Min and Max with two constraints I'm asked to find the minimum and maximum values of $f(x, y, z) = x^2+y^2+z^2$ given the constraints $x+2y+z=3$ and $x-y=7$. 
I'm pretty sure I need to set up Lagrange equations, giving:
$$2x = A+B$$
$$2y = 2A-B$$
$$2z = A$$
From here, I'm not sure where to go.

Here's how to do it with Lagrange multipliers.
Put $g(x,y,z) = x + 2y + z -3 = 0$ and $h(x,y) = x - y - 7 =0$ as constraints.
Then we have 5 equations


*

*$f_x = 2x = \lambda g_x + \mu h_x = \lambda + \mu$

*$f_y = 2y = \lambda g_y + \mu h_y = 2\lambda  - \mu$

*$f_z = 2z = \lambda g_z + \mu h_z = \lambda + 0 = \lambda$

*$g(x,y,z) = x + 2y + z -3 = 0$

*$h(x,y) = x - y - 7 =0$


Now, from (3) we have $\frac{\lambda}{2} = z$. Using our expressions for $x$ and $y$ from (1) and (2) and substituting them in in (4) and (5), 
we have now


*

*$\frac{\lambda + \mu}{2} + 2\lambda - \mu + \frac{\lambda}{2} -3 = 3\lambda - \frac{\mu}{2} -3 = 0$

*$\frac{\lambda + \mu}{2} - \lambda + \frac{\mu}{2} - 7 = -\frac{\lambda}{2} + \mu -7 = 0$.


Now we have two linear equations, two unknowns. Multiply the first by two and add to the second two get


*

*$\frac{11 \lambda}{2} -13 =0 \rightarrow \lambda = \frac{26}{11}$.


Plugging this into the first equation above, $\mu = \frac{90}{11}$.
Now from (1), (2), (3), we see that our only critical point occurs at the point where $x = \frac{58}{11}, y = \frac{-19}{11}, z= \frac{13}{11}$. 
Plugging this into our equation for $f$, this comes out to $f(\frac{58}{11}, \frac{-19}{11}, \frac{13}{11}) = \frac{3894}{121} = \frac{354}{11}$.
Now, how do we know whether this is the minimum or maximum? To figure it out we just have to produce a point whose function value is less than or greater than $\frac{354}{11}$. If there's one with a higher function value, than this must be a minimum, and if there's one with a lower function value it's a maximum. 
We want points satisfying (4) and (5). For ease, let $z=0$. Then we just pick the point of intersection of the lines $ y = \frac{-x}{2} + \frac{3}{2}$ and $y = x -7$. This point is $(\frac{17}{3}, \frac{-4}{3})$.
Plugging this into $f$, we see $f(\frac{17}{3}, \frac{-4}{3}, 0)= \frac{305}{9} \gt \frac{354}{11} = f(\frac{58}{11}, \frac{-19}{11}, \frac{13}{11})$. So our point must be a minimum.
Then we must have the minimum of $f$ is $f(\frac{58}{11}, \frac{-19}{11}, \frac{13}{11}) = \frac{3894}{121} = \frac{354}{11}$.
