Linear Programming Question 131
The solution set of the following system of inequalities: $ x+2y\le 3, $
$ 3x+4y\ge 12, $
$ x\ge 0, $
$ y\ge 1, $ is
Options:
A) Bounded region
B) Unbounded region
C) Only one point is valid
D) Empty set
Show Answer
Answer:
Correct Answer: D
Solution:
The solution region is bounded by the straight lines $ x+2y=3 $ (1) $ 3x+4y=12 $ (2) $ x=0 $ (3) $ y=1 $ (4) The straight lines (1) and (2) meet the x-axis in (1.5, 0) and (4, 0) and for (0, 0), $ x+2y\le 3\Rightarrow 0\le 3 $ which is true.
Hence (0, 0) lies in the half plane $ x+2y\le 3. $ Also the lines (1) and (2) meet the y-axis in (0, 3/2) and (0, 3) and for (0, 0) $ 3x+4y\ge 12\Rightarrow 0\ge 12 $ which is not true.
Hence (0, 0) belongs to the half plane $ 3x+4y\ge 0. $
Also $ x\ge 0,y\ge 1\Rightarrow $ that the solution set belongs to the first quadrant.
Moreover all the boundary lines are part of the solution.
From the shaded region, we find that there is no solution of the given system.
Hence the solution set is an empty set.
BETA
