How to minimize $\max(x_1, x_2)$ and $x_1^2 + 9x_2^2$ subject to constraints? My textbook came up with a solution without explanation. I'm looking for a systematic way of solving the following optimization problems and similar ones (by hand), because I'm drawing a blank:
minimize 

$$\max(x_1, x_2)$$ and after, $$x_1^2 + 9x_2^2$$
subject to 

$$2x_1 + x_2 \ge 1$$
$$x_1 + 3x_2 \ge1$$
$$x_1 \ge 0, x_2 \ge 0$$
BTW the answer is $\left(\dfrac{1}{3}, \dfrac{1}{3}\right)$.
 A: Let us enter in the Karush–Kuhn–Tucker framework, 
with
$$
f(x,y) = -\max(x,y);\\
g_1(x,y) = 2x+y-1;\\
g_2(x,y) = x+3y-1.
$$
The Lagrangian is
$$
L(x,y,\lambda) =
-\max(x,y) + \lambda\cdot g(x,y)
$$
and the solution is, in the region $x\neq y$ (where $L$ is smooth):
$$
\begin{cases}
0&=&-1_{x>y} + 2\lambda_1 + \lambda_2  \\
0&=&-1_{x<y} + \lambda_1 + 3\lambda_2 \\
0&=&\min(\lambda_{i} , g_{i}(x,y)), & (i=1,2).
\end{cases}
$$
Hence, both $\lambda_i =0$
and the system has no solution.
In the region $x = y$,
$\max (x,y) = x$.
$$L(x,y,\lambda) =
-x + \lambda\cdot g(x,y) =: M(x,\lambda).
$$The new system for $M$ is:
$$
\begin{cases}
0&=&-1+ 3\lambda_1 +4\lambda_2\\
0&=&\min(\lambda_{i} , g_{i}(x,x)), & (i=1,2).
\end{cases}
$$
As we can't have $
g_1(x,x)=0=g_2(x,x),
$ either $\lambda_1=0$ or $\lambda_2=0$, then 
$$
x\in\{1/3, 1/4\}.
$$
But if $x=y=1/4$, $2x+y<1$ and the point is not an admissible solution.
Then the solution is
$$
x=y=\frac 13;\\
\min_{g_{1,2}(x,y)\ge 0} \max(x,y)
= -\max_{g_{1,2}(x,y)\ge 0} (- \max(x,y))
\\ = \max(1/3,1/3) = 1/3.
$$

For the second problem,
the equations are 
$$
\begin{cases}
0&=&-2x + 2\lambda_1 + \lambda_2  \\
0&=&-18y + \lambda_1 + 3\lambda_2 \\
0&=&\min(\lambda_{i} , g_{i}(x,y)), & (i=1,2),
\end{cases}
$$
hence  $(\lambda_1,\lambda_2)\neq (0,0).$
If $\lambda_2 = 0$:
$$
\begin{cases}
0&=&-2x + 2\lambda_1   \\
0&=&-18y + \lambda_1  \\
1 &=& 2x+y
\end{cases}
$$
then $x=18y, y=1/37, x = 18/37$. Then $x+3y = 21/37<1$: this is not the solution.
The solution is such as
$$
\begin{cases}
0&=&-2x + \lambda_2   \\
0&=&-18y + 3\lambda_2  \\
1 &=& x+3y
\end{cases}
$$
then $x=3y, y=1/6, x=1/2$.
