Answer
The point $\left( { - 1,0,2} \right)$ on the ellipse is farthest from the origin.
Work Step by Step
We are given a cylinder ${x^2} + {y^2} = 1$ and a plane $x + z = 1$. The cylinder ${x^2} + {y^2} = 1$ intersects the plane $x + z = 1$ in an ellipse. Our task is to find a point $\left( {x,y,z} \right)$ on the ellipse such that the distance $d = \sqrt {{x^2} + {y^2} + {z^2}} $ is maximum.
Since finding the maximum $d$ is the same as finding the maximum square of the distance ${d^2}$, our problem is to maximize $f\left( {x,y,z} \right) = {x^2} + {y^2} + {z^2}$ subject to the two constraints $g\left( {x,y,z} \right) = {x^2} + {y^2} - 1 = 0$ and $h\left( {x,y,z} \right) = x + z - 1 = 0$.
The Lagrange condition is
$\nabla f = \lambda \nabla g + \mu \nabla h$
$\left( {2x,2y,2z} \right) = \lambda \left( {2x,2y,0} \right) + \mu \left( {1,0,1} \right)$
From the Lagrange condition we obtain three equations:
(1) ${\ \ \ \ }$ $2x = 2\lambda x + \mu $, ${\ \ \ }$ $2y = 2\lambda y$, ${\ \ \ }$ $2z = \mu $
The second equation suggests that $\lambda=1$ or $y=0$. Therefore, we may have the following cases:
Case 1. $\lambda=1$
Substituting $\lambda=1$ in the first equation of (1) gives $\mu = 0$.
Since $\mu = 0$, the third equation of (1) gives $z=0$.
Substituting $z=0$ in the second constraint $h\left( {x,y,z} \right) = x + z - 1 = 0$ gives $x=1$.
Substituting $x=1$ in the first constraint $g\left( {x,y,z} \right) = {x^2} + {y^2} - 1 = 0$ gives $y=0$.
So, the critical point is $\left( {1,0,0} \right)$.
Case 2. $y=0$
Substituting $y=0$ in the first constraint $g\left( {x,y,z} \right) = {x^2} + {y^2} - 1 = 0$ gives $x = \pm 1$.
Substituting $x=1$ and $x=-1$ in the second constraint $h\left( {x,y,z} \right) = x + z - 1 = 0$ gives $z=0$ and $z=2$, respectively.
So, the critical points are $\left( {1,0,0} \right)$ and $\left( { - 1,0,2} \right)$. Notice that the first critical point has been accounted for in the first case above.
The extreme values corresponds to the two critical points are $f\left( {1,0,0} \right) = 1$ and $f\left( { - 1,0,2} \right) = 5$.
From these results we conclude that the maximum of $f$ is $f\left( { - 1,0,2} \right) = 5$ subject to the two constraints $g\left( {x,y,z} \right) = {x^2} + {y^2} - 1 = 0$ and $h\left( {x,y,z} \right) = x + z - 1 = 0$.
Hence, the point $\left( { - 1,0,2} \right)$ on the ellipse is farthest from the origin. Since $f$ is the square of distance, the distance of the point $\left( { - 1,0,2} \right)$ from the origin is $\sqrt 5 $.
