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?

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.
