Question

Match List I with List II

List I	List II
A. The region represented by \(x \geq 0, y \geq 0\)	I. no feasible region
B. The region represented by the inequalities \(2x + y \geq 3, x + 2y \geq 6, x,y \geq 0\)	II. 1st quadrant
C. The region represented by the inequalities \(x + 2y \leq 8, 3x + 2y \leq 12, x,y \geq 0\)	III. unbounded
D. The region represented by the inequalities \(x + y \leq 2, 3x + 5y \geq 15, x,y \geq 0\)	IV. bounded

Choose the correct answer from the options given below:

• A. A-I, B-II, C-III, D-IV
• B. A-IV, B-I, C-II, D-III
• C. А-II, В-III, C-IV, D-I
• D. A-III, B-IV, C-I, D-II

Answer 1

The Correct Option is C

To match the regions given in list I with the descriptions in list II, we analyze each set of inequalities to determine the nature of the feasible region.

1. A. The region represented by \(x \geq 0, y \geq 0\):
   • These inequalities represent the first quadrant of the Cartesian plane, where both \(x\) and \(y\) are non-negative.
   Match: II. 1st quadrant
2. B. The region represented by \(2x + y \geq 3, x + 2y \geq 6, x,y \geq 0\):
   • Graphically, these lines may not have a common intersection region that is closed or bounded.
   Match: III. unbounded
3. C. The region represented by \(x + 2y \leq 8, 3x + 2y \leq 12, x,y \geq 0\):
   • These inequalities create a polygon in the first quadrant, giving rise to a bounded solution region.
   Match: IV. bounded
4. D. The region represented by \(x + y \leq 2, 3x + 5y \geq 15, x,y \geq 0\):
   • These constraints do not provide any feasible region, as the inequalities are contradictory.
   Match: I. no feasible region

Thus, the correct option is: А-II, В-III, C-IV, D-I.