Question

For the given linear programming problem,
Minimum Z = 6x + 10y
subject to the constraints
\( x \ge 6 \); \( y \ge 2 \); \( 2x+y \ge 10 \); \( x,y \ge 0 \),
the redundant constraints are:

• A. \( x \ge 6, 2x+y \ge 10 \)
• B. \( 2x+y \ge 10 \)
• C. \( x \ge 6, y \ge 2, x \ge 0, y \ge 0 \)
• D. \( y \ge 2, x \ge 0 \)

Hint:

To spot redundant constraints, look for implications. If one constraint (e.g., \(x \ge 6\)) is stricter than another (e.g., \(x \ge 0\)), the less strict one is redundant. Also, check combinations of constraints. If two constraints \(A\) and \(B\) together imply a third constraint \(C\), then \(C\) is redundant.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
A redundant constraint is a constraint in a linear programming problem that does not affect the feasible region. In other words, if we remove a redundant constraint, the set of feasible solutions remains the same. We can identify redundant constraints by checking if they are automatically satisfied whenever the other constraints are met.

Step 2: Detailed Explanation:
Let's analyze the given constraints: 1. \( x \ge 6 \) 2. \( y \ge 2 \) 3. \( 2x+y \ge 10 \) 4. \( x \ge 0 \) 5. \( y \ge 0 \) Let's check for redundancy: - Consider constraint \( x \ge 0 \). If \( x \ge 6 \) is satisfied, then \(x\) is certainly greater than 0. So, \( x \ge 0 \) is redundant. - Consider constraint \( y \ge 0 \). If \( y \ge 2 \) is satisfied, then \(y\) is certainly greater than 0. So, \( y \ge 0 \) is redundant. - Consider constraint \( 2x+y \ge 10 \). Let's see if the other "stronger" constraints \( x \ge 6 \) and \( y \ge 2 \) imply this one. If \( x \ge 6 \) and \( y \ge 2 \), then the smallest value the expression \( 2x+y \) can take is when \(x\) and \(y\) are at their minimums: \[ 2x+y \ge 2(6) + 2 = 12+2 = 14 \] Since \( 14>10 \), any point \( (x,y) \) that satisfies \( x \ge 6 \) and \( y \ge 2 \) will automatically satisfy \( 2x+y \ge 14 \), which means it will definitely satisfy \( 2x+y \ge 10 \). Therefore, the constraint \( 2x+y \ge 10 \) is made redundant by the constraints \( x \ge 6 \) and \( y \ge 2 \). The set of non-redundant constraints is just \(x \ge 6\) and \(y \ge 2\). The constraints \(x \ge 0\), \(y \ge 0\), and \(2x+y \ge 10\) are all redundant. The question asks for "the redundant constraints" from the options. Let's evaluate the options: (A) \( x \ge 6, 2x+y \ge 10 \): \(x \ge 6\) is a defining constraint of the final feasible region, not redundant. (B) \( 2x+y \ge 10 \): As shown above, 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 question is slightly ambiguous. 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 among the given options.

Step 3: Final Answer:
The constraints \( x \ge 0 \) and \( y \ge 0 \) are made redundant by \( x \ge 6 \) and \( y \ge 2 \). The constraint \( 2x+y \ge 10 \) is also made redundant by \( x \ge 6 \) and \( y \ge 2 \). Therefore, \( 2x+y \ge 10 \) is a redundant constraint.