Lagrange multiplers Method:

\(\displaystyle{F}{\left({x},{y},{z}\right)}={x}^{{2}}+{y}^{{2}}+{z}^{{2}},{g{{\left({x},{y},{z}\right)}}}-{x}+{y}+{z}={12}\)

On finding the corresponding partial derivatives of F and g and equating,

\(\displaystyle{2}{x}=\lambda\)

\(\displaystyle{2}{y}=\lambda\)

\(\displaystyle{2}{z}=\lambda\)

Therefore, \(\displaystyle{x}={y}={z}={\frac{{\lambda}}{{{2}}}}\)

Hence, \(\displaystyle{x}+{y}+{z}={12}\to{3}{x}={12}\to{x}={4}\)

Hence, \(\displaystyle{F}{\left({x},{y},{z}\right)}\) has a minimum at x=y=z=4

Minimum value is \(\displaystyle{x}^{{2}}+{y}^{{2}}+{z}^{{2}}={4}^{{2}}+{4}^{{2}}+{4}^{{2}}\)

\(\displaystyle={16}+{16}+{16}={48}\)

\(\displaystyle{F}{\left({x},{y},{z}\right)}={x}^{{2}}+{y}^{{2}}+{z}^{{2}},{g{{\left({x},{y},{z}\right)}}}-{x}+{y}+{z}={12}\)

On finding the corresponding partial derivatives of F and g and equating,

\(\displaystyle{2}{x}=\lambda\)

\(\displaystyle{2}{y}=\lambda\)

\(\displaystyle{2}{z}=\lambda\)

Therefore, \(\displaystyle{x}={y}={z}={\frac{{\lambda}}{{{2}}}}\)

Hence, \(\displaystyle{x}+{y}+{z}={12}\to{3}{x}={12}\to{x}={4}\)

Hence, \(\displaystyle{F}{\left({x},{y},{z}\right)}\) has a minimum at x=y=z=4

Minimum value is \(\displaystyle{x}^{{2}}+{y}^{{2}}+{z}^{{2}}={4}^{{2}}+{4}^{{2}}+{4}^{{2}}\)

\(\displaystyle={16}+{16}+{16}={48}\)