something,\frac{1}{16}]$

Can you help me figure out if I missed something? and if I got it right, is there a reason why there is no minimum?

Answer 1

The constraint $\ x^2 - y^2 - z^2 + 16 \le 0\ $ is not satisfied by $\ \left(\pm 4, 0, 0\right)\ $, so it's not even a feasible point, let alone an extremum.

All points of the form $\ \left(\,0, y, z\,\right)\ $ with $\ y^2 + z^2 = 16\ $ satisfy the Lagrange extremality conditions, and are all maxima, since $\ f\left(\,0,y,z\,\right)\ = \frac{1}{16}\ $ for such points, and the constraint $\ x^2 - y^2 - z^2 \le 16\ $ implies that its value cannot be any larger than that.

There are also arbitrarily large values of $\ x, y,\ \mbox{and } z\ $ that can satisfy the constraint $\ x^2 - y^2 - z^2 \le 16\ $. If $\ x=a^2 + 1, y=2a,\ $ and $ z=a^2 - 1\ $, for instance, then $\ x^2 - y^2 - z^2 = 0 \le 16\ $ for any value of $\ a\ $, and $\ a\ $ can be made aribitrarily large. Thus, although the value of the objective function is always strictly positive, it can nevertheless be made arbitrarily close to zero, which is therefore its infimum, and $\ f\left(R\right) = \left(0, \frac{1}{16}\,\right]\ $.

Answer 2

The Lagrange equations are not all consistent, as we obtain $$ \frac{-2x}{(x^2+y^2+z^2)^2} \ = \ 2x· \lambda \ \ \Rightarrow \ \ -2x \ · \left[ \ \lambda \ + \ \frac{1}{(x^2+y^2+z^2)^2} \ \right] \ = \ 0 \ \ , $$ $$ \frac{-2y}{(x^2+y^2+z^2)^2} \ = \ -2y· \lambda \ \ \Rightarrow \ \ 2y \ · \left[ \ \lambda \ - \ \frac{1}{(x^2+y^2+z^2)^2} \ \right] \ = \ 0 \ \ , $$ $$ \frac{-2z}{(x^2+y^2+z^2)^2} \ = \ -2z· \lambda \ \ \Rightarrow \ \ 2z \ · \left[ \ \lambda \ - \ \frac{1}{(x^2+y^2+z^2)^2} \ \right] \ = \ 0 \ \ . $$

[The denominator in the derivatives given in the post is inaccurate, but does not materially affect the conclusion.]

The expression in brackets in the latter two equations doesn't tell us much. We can extract that $ \ x \ , \ y \ , \ \text{or} \ \ z = 0 \ \ , $ but we will see that this is not giving us a complete description of the solution.

We can arrive at a geometrical interpretation of the situation by treating the function $ \ f(x,y,z) \ = \ \frac{1}{x^2 + y^2 + z^2} \ $ as the "reciprocal distance-squared" for a point $ \ (x,y,z) \ $ measured from the origin. Expressed in "standard form" for a quadric surface, the region $ \ R \ $ is $ \ -x^2+y^2+z^2 \ \ge \ 16 \ $ with the boundary $ \ \partial R \ $ being $ \ -x^2+y^2+z^2 \ = \ 16 \ \ . $ We see then that this boundary is an "inner surface" which is a hyperboloid of one sheet with its symmetry axis being the $ \ x-$axis [depicted in the graph below]. The region $ \ R \ $ comprises all points "outside" of this surface, extending infinitely.

Writing the hyperboloid equation as $ \ y^2+z^2 \ = \ 16 + x^2 \ \ , $ we observe that the "narrowest" cross-section occurs at $ \ x = 0 \ \ , $ giving us a circle in the $ \ yz-$plane with a radius of $ \ 4 \ $ centered on the $ \ x-$ axis. So the critical points $ \ ( 0 \ , \ \pm4 \ , \ 0) \ $ and $ \ ( 0 \ , \ 0 \ , \ \pm4) \ $ are "degenerate" and belong to the minimal-distance "ring" $ \ y^2+z^2 \ = \ 16 \ \ , $ as lonza leggiera also describes. [This also indicates that the specified exclusion of the origin from $ \ R \ $ and $ \ \partial R \ $ in the problem statement is unnecessary, as the given inequality already prevents it from being included.]

Since no point in $ \ R \ $ is closer to the origin than $ \ 4 \ $ units, the global maximum for $ \ f(x,y,z) \ $ is $ \ \frac{1}{16} \ \ . $ There is no upper limit to distances from the origin for points in the region, so our function is bounded below by zero, but does not possess a global minimum.