Quickie

W

WTF?

Junior Member
Joined
Sep 16, 2005
Messages
95
Hi, how are you?

Okay, I got this problem, that to me, it makes little sense.

01231321.jpg


On #28; What does it mean by boundary? Does it mean if the lines intersect? Because if it did, wouldn't the feasible set be two lines?

On #29; THis one is confusing, because in my opinion, they could all work. However, the book says a and c. Can you help please?
 
Hello, WTF?!

I too wondered about #29 . . . then I woke up.

I have a vague memory of "convex sets".
. . That is, the shaded region must be a <u>convex</u> polygon.
The region must not be concave, where a side "caves in"
. . which happens in (b) and (d).


It's hard to explain without drawing pictures, so I'll give it a try.

Suppose we graphed this system:

. . \(\displaystyle 2x\,+\,y\:\leq\:4\)
. . \(\displaystyle x\,+\,2y\:\leq\:4\)
Code: 
      *
      |*
      | *
      |  *
      o   *                 The graph looks like this.
      |::o *
      |:::::o              The shaded region is convex.
      |::::::*  o 
    --+-------*-----o-

Code: 
      *
      |*                    We cannot construct inequalities
      |:*
      |::*                    that give us a concave region.
      |:::*
      |::::*                             (try it!)
      |:::::o          
      |:::::::::o 
    --+-------------o-
 
I would conjecture that one boundry is y=2 but since it says > it would not be included.
 
Y = 2 would be part of the feasible area. Since it says greater than 2, the boundry, y = 2 is not part of the feasible area.
Just as on a line 2<x<4 would be all points between 2 and 4 but NOT including 2 nor 4. They would be the boundry.
 
Y = 2 would be part of the feasible area. Since it says greater than 2, the boundry, y = 2 is not part of the feasible area.
Just as on a line 2<x<4 would be all points between 2 and 4 but NOT including 2 nor 4. They would be the boundry.
 
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