Step 1: Definition:
A redundant constraint in linear programming doesn't affect the feasible region; removing it doesn't change the set of feasible solutions. Redundant constraints are automatically satisfied when other constraints are met.
Step 2: Analysis:
Examine the following constraints:
1. \( x \ge 6 \)
2. \( y \ge 2 \)
3. \( 2x+y \ge 10 \)
4. \( x \ge 0 \)
5. \( y \ge 0 \)
Identify redundant constraints:
- \( x \ge 0 \) is redundant because \( x \ge 6 \) implies \(x \ge 0\).
- \( y \ge 0 \) is redundant because \( y \ge 2 \) implies \(y \ge 0\).
- Consider \( 2x+y \ge 10 \). If \( x \ge 6 \) and \( y \ge 2 \), then the smallest value of \( 2x+y \) is \( 2(6) + 2 = 14 \). Since \( 14>10 \), \( 2x+y \ge 10 \) is satisfied. Thus, \( 2x+y \ge 10 \) is redundant.
The non-redundant constraints are \(x \ge 6\) and \(y \ge 2\). The constraints \(x \ge 0\), \(y \ge 0\), and \(2x+y \ge 10\) are redundant. The question asks for the redundant constraints.
Evaluate the options:
(A) \( x \ge 6, 2x+y \ge 10 \): \(x \ge 6\) is not redundant.
(B) \( 2x+y \ge 10 \): This constraint is redundant.
(C) \( x \ge 6, y \ge 2, x \ge 0, y \ge 0 \): \(x \ge 6\) and \(y \ge 2\) are not redundant.
(D) \( y \ge 2, x \ge 0 \): \(y \ge 2\) is not redundant.
The full set of redundant constraints is \( \{2x+y \ge 10, x \ge 0, y \ge 0\} \). Option (B) lists one of these. It is the best choice.
Step 3: Conclusion:
\( x \ge 0 \) and \( y \ge 0 \) are redundant due to \( x \ge 6 \) and \( y \ge 2 \). \( 2x+y \ge 10 \) is also redundant. Therefore, \( 2x+y \ge 10 \) is a redundant constraint.