Consider the following linear programming problem:
Maximize 12X + 10Y
Subject to: 4X + 3Y ≤ 480
2X + 3Y ≤ 360
All variable ≥ 0
The maximum possible value for the objective function is
A)360.
B)480.
C)1520.
D)1560.

Question

Consider the following linear programming problem: 
Maximize 12X + 10Y
Subject to: 4X + 3Y ≤ 480
2X + 3Y ≤ 360
All variable ≥ 0
The maximum possible value for the objective function is
A)360.
B)480.
C)1520.
D)1560.

Quizplus · Accepted Answer

We can solve this by using the graphical method. Plotting the two equations we get:

Now, we need to find the intersection points. Solving the equations 4X + 3Y = 480 and 2X + 3Y = 360, we get X = 120 and Y = 80. Thus, the intersection point is (120, 80).

Now, we need to determine the feasible region. This region is shaded in the above graph. The vertices of this region are (0, 160), (120, 80), and (144, 0).

Now, we need to find the value of the objective function at each of these vertices:

At (0, 160), the value of the objective function is 10(160) = 1600.
At (120, 80), the value of the objective function is 12(120) + 10(80) = 1520.
At (144, 0), the value of the objective function is 12(144) = 1728.

Thus, the maximum possible value for the objective function is 1728, which occurs at (144, 0). However, this point is not feasible as it violates the constraint 2X + 3Y ≤ 360.

Therefore, the maximum possible value for the objective function within the feasible region is 1520, which occurs at (120, 80).

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