Question

The maximum value of \( Z = 3x + 2y \) subject to the constraints: \[ 3x + y \leq 15, \quad x \geq 0, \quad y \geq 0 \] is:

• A. 30
• B. 15
• C. 10
• 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 A

We are given the linear programming problem with the objective function: \[ Z = 3x + 2y \] subject to the constraints: \[ 3x + y \leq 15, \quad x \geq 0, \quad y \geq 0 \] Step 1: Plot the constraints - The line \( 3x + y = 15 \) is the boundary of the constraint \( 3x + y \leq 15 \). - 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 line \( 3x + y = 15 \) and the coordinate axes. We can find the intercepts by setting \( x = 0 \) and \( y = 0 \): - When \( x = 0 \), \( y = 15 \), so the point is \( (0, 15) \). - When \( y = 0 \), \( 3x = 15 \), so \( x = 5 \), and the point is \( (5, 0) \). Thus, the feasible region is a triangle with vertices at: - \( (0, 0) \), - \( (5, 0) \), - \( (0, 15) \). Step 3: Evaluate the objective function at the vertices We now calculate the value of \( Z = 3x + 2y \) at each vertex of the feasible region: - At \( (0, 0) \): \[ Z = 3(0) + 2(0) = 0 \] - At \( (5, 0) \): \[ Z = 3(5) + 2(0) = 15 \] - At \( (0, 15) \): \[ Z = 3(0) + 2(15) = 30 \] Step 4: Conclusion The maximum value of \( Z \) occurs at the vertex \( (0, 15) \), where \( Z = 30 \). Thus, the correct answer is: \[ \boxed{30} \]