Answer
$\frac{{138}}{{11}}$ is the minimum value of $f\left( {x,y,z} \right) = {x^2} + {y^2} + {z^2}$ subject to the two constraints, $g\left( {x,y,z} \right) = x + 2y + z - 3 = 0$ and $h\left( {x,y,z} \right) = x - y - 4 = 0$. This is illustrated in the figure attached.
Work Step by Step
Our task is find the minimum value of $f\left( {x,y,z} \right) = {x^2} + {y^2} + {z^2}$ subject to two constraints, $g\left( {x,y,z} \right) = x + 2y + z - 3 = 0$ and $h\left( {x,y,z} \right) = x - y - 4 = 0$.
The Lagrange condition is
$\nabla f = \lambda \nabla g + \mu \nabla h$
$\left( {2x,2y,2z} \right) = \lambda \left( {1,2,1} \right) + \mu \left( {1, - 1,0} \right)$
From the Lagrange condition we obtain three equations:
(1) ${\ \ \ \ }$ $2x = \lambda + \mu $, ${\ \ \ }$ $2y = 2\lambda - \mu $, ${\ \ \ }$ $2z = \lambda $
Adding the first equation and the second equation of (1) gives
$2x + 2y = 3\lambda $
$\lambda = \frac{{2x + 2y}}{3}$
Using the third equation of (1), we get
$z = \frac{{x + y}}{3}$
Substituting $z = \frac{{x + y}}{3}$ in $g$ gives
$x + 2y + \frac{{x + y}}{3} - 3 = 0$
$\frac{4}{3}x + \frac{7}{3}y - 3 = 0$
(2) ${\ \ \ \ }$ $4x + 7y - 9 = 0$
We have from the second constraint:
(3) ${\ \ \ \ }$ $x - y - 4 = 0$
Solving equations (2) and (3) yields $x = \frac{{37}}{{11}}$, $y = - \frac{7}{{11}}$.
Using $z = \frac{{x + y}}{3}$, we obtain $z = \frac{{10}}{{11}}$.
So, the critical point is $\left( {\frac{{37}}{{11}}, - \frac{7}{{11}},\frac{{10}}{{11}}} \right)$.
The extreme value corresponds to this critical point is $f\left( {\frac{{37}}{{11}}, - \frac{7}{{11}},\frac{{10}}{{11}}} \right) = \frac{{138}}{{11}}$.
Since the function $f\left( {x,y,z} \right) = {x^2} + {y^2} + {z^2}$ is increasing, we conclude that $\frac{{138}}{{11}}$ is the minimum value of $f$ subject to the two constraints, $g\left( {x,y,z} \right) = x + 2y + z - 3 = 0$ and $h\left( {x,y,z} \right) = x - y - 4 = 0$. This is illustrated in the figure attached.
