# Get the set of every direction of S

• Oct 21st 2011, 08:13 AM
Lolyta
Get the set of every direction of S
Well, d is a direction of a set S if for all $\displaystyle x \in{S}, \lambda \geq{0}, x+\lambda d \in{S}$.
Let $\displaystyle S=\left\{{(x_1,x_2): x_1-2x_2 \geq{-6}, x_1-x_2\geq{-2}, x_1\geq{0}, x_2 \geq{1}}\right\}$ I have to get the set of every direction of S.

And I have the answer but I don't know how to reach it. The answer is: d has two components $\displaystyle d=(d_1,d_2), d_1 = 2\alpha + \beta; d_2=\alpha: \alpha, \beta \geq{0}$. Could anyone help me?
Thanks.
• Oct 21st 2011, 02:40 PM
Deveno
Re: Get the set of every direction of S
a direction is not a number. it could be a vector (like say, a unit vector).

your set S is a subset of the first quadrant. it is bounded below by $\displaystyle x_2 = 1$, bounded on the left by $\displaystyle x_1 = 0$ and bounded above by

$\displaystyle x_2 = x_1 + 2$ and $\displaystyle x_2 = \dfrac{1}{2}x_1 + 3$.

of these two lines, the second one has the lesser slope, and eventually is the more restrictive.

so for $\displaystyle (x_1,x_2) + \lambda(d_1,d_2) \in S$, whenever $\displaystyle (x_1,x_2) \in S$ for all $\displaystyle \lambda > 0$ we have to have:

$\displaystyle d_1 \geq 0,\ d_2 \leq \dfrac{1}{2}d_1$

that is, $\displaystyle (d_1,d_2)$ must lie between the rays$\displaystyle (\alpha,0)$ and $\displaystyle (\alpha,\alpha/2), \alpha \geq 0$.
• Oct 23rd 2011, 07:46 AM
Lolyta
Re: Get the set of every direction of S
Quote:

Originally Posted by Deveno
a direction is not a number. it could be a vector (like say, a unit vector).

your set S is a subset of the first quadrant. it is bounded below by $\displaystyle x_2 = 1$, bounded on the left by $\displaystyle x_1 = 0$ and bounded above by

$\displaystyle x_2 = x_1 + 2$ and $\displaystyle x_2 = \dfrac{1}{2}x_1 + 3$.

of these two lines, the second one has the lesser slope, and eventually is the more restrictive.

so for $\displaystyle (x_1,x_2) + \lambda(d_1,d_2) \in S$, whenever $\displaystyle (x_1,x_2) \in S$ for all $\displaystyle \lambda > 0$ we have to have:

$\displaystyle d_1 \geq 0,\ d_2 \leq \dfrac{1}{2}d_1$

that is, $\displaystyle (d_1,d_2)$ must lie between the rays$\displaystyle (\alpha,0)$ and $\displaystyle (\alpha,\alpha/2), \alpha \geq 0$.

Thanks for answering but I don't see how do you get those inequalities: $\displaystyle d_1 \geq 0,\ d_2 \leq \dfrac{1}{2}d_1$.

Thank you again.
• Oct 23rd 2011, 08:24 AM
Deveno
Re: Get the set of every direction of S
it might help if you draw a picture of S.

what happens if we add $\displaystyle (d_1,d_2)$ to the point (0,1) (which is in the lower left corner of S), if $\displaystyle d_1 < 0$? isn't the first coordinate going to shift to the left? that will put us outside of S. so we know we can only move to the right.

that leaves up-and-down. now, i should have included the condition $\displaystyle d_2 \geq 0$ as well, because if we go down, the second coordinate could drop below 1, and that will cause us to leave S as well.

so, we can only go up and to the right. well that's 3/4ths of the possible directions eliminated. you can think of this as all slopes between 0 and $\displaystyle \infty$ going up and to the right.

but now, if the slope of our ray $\displaystyle \lambda(d_1,d_2)$ is greater than 1/2, we'll eventually cross the line:

$\displaystyle x_2 = \frac{1}{2}x_1 + 3$, which will take us outside of S. at best, we can only go parallel to it. this limits which directions we can go, for $\displaystyle (d_1,d_2)$ to work,

$\displaystyle d_2$ has to be less than or equal to 1/2 of $\displaystyle d_1$.

it's a lot easier to understand all this, if you draw S, because then you can SEE it.