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 \( 162 \).
If the value of \( \cos \alpha \) is \( \frac{\sqrt{3}}{2} \), then \( A + A = I \), where \[ A = \begin{bmatrix} \sin\alpha & -\cos\alpha \\ \cos\alpha & \sin\alpha \end{bmatrix}. \]