Problem 1 - Applied Mathematics: For the Managerial, Life, and Social Sciences

Question

A company manufactures two products, \(A\) and \(B\), on two machines, \(\bar{I}\) and II. It has been determined that the company will realize a profit of $$\$ 3$$ on each unit of product \(A\) and a profit of $$\$ 4$$ on each unit of product \(\mathrm{B}\). To manufacture a unit of product A requires \(6 \mathrm{~min}\) on machine \(\mathrm{I}\) and \(5 \mathrm{~min}\) on machine II. To manufacture a unit of product B requires 9 min on machine \(\mathrm{I}\) and \(4 \mathrm{~min}\) on machine \(\mathrm{II}\). There are \(5 \mathrm{hr}\) of machine time available on machine \(\mathrm{I}\) 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?

Step-by-Step Solution

Answer

The company should produce 0 units of product A and 45 units of product B in each shift to maximize their profit, which will be 180 dollars.

Step 1: Define Variables

Let's denote x as the number of units of product A and y as the number of units of product B.

Step 2: Determine Constraints

We have two machines and the processing time allotted for products A and B are given. To write the inequalities, we will consider the total time consumption of machines I and II for both products, which must not exceed their available time in each shift. Machine I: \(6x + 9y \leq 5(hour)\cdot60(minutes) \) (time used by products A and B should be under 5 hours) Machine II: \(5x + 4y \leq 3(hour)\cdot60(minutes)\) (time used by products A and B should be under 3 hours) We also have non-negativity constraints, since the number of products cannot be negative: \(x \geq 0\) \(y \geq 0\)

Step 3: Define the Objective Function

The objective is to maximize the profit. The profit function is given by: \( P(x, y) = 3x + 4y\)

Step 4: Graph the Constraints and Find the Feasible Region

Graph the inequalities: Machine I: \(6x + 9y \leq 300\) Machine II: \(5x + 4y \leq 180\) Along with non-negativity constraints: \(x \geq 0\) \(y \geq 0\) The feasible region is the area where all constraints are satisfied.

Step 5: Find the Vertices of the Feasible Region

Now, we need to find the vertices of the feasible region, which are the points where the constraints intersect. It is where the optimal solution will be located. Vertices: 1. Intersection of Machine I and the x-axis (y = 0): \(6x = 300\), x = 50 2. Intersection of Machine II and the x-axis (y = 0): \(5x = 180\), x = 36 3. Intersection of Machine I and the y-axis (x = 0): \(9y = 300\), y = 100/3 4. Intersection of Machine II and the y-axis (x = 0): \(4y = 180\), y = 45 5. Intersection of Machine I and Machine II: Solve the system of equations \(6x + 9y = 300\) \(5x + 4y = 180\) After solving the system of equations, we get x = 20 and y = 20. Our vertices are: 1. (50,0): \(P(50,0) = 3(50) + 4(0) = 150\) 2. (36,0): \(P(36,0) = 3(36) + 4(0) = 108\) 3. (0,100/3): \(P(0,100/3) = 3(0) + 4(100/3) = 400/3\) 4. (0,45): \(P(0,45) = 3(0) + 4(45) = 180\) 5. (20,20): \(P(20,20) = 3(20) + 4(20) = 140\)

Step 6: Determine the Optimal Solution

Compare the profit function values at the feasible region's vertices and select the one that maximizes profit. Maximized profit is achieved at the point (0,45), which gives a profit of 180 dollars. Therefore, the company should produce 0 units of product A and 45 units of product B in each shift to maximize their profit.

Key Concepts

Optimization

Optimization refers to the process of finding the best possible solution to a problem, specifically focusing on maximizing or minimizing a particular outcome. In the context of linear programming, which deals with linear relationships, optimization aims to find the maximum profit, minimum cost, or most efficient use of resources.

In our exercise, the company wishes to optimize their profit from products A and B. The optimization process involves choosing the best combination of products A and B that leads to the highest profit, given the available machine time. This requires setting up an objective function and a set of constraints, then using mathematical or graphical methods to find the optimal values for the decision variables, here being the number of units for each product.

Optimization in linear programming is particularly useful in many fields like manufacturing, finance, and operations research because it provides clear guidelines on resource allocation to achieve specific objectives.

Feasibility Region

The feasibility region is a key concept in linear programming. It is the set of all possible points that satisfy the problem's constraints, representing all potential solutions. Graphically, these points typically form a polygon in two-dimensional space when dealing with two variables, like our products A and B.

To determine the feasible region, we plot each inequality constraint on a graph. In our case, constraints related to machine time usage for products A and B form two inequalities. Additional non-negativity constraints (i.e., the number of products cannot be negative) further define the feasible region.

The feasible region is crucial because it visually represents the boundaries within which optimal solutions can be found. If a point is outside this region, it does not satisfy one or more constraints, hence cannot be considered a valid solution. In our exercise, the feasible region is where all the constraints overlap, allowing us to find out the best possible combination of products that can be manufactured within the given limits.

Objective Function

The objective function is a mathematical expression that defines the goal of the optimization problem, such as maximizing profit or minimizing cost. This function is based on the problem’s key variables, often centered on decision-making to reach a particular outcome.

In the exercise provided, the objective function is established to maximize profits from producing products A and B. It is given by \( P(x, y) = 3x + 4y \), where \( x \) and \( y \) are the units of products A and B, respectively. Here, the coefficients 3 and 4 represent the profit obtained from each unit of product A and product B.

Evaluating the objective function involves substituting different combinations of \( x \) and \( y \) (within the feasibility region) into the function to determine which combination yields the highest profit. Solving the exercise involves finding the point within the feasibility region that maximizes this function, illustrating how critical the objective function is to the decision-making process in optimization problems.

Constraints

Constraints in linear programming are the conditions that the solution to an optimization problem must satisfy. They are written as inequalities that limit the values of the decision variables, dictating what combinations are possible.

In this exercise, constraints are derived from the time available on machines I and II for producing products A and B. They are as follows:

• Machine I: \(6x + 9y \leq 300\), since machine I has a total available time of 300 minutes per shift.
• Machine II: \(5x + 4y \leq 180\), since machine II has a total available time of 180 minutes per shift.

Additionally, non-negativity constraints \(x \geq 0\) and \(y \geq 0\) ensure that the number of products cannot be negative.

Constraints are essential components of linear programming as they restrict the set of possible solutions, defining the feasible region. Solving an optimization problem entails analyzing these constraints to find the best solution within these limits. They reflect real-world limitations, such as resource availability and production capacity, making them key considerations in planning and decision-making processes.