Problem 30 - Applied Mathematics: For the Managerial, Life, and Social Sciences

Question

National Business Machines manufactures two models of fax machines: A and B. Each model A costs $$\$ 100$$ to make, and each model \(\mathrm{B}\) costs $$\$ 150$$. The profits are $$\$ 30$$ for each model \(\mathrm{A}\) and $$\$ 40$$ for each model B fax machine. If the total number of fax machines demanded per month does not exceed 2500 and the company has earmarked no more than $$\$ 600,000 $$ month for manufacturing costs, how many units of each model should National make each month in order to maximize its monthly profit? What is the optimal profit?

Step-by-Step Solution

Answer

To maximize the monthly profit, National Business Machines should manufacture 1000 units of model A and 1500 units of model B each month. The optimal profit in this case is $$\$ 90,000$$.

Step 1: Define the variables

Let x represent the number of model A fax machines, and y represent the number of model B fax machines.

Step 2: Formulate the objective function

The objective function represents the total profit. As the company wants to maximize its profit, we will use the given profits for each model and multiply them by the number of fax machines to derive the objective function: Objective function: \(P(x, y) = 30x + 40y\)

Step 3: Formulate the constraints

There are two constraints in this problem: the total number of fax machines and the budget for manufacturing costs. Constraint 1 (total number of fax machines): \(x + y \leq 2500\) Constraint 2 (total manufacturing costs): \(100x + 150y \leq 600,000\) Additionally, we have non-negativity constraints as the number of fax machines can't be negative: \(x \geq 0\) \(y \geq 0\)

Step 4: Find the feasible region

By graphing the constraints on a coordinate plane, we can find the feasible region representing the possible values for the number of model A and model B fax machines the company can manufacture. 1. \(x + y \leq 2500\): This divides the plane into two regions. The feasible region is below the line. 2. \(100x + 150y \leq 600,000\): This also divides the plane into two regions. The feasible region is below the line. 3. \(x \geq 0\) and \(y \geq 0\): These are the non-negativity constraints, which restrict the feasible region to the first quadrant. After graphing, the feasible region is the intersection of these regions in the first quadrant.

Step 5: Determine the optimal solution

To find the optimal solution, we must look at the vertices of the feasible region. Since the objective function is linear, the optimal solution must be on one of these vertices. Calculate the objective function value (profit) for each vertex, and choose the one that gives the maximum profit (objective function value). Vertices: 1. \((0, 0)\): Profit: \(P(0, 0) = 0\) 2. \((2500, 0)\): Profit: \(P(2500, 0) = 30(2500) = 75,000\) 3. \((0, 4000)\): Not in the feasible region as it violates the constraint \(x + y \leq 2500\) 4. Intersection point of the two constraint lines: To find the intersection point, solve the system of equations: \(\begin{cases} x + y = 2500 \\ 100x + 150y = 600,000 \end{cases}\) Solving this system, we get \(x = 1000\) and \(y = 1500\). Profit: \(P(1000, 1500) = 30(1000) + 40(1500) = 90,000\) The vertex with the maximum profit is the intersection point \((1000, 1500)\) with a profit of $$\$ 90,000$$. So, the company should manufacture 1000 units of model A and 1500 units of model B each month in order to maximize the monthly profit, with an optimal profit of $$\$ 90,000$$.

Key Concepts

Objective Function

In linear programming, the **objective function** is a mathematical expression that describes the problem's goal—typically to maximize or minimize some quantity. In the exercise, National Business Machines' goal is to maximize its profit from manufacturing two fax machine models, A and B.

The profit for each model is given as follows:

• Model A: Profit of \(30
• Model B: Profit of \)40

To formulate the objective function, we consider how many units (\(x\) for model A and \(y\) for model B) are produced and multiply them by their respective profits.

Thus, the objective function becomes:\[ P(x, y) = 30x + 40y \]where \(P(x, y)\) represents the total profit.
This function is crucial as it directs us towards the solution that achieves the highest possible profit within the given constraints.

Constraints

**Constraints** are the restrictions or limits on the decision variables given within the problem. In the exercise, the constraints ensure that the manufacturing process stays within given limits. They include:

• Total number of fax machines manufactured cannot exceed 2500:

\[ x + y \leq 2500 \]

• Budget for manufacturing models cannot exceed $600,000:

\[ 100x + 150y \leq 600,000 \]These constraints represent physical and economic limits that must be respected when determining the number of fax machines to produce. Additional non-negativity constraints:

• \(x \geq 0\)
• \(y \geq 0\)

These ensure that the number of products is feasible (non-negative numbers). Each constraint is a linear equation or inequality and when graphically plotted, they help define the boundaries within which the solution must lie.

Feasible Region

The **feasible region** in linear programming is the area on a graph where all constraints are satisfied. This region is formed by overlapping the constraints' boundaries. In our exercise, to locate this region, we graph the constraints:

• \(x + y \leq 2500\)
• \(100x + 150y \leq 600,000\)

Along with non-negativity constraints, these form a closed shape—usually a polygon in the first quadrant of the coordinate plane.

The feasible region is critical because it contains all potential solutions that comply with the problem's limitations. By checking the area of intersection of these plotted lines, we find the region where all conditions are true. This is the only region where one can find the feasible solutions to the problem.

Optimal Solution

The **optimal solution** is the best possible outcome, maximizing or minimizing the objective function within the feasible region. Here, it means maximizing the profit function:
\[ P(x, y) = 30x + 40y \]To find this, we examine the vertices (corners) of the feasible region. Experience tells us that linear programming problems involving polygons tend to have optimal solutions at these vertices.

In the exercise, the vertices of interest are:

• \((0, 0)\) – Profit: 0
• \((2500, 0)\) – Profit: 75,000
• Intersection of constraints (where \(x=1000\) and \(y=1500\)) – Profit: 90,000

Calculating the profits at these points, the intersection point \((1000, 1500)\) gives the maximum profit of \$90,000. Hence, producing 1000 units of Model A and 1500 units of Model B each month aligns with the company's goal to maximize its profit. This demonstrates how linear programming can effectively determine optimal decisions in resource allocation scenarios.