Problem 29 - Applied Mathematics: For the Managerial, Life, and Social Sciences

Question

A company manufactures two products, \(\mathrm{A}\) and \(\mathrm{B}\), on two machines, 1 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 a unit of product A requires 6 min on machine \(\mathrm{I}\) and 5 min on machine II. To manufacture a unit of product \(\mathrm{B}\) requires \(9 \mathrm{~min}\) on machine \(\mathrm{I}\) and 4 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 optimal profit?

Step-by-Step Solution

Answer

The company should produce 20 units of product A and 15 units of product B per shift to maximize its profit. The optimal profit is \(\$120\) per shift.

Step 1: Define the variables

Let \(x\) be the number of units of product A, and \(y\) be the number of units of product B produced per shift.

Step 2: Create the objective function

The objective is to maximize the profit. Since the company earns \(\$3\) for each unit of product A and \(\$4\) for each unit of product B, the profit can be represented by the function: $$ P(x, y) = 3x + 4y $$

Step 3: Determine the constraints

The constraints of this problem are based on the time required to produce the products on each machine and the available machine time. The constraints can be written as the following inequalities: Machine I constraint: \(6x + 9y \leq 300\) (because there are 5 hours available on machine I, and 5 hours * 60 minutes = 300 minutes) Machine II constraint: \(5x + 4y \leq 180\) (because there are 3 hours available on machine II, and 3 hours * 60 minutes = 180 minutes) Additionally, the number of units produced cannot be negative. Therefore: \(x \geq 0\) \(y \geq 0\)

Step 4: Solve the linear programming problem

To solve this linear programming problem, we can graph the constraints and identify the region where all constraints are satisfied. Then, we will find the vertices of the feasible region and evaluate the objective function at each vertex to find the maximum profit. Graph the inequalities to find the feasible region. You will find that the vertices of the feasible region are: 1. \((0, 0)\) 2. \((0, 20)\) 3. \((30, 0)\) 4. \((20, 15)\) Now, evaluate the profit function at each vertex. 1. \(P(0,0) = 0\) 2. \(P(0,20) = 80\) 3. \(P(30,0) = 90\) 4. \(P(20,15) = 120\) The maximum profit is achieved at \((20, 15)\), which means producing 20 units of product A and 15 units of product B.

Step 5: Determine the optimal profit and solution

The maximum profit is \(P(20,15) = 120\) dollars per shift. So, the company should produce 20 units of product A and 15 units of product B per shift to maximize its profit.

Key Concepts

Objective Function

In the world of linear programming, the objective function plays a crucial role. It is a mathematical expression that defines the goal of the optimization problem. In this exercise, the objective function is designed to maximize the company's profit based on the number of products A and B produced. The objective function is expressed as \( P(x, y) = 3x + 4y \), where:

• \( x \) is the number of units of product A.
• \( y \) is the number of units of product B.
• The coefficients 3 and 4 represent the profit obtained from each unit of products A and B, respectively.

To maximize the profit, the objective is to find the values of \( x \) and \( y \) that make \( P(x, y) \) as large as possible, within the given constraints.

Constraints

Constraints are the mathematical expressions that represent the limitations within which an optimization problem must be solved. In this exercise, the constraints are based on the machine time available per shift for producing products A and B.For machine I, the constraint is:\[ 6x + 9y \leq 300 \]This means that the total time taken by products A and B on machine I should not exceed 300 minutes (5 hours).For machine II, the constraint is:\[ 5x + 4y \leq 180 \]Here, the total usage of machine II must remain within 180 minutes (3 hours).Additionally, we consider non-negativity constraints to ensure the practical meaning:

• \( x \geq 0 \)
• \( y \geq 0 \)

These constraints define a set of possible solutions, known as the feasible region, where the objective function will be evaluated.

Feasible Region

The feasible region in linear programming is the set of all possible points that satisfy all the constraints. It is typically represented as an area on a graph where the constraints overlap. In this problem, the feasible region is determined by the intersection of the constraints:

• \( 6x + 9y \leq 300 \)
• \( 5x + 4y \leq 180 \)
• \( x \geq 0 \)
• \( y \geq 0 \)

When plotted on a graph, these constraints create a polygonal area. The corners, or vertices, of the feasible region are crucial because they are potential solutions to check for the maximum profit. In this exercise, the vertices were found to be

• (0, 0)
• (0, 20)
• (30, 0)
• (20, 15)

Evaluating the objective function at these points helps identify the optimal solution.

Profit Maximization

Profit maximization is the ultimate goal of the linear programming problem presented in this exercise. After graphing the constraints and identifying the feasible region, the next step is to apply the objective function to each vertex of the feasible region.The profit function \( P(x, y) = 3x + 4y \) is calculated at each vertex:

• \( P(0, 0) = 0 \)
• \( P(0, 20) = 80 \)
• \( P(30, 0) = 90 \)
• \( P(20, 15) = 120 \)

The maximum profit of 120 dollars is identified at the point (20, 15). Thus, to achieve optimal profitability, the company should produce 20 units of product A and 15 units of product B per shift. This solution not only respects all constraints but also ensures the highest possible profit.