extreme points and representing

Asked Feb 11 '15 at 21:39

Active Feb 11 '15 at 22:52

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$$X=\{(x_1,x_2)^T : x_1-x_2\leq3, 2x_1+x_2\leq 4, x_1\geq -3\}$$ Find all extreme points of $X$, and represent $x^*= {0\choose1}$ as a convex combination of those extreme points.

I sketched it out with $x_2$ being the vertical axis and $x_1$ being the horizontal and found that the intersections or 'corners' were $(-3,10)^T$, $(-3,-6)^T$ and $(\frac73,\frac{-2}3)^T$. Are these the extreme points? How do I exactly answer the question?

Am I supposed to find a linear combination of $x^*= {0\choose1}$ using these points that I found? As in let the points I found be $p$, $q$ and $r$ respectively. So am I meant to find $Ap+Bq+Cr=x^*$, where $A,B,C\geq0$?

edited Feb 11 '15 at 22:52

asked Feb 11 '15 at 21:39

snowman

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Almost yes, but you have one more condition on $A,B,C$: $A+B+C=1$. – Stefan Gyürki Feb 13 '15 at 08:25
Why is that? Is that included iv the definition? – snowman Feb 13 '15 at 09:28
Yes, please check how "convex combination" is defined. – Stefan Gyürki Feb 13 '15 at 11:28
So all I have to find is A, B and C with that other condition right? – snowman Feb 16 '15 at 19:14
Yes, you have to find a convex combination. Since that point is inside the triangle given by those three points, it has a solution. – Stefan Gyürki Feb 16 '15 at 21:55

extreme points and representing

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