Answer
Maximum:$f(1, \pm\sqrt2,1)=f(-1, \pm\sqrt2,-1)=2$, Minimum: $f(1, \pm\sqrt2,-1)=f(-1, \pm\sqrt2,1)=-2$
Work Step by Step
Our aim is to calculate the extreme values with the help of Lagrange Multipliers Method subject to the given constraints. For this, we have:$\nabla f(x,y)=\lambda \nabla g(x,y)$
This yields $\nabla f=\lt y^2z,2xyz,xy^2 \gt$ and $\lambda \nabla g=\lambda \lt 2x,2y,2z\gt$
From the given constraint condition $x^2+y^2+z^2=4$ we get, $y^2z=\lambda 2x, 2xyz=\lambda 2y,xy^2=\lambda 2z$
Simplify. Thus, $x^2=z^2$ and $y^2=2z^2$
Since, $g(x,y)=x^2+y^2+z^2=4 \implies$ $z=\pm 1$
and $x=\pm 1$ and $y=\pm \sqrt2$
Hence, Maximum value is $f(1, \pm\sqrt2,1)=f(-1, \pm\sqrt2,-1)=2$, Minimum value is $f(1, \pm\sqrt2,-1)=f(-1, \pm\sqrt2,1)=-2$
