Answer :
To solve this problem using the graphical method and verify the optimal solution, let's go through the detailed steps.
### Part A: Graph the Model
1. Define the constraints:
- Time constraint: [tex]\(20X_1 + 15X_2 \leq 1200\)[/tex]
- Materials constraint: [tex]\(8X_1 + 12X_2 \leq 600\)[/tex]
- Non-negativity constraints: [tex]\(X_1 \geq 0\)[/tex], [tex]\(X_2 \geq 0\)[/tex]
2. Plot the constraints on a graph:
To graph the inequalities, we first convert them to equalities and then find their intercepts.
- For the time constraint [tex]\(20X_1 + 15X_2 = 1200\)[/tex]:
- [tex]\(X_1\)[/tex]-intercept (set [tex]\(X_2 = 0\)[/tex]): [tex]\(20X_1 = 1200 \Rightarrow X_1 = 60\)[/tex]
- [tex]\(X_2\)[/tex]-intercept (set [tex]\(X_1 = 0\)[/tex]): [tex]\(15X_2 = 1200 \Rightarrow X_2 = 80\)[/tex]
- For the materials constraint [tex]\(8X_1 + 12X_2 = 600\)[/tex]:
- [tex]\(X_1\)[/tex]-intercept (set [tex]\(X_2 = 0\)[/tex]): [tex]\(8X_1 = 600 \Rightarrow X_1 = 75\)[/tex]
- [tex]\(X_2\)[/tex]-intercept (set [tex]\(X_1 = 0\)[/tex]): [tex]\(12X_2 = 600 \Rightarrow X_2 = 50\)[/tex]
3. Draw the feasible region:
- Plot the constraint lines on the graph based on the intercepts:
- Time constraint line: Connect points (60, 0) and (0, 80)
- Materials constraint line: Connect points (75, 0) and (0, 50)
- Shade the feasible region that satisfies both inequalities.
4. Identify the vertices of the feasible region:
The feasible region is typically a polygon (or unbounded) area where all constraints overlap. The vertices of this region are potential candidates for the optimal solution. Identify these vertices by solving the equations of the lines that intersect.
### Part B: Determine the Optimal Value
1. Calculating the vertices:
- Let's identify the points of intersection:
- The intercepts with X and Y axes provide points (60,0), (0,80), (75,0), and (0,50).
- Calculate the intersection of the lines [tex]\(20X_1 + 15X_2 = 1200\)[/tex] and [tex]\(8X_1 + 12X_2 = 600\)[/tex]:
Solve the systems of equations:
[tex]\(20X_1 + 15X_2 = 1200\)[/tex] and [tex]\(8X_1 + 12X_2 = 600\)[/tex]
We can scale the second equation for easier solving:
[tex]\(\frac{8}{8}X_1 + \frac{12}{8}X_2 = \frac{600}{8} \Rightarrow X_1 + \frac{3}{2}X_2 = 75\)[/tex]
Now we have:
[tex]\(20X_1 + 15X_2 = 1200\)[/tex]
[tex]\(8X_1 + 12X_2 = 600 (\Rightarrow 4X_1 + 6X_2 = 300)\)[/tex]
From the above we can form:
[tex]\(10X_1 + 15X_2 = 600\)[/tex]
Solving | For optimal:
X_1 = 45, X_2 = 20
2. The Objective Function:
To find the optimal solution, we need to evaluate the objective function [tex]\(25X_1 + 30X_2\)[/tex] at each vertex of the feasible region.
- (60, 0): [tex]\(25(60) + 30(0) = 1500\)[/tex]
- (0, 80): [tex]\(25(0) + 30(80) = 2400\)[/tex]
- (75, 0): [tex]\(25(75) + 30(0) = 1875\)[/tex]
- (0, 50): [tex]\(25(0) + 30(50) = 1500\)[/tex]
- (45, 20): [tex]\(25(45) + 30(20) = 1725\)[/tex]
3. Identify the Maximum Profit:
The maximum profit occurs at the vertex that gives the highest value of the objective function. In this case, we have calculated:
- The point [tex]\((45, 20)\)[/tex] yields the maximum profit of [tex]\(1725\)[/tex].
Thus, the optimal values for the decision variables are:
- [tex]\(X_1 = 45\)[/tex]
- [tex]\(X_2 = 20\)[/tex]
The maximum profit is 1725 units.
### Part A: Graph the Model
1. Define the constraints:
- Time constraint: [tex]\(20X_1 + 15X_2 \leq 1200\)[/tex]
- Materials constraint: [tex]\(8X_1 + 12X_2 \leq 600\)[/tex]
- Non-negativity constraints: [tex]\(X_1 \geq 0\)[/tex], [tex]\(X_2 \geq 0\)[/tex]
2. Plot the constraints on a graph:
To graph the inequalities, we first convert them to equalities and then find their intercepts.
- For the time constraint [tex]\(20X_1 + 15X_2 = 1200\)[/tex]:
- [tex]\(X_1\)[/tex]-intercept (set [tex]\(X_2 = 0\)[/tex]): [tex]\(20X_1 = 1200 \Rightarrow X_1 = 60\)[/tex]
- [tex]\(X_2\)[/tex]-intercept (set [tex]\(X_1 = 0\)[/tex]): [tex]\(15X_2 = 1200 \Rightarrow X_2 = 80\)[/tex]
- For the materials constraint [tex]\(8X_1 + 12X_2 = 600\)[/tex]:
- [tex]\(X_1\)[/tex]-intercept (set [tex]\(X_2 = 0\)[/tex]): [tex]\(8X_1 = 600 \Rightarrow X_1 = 75\)[/tex]
- [tex]\(X_2\)[/tex]-intercept (set [tex]\(X_1 = 0\)[/tex]): [tex]\(12X_2 = 600 \Rightarrow X_2 = 50\)[/tex]
3. Draw the feasible region:
- Plot the constraint lines on the graph based on the intercepts:
- Time constraint line: Connect points (60, 0) and (0, 80)
- Materials constraint line: Connect points (75, 0) and (0, 50)
- Shade the feasible region that satisfies both inequalities.
4. Identify the vertices of the feasible region:
The feasible region is typically a polygon (or unbounded) area where all constraints overlap. The vertices of this region are potential candidates for the optimal solution. Identify these vertices by solving the equations of the lines that intersect.
### Part B: Determine the Optimal Value
1. Calculating the vertices:
- Let's identify the points of intersection:
- The intercepts with X and Y axes provide points (60,0), (0,80), (75,0), and (0,50).
- Calculate the intersection of the lines [tex]\(20X_1 + 15X_2 = 1200\)[/tex] and [tex]\(8X_1 + 12X_2 = 600\)[/tex]:
Solve the systems of equations:
[tex]\(20X_1 + 15X_2 = 1200\)[/tex] and [tex]\(8X_1 + 12X_2 = 600\)[/tex]
We can scale the second equation for easier solving:
[tex]\(\frac{8}{8}X_1 + \frac{12}{8}X_2 = \frac{600}{8} \Rightarrow X_1 + \frac{3}{2}X_2 = 75\)[/tex]
Now we have:
[tex]\(20X_1 + 15X_2 = 1200\)[/tex]
[tex]\(8X_1 + 12X_2 = 600 (\Rightarrow 4X_1 + 6X_2 = 300)\)[/tex]
From the above we can form:
[tex]\(10X_1 + 15X_2 = 600\)[/tex]
Solving | For optimal:
X_1 = 45, X_2 = 20
2. The Objective Function:
To find the optimal solution, we need to evaluate the objective function [tex]\(25X_1 + 30X_2\)[/tex] at each vertex of the feasible region.
- (60, 0): [tex]\(25(60) + 30(0) = 1500\)[/tex]
- (0, 80): [tex]\(25(0) + 30(80) = 2400\)[/tex]
- (75, 0): [tex]\(25(75) + 30(0) = 1875\)[/tex]
- (0, 50): [tex]\(25(0) + 30(50) = 1500\)[/tex]
- (45, 20): [tex]\(25(45) + 30(20) = 1725\)[/tex]
3. Identify the Maximum Profit:
The maximum profit occurs at the vertex that gives the highest value of the objective function. In this case, we have calculated:
- The point [tex]\((45, 20)\)[/tex] yields the maximum profit of [tex]\(1725\)[/tex].
Thus, the optimal values for the decision variables are:
- [tex]\(X_1 = 45\)[/tex]
- [tex]\(X_2 = 20\)[/tex]
The maximum profit is 1725 units.
