LP graphical method true/false Q

[csc2iffy]

Homework Statement

Label each of the following statements as True or False, and then justify your answer based on the graphical method. In each case, give an example of an objective function that illustrates your answer.

(a) If (3,3) produces a larger value of the objective function than (0,2) and (6,3), then (3,3) must be an optimal solution.

(b) If (3, 3) is an optimal solution and multiple optimal solutions exist, then either (0,2) or (6,3) must also be an optimal solution.

(c) The point (0,0) cannot be an optimal solution.

Homework Equations

I posted the attachment of the picture

The Attempt at a Solution

I said...

(a) False, (3,3) cannot be an optimal solution because ? I just think it's because (6,3) would always be greater, I can't think of an example where it wouldn't be

(b) False, either (3,3) and (0,2) are optimal, or (3,3) and (6,3) are optimal.

(c) True

I need a little help justifying my answers and finding an example of an objective function illustrating my answers.

 

[mighty2000]

(a) False. The graphical method involves plotting the constraints and finding the feasible region, which is the area where all constraints are satisfied. The optimal solution is the point within the feasible region that maximizes or minimizes the objective function. Just because (3,3) produces a larger value of the objective function than (0,2) and (6,3) does not necessarily mean it is the optimal solution. It could be that there is another point within the feasible region that produces an even larger value of the objective function. For example, if the objective function is to maximize profit and the constraints are x + y ≤ 4 and 2x + y ≤ 6, the feasible region is the shaded area bounded by the two lines and the coordinate axes. The point (3,3) is within the feasible region and produces a profit of 6, but the point (2,4) also satisfies the constraints and produces a profit of 8, which is larger. Therefore, (3,3) is not necessarily the optimal solution.

(b) False. If (3,3) is an optimal solution, it means that it is the best possible solution within the feasible region. If there are multiple optimal solutions, it means that they all produce the same optimal value of the objective function. Therefore, if (0,2) or (6,3) were also optimal solutions, they would produce the same optimal value as (3,3). This would mean that the point (3,3) is not the best possible solution, which contradicts the definition of an optimal solution.

(c) True. The point (0,0) is the origin and is outside of the feasible region. Since the optimal solution must be within the feasible region, (0,0) cannot be an optimal solution. An example of an objective function that illustrates this is to maximize revenue with the constraints x ≥ 0 and y ≥ 0. The feasible region is the first quadrant and the origin is outside of it. Therefore, (0,0) cannot be an optimal solution.