Use Lagrange multipliers to find the maximum and minimum values of the function subject to the given constraints. (If a value does not exist, enter NONE.) f(x,y,z) = 3x – y – 3z; x + y – z = 0, x2 + 2z2 = 1

The Lagrangian is[tex]L(x,y,z,\lambda,\mu)=3x-y-3z+\lambda(x+y-z)+\mu(x^2+2z^2-1)[/tex]with critical points where its partial derivatives vanish:[tex]L_x=3+\lambda+2\mu x=0[/tex][tex]L_y=-1+\lambda=0\implies\lambda=1[/tex][tex]L_z=-3-\lambda+4\mu z=0[/tex][tex]L_\lambda=x+y-z=0[/tex][tex]L_\mu=x^2+2z^2-1=0\implies x^2+2z^2=1[/tex]Since we know [tex]\lambda=1[/tex], we have[tex]L_x=0\implies4+2\mu x=0\implies x=-\dfrac2\mu[/tex][tex]L_z=0\implies-4+4\mu z=0\implies z=\dfrac1\mu[/tex][tex]L_\mu=0\implies\dfrac6{\mu^2}=1\implies\mu=\pm\sqrt6[/tex], so that [tex]x=\mp\dfrac2{\sqrt6}[/tex] and [tex]z=\pm\dfrac1{\sqrt6}[/tex][tex]L_\lambda=0\implies y=z-x\implies y=\pm\dfrac3{\sqrt6}[/tex]So there are two critical points, [tex]\left(-\dfrac2{\sqrt6},\dfrac3{\sqrt6},\dfrac1{\sqrt6}\right)[/tex] and [tex]\left(\dfrac2{\sqrt6},-\dfrac3{\sqrt6},-\dfrac1{\sqrt6}\right)[/tex], which give a minimum value of [tex]-2\sqrt6[/tex] and a maximum value of [tex]2\sqrt6[/tex], respectively.