Final answer:
The optimal solution to the linear program is a = 0, b = 0, with a maximum objective function value of 0.
Step-by-step explanation:
To solve this linear program, we'll use the constraints -4a + 3b ≤ 9 and 1a - 1b ≤ 4 along with the non-negativity constraints a ≥ 0 and b ≥ 0. First, let's graph the constraints on a coordinate plane. The line -4a + 3b = 9 intersects the axes at (0, 3) and (9/4, 0). The line 1a - 1b = 4 intersects the axes at (0, -4) and (4, 0). The feasible region is the area where both constraints are satisfied, which is the intersection of these lines in the first quadrant.
Upon evaluating the objective function 1a - 2b at the corner points of the feasible region, we find the values at the vertices: (0, 0), (0, 3), and (4, 0). Plugging these into the objective function gives us values of 0, -6, and 4 respectively. The maximum value occurs at (4, 0) with a = 4 and b = 0, resulting in a maximum objective function value of 4. However, this violates the constraint -4a + 3b ≤ 9.
Therefore, the maximum value within the feasible region is at the point (0, 0), where a = 0 and b = 0, yielding an objective function value of 0. This means that the optimal solution occurs at the origin, and the maximum value of the objective function is 0 within the given constraints.