Csdrhrt

I want to calculate the minimum and maximum values the function $f:z^2-2x^2-y^2-4xy-2xz-z+x$ takes on the boundary of the compact set $$begin{cases}6x^2+y^2+z^2+4xy-2zxleq1 spacespace(1) \ 4x^2+y^2+4xy+2xzleq z^2space space spacespacespacespacespacespacespace(2)end{cases}$$
To do so, I use the Lagrange multiplier method. From $(1)$ and $(2)$ I get $$g(x,y)=10x^2+2y^2+8xy=1$$
Therefore I need to solve the system of equations $$begin{cases}-4x-4y-2z+1+lambda(20x+8y)=0 \
-2y-4x+lambda(4y + 8x)=0\
2z-2x-1=0\
10x^2+2y^2+8xy=1end{cases}$$
Using substitution, I solved this system of equations, the solutions are $x=0,y=frac{1}{sqrt{2}},z=frac{1}{2},lambda=frac{1}{2}$ and $x=frac{1}{sqrt{2}},y=frac{-2}{sqrt{2}},z=frac{1+frac{2}{sqrt{2}}}{2},lambda=frac{1}{2}$

Which gives a minimum: $frac{-3}{4}$ and a maximum $frac{-15}{4}+frac{8}{sqrt{2}}$

Is this correct?
I am told there are more minimums and maximums, maybe I missed more solutions of the system of equations? I don't have much experience solving systems of non linear equations.

Calling

$$
f(x,y,z) = z^2 - 2 x^2 - y^2 - 4 x y - 2 x z - z + x\
g_1(x,y,z) = 6 x^2 + y^2 + z^2 + 4 x y - 2 z x - 1\
g_2(x,y,z) = 4 x^2 + y^2 + 4 x y + 2 x z - z^2
$$

and introducing the auxiliary slack variables $epsilon_i$ to transform the inequalities into equalities, we have the lagrangian

$$
L(X,lambda,epsilon) = f(x,y,z)+lambda_1(g_1(x,y,z)+epsilon_1^2)+lambda_2(g_2(x,y,z)+epsilon_2^2)
$$

The stationary points are obtained by solving

$$
left{
begin{array}{rcl}
-4 x-4 y+lambda_1 (12 x+4 y-2 z)-2 z+lambda_2(8 x+4 y+2 z)+1&=&0 \
-4 x-2 y+lambda_1(4 x+2 y)+lambda_2(4 x+2 y)&=&0 \
-2 x+lambda_2 (2 x-2 z)+2 z+lambda_1(2 z-2 x)-1&=&0 \
epsilon_1^2+6 x^2+y^2+z^2+4 x y-2 x z-1&=&0 \
epsilon_2^2+4 x^2+y^2-z^2+4 x y+2 x z&=&0 \
2 epsilon_1lambda_1&=&0 \
2 epsilon_2 lambda_2&=&0 \
end{array}
right.
$$

with the solutions

$$
begin{array}{cccccccc}
x & y & z & lambda_1 & lambda_2 & epsilon_1 & epsilon_2 & f(x,y, z) \
-frac{1}{4} & frac{1}{2} & 0 & 0 & -1 & frac{sqrt{frac{7}{2}}}{2} & 0 & -frac{1}{8} \
0 & 0 & -1 & -frac{3}{2} & 0 & 0 & 1 & 2 \
0 & 0 & frac{1}{2} & 0 & 0 & frac{sqrt{3}}{2} & frac{1}{2} & -frac{1}{4} \
0 & 0 & 1 & -frac{1}{2} & 0 & 0 & 1 & 0 \
0 & -frac{1}{sqrt{2}} & -frac{1}{sqrt{2}} & -frac{1}{2 sqrt{2}} & frac{1}{2} left(2+frac{1}{sqrt{2}}right) & 0 & 0 & frac{1}{sqrt{2}}
\
0 & frac{1}{sqrt{2}} & -frac{1}{sqrt{2}} & -frac{1}{2 sqrt{2}} & frac{1}{2} left(2+frac{1}{sqrt{2}}right) & 0 & 0 & frac{1}{sqrt{2}} \
0 & -frac{1}{sqrt{2}} & frac{1}{sqrt{2}} & frac{1}{2 sqrt{2}} & frac{1}{2} left(2-frac{1}{sqrt{2}}right) & 0 & 0 & -frac{1}{sqrt{2}} \
0 & frac{1}{sqrt{2}} & frac{1}{sqrt{2}} & frac{1}{2 sqrt{2}} & frac{1}{2} left(2-frac{1}{sqrt{2}}right) & 0 & 0 & -frac{1}{sqrt{2}} \
frac{1}{4} & -frac{1}{2} & frac{1}{2} & 0 & -1 & frac{sqrt{frac{7}{2}}}{2} & 0 & -frac{1}{8} \
-frac{1}{sqrt{2}} & sqrt{2} & 0 & frac{1}{2} left(-2+frac{1}{sqrt{2}}right) & -frac{1}{2 sqrt{2}} & 0 & 0 & 1-frac{1}{sqrt{2}} \
-frac{1}{sqrt{2}} & sqrt{2} & -sqrt{2} & frac{1}{2} left(-2-frac{1}{sqrt{2}}right) & frac{1}{2 sqrt{2}} & 0 & 0 &
1+frac{1}{sqrt{2}} \
frac{1}{sqrt{2}} & -sqrt{2} & 0 & frac{1}{2} left(-2-frac{1}{sqrt{2}}right) & frac{1}{2 sqrt{2}} & 0 & 0 & 1+frac{1}{sqrt{2}} \
frac{1}{sqrt{2}} & -sqrt{2} & sqrt{2} & frac{1}{2} left(-2+frac{1}{sqrt{2}}right) & -frac{1}{2 sqrt{2}} & 0 & 0 &
1-frac{1}{sqrt{2}}\
end{array}
$$

as we can observe, there are a lot of stationary points. Stationary points with at least one $epsilon_i = 0$ are located at the feasible region boundary. Solutions with all $epsilon_i ne 0$ are internal stationary points.