MeanBeats is a company that manufactures and sells electronic musical equipment. Their
headphone product line consists of two products, Reverbia and Bscape. MeanBeats sells
the headphones to retailers at $600 and $300 per pack for Reverbia and Bscape,
respectively. The company needs to know how many packs of each headphone to make to maximize potential revenue. They know that there will be at least twice as many
Bscape orders as there are for Reverbia. Additionally, below are business rules that the
company must follow on ordering packs of headphones:
i) the Bscape orders plus twice the Reverbia orders cannot exceed 20 packs.
ii) the Reverbia orders are no more than 10 more than the Bscape orders, and
iii) the Bscape orders are no more than 3 more than the Reverbia orders.
a. Formulate a linear programming model (that is, define the variables, and write down
the objective function and all constraints mathematically).
b. Solve this problem using graphical analysis. Please label each constraint and identify
the feasible region on the graph. Plot several level curves of the objective function
line and clearly identify the optimal point.
c. What is the optimal solution? What is the optimal value?
The answer is one that you must learn for yourself my friend...
a. To formulate the linear programming model, let's define the following variables:
Let x be the number of packs of Reverbia headphones to produce.
Let y be the number of packs of Bscape headphones to produce.
The objective is to maximize the potential revenue, which can be expressed as the total cost of packs sold:
Revenue = (Number of Reverbia packs sold) * (Price per Reverbia pack) + (Number of Bscape packs sold) * (Price per Bscape pack)
Revenue = 600x + 300y
Now, let's write down the constraints mathematically:
Constraint 1: The Bscape orders plus twice the Reverbia orders cannot exceed 20 packs.
y + 2x <= 20
Constraint 2: The Reverbia orders are no more than 10 more than the Bscape orders.
x <= y + 10
Constraint 3: The Bscape orders are no more than 3 more than the Reverbia orders.
y <= x + 3
Since we need non-negative quantities, x and y are greater than or equal to zero:
x >= 0
y >= 0
b. To solve this problem using graphical analysis, we need to plot the feasible region determined by the constraints.
Let's plot the three constraints on a graph:
1. Constraint 1: y + 2x <= 20
This is a straight line with intercepts (0, 20) and (10, 0).
2. Constraint 2: x <= y + 10
This is a straight line with intercepts (0, 10) and (20, 20).
3. Constraint 3: y <= x + 3
This is a straight line with intercepts (0, 3) and (20, 23).
We also need to plot the non-negativity constraints:
x >= 0 (Vertical line passing through (0, 0))
y >= 0 (Horizontal line passing through (0, 0))
The feasible region is the intersection of all the shaded regions formed by these constraints.
c. The optimal solution can be found by evaluating the objective function at the vertices of the feasible region.
To find the vertices, we can solve the system of equations formed by the intersecting lines:
1. Solve the system of equations for constraints 1 and 3:
y + 2x = 20
y = x + 3
2. Solve the system of equations for constraints 2 and 3:
x = y + 10
y = x + 3
Solving these equations will give us the intersection points.
Once we have the vertices, we can substitute their values into the objective function to find the maximum potential revenue.
After evaluating the objective function at the vertices, we can determine the optimal solution, which will be the vertex with the highest revenue.
The optimal value is the maximum potential revenue obtained from the objective function at the optimal solution point.