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Question 1186203: A food factory is making a beverage for a customer from mixing two different existing products A and B. The compositions of A and B and prices ($/L) are given as follows,
Amount (L) in /100 L of A and B
Lime Orange Mango Cost ($/L)
A 3 6 4 3
B 7 4 8 11
The customer requires that there must be at least 5 Litres (L) Orange and at least 5Litres of Mango concentrate per 100 Litres of the beverage respectively, but no morethan 6 Litres of Lime concentrate per 100 Litres of beverage. The customer needs atleast 100 Litres of the beverage per week.
(a) Explain why a linear programming model would be suitable for this case study
(b) Formulate a Linear Programming (LP) model for the factory that minimises the total cost of producing the beverage while satisfying all constraints.
(c) Use the graphical method to nd the optimal solution. Show the feasible region andthe optimal solution on the graph. Annotate all lines on your graph. What is the mini-mal cost for the product?
Note: you can use graphical solvers available online but make sure that your graph isclear, all variables involved are clearly represented and annotated, and each line is clearlymarked and related to the corresponding equation.
(d) Is there a range for the cost ($) of A that can be changed without a ecting the opti-mum solution obtained above
Answer by CPhill(1959)   (Show Source):
You can put this solution on YOUR website! Here's how to approach this linear programming problem:
**(a) Suitability of Linear Programming:**
A linear programming model is suitable for this case study because:
1. **Linearity:** The relationships between the amounts of products A and B and the amounts of lime, orange, and mango concentrates are linear. The cost function is also linear.
2. **Optimization:** The problem seeks to minimize the cost, which is a typical objective of linear programming.
3. **Constraints:** There are specific constraints on the amounts of each concentrate and the total volume of the beverage, which can be expressed as linear inequalities.
4. **Divisibility:** The amounts of products A and B can be considered divisible (fractions of liters are possible).
5. **Non-negativity:** The amounts of products A and B cannot be negative.
**(b) Linear Programming Model:**
**Decision Variables:**
* x = Liters of product A per 100L of beverage
* y = Liters of product B per 100L of beverage
**Objective Function (Minimize Cost):**
Cost (C) = 3x + 11y
**Constraints:**
* Orange: 6x + 4y ≥ 5
* Mango: 4x + 8y ≥ 5
* Lime: 3x + 7y ≤ 6
* Total Volume (per 100L): x + y = 100
* Non-negativity: x ≥ 0, y ≥ 0
Since the total volume is *exactly* 100L, the x and y values will directly represent the liters of each product per 100L of beverage.
**(c) Graphical Solution:**
1. **Graph the constraints:** Plot each constraint on a graph with x and y axes. Treat each inequality as an equality to draw the lines, and then shade the appropriate region.
* 6x + 4y = 5 => y = (5 - 6x) / 4
* 4x + 8y = 5 => y = (5 - 4x) / 8
* 3x + 7y = 6 => y = (6 - 3x) / 7
* x + y = 100
2. **Identify the feasible region:** The feasible region is the area where all constraints are satisfied. It will be a polygon bounded by the constraint lines.
3. **Find the vertices:** The vertices are the intersection points of the constraint lines that bound the feasible region. Solve the systems of equations for each intersecting pair of lines. The relevant vertices will be where the orange, mango, and lime constraints intersect with each other and the x+y=100 line.
4. **Evaluate the objective function at each vertex:** Substitute the x and y values of each vertex into the cost function C = 3x + 11y.
5. **Optimal Solution:** The vertex that yields the lowest cost is the optimal solution.
*Intersection points and cost calculations:*
*6x+4y=5 and 4x+8y=5:* solving gives x=0.25, y=0.875, Cost = 3(0.25) + 11(0.875) = $10.4375
*6x+4y=5 and 3x+7y=6:* solving gives x=0.45, y=0.55, Cost = 3(0.45) + 11(0.55) = $7.4
*4x+8y=5 and 3x+7y=6:* solving gives x=1.1, y=0.075, Cost = 3(1.1) + 11(0.075) = $4.025
The minimum cost is $4.025.
**(d) Range for Cost of A:**
To determine the range for the cost of A without affecting the optimal solution, we need to consider the slopes of the objective function and the binding constraints (the constraints that determine the optimal solution).
1. **Binding Constraints:** The binding constraints are the ones that intersect at the optimal solution. In this case, it will be the intersection of 4x+8y=5 and 3x+7y=6.
2. **Slope of the Objective Function:** The slope of the objective function C = 3x + 11y is -3/11.
3. **Slopes of the Binding Constraints:** The slopes of the binding constraints are -4/8 = -1/2 and -3/7.
4. **Range:** The cost coefficient of A can vary as long as the slope of the objective function stays between the slopes of the binding constraints. So, the range of the cost of A (let's call it 'a') will be:
-1/2 <= -a/11 <= -3/7
Multiplying by -11 and flipping the inequalities:
11/2 >= a >= 33/7
5.5 >= a >= 4.71
So, the cost of A can range from $4.71 to $5.5 without changing the optimal solution.
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