The feasible region determined by the constraints \(x + 2y \le 10\), \(2x + y \le 11\), \(x + y \ge k\), \(x \ge 0\), \(y \ge 0\) is non-empty if and only if
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \(k \le 7\)
The constraint \(x + y \ge k\) can be met by some point of the region R given by the other constraints exactly when the largest value of \(x + y\) on R is at least k.
The lines \(x + 2y = 10\) and \(2x + y = 11\) meet where
\(3x + 3y = 21\), so \(x + y = 7\)
\(x - y = 1\), so \(x = 4, \; y = 3\)
The corners of R are (0, 0), \(\left(\frac{11}{2}, 0\right)\), (4, 3) and (0, 5). Values of \(x + y\):
\(0, \; \frac{11}{2}, \; 7, \; 5\)
So the largest value of \(x + y\) on R is 7, at (4, 3). The region is non-empty if and only if \(k \le 7\); for \(k = 7\) it is the single point (4, 3), and for \(k \gt 7\) it is empty.
Why the other options are incorrect
Option (1) \(k \le 5\) — Incorrect. For \(k = 6\) the point (4, 3) is still feasible, as \(4 + 3 = 7 \ge 6\).
Option (2) \(k \le \displaystyle\frac{11}{2}\) — Incorrect. This uses only the corner on the x-axis; (4, 3) is feasible for k up to 7.
Option (4) \(k \le 10\) — Incorrect. For \(k = 8\), no point of R has \(x + y \ge 8\), so the region is empty.