Question

Objective function \(z=30x-30y \) is subject to which combination of constraints, with feasible solution shown in the figure.

(A)\(x \geq 0, \quad y \geq 0, \quad x \leq 15\)
(B)\(y \leq 20, \quad x + y \leq 30\)
(C)\(x + y \leq 30, \quad x + y \leq 15, \quad 2x - y \leq 5\)
(D)\(2x + y \leq 30, \quad x + y \leq 15, \quad x > 15\)
(E)\(3x + y \leq 30, \quad x + 3y \leq 15, \quad y \geq 20\)
Choose the correct answer from the options given below:
 

• A. (A), (B) and (C) Only
• B. (A) and (B) Only
• C. (A) and (D) Only
• D. (A) and (E) Only

Answer 1

The Correct Option is B

To determine which constraints allow a feasible solution for the objective function \( z = 30x - 30y \), we must analyze each option's constraints to see whether they define a valid, non-empty region in the coordinate plane.

1. Option (A): \( x \geq 0 \), \( y \geq 0 \), \( x \leq 15 \)
   This defines a vertical strip in the first quadrant, bounded on the right by \( x = 15 \). It is a feasible, non-empty region.
2. Option (B): \( y \leq 20 \), \( x + y \leq 30 \)
   This defines a region below the line \( y = 20 \) and below the line \( x + y = 30 \). This region is also feasible and bounded.
3. Option (C): \( x + y \leq 30 \), \( x + y \leq 15 \), \( 2x - y \leq 5 \)
   The first two constraints both bound \( x + y \), with \( x + y \leq 15 \) being the tighter constraint. Now, testing a point like \( (x, y) = (5, 10) \) gives \( x + y = 15 \) and \( 2x - y = 0 \leq 5 \), which works. So this region is feasible.
4. Option (D): \( 2x + y \leq 30 \), \( x + y \leq 15 \), \( x > 15 \)
   The condition \( x > 15 \) contradicts \( x + y \leq 15 \) because even if \( y = 0 \), \( x \) would need to be at most 15. Thus, this region is infeasible.
5. Option (E): \( 3x + y \leq 30 \), \( x + 3y \leq 15 \), \( y \geq 20 \)
   The constraint \( y \geq 20 \) forces a large value of \( y \), but then \( x + 3y \leq 15 \) becomes impossible. Even for \( y = 20 \), we get \( x + 60 \leq 15 \Rightarrow x \leq -45 \), which is not feasible. So this region is also infeasible.

✅ Final Answer: The objective function has feasible solutions under constraint sets:
(A), (B), and (C)