Answer
The minimum of $f\left( {x,y,z} \right) = x + 2y + 3z$ is $2 - \frac{2}{{\sqrt 3 }}$ and the maximum is $2 + \frac{2}{{\sqrt 3 }}$ subject to the two constraints $x + y + z = 1$ and ${x^2} + {y^2} + {z^2} = 1$.
Work Step by Step
Our task is to find the extreme values of $f\left( {x,y,z} \right) = x + 2y + 3z$ subject to two constraints, $g\left( {x,y,z} \right) = x + y + z - 1 = 0$ and $h\left( {x,y,z} \right) = {x^2} + {y^2} + {z^2} - 1 = 0$.
The Lagrange condition is
$\nabla f = \lambda \nabla g + \mu \nabla h$
$\left( {1,2,3} \right) = \lambda \left( {1,1,1} \right) + \mu \left( {2x,2y,2z} \right)$
From the Lagrange condition we obtain three equations:
(1) ${\ \ \ }$ $1 = \lambda + 2\mu x$, ${\ \ }$ $2 = \lambda + 2\mu y$, ${\ \ }$ $3 = \lambda + 2\mu z$
So, $\lambda = 1 - 2\mu x = 2 - 2\mu y = 3 - 2\mu z$.
From here, we obtain
$1 - 2\mu x = 2 - 2\mu y$, ${\ \ }$ $2 - 2\mu y = 3 - 2\mu z$, ${\ \ }$ $1 - 2\mu x = 3 - 2\mu z$
$2\mu \left( {y - x} \right) = 1$, ${\ \ \ }$ $2\mu \left( {z - y} \right) = 1$, ${\ \ \ }$ $2\mu \left( {z - x} \right) = 2$
Notice that $\mu $ cannot be zero, otherwise equation (1) does not have solutions. Thus, we can write
$\mu = \frac{1}{{2\left( {y - x} \right)}} = \frac{1}{{2\left( {z - y} \right)}} = \frac{1}{{z - x}}$
From the first two equations on the right-hand side of $\mu $, we get
$2\left( {y - x} \right) = 2\left( {z - y} \right)$
$y - x = z - y$
(2) ${\ \ \ \ }$ $x - 2y + z = 0$
We have from the constraint:
(3) ${\ \ \ \ }$ $x + y + z = 1$
Subtracting equation (2) from equation (3) gives $3y = 1$.
So, $y = \frac{1}{3}$.
Substituting $y = \frac{1}{3}$ in equation (3) gives
$x + \frac{1}{3} + z = 1$
$z = - x + \frac{2}{3}$
Substituting $y = \frac{1}{3}$ and $z = - x + \frac{2}{3}$ in $h\left( {x,y,z} \right) = {x^2} + {y^2} + {z^2} - 1 = 0$ gives
${x^2} + {\left( {\frac{1}{3}} \right)^2} + {\left( { - x + \frac{2}{3}} \right)^2} - 1 = 0$
$2{x^2} - \frac{{4x}}{3} - \frac{4}{9} = 0$
$9{x^2} - 6x - 2 = 0$
$x = \frac{{6 \pm \sqrt {{{\left( { - 6} \right)}^2} - 4\cdot9\cdot\left( { - 2} \right)} }}{{2\cdot9}} = \frac{{6 \pm \sqrt {108} }}{{18}} = \frac{{1 \pm \sqrt 3 }}{3}$
Using $y = \frac{1}{3}$ and $z = - x + \frac{2}{3}$, we obtain the critical points: $\left( {\frac{{1 + \sqrt 3 }}{3},\frac{1}{3},\frac{{1 - \sqrt 3 }}{3}} \right)$ and $\left( {\frac{{1 - \sqrt 3 }}{3},\frac{1}{3},\frac{{1 + \sqrt 3 }}{3}} \right)$.
The extreme values corresponds to these critical points are listed in the following table:
$\begin{array}{*{20}{c}}
{{\rm{Critical{\ }point}}}&{f\left( {x,y,z} \right)}\\
{\left( {\frac{{1 + \sqrt 3 }}{3},\frac{1}{3},\frac{{1 - \sqrt 3 }}{3}} \right)}&{2 - \frac{2}{{\sqrt 3 }} \simeq 0.85}\\
{\left( {\frac{{1 - \sqrt 3 }}{3},\frac{1}{3},\frac{{1 + \sqrt 3 }}{3}} \right)}&{2 + \frac{2}{{\sqrt 3 }} \simeq 3.15}
\end{array}$
From these results we conclude that the minimum of $f\left( {x,y,z} \right) = x + 2y + 3z$ is $2 - \frac{2}{{\sqrt 3 }}$ and the maximum is $2 + \frac{2}{{\sqrt 3 }}$ subject to the two constraints $x + y + z = 1$ and ${x^2} + {y^2} + {z^2} = 1$.
