Question

The minimum value of \( Z = 3x + 5y \) subject to the constraints: \[ x + y \leq 2, \quad x \geq 0, \quad y \geq 0 \] is:

• A. 16
• B. 15
• C. 0
• D. none of these

Hint:

In linear programming problems, always evaluate the objective function at the vertices of the feasible region to find the minimum or maximum value.

Answer 1

The Correct Option is C

We are given the linear programming problem with the objective function: \[ Z = 3x + 5y \] subject to the constraints: \[ x + y \leq 2, \quad x \geq 0, \quad y \geq 0 \] Step 1: Plot the constraints - The line \( x + y = 2 \) is the boundary of the constraint \( x + y \leq 2 \). - The constraints \( x \geq 0 \) and \( y \geq 0 \) restrict the solution to the first quadrant. Step 2: Find the feasible region The feasible region is the area bounded by the lines: - \( x + y = 2 \), - \( x = 0 \), - \( y = 0 \). This forms a triangle with vertices at: - \( (0, 0) \), - \( (2, 0) \), - \( (0, 2) \). Step 3: Evaluate the objective function at the vertices We now calculate the value of \( Z = 3x + 5y \) at each vertex of the feasible region: - At \( (0, 0) \): \[ Z = 3(0) + 5(0) = 0 \] - At \( (2, 0) \): \[ Z = 3(2) + 5(0) = 6 \] - At \( (0, 2) \): \[ Z = 3(0) + 5(2) = 10 \] Step 4: Conclusion The minimum value of \( Z \) occurs at the vertex \( (0, 0) \), where \( Z = 0 \). Thus, the correct answer is: \[ \boxed{0} \]