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Answer :
The linear programming model aims to maximize profit contribution for RMC, where [tex]\(X_1\)[/tex] represents fuel additive tons, and [tex]\(X_2\)[/tex] represents solvent base tons, subject to material constraints. The graphical solution procedure identifies the optimal solution within the feasible region to maximize [tex]\(Z = 40X_1 + 30X_2\)[/tex].
To solve this linear programming problem step by step, we'll use the graphical solution procedure. We'll start by graphing the constraints and then identify the optimal solution within the feasible region.
We have the following constraints:
1. Material 1 constraint: [tex]\(2/5X_1 + 1/2X_2 \leq 20\)[/tex]
2. Material 2 constraint: [tex]\(1/5X_2 \leq 5\)[/tex]
3. Material 3 constraint: [tex]\(3/5X_1 + 3/10X_2 \leq 21\)[/tex]
Let's plot these constraints on a graph:
Constraint 1: [tex]\(2/5X_1 + 1/2X_2 \leq 20\)[/tex]
- When [tex]\(X_1 = 0\), \(X_2 = 40\)[/tex]
- When [tex]\(X_2 = 0\), \(X_1 = 50\)[/tex]
Plotting these two points and drawing the line between them.
Constraint 2: [tex]\(1/5X_2 \leq 5\)[/tex]
- When [tex]\(X_2 = 0\), \(X_1 = 0\)[/tex]
- When [tex]\(X_1 = 50\), \(X_2 = 5\)[/tex]
Plotting these two points and drawing the line between them.
Constraint 3: [tex]\(3/5X_1 + 3/10X_2 \leq 21\)[/tex]
- When [tex]\(X_1 = 0\), \(X_2 = 70\)[/tex]
- When [tex]\(X_2 = 0\), \(X_1 = 35\)[/tex]
Plotting these two points and drawing the line between them.
The feasible region is the area where all constraints overlap. In this case, it is the region enclosed by the three constraint lines.
We want to maximize [tex]\(Z = 40X_1 + 30X_2\)[/tex]. To do this, we'll calculate the value of Z at the corner points (vertices) of the feasible region.
- Vertex A: (0, 0) -> Z = 0
- Vertex B: (0, 5) -> Z = 150
- Vertex C: (35, 5) -> Z = 1,250
- Vertex D: (50, 0) -> Z = 2,000
The optimal solution is the one that maximizes [tex]\(Z\)[/tex]. In this case, the highest [tex]\(Z\)[/tex] is achieved at Vertex D: [tex]\((50, 0)\) with \(Z = 2,000\)[/tex].
The optimal solution for RMC is to produce 50 tons of fuel additive (\[tex](X_1\))[/tex] and 0 tons of solvent base [tex](\(X_2\))[/tex] to maximize the profit contribution, resulting in a profit of $2,000.
To learn more about linear programming, click here.
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