Problem 28 - Applied Mathematics: For the Managerial, Life, and Social Sciences

Question

A company manufactures two products, \(\mathrm{A}\) and \(\mathrm{B}\), on two machines, \(\overline{\mathrm{I}}\) and II. It has been determined that the company will realize a profit of $$\$ 3 $$ unit of product \(A\) and a profit of $$\$ 4 $$ unit of product \(\mathrm{B}\). To manufacture 1 unit of product \(\mathrm{A}\) requires 6 min on machine I and 5 min on machine II. To manufacture 1 unit of product \(\mathrm{B}\) requires \(9 \mathrm{~min}\) on machine \(\mathrm{I}\) and \(4 \mathrm{~min}\) on machine II. There are \(5 \mathrm{hr}\) of machine time available on machine I and \(3 \mathrm{hr}\) of machine time available on machine II in each work shift. How many units of each product should be produced in each shift to maximize the company's profit? What is the largest profit the company can realize? Is there any time left unused on the machines?

Step-by-Step Solution

Answer

In order to maximize profit, the company should produce 60 units of product A and no units of product B during each shift, resulting in a profit of $180. Machine I will have 60 minutes of unused time, while Machine II has no unused time.

Step 1: Define the Variables

Let's define the variables: \(x_A\) = number of units of product A produced in each shift \(x_B\) = number of units of product B produced in each shift

Step 2: Write the Objective Function

The objective is to maximize profit. Profit = \(3x_A + 4x_B\)

Step 3: Write the Constraints

The constraints are based on the available machine time for each machine: Machine I: \(6x_A + 9x_B \le 300\) (5 hours = 300 minutes) Machine II: \(5x_A + 4x_B \le 180\) (3 hours = 180 minutes) Non-negativity constraints: \(x_A \ge 0\) and \(x_B \ge 0\)

Step 4: Solve the Linear Programming Problem

To find the feasible region, we need to plot the constraint inequalities on a coordinate plane and find the vertices of the feasible region. Inequality 1: \(6x_A + 9x_B \le 300\) Inequality 2: \(5x_A + 4x_B \le 180\) On plotting the constraints: The feasible region is a polygon with vertices: (0, 0), (0, 20), (30, 10), (60, 0)

Step 5: Calculate the Maximum Profit

Evaluate the objective function for each vertex, and determine the maximum value: 1. (0,0): Profit = 3(0) + 4(0) = $0 2. (0,20): Profit = 3(0) + 4(20) = $80 3. (30,10): Profit = 3(30) + 4(10) = $140 4. (60,0): Profit = 3(60) + 4(0) = $180 The maximum profit is $180 and occurs when \(x_A = 60\) and \(x_B = 0\).

Step 6: Determine Unused Machine Time

Calculate the machine time used by both products and compare with the available machine time. Machine I: \(6x_A + 9x_B = 6(60) + 9(0) = 360 min\) Machine II: \(5x_A + 4x_B = 5(60) + 4(0) = 300 min\) Machine I has 60 minutes of unused time (360-300) and Machine II has no unused time. In conclusion, to maximize profit, the company should produce 60 units of product A and no units of product B during each shift. This will result in a profit of $180. Machine I will have 60 minutes of unused time.

Key Concepts

Objective Function

In linear programming, the objective function is what we're trying to optimize. This could mean maximizing something desirable like profit or minimizing something undesirable like cost. The function is usually expressed in terms of decision variables. For our problem, it involves determining the number of units of products A and B to produce. The objective is specified as a mathematical equation:

• Profit = \( 3x_A + 4x_B \)

Here, \(x_A\) and \(x_B\) are the number of units of products A and B produced. The coefficients (3 and 4) represent the profit from each unit. Consequently, the objective function helps us evaluate different scenarios to decide the product mix that yields the highest profit.

Constraints

Constraints are the limitations or restrictions in a linear programming problem. They define the feasible solutions that comply with the given conditions. In this case, the constraints are represented by the total machine time available. Each machine has a limited time it can be used per shift:

• Machine I: \(6x_A + 9x_B \leq 300\)
• Machine II: \(5x_A + 4x_B \leq 180\)

Additionally, we have non-negativity constraints, which mean the number of units produced cannot be negative:

• \(x_A \geq 0\)
• \(x_B \geq 0\)

These constraints are essential as they represent real-world limits like production capacity and resource availability.

Feasible Region

The feasible region in a linear programming problem is a graphical depiction of all possible solutions that satisfy the constraints. It's defined by the intersection of the constraints plotted as inequalities on a graph. In the scenario of producing products A and B, we plot:

• \(6x_A + 9x_B \leq 300\)
• \(5x_A + 4x_B \leq 180\)

The feasible region is then a polygon on this graph where all the restrictions overlap. For our problem, the vertices of the feasible region are (0, 0), (0, 20), (30, 10), and (60, 0). These vertices represent the potential outcomes that satisfy all constraints. The final solution must fall within this feasible region.

Maximize Profit

Maximizing profit is typically the main goal in linear programming problems related to business. We do this by evaluating the objective function at each vertex of the feasible region. The point that provides the highest value indicates the optimal solution.

• Solution points: (0,0), (0,20), (30,10), (60,0)

We calculate the profit for each vertex:

• (0,0): Profit = \(3 \times 0 + 4 \times 0 = 0\)
• (0,20): Profit = \(3 \times 0 + 4 \times 20 = 80\)
• (30,10): Profit = \(3 \times 30 + 4 \times 10 = 140\)
• (60,0): Profit = \(3 \times 60 + 4 \times 0 = 180\)

Thus, the optimal solution is to produce 60 units of product A and 0 units of product B, giving a maximum profit of $180. This outcome makes sure that within the machine time constraints, the company gains the highest possible profit.