Minimize $x^2+6y^2+4z^2$ subject to $x+2y+z-4=0$ and $2x^2+y^2=16$ Minimize $x^2+6y^2+4z^2$ subject to $x+2y+z-4=0$ and $2x^2+y^2=16$
My try: 
By Lagrange Multiplier method we have
$$L(x,y,z,\lambda, \mu)=(x^2+6y^2+4z^2)+\lambda(x+2y+z-4)+\mu(2x^2+y^2-16)$$
For $$L_x=0$$ we get
$$2x+\lambda+4\mu x=0 \tag{1}$$
For $$L_y=0$$ we get
$$12y+2\lambda+2\mu y=0 \tag{2}$$
For $$L_z=0$$ we get
$$8z+\lambda=0 \tag{3}$$
From $(1)$ and $(2)$ we get
$$x=\frac{4 \lambda}{1-2\mu}$$
$$y=\frac{8 \lambda}{6-\mu}$$
Substituting $x$ , $y$ and $z$ above in constrainst we get
$$2 \frac{\lambda^2}{(1-2\mu)^2}+4 \frac{\lambda^2}{(6-\mu)^2}=1 \tag{4}$$
$$\frac{4 \lambda}{1-2\mu}+\frac{16 \lambda}{6-\mu}+\frac{\lambda}{8}=4 \tag{5}$$
But its tedious to solve above equations for $\lambda$ and $\mu$
Any other approach?
 A: Since you are assuming that $2x^2+y^2=16$, your problem is equivalent to the problem of minimizing $\frac{11}2y^2+4z^2$. But then your first equation becomes $\lambda+4\mu x=0$, which is much simpler.
A: Hint:
Let $u=x^2+6y^2+4(4-2y-x)^2$
$\iff5x^2-16(2-y)x+22y^2-16y+16-u=0$
As $x$ is real, the discriminant will be $\ge0$
$$\implies256(y-2)^2\ge20(22y^2-16y+16-u)$$
$$\implies64(y-2)^2\ge5(22y^2-16y+16-u)$$
$$\implies5u\ge46y^2-176y+176=46\left(y-\dfrac{44}{23}\right)^2+176-46\left(\dfrac{44}{23}\right)^2$$
Now $y^2=16-2x^2\le16\implies-4\le y\le4\iff-4-\dfrac{44}{23}\le y-\dfrac{44}{23}\le4-\dfrac{44}{23}$
A: With computer I get numeric solution
$$f_{min}=9.445967377367024$$
if
$$x=2.811441051354938\\
y=0.4377195786259503\\
z=0.3131197913931612$$ 
Exact values of  $x,y,z$ is solutions of equations:
$$2033x^4-800x^3-13960x^2-1792x+6144=0,\\
2033y^4-6016y^3-27920y^2+88064y-32768=0,\\
2033z^4-19696z^3-84456z^2+176704z-46464=0.$$
We can solve these equations exact and get exact solutions. 
However, there are too big expressions.
