The area (in square units) of the feasible region determined by the constraints \(x + y \le 6\), \(x \ge 2\), \(y \ge 1\) is
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \(\displaystyle\frac{9}{2}\)
The region lies to the right of \(x = 2\), above \(y = 1\) and below \(x + y = 6\). The line \(x + y = 6\) meets the other two lines at
\(y = 1\): \(x = 5\), point (5, 1)
\(x = 2\): \(y = 4\), point (2, 4)
So the feasible region is the right-angled triangle with corner points (2, 1), (5, 1) and (2, 4). Its two shorter sides are
\(5 - 2 = 3\) and \(4 - 1 = 3\)
\( \displaystyle \text{Area} = \frac{1}{2} \times 3 \times 3 = \frac{9}{2} \)
Why the other options are incorrect
Option (2) 9 — Incorrect. This is \(3 \times 3\), the area of the square on the two sides, without the factor \(\frac{1}{2}\).
Option (3) 18 — Incorrect. This is the area of the triangle cut off by \(x + y = 6\) from the first quadrant, which ignores \(x \ge 2\) and \(y \ge 1\).
Option (4) 6 — Incorrect. The triangle with sides 3 and 3 has area \(\frac{9}{2}\), not 6.