This extreme value problem has a solution with both a maximum value and a minimum value. use lagrange multipliers to find the extreme values of the function subject to the given constraint. f(x, y, z) = 8x + 8y + 4z; 4x2 + 4y2 + 4z2 = 36

[tex]L(x,y,z,\lambda)=8x+8y+4z+\lambda(4x^2+4y^2+4z^2-36)[/tex]

[tex]L_x=8+8\lambda x=0\implies 1+\lambda x=0[/tex]
[tex]L_y=8+8\lambda y=0\implies 1+\lambda y=0[/tex]
[tex]L_z=4+8\lambda z=0\implies 1+2\lambda z=0[/tex]
[tex]L_\lambda=4x^2+4y^2+4z^2-36=0\implies x^2+y^2+z^2=9[/tex]

[tex]yL_x=y+\lambda xy=0[/tex]
[tex]xL_y=x+\lambda xy=0[/tex]
[tex]\implies yL_x-xL_y=y-x=0\implies y=x[/tex]

[tex]2zL_x=2z+2\lambda xz=0[/tex]
[tex]xL_z=x+2\lambda xz=0[/tex]
[tex]\implies 2zL_x-xL_z=2z-x=0\implies x=2z[/tex]

[tex]2zL_y=2z+2\lambda yz=0[/tex]
[tex]yL_z=y+2\lambda yz=0[/tex]
[tex]\implies 2zL_y-yL_z=2z-y=0\implies y=2z[/tex]

[tex]x=y=2z\implies x^2+y^2+z^2=9\iff 4z^2+4z^2+z^2=9z^2=9\implies z=\pm1[/tex]

[tex]z=\pm1\implies y=x=\pm2[/tex]

So we have two critical points, (2, 2, 1) and (-2, -2, -1), which respectively give a max of 36 and a min of -36.