Answer
Maximum:$f( \sqrt2,1,\sqrt{\dfrac{2}{3}})=f( \sqrt2,-1,-\sqrt{\dfrac{2}{3}})=f(- \sqrt2,1,-\sqrt{\dfrac{2}{3}})=f( \sqrt2,-1,-\sqrt{\dfrac{2}{3}})= {\dfrac{2}{\sqrt 3}}$, Minimum: $f( \sqrt2,1,-\sqrt{\dfrac{2}{3}})=f( \sqrt2,-1,\sqrt{\dfrac{2}{3}})=f(- \sqrt2,1,\sqrt{\dfrac{2}{3}})=f( -\sqrt2,-1,-\sqrt{\dfrac{2}{3}})=- {\dfrac{2}{\sqrt 3}}$
or,
Maximum: $ {\dfrac{2}{\sqrt 3}}$ and minimum: $- {\dfrac{2}{\sqrt 3}}$
Work Step by Step
Use Lagrange Multipliers Method:
$\nabla f(x,y,z)=\lambda \nabla g(x,y,z)$
This yields $\nabla f=\lt y^2z,2xyz,xy^2 \gt$ and $\lambda \nabla g=\lambda \lt 2x,2y,2z\gt$
Using the constraint condition $x^2+y^2+z^2=4$ we get, $y^2z=\lambda 2x, 2xyz=\lambda 2y,xy^2=\lambda 2z$
After solving, we get $x^2=z^2$ and $y^2=2z^2$
Since, $g(x,y)=x^2+y^2+z^2=4$ gives $z=\pm 1$
Thus, $x=\pm 1$ and $y=\pm \sqrt2$
Hence, Maximum:$f( \sqrt2,1,\sqrt{\dfrac{2}{3}})=f( \sqrt2,-1,-\sqrt{\dfrac{2}{3}})=f(- \sqrt2,1,-\sqrt{\dfrac{2}{3}})=f( \sqrt2,-1,-\sqrt{\dfrac{2}{3}})= {\dfrac{2}{\sqrt 3}}$, Minimum: $f( \sqrt2,1,-\sqrt{\dfrac{2}{3}})=f( \sqrt2,-1,\sqrt{\dfrac{2}{3}})=f(- \sqrt2,1,\sqrt{\dfrac{2}{3}})=f( -\sqrt2,-1,-\sqrt{\dfrac{2}{3}})=- {\dfrac{2}{\sqrt 3}}$
or,
Maximum: $ {\dfrac{2}{\sqrt 3}}$ and minimum: $- {\dfrac{2}{\sqrt 3}}$
