I am trying to solve this problem, but I am doing something wrong: $$f(x,y,z)=x^2-y^2,M=\{[x,y,z]\in\mathbb{R}^3:x^2+y^2+z^2=9,x+z\ge1\}$$
And let $g(x,y,z)=x^2+y^2+z^2-9$. Set M is closed and bounded ($M\subset K(0,3)) $. M is therefore compact and $f$ is continuous $\Rightarrow$ $f$ has max/min.
I use Lagrangian multiplier:
1) $\nabla g(x,y,z)=(2x,2y,2z)=\vec{0}\leftrightarrow x=y=z=0$, but $[0,0,0]\not \in M$
2)$\nabla f-\lambda \nabla g=0$ $$(2x-\lambda2x=0$$ $$-2y-\lambda 2y=0$$ $$0-\lambda 2z=0$$ $$x^2+y^2+z^2=9$$
A) $\lambda=0$:First critical point$[0,0,3]$$$x=y=0\rightarrow z=3$$ B) $\lambda\not=0,z=0$
$\lambda =\pm 1:x=y=0,[0,0,0]\not \in M$
So I have only point that could be extrema, but that is entirely wrong. Can you please identify the mistake?
edit: I found that for $\lambda = 1 $ or $y=0$ we have $[3,0,0]\in M$, which is maxima, but I still can't determine minima.
The second constraint given turns out to be of some importance in locating the minimum value for $ \ f(x,y,z) \ $ . We will need to "turn it on and off" as we proceed, since an inequality is used in describing the region $ \ M \ $ . Here is a graph of the geometrical arrangement:
As you already found from the Lagrange equations,
$$ \ 2x \ ( \ \lambda \ - \ 1 \ ) \ = \ 0 \ \ , \ \ 2y \ ( \ \lambda \ + \ 1 \ ) \ = \ 0 \ \ , \ \ 2z \ \lambda \ = \ 0 \ \ , \ \ $$
we may have:
I : from the third equation, $ \ \lambda \ = \ 0 \ , \ z \ $ unspecified
$$ \Rightarrow \quad \ 2x \ (-1) \ = \ 0 \ \ \Rightarrow \ x \ = \ 0 \ \ , \ \ 2y \ (1) \ = \ 0 \ \ \Rightarrow \ y \ = \ 0 \ \ \Rightarrow \ \ z \ = \ \pm 3 \ \ , $$
from the equation for the sphere; $ \ x \ + \ z \ \ge \ 1 \ $ requires that $ \ z \ = \ 3 \ $ ; and $ \ f(0, 0, 3) \ = \ 0 \ $ , which proves not to be extremal
II : from the first equation, $ \ \lambda \ = \ 1 \ , \ x \ $ unspecified
$$ \Rightarrow \quad \ 2y \ (2) \ = \ 0 \ \ \Rightarrow \ y \ = \ 0 \ \ , \ \ 2z \ (1) \ = \ 0 \ \ \Rightarrow \ z \ = \ 0 \ \ \Rightarrow \ \ x \ = \ \pm 3 \ \ , $$
from the equation for the sphere; $ \ x \ + \ z \ \ge \ 1 \ $ requires that $ \ x \ = \ 3 \ $ ; and $ \ f(3, 0, 0) \ = \ 9 \ , $
which is the maximal value [more on this shortly]
III : from the second equation, $ \ \lambda \ = \ -1 \ , \ y \ $ unspecified
$$ \Rightarrow \quad \ 2x \ (-2) \ = \ 0 \ \ \Rightarrow \ x \ = \ 0 \ \ , \ \ 2z \ (-1) \ = \ 0 \ \ \Rightarrow \ z \ = \ 0 \ \ \Rightarrow \ \ y \ = \ \pm 3 \ \ , $$
from the equation for the sphere; but these points lies below the plane $ \ x \ + \ z \ \ge \ 1 \ $ , so they are not in the region $ \ M \ $ .
Case I above corresponds to the "level surface" for $ \ f \ , \ x^2 \ - \ y^2 \ = \ 0 \ $ , which is the pair of "crossed planes" $ \ y \ = \ \pm x \ $ intersecting on the $ \ z-$ axis. Case II corresponds to the hyperbolic cylinder $ \ x^2 \ - \ y^2 \ = \ 9 \ $ , which is just tangent to the sphere at $ \ (3, \ 0, \ 0) \ $ as seen in the graph below. (The constant for the hyperbolic cylinder is made a little "too small" in order to make the tangent point visible.)
We need to explore the boundary of the region in order to locate the minimal value for $ \ f \ $ . If we insert the equation for the plane, $ \ x \ + \ z \ = \ 1 \ $ into the equation for the sphere, we obtain
$$ x^2 \ + \ y^2 \ + \ (1-z)^2 \ = \ 1 \ \ \Rightarrow \ \ 2x^2 \ - \ 2x \ + \ y^2 \ = \ 8 \ \ . $$
We can use this to construct a Lagrange equation, which we can then put together with the one for $ \ y \ $ :
$$ 2x \ = \ \lambda \ ( \ 4x \ - \ 2 \ ) \ \ , \ \ 2y \ ( \ \lambda \ + \ 1 \ ) \ = \ 0 \ \ , $$
giving us two cases to examine:
A -- $ \ y \ = \ 0 \ \ \Rightarrow \ \ 2x^2 \ - \ 2x \ - \ 8 \ = \ 0 \ \ $
$$ \Rightarrow \ \ x \ = \ \frac{1 \ + \ \sqrt{17}}{2} \ , \ z \ = \ \frac{1 \ - \ \sqrt{17}}{2} $$ $$ f \left(\frac{1 \ + \ \sqrt{17}}{2} \ , \ 0 \ , \ \frac{1 \ - \ \sqrt{17}}{2} \right) \ = \ \frac{9 \ + \ \sqrt{17}}{2} \ \approx \ 6.56 $$
or
$$ \Rightarrow \ \ x \ = \ \frac{1 \ - \ \sqrt{17}}{2} \ , \ z \ = \ \frac{1 \ + \ \sqrt{17}}{2} $$ $$ f \left(\frac{1 \ - \ \sqrt{17}}{2} \ , \ 0 \ , \ \frac{1 \ + \ \sqrt{17}}{2} \right) \ = \ \frac{9 \ - \ \sqrt{17}}{2} \ \approx \ 2.44 \ \ , $$
neither of which is extremal
B -- $ \ \lambda \ = \ -1 \ \ \Rightarrow \ y \ $ unspecified, $ \ 2x \ = \ (-1) \ ( \ 4x \ - \ 2 \ ) \ \ \Rightarrow \ \ 6x \ = \ 2 \ \ \Rightarrow \ \ x \ = \ \frac{1}{3} \ $ ;
from the equation for the plane, $ \ z \ = \ 1 \ - \ x \ = \ \frac{2}{3} \ $ ;
from the equation for the sphere, $ \ y^2 \ = \ 9 \ - \ \left(\frac{1}{3} \right)^2 \ - \ \left(\frac{2}{3} \right)^2 \ = \ \frac{76}{9} \ \ \Rightarrow \ \ y \ = \ \pm \frac{2 \ \sqrt{19}}{3} $
$$ \Rightarrow \ \ f \left(\frac{1}{3} \ , \ \pm \frac{2 \ \sqrt{19}}{3} \ , \ \frac{2}{3} \right) \ = \ -\frac{75}{9} \ \approx \ -8.33 \ \ . $$
This is in fact the minimal value for the function, its level surface represented by the hyperbolic cylinder $ \ x^2 \ - \ y^2 \ = \ -\frac{75}{9} \ $ or $ \ y^2 \ - \ x^2 \ = \ \frac{75}{9} \ , $ which is rotated 90º with respect to the maximal level surface. (In the graph below, its constant is again taken a trifle smaller in absolute value in order to make a tangent point visible; the other point is on the "far side" of the figure.)
This works out a bit more directly if we use Lagrange multipliers for both constraints. We use the constraint functions $ \ g(x,y,z) \ = \ x^2 \ + \ y^2 \ + \ z^2 \ - 9 \ $ and $ h(x,y,z) \ = \ x \ + \ z \ - \ 1 \ $ , together with the gradient equation $ \ \nabla f \ = \ \lambda \ \nabla g \ + \ \mu \ \nabla h \ $ to produce
$$ 2x \ = \ \lambda \cdot 2x \ + \ \mu \cdot 1 \ \ , \ \ -2y \ = \ \lambda \cdot 2y \ \ , \ \ 0 \ = \ \lambda \cdot 2z \ + \ \mu \cdot 1 \ \ . $$
We have already seen from Cases II and A above that we will need to "relax" the constraint from the plane, and instead make use of the inequality $ \ x \ + \ z \ \ge \ 1 \ $ to locate the tangent point $ \ ( 3, \ 0, \ 0) \ $ on the surface of the sphere, where $ \ f(x,y,z) \ $ is maximized.
On the other hand, if we use $ \ \lambda \ = \ -1 \ $ with $ \ y \ $ being unspecified in the second Lagrange equation immediately above, the third equation now gives us $ \ 2 z \ = \ \mu \ $ . We then have from the first equation $ \ 4x \ = \ 2z \ \Rightarrow \ z \ = \ 2x \ $ . The equation for the plane yields $ \ x \ + \ z \ = \ x \ + \ 2x \ = \ 1 \ \Rightarrow \ x \ = \ \frac{1}{3} \ $ , from which we may now derive the information for the points where $ \ f(x,y,z) \ $ is minimal, in the manner we carried out above.