Answer
The maximum of $f\left( {x,y,z} \right) = z$ subject to the two constraints ${x^2} + {y^2} = 1$ and $x + y + z = 1$ is $1 + \sqrt 2 $.
Work Step by Step
In this exercise our task is to maximize the function $f\left( {x,y,z} \right) = z$ subject to the two constraints $g\left( {x,y,z} \right) = {x^2} + {y^2} - 1 = 0$ and $h\left( {x,y,z} \right) = x + y + z - 1 = 0$.
The Lagrange condition is
$\nabla f = \lambda \nabla g + \mu \nabla h$
$\left( {0,0,1} \right) = \lambda \left( {2x,2y,0} \right) + \mu \left( {1,1,1} \right)$
From the Lagrange condition we obtain three equations:
$0 = 2\lambda x + \mu $, ${\ \ }$ $0 = 2\lambda y + \mu $, ${\ \ }$ $1 = \mu $
Substituting $\mu = 1$ in the first and the second equation gives
$2\lambda x + 1 = 0$, ${\ \ \ }$ $2\lambda y + 1 = 0$
These equations imply that $x \ne 0$, $y \ne 0$ and $\lambda \ne 0$. So,
$\lambda = - \frac{1}{{2x}} = - \frac{1}{{2y}}$
So, $y=x$.
Substituting $y=x$ in $g$ gives
$2{x^2} - 1 = 0$
$x = \pm \frac{1}{{\sqrt 2 }}$
Using $y=x$, we obtain $y = \pm \frac{1}{{\sqrt 2 }}$.
Substituting $x = \frac{1}{{\sqrt 2 }}$ and $y = \frac{1}{{\sqrt 2 }}$ in $h$ gives
$\sqrt 2 + z - 1 = 0$
$z = 1 - \sqrt 2 $
Substituting $x = - \frac{1}{{\sqrt 2 }}$ and $y = - \frac{1}{{\sqrt 2 }}$ in $h$ gives
$ - \sqrt 2 + z - 1 = 0$
$z = 1 + \sqrt 2 $
So, the critical points are $\left( {\frac{1}{{\sqrt 2 }},\frac{1}{{\sqrt 2 }},1 - \sqrt 2 } \right)$ and $\left( { - \frac{1}{{\sqrt 2 }}, - \frac{1}{{\sqrt 2 }},1 + \sqrt 2 } \right)$.
The extreme values corresponds to the two critical points are
$f\left( {\frac{1}{{\sqrt 2 }},\frac{1}{{\sqrt 2 }},1 - \sqrt 2 } \right) = 1 - \sqrt 2 $
$f\left( { - \frac{1}{{\sqrt 2 }}, - \frac{1}{{\sqrt 2 }},1 + \sqrt 2 } \right) = 1 + \sqrt 2 $
From these results we conclude that the maximum of $f\left( {x,y,z} \right) = z$ subject to the two constraints ${x^2} + {y^2} = 1$ and $x + y + z = 1$ is $1 + \sqrt 2 $.
