Find the highest and lowest points on the ellipse of intersection of the cylinder $x^2+y^2 = 1$ and the plane $x+y+z=1$ 
Find the highest and lowest points on the ellipse of intersection of the cylinder $x^2+y^2 = 1$ and the plain $x+y+z=1$

Hi i was doing this question but i'm not sure i was right. Does this make sense?
Let
$$f(x) = x^2+y^2 = 1$$
$$f(y) = x+y+z=1, \quad \text{our constraint}$$
$$ \nabla f(x) = \lambda \nabla g(x) $$
$f_x = 2x = \lambda$ ..... (1)
$f_y = 2y = \lambda$ ...... (2)
$f_z = 0 = \lambda$
(1) = (2)
$2x = 2y = \lambda = 0 $
$x=y$
Sub into f(x) 
$$f(x) = x^2 + x^2 = 1$$ $$f(x) = 2x^2 = 1$$ $$f(x) = x = \pm \frac{1}{\sqrt{2}} $$ $$ \therefore \, y =  \pm \frac{1}{\sqrt{2}}$$
Points: $\left(\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}} 
\right), \quad \left(-\frac{1}{\sqrt{2}}, -\frac{1}{\sqrt{2}} \right)$ 
Sub these points into g
$$g\left(\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}, z \right) = \frac{2}{\sqrt{2}} + z = 10 $$ 
$$ z = 1 - \frac{2}{\sqrt{2}} $$
$$g\left(-\frac{1}{\sqrt{2}}, -\frac{1}{\sqrt{2}}, z \right) = -\frac{2}{\sqrt{2}} + z = 10 $$ 
$$ z = 1 + \frac{2}{\sqrt{2}} $$
So highest and lowest points are....
$$\left(\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}, 1 - \frac{2}{\sqrt{2}}  \right) \quad lowest$$
$$\left(-\frac{1}{\sqrt{2}}, -\frac{1}{\sqrt{2}}, 1 + \frac{2}{\sqrt{2}}  \right) \quad highest$$

By inequality $2x^2+2y^2\geq (x+y)^2$  we indeed have $-\sqrt2 \leq x+y \leq \sqrt{2}$ and to maximize/minimize $z$ we need to minimize/maximize $x+y$ so your solution until $x=y=\pm{1\over\sqrt2}$ is correct.
However how do you get $z$ looks unreasonable as $x+y+z=1$ but your solution, for example ${1\over\sqrt2}+{1\over\sqrt2}+{1\over\sqrt8}$ does not equal $1$.
You should simply calculate $z=1-x-y=1+{\sqrt2}$ for maximum and $1-\sqrt2$ for minimum.
There are two major mistakes after a look at your way of finding z:
(1) $g(x,y,z)=1\neq10$ as stated in the question. I do not know where the $10$ comes from.
(2) $10-{2\over \sqrt2}$ is not $8\over\sqrt2$. You cannot calculate irrational numbers like this.
As to answer the comment, what Matthew is talking about is the general way of solving multiple constraints. 
We want to maximize $h(x,y,z)=z$ using constraint $g(x,y,z)=x+y+z=1$ and $f(x,y,z)=s^2+y^2=1$. Let the first multiplier be $\lambda$ and the second be $u$
Then we have the system:
$0=\lambda+2ux=\lambda+2uy$
$1=\lambda$
$x^2+y^2=1$
$x+y+z=1$
By solving the equations we will find the value of $z$.
