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 $$\begin{cases}6x^2+y^2+z^2+4xy-2zx\leq1 \space\space(1) \\ 4x^2+y^2+4xy+2xz\leq z^2\space \space \space\space\space\space\space\space\space(2)\end{cases}$$ 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 $$\begin{cases}-4x-4y-2z+1+\lambda(20x+8y)=0 \\ -2y-4x+\lambda(4y + 8x)=0\\ 2z-2x-1=0\\ 10x^2+2y^2+8xy=1\end{cases}$$ 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
$$ \left\{ \begin{array}{rcl} -4 x-4 y+\lambda_1 (12 x+4 y-2 z)-2 z+\lambda_2(8 x+4 y+2 z)+1&=&0 \\ -4 x-2 y+\lambda_1(4 x+2 y)+\lambda_2(4 x+2 y)&=&0 \\ -2 x+\lambda_2 (2 x-2 z)+2 z+\lambda_1(2 z-2 x)-1&=&0 \\ \epsilon_1^2+6 x^2+y^2+z^2+4 x y-2 x z-1&=&0 \\ \epsilon_2^2+4 x^2+y^2-z^2+4 x y+2 x z&=&0 \\ 2 \epsilon_1\lambda_1&=&0 \\ 2 \epsilon_2 \lambda_2&=&0 \\ \end{array} \right. $$
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.