Using Lagrange multipliers, we have the Lagrangian
[tex]L(x,y,z,\lambda)=x+y-z+\lambda(x^2+y^2+z^2-81)[/tex]
with partial derivatives (set equal to 0)
[tex]L_x=1+2\lambda x=0\implies x=-\dfrac1{2\lambda}[/tex]
[tex]L_y=1+2\lambda y=0\implies y=-\dfrac1{2\lambda}[/tex]
[tex]L_z=-1+2\lambda z=0\implies z=\dfrac1{2\lambda}[/tex]
[tex]L_\lambda=x^2+y^2+z^2-81=0\implies x^2+y^2+z^2=81[/tex]
Substituting the first three equations into the fourth allows us to solve for [tex]\lambda[/tex]:
[tex]x^2+y^2+z^2=\dfrac1{4\lambda^2}+\dfrac1{4\lambda^2}+\dfrac1{4\lambda^2}=81\implies\lambda=\pm\dfrac1{6\sqrt3}[/tex]
For each possible value of [tex]\lambda[/tex], we get two corresponding critical points at [tex](\mp3\sqrt3,\mp3\sqrt3,\pm3\sqrt3)[/tex].
At these points, respectively, we get a maximum value of [tex]f(3\sqrt3,3\sqrt3,-3\sqrt3)=9\sqrt3[/tex] and a minimum value of [tex]f(-3\sqrt3,-3\sqrt3,3\sqrt3)=-9\sqrt3[/tex].
