Answer
The maximum of $f\left( {x,y,z} \right) = x + y + z$ is $\frac{{3 + 6\sqrt 2 }}{{\sqrt 5 }} \simeq 5.14$ subject to the two constraints ${x^2} + {y^2} + {z^2} = 9$ and $\frac{1}{4}{x^2} + \frac{1}{4}{y^2} + 4{z^2} = 9$.
Work Step by Step
Our task is to maximize the function $f\left( {x,y,z} \right) = x + y + z$ subject to the two constraints $g\left( {x,y,z} \right) = {x^2} + {y^2} + {z^2} - 9 = 0$ and $h\left( {x,y,z} \right) = \frac{1}{4}{x^2} + \frac{1}{4}{y^2} + 4{z^2} - 9 = 0$.
The Lagrange condition is
$\nabla f = \lambda \nabla g + \mu \nabla h$
$\left( {1,1,1} \right) = \lambda \left( {2x,2y,2z} \right) + \mu \left( {\frac{1}{2}x,\frac{1}{2}y,8z} \right)$
From the Lagrange condition we obtain three equations:
$1 = 2\lambda x + \frac{1}{2}\mu x$, ${\ \ }$ $1 = 2\lambda y + \frac{1}{2}\mu y$, ${\ \ }$ $1 = 2\lambda z + 8\mu z$
Since the left-hand sides are 1 (nonzero), we have $x \ne 0$, $y \ne 0$ and $z \ne 0$.
Write
$\lambda = \frac{1}{{2x}}\left( {1 - \frac{1}{2}\mu x} \right) = \frac{1}{{2y}}\left( {1 - \frac{1}{2}\mu y} \right) = \frac{1}{{2z}}\left( {1 - 8\mu z} \right)$
$\frac{1}{{2x}} - \frac{\mu }{4} = \frac{1}{{2y}} - \frac{\mu }{4}$, ${\ \ }$ $\frac{1}{{2x}} - \frac{\mu }{4} = \frac{1}{{2z}} - 4\mu $
$\frac{1}{{2x}} = \frac{1}{{2y}}$
So, $y=x$.
Substituting $y=x$ in the constraint $g$ gives
$2{x^2} + {z^2} - 9 = 0$
$z = \pm \sqrt {9 - 2{x^2}} $
Substituting $y=x$ and $z = \pm \sqrt {9 - 2{x^2}} $ in $h$ gives
$\frac{1}{2}{x^2} + 4\left( {9 - 2{x^2}} \right) - 9 = 0$
$27 - \frac{{15}}{2}{x^2} = 0$
So, $x = \pm 3\sqrt {\frac{2}{5}} $.
Using $y=x$ and $z = \pm \sqrt {9 - 2{x^2}} $, we obtain the critical points: $\left( {3\sqrt {\frac{2}{5}} ,3\sqrt {\frac{2}{5}} ,\frac{3}{{\sqrt 5 }}} \right)$, $\left( {3\sqrt {\frac{2}{5}} ,3\sqrt {\frac{2}{5}} , - \frac{3}{{\sqrt 5 }}} \right)$, $\left( { - 3\sqrt {\frac{2}{5}} , - 3\sqrt {\frac{2}{5}} ,\frac{3}{{\sqrt 5 }}} \right)$ and $\left( { - 3\sqrt {\frac{2}{5}} , - 3\sqrt {\frac{2}{5}} , - \frac{3}{{\sqrt 5 }}} \right)$.
The extreme values corresponds to the four critical points are listed in the following table
$\begin{array}{*{20}{c}}
{{\rm{Critical{\ }point}}}&{f\left( {x,y,z} \right)}\\
{\left( {3\sqrt {\frac{2}{5}} ,3\sqrt {\frac{2}{5}} ,\frac{3}{{\sqrt 5 }}} \right)}&{\frac{{3 + 6\sqrt 2 }}{{\sqrt 5 }} \simeq 5.14}\\
{\left( {3\sqrt {\frac{2}{5}} ,3\sqrt {\frac{2}{5}} , - \frac{3}{{\sqrt 5 }}} \right)}&{\frac{{ - 3 + 6\sqrt 2 }}{{\sqrt 5 }} \simeq 2.45}\\
{\left( { - 3\sqrt {\frac{2}{5}} , - 3\sqrt {\frac{2}{5}} ,\frac{3}{{\sqrt 5 }}} \right)}&{\frac{{3 - 6\sqrt 2 }}{{\sqrt 5 }} \simeq - 2.45}\\
{\left( { - 3\sqrt {\frac{2}{5}} , - 3\sqrt {\frac{2}{5}} , - \frac{3}{{\sqrt 5 }}} \right)}&{ - \frac{{3 + 6\sqrt 2 }}{{\sqrt 5 }} \simeq - 5.14}
\end{array}$
From these results we conclude that the maximum of $f\left( {x,y,z} \right) = x + y + z$ is $\frac{{3 + 6\sqrt 2 }}{{\sqrt 5 }} \simeq 5.14$ subject to the two constraints ${x^2} + {y^2} + {z^2} = 9$ and $\frac{1}{4}{x^2} + \frac{1}{4}{y^2} + 4{z^2} = 9$.
