The feasible region for the constrains x ≥ 0,\ x+y ≤ 1 and x-y ≤ 1 ,…

The correct answer is Option (2) → (B) and (D) Only

Write each constraint as a bound on y:

    \(x + y \le 1 \Rightarrow y \le 1 - x\)
    \(x - y \le 1 \Rightarrow y \ge x - 1\)

So for each x, y lies between \(x - 1\) and \(1 - x\). This needs \(x - 1 \le 1 - x\), that is \(x \le 1\). Together with \(x \ge 0\):

    \(0 \le x \le 1,\quad x - 1 \le y \le 1 - x\)

The feasible region is the triangle with corners (0, 1), (1, 0) and (0, −1). Its part above the x-axis (y > 0) lies in the first quadrant, and its part below the x-axis (y < 0) lies in the fourth quadrant; since \(x \ge 0\), no point is on the left of the y-axis.

So the region lies in the I and IV quadrants and has no point in the II or III quadrant.

Checking the statements

(A) I and II Quadrant — Incorrect. The II quadrant needs x < 0, but \(x \ge 0\).

(B) I Quadrant — Correct. For example (0.5, 0.25) satisfies all three constraints.

(C) II and IV Quadrant — Incorrect. No point of the region has x < 0, so none is in the II quadrant.

(D) IV Quadrant — Correct. For example (0.5, −0.25) satisfies all three constraints.

(E) I, II, III and IV Quadrant — Incorrect. The II and III quadrants need x < 0, which \(x \ge 0\) rules out.