find absolute maximum and minimum with explanation

Question

if $f(x,y,z)=x+y-z$ and constraint is sphere $g(x,y,z)=x^2+y^2+z^2=81$. I have to find the absolute maximum and minimum. And I have to explain how both an absolute maximum and minimum must be attained.

my solution: I obtained critical points $(-3\sqrt{3} ,-3\sqrt{3},3\sqrt{3}) and (3\sqrt{3} ,3\sqrt{3},-3\sqrt{3})$ by solving the system $\nabla f= \lambda \nabla g$ and constraint equation.

Then I used the equation $L(l;x,y,z) =x+y-z-l(x^2+y^2+z^2-81)$ and found the matrix $\begin{bmatrix} 0 & -2x & -2y & -2z\\ -2x & -2l & 0 & 0\\ -2y & 0 & -2l & 0\\ -2z & 0 & 0 & -2l \end{bmatrix}$.

I plugged in the critical points, found the sequence of determinants and found that $(-3\sqrt{3} ,-3\sqrt{3},3\sqrt{3})$ is local minimum and $(3\sqrt{3} ,3\sqrt{3},-3\sqrt{3})$ is local maximum.

How do I say that they are absolute maximum and minimum, as well as that they both must be attained?

Answer 1

Hint (for the 1st part): using the RMS-AM inequality, and also that $|a| \ge a$ and $|a| \ge -a$:

$$ 3 \sqrt{3} = \sqrt{\frac{81}{3}} = \sqrt{\frac{x^2+y^2+z^2}{3}} \ge \frac{|x|+|y|+|z|}{3} \ge \frac{x+y-z}{3} \\ \iff\quad x+y -z \le 9 \sqrt{3} $$

Equality is attained iff $|x|=|y|=|z|=3 \sqrt{3}$ and $x,y \ge 0, z \le 0$, which gives $x=y=3 \sqrt{3}$ and $z = -3 \sqrt{3}$, so the upper bound of $9 \sqrt{3}$ is indeed a global maximum.

Answer 2

By the Cauchy-Schwarz inequality one has $f^2 \le (1^2+1^2+(-1)^2)(x^2+y^2+z^2)=243$ and equality only occurs when $(x,y,z)$ is a multiple of $(1,1,-1)$, so the extrema are at $(t,t,-t)$ with $3t^2=81$ and therefore $t=\pm 3\sqrt 3$.

Answer 3

The problem can also be interpreted geometrically. We have the sphere $ \ x^2 \ + \ y^2 \ + \ z^2 \ = \ 81 \ $ of radius $ \ 9 \ $ centered on the origin. The plane $ \ x \ + \ y \ - \ z \ = \ 0 \ $ passes through the origin and all planes $ \ x \ + \ y \ - \ z \ = \ d \ $ are parallel to it.

It is a reasonably well-known result (and not difficult to prove) that the perpendicular distance between parallel planes $ \ ax \ + \ by \ + \ cz \ = \ d_1 \ $ and $ \ ax \ + \ by \ + \ cz \ = \ d_2 \ $ is $ \ D \ = \ \frac{| \ d_2 \ - \ d_1 \ |}{\sqrt{a^2 \ + \ b^2 \ + \ c^2}} \ $ . The planes parallel to $ \ x \ + \ y \ - \ z \ = \ 0 \ $ are displaced along a line with direction vector $ \ \langle \ 1 \ , \ 1 \ , \ -1 \ \rangle \ $ .

We will want the tangent planes to the sphere which are the result of "parallel displacement" along the radii lying on this line. Those tangent planes then lie at a distance $ \ D \ = \ 9 \ $ from the plane $ \ x \ + \ y \ - \ z \ = \ 0 \ $ , which tells us that $$ \ 9 \ = \ \frac{| \ d \ - \ 0 \ |}{\sqrt{1^2 \ + \ 1^2 \ + \ (-1)^2}} \ \ \Rightarrow \ \ |d| \ = \ 9 \sqrt{3} \ \ . $$

(The line is a diameter of the sphere, so we would expect the symmetry of the sphere to produce two such planes.)

These planes are thus $ \ x \ + \ y \ - \ z \ = \ 9 \sqrt{3} \ $ and $ \ x \ + \ y \ - \ z \ = \ -9 \sqrt{3} \ \ , $ giving us the maximum and minimum values of $ \ f (x,y,z) \ $ , subject to the "spherical" constraint.

Examples of this sort were likely involved in the reasoning that led to the "Lagrange method".