In each system of List-I, the variables also satisfy \(x \ge 0,\ y \ge 0\). Match each system with the corner points of its feasible region in List-II.
| List-I | List-II | ||
|---|---|---|---|
| Constraints | Corner points | ||
| (A) | \(x + y \le 4,\ x \le 3\) | (I) | (0, 0), (2, 0), (3, 1), (0, 4) |
| (B) | \(x + y \le 4,\ y \le 3\) | (II) | (0, 0), (4, 0), (1, 3), (0, 3) |
| (C) | \(x + y \le 4,\ x - y \le 2\) | (III) | (0, 0), (4, 0), (1, 3), (0, 2) |
| (D) | \(x + y \le 4,\ y - x \le 2\) | (IV) | (0, 0), (3, 0), (3, 1), (0, 4) |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → (A)-(IV), (B)-(II), (C)-(I), (D)-(III)
Each feasible region lies under the line \(x + y = 4\) in the first quadrant, with one more line cutting off a part of it. The corner points are the points where two boundary lines meet inside the region.
(A) \(x + y \le 4,\ x \le 3\)
This matches (IV). The line \(x = 3\) cuts the x-axis at (3, 0) and meets \(x + y = 4\) at
\(y = 4 - 3 = 1\), the point (3, 1)
so the corners are (0, 0), (3, 0), (3, 1), (0, 4).
(B) \(x + y \le 4,\ y \le 3\)
This matches (II). The line \(y = 3\) cuts the y-axis at (0, 3) and meets \(x + y = 4\) at
\(x = 4 - 3 = 1\), the point (1, 3)
so the corners are (0, 0), (4, 0), (1, 3), (0, 3).
(C) \(x + y \le 4,\ x - y \le 2\)
This matches (I). The line \(x - y = 2\) cuts the x-axis at (2, 0). Adding \(x + y = 4\) and \(x - y = 2\):
\(2x = 6\)
\(x = 3,\ y = 1\)
so the corners are (0, 0), (2, 0), (3, 1), (0, 4).
(D) \(x + y \le 4,\ y - x \le 2\)
This matches (III). The line \(y - x = 2\) cuts the y-axis at (0, 2). Adding \(x + y = 4\) and \(y - x = 2\):
\(2y = 6\)
\(y = 3,\ x = 1\)
so the corners are (0, 0), (4, 0), (1, 3), (0, 2).