Extreme directions of $S=\{x:-x_1+2x_2\le 4,x_1-3x_2\le 3,x_1,x_2\ge 0\}$

Question

Consider the set $S=\{x:-x_1+2x_2\le 4,x_1-3x_2\le 3,x_1,x_2\ge 0\}$. Identify all extreme points and extreme directions of $S$. Represent the point $(4,1)^t$ as a convex combination of the extreme points plus a nonnegative combination of the extreme directions.

Recall:

Definition: $d\in \mathbb R^n$ is a extreme direction of $S$ if $d=\lambda_1d_1+\lambda_2d_2,$ then $\lambda d_1=d_2.$, where $d_1,d_2$ are directions of $S.$ and $\lambda>0,\lambda_i>0$

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This is what I've done:

I draw the region and I found 3 extreme points $(0,0),(0,2),(3,0).$

And I don't know how to identify the extreme directions, I don't even see it geometrically.

How can I see it geometrically?

For instance the extreme points are "the vertex of the figure". So the extreme directions are ... ?

And finally, how do I prove it?

Answer 1

Check out the graph at https://www.desmos.com/calculator/sarf5jl1cq

Extreme directions are $(2,1)$ and $(3,1)$, which are the direction vectors emanating from the two nonzero extreme points in the boundary.