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Sagot :
Explanation:
To determine the optimal way to meet the demand of 80 units given the constraints and costs, we need to solve a linear programming problem. Here's a step-by-step outline:
### Define the Variables:
- Let \( x_1 \) be the number of hours Machine 1 operates.
- Let \( x_2 \) be the number of hours Machine 2 operates.
### Objective Function:
Minimize the total cost:
\[ \text{Minimize Cost} = 4x_1 + 5x_2 \]
### Constraints:
1. **Demand Constraint**: The total units produced should be at least 80 units.
\[ 2x_1 + 3x_2 \geq 80 \]
2. **Regular Time Constraint**: The total hours should not exceed 40 hours.
\[ x_1 + x_2 \leq 40 \]
3. **Non-negativity Constraint**:
\[ x_1 \geq 0 \]
\[ x_2 \geq 0 \]
### Steps to Solve the Linear Programming Problem:
1. **Formulate the Problem**: Write down the objective function and constraints clearly.
2. **Graph the Constraints**: Plot the constraints on a graph to find the feasible region.
3. **Find the Intersection Points**: Determine where the constraint lines intersect.
4. **Evaluate the Objective Function at Each Vertex of the Feasible Region**: The minimum cost will occur at one of these vertices.
Let's proceed with these steps.
### Graphing the Constraints:
The two main constraints are:
1. \( 2x_1 + 3x_2 \geq 80 \)
2. \( x_1 + x_2 \leq 40 \)
First, we need to find the intersection points by solving these equations simultaneously.
### Solving for Intersections:
1. Solve for the point where \( 2x_1 + 3x_2 = 80 \) intersects \( x_1 + x_2 = 40 \):
\[ x_1 + x_2 = 40 \]
\[ 2x_1 + 3x_2 = 80 \]
Multiply the first equation by 2:
\[ 2x_1 + 2x_2 = 80 \]
Subtract this from the second equation:
\[ (2x_1 + 3x_2) - (2x_1 + 2x_2) = 80 - 80 \]
\[ x_2 = 0 \]
Substitute \( x_2 = 0 \) into \( x_1 + x_2 = 40 \):
\[ x_1 + 0 = 40 \]
\[ x_1 = 40 \]
So, one intersection point is (40, 0).
2. Find the x-intercept of \( 2x_1 + 3x_2 = 80 \):
Set \( x_2 = 0 \):
\[ 2x_1 = 80 \]
\[ x_1 = 40 \]
3. Find the y-intercept of \( 2x_1 + 3x_2 = 80 \):
Set \( x_1 = 0 \):
\[ 3x_2 = 80 \]
\[ x_2 = \frac{80}{3} \approx 26.67 \]
4. Intersection of \( x_1 + x_2 = 40 \) with the x and y axes:
\[ x_1 = 0 \Rightarrow x_2 = 40 \]
\[ x_2 = 0 \Rightarrow x_1 = 40 \]
Now we can evaluate the objective function at these points:
- (40, 0)
- (0, 40)
- (0, 26.67)
### Evaluate the Objective Function at Each Point:
1. At (40, 0):
\[ 4x_1 + 5x_2 = 4(40) + 5(0) = 160 \]
2. At (0, 40):
\[ 4x_1 + 5x_2 = 4(0) + 5(40) = 200 \]
3. At (0, 26.67):
\[ 4x_1 + 5x_2 = 4(0) + 5(26.67) = 133.35 \]
The minimum cost occurs at (0, 26.67) with a cost of 133.35.
### Conclusion:
To meet the demand of 80 units at the minimum cost:
- Run Machine 1 for 0 hours.
- Run Machine 2 for approximately 26.67 hours.
The minimum cost will be $133.35.
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