Answer
The minimum of $f\left( {x,y,z} \right) = y + 2z$ is $8 - \sqrt {17} $ and the maximum is $8 + \sqrt {17} $ subject to the two constraints $2x+z=4$ and ${x^2} + {y^2} = 1$.
Work Step by Step
Our task is to find the minimum and maximum of $f\left( {x,y,z} \right) = y + 2z$ subject to two constraints, $g\left( {x,y,z} \right) = 2x + z - 4 = 0$ and $h\left( {x,y,z} \right) = {x^2} + {y^2} - 1 = 0$.
The Lagrange condition is
$\nabla f = \lambda \nabla g + \mu \nabla h$
$\left( {0,1,2} \right) = \lambda \left( {2,0,1} \right) + \mu \left( {2x,2y,0} \right)$
From the Lagrange condition we obtain three equations:
(1) ${\ \ \ \ }$ $0 = 2\lambda + 2\mu x$, ${\ \ \ }$ $1 = 2\mu y$, ${\ \ \ }$ $2 = \lambda $
Since $\lambda = 2$, these equations imply that $x \ne 0$, $y \ne 0$, $z \ne 0$ and $\mu \ne 0$.
Using $\lambda = 2$, equation (1) becomes
$0 = 4 + 2\mu x$, ${\ \ \ }$ $1 = 2\mu y$
$\mu = - \frac{2}{x} = \frac{1}{{2y}}$
So, $y = - \frac{1}{4}x$.
Substituting $y = - \frac{1}{4}x$ in the constraint $h$ gives
${x^2} + \frac{1}{{16}}{x^2} - 1 = 0$
$\frac{{17}}{{16}}{x^2} = 1$
So, $x = \pm \frac{4}{{\sqrt {17} }}$.
The first constraint $g\left( {x,y,z} \right) = 2x + z - 4 = 0$ gives $z=4-2x$.
Using $y = - \frac{1}{4}x$ and $z=4-2x$, we obtain the critical points: $\left( {\frac{4}{{\sqrt {17} }}, - \frac{1}{{\sqrt {17} }},4 - \frac{8}{{\sqrt {17} }}} \right)$ and $\left( { - \frac{4}{{\sqrt {17} }},\frac{1}{{\sqrt {17} }},4 + \frac{8}{{\sqrt {17} }}} \right)$.
The extreme values corresponds to the two critical points are listed in the following table
$\begin{array}{*{20}{c}}
{{\rm{Critical{\ }point}}}&{f\left( {x,y,z} \right)}\\
{\left( {\frac{4}{{\sqrt {17} }}, - \frac{1}{{\sqrt {17} }},4 - \frac{8}{{\sqrt {17} }}} \right)}&{8 - \sqrt {17} \simeq 3.88}\\
{\left( { - \frac{4}{{\sqrt {17} }},\frac{1}{{\sqrt {17} }},4 + \frac{8}{{\sqrt {17} }}} \right)}&{8 + \sqrt {17} \simeq 12.12}
\end{array}$
From these results we conclude that the minimum of $f\left( {x,y,z} \right) = y + 2z$ is $8 - \sqrt {17} $ and the maximum is $8 + \sqrt {17} $ subject to the two constraints $2x+z=4$ and ${x^2} + {y^2} = 1$.
