Question

The maximum value of \[ Z = 3x + 5y, \] subject to \[ x + 4y \leq 24, \quad y \leq 4, \quad x \geq 0, \quad y \geq 0 \] is

• A. 20
• B. 120
• C. 72
• D. 44

Hint:

In linear programming problems, always evaluate the objective function at the corner points of the feasible region.

Answer 1

The Correct Option is C

Step 1: Identify the constraints.
The constraints are \[ x + 4y \leq 24, \quad y \leq 4, \quad x \geq 0, \quad y \geq 0 \] Step 2: Graph the constraints.
Graph the inequalities to find the feasible region. The intersection points of the boundary lines will be the possible solutions.
Step 3: Evaluate \( Z \) at the corner points.
The corner points of the feasible region are \( (0, 4) \), \( (12, 0) \), and \( (6, 3) \). Substituting these into the objective function \( Z = 3x + 5y \), we get: \[ Z(0, 4) = 3(0) + 5(4) = 20 \] \[ Z(12, 0) = 3(12) + 5(0) = 36 \] \[ Z(6, 3) = 3(6) + 5(3) = 18 + 15 = 33 \] Step 4: Conclusion.
The maximum value of \( Z \) is 72 at the point \( (6, 3) \).