Let $S=\{x:x_1+2x_2\le 4\}$. Find the extreme points and directions of S. Can you represent any point in $S$ as a convex combination of its extreme points plus a nonnegative linear combination of its extreme directions?
There is no extreme points. (Seen in another question)
Now I am having trouble finding the directions of $S$ which by definition is $d\in\mathbb R^n$ such that $\forall x\in S ,\ \forall \lambda\ge 0:x+\lambda d\in S$.
So take $d=(x,y)\in\mathbb R^2,\lambda\ge0,\ s=(x_1,x_2)\in S.$
Then we should get $d\lambda+s\in S.$
Thus do we try to solve this $(\lambda x,\lambda y)+(x_1,x_2)\le2-x_1/2$ $?\ \ ?$
I feel lost, help please!
Hint: The domain $S$ is a half-plane bounded by a line, whose equation is $x_1+2x_2=4$ (draw a picture).
If you start at a point in $S$ and move towards the boundary, you will eventually hit it and get out of $S$. That means you're not following a "direction".
If you move parallel to the boundary, of away from it, then you will never get out of $S$. This means you are now following a direction.
So you need to understand which vectors "look towards the boundary" and which don't.