The equations that must be solved for maximum or minimum values of a differentiable function w=​f(x,y,z) subject to two constraints ​g(x,y,z)=0 and ​h(x,y,z)=​0, where g and h are also​ differentiable, are gradientf=lambdagradientg+mugradient​h, ​g(x,y,z)=​0, and ​h(x,y,z)=​0, where lambda and mu ​(the Lagrange​ multipliers) are real numbers. Use this result to find the maximum and minimum values of ​f(x,y,z)=xsquared+ysquared+zsquared on the intersection between the cone zsquared=4xsquared+4ysquared and the plane 2x+4z=2.

The Lagrangian is[tex]L(x,y,z,\lambda,\mu)=x^2+y^2+z^2+\lambda(4x^2+4y^2-z^2)+\mu(2x+4z-2)[/tex]with partial derivatives (set equal to 0)[tex]L_x=2x+8\lambda x+2\mu=0\implies x(1+4\lambda)+\mu=0[/tex][tex]L_y=2y+8\lambda y=0\implies y(1+4\lambda)=0[/tex][tex]L_z=2z-2\lambda z+4\mu=0\implies z(1-\lambda)+2\mu=0[/tex][tex]L_\lambda=4x^2+4y^2-z^2=0[/tex][tex]L_\mu=2x+4z-2=0\implies x+2z=1[/tex]Case 1: If [tex]y=0[/tex], then[tex]4x^2-z^2=0\implies4x^2=z^2\implies2|x|=|z|[/tex]Then[tex]x+2z=1\implies x=1-2z\implies2|1-2z|=|z|\implies z=\dfrac25\text{ or }z=\dfrac23[/tex][tex]\implies x=\dfrac15\text{ or }x=-\dfrac13[/tex]So we have two critical points, [tex]\left(\dfrac15,0,\dfrac25\right)[/tex] and [tex]\left(-\dfrac13,0,\dfrac23\right)[/tex]Case 2: If [tex]\lambda=-\dfrac14[/tex], then in the first equation we get[tex]x(1+4\lambda)+\mu=\mu=0[/tex]and from the third equation,[tex]z(1-\lambda)+2\mu=\dfrac54z=0\implies z=0[/tex]Then[tex]x+2z=1\implies x=1[/tex][tex]4x^2+4y^2-z^2=0\implies1+y^2=0[/tex]but there are no real solutions for [tex]y[/tex], so this case yields no additional critical points.So at the two critical points we've found, we get extreme values of[tex]f\left(\dfrac15,0,\dfrac25\right)=\dfrac15[/tex] (min)and[tex]f\left(-\dfrac13,0,\dfrac23\right)=\dfrac59[/tex] (max)