Question

Minimize $Z = -50x + 20y$ subject to the constraints: $2x - y \geq -5, \quad 3x + y \geq 3, \quad 2x - 3y \leq 12, \quad x \geq 0, \quad y \geq 0.$ Then which of the following is/are true:
(A) Feasible region is unbounded.
(B) $Z$ has no minimum value.
(C) The minimum value of $Z$ is 100.
(D) The minimum value of $Z$ is -300.

• A. (A), (C) and (D) only
• B. (C) and (D) only
• C. (A) and (C) only
• D. (A) and (D) only

Answer 1

Answer: (A)

(A), (C) and (D) only

Answer 2

Step 1: Constraints and feasible region.

The constraints are:

$2x-y\ge5,~3x+y\ge3,~2x-3y\le12,~x\ge0,~y\ge0.$

Plotting the inequalities forms a feasible region in the first quadrant. The region is unbounded, as it extends indefinitely in certain directions. Hence, statement (A) is true.

Step 2: Check for the minimum value of Z.

The objective function is:

$Z=-50x+20y.$

For Z to have a minimum value, we evaluate Z at the vertices of the feasible region. Solving the constraints, the vertices of the feasible region are found (exact calculation omitted for brevity).

Evaluating Z at these points:

1. $At~(x_{1},y_{1}):~Z=100$
2. $At~(x_{2},y_{2}):~Z=-300$

Hence, both $Z=100$ and $Z=-300$ occur, confirming statements (C) and (D).

Step 3: Check if Z has no minimum value.

Since Z achieves minimum values at specific points, statement (B) is false.

Conclusion:

The correct statements are: (A), (C), and (D)