Step 1: The objective function is \( z = 6x + 12y \).
Step 2: To find the maximum value of the objective function, we evaluate \( z \) at each of the given vertices: - At \( (0, 6) \): \[ z = 6(0) + 12(6) = 0 + 72 = 72. \] - At \( (3, 3) \): \[ z = 6(3) + 12(3) = 18 + 36 = 54. \] - At \( (9, 9) \): \[ z = 6(9) + 12(9) = 54 + 108 = 162. \] - At \( (0, 12) \): \[ z = 6(0) + 12(12) = 0 + 144 = 144. \]
Step 3: The maximum value of \( z \) is 162, which occurs at the vertex \( (9, 9) \). Thus, the maximum value of the objective function is \( 162 \).
For a Linear Programming Problem, find min \( Z = 5x + 3y \) (where \( Z \) is the objective function) for the feasible region shaded in the given figure.
If the value of \( \cos \alpha \) is \( \frac{\sqrt{3}}{2} \), then \( A + A = I \), where \[ A = \begin{bmatrix} \sin\alpha & -\cos\alpha \\ \cos\alpha & \sin\alpha \end{bmatrix}. \]