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

Question

Kane Manufacturing has a division that produces two models of hibachis, model A and model B. To produce each model-A hibachi requires \(3 \mathrm{lb}\) of cast iron and \(6 \mathrm{~min}\) of labor. To produce each model-B hibachi requires \(4 \mathrm{lb}\) of cast iron and \(3 \mathrm{~min}\) of labor. The profit for each modelA hibachi is $$\$ 2$$, and the profit for each model-B hibachi is $$\$ 1.50 .$$ If \(1000 \mathrm{lb}\) of cast iron and 20 labor-hours are available for the production of hibachis each day, how many hibachis of each model should the division produce in order to maximize Kane's profit? What is the largest profit the company can realize? Is there any raw material left over?

Step-by-Step Solution

Answer

The company should produce 200 hibachis of model A and none of model B in order to maximize the profit, resulting in a largest profit of \(\$400\). There will be 400 lb of cast iron left over after production.

Step 1: Define the variables and the objective function

Let's define \(x\) as the number of model A hibachis produced, and \(y\) as the number of model B hibachis produced. The profit from producing hibachis is given by \(\$2\) per model A hibachi and \(\$1.50\) per model B hibachi. Therefore, the objective function to maximize the profit, denoted by \(P\), can be defined as: \[P(x, y) = 2x + 1.5y\]

Step 2: Develop the constraints

In order to develop the constraints for this linear programming problem, we have to consider the amount of cast iron and labor hours available for production. For cast iron, we have: \[3x + 4y \leq 1000\] For labor hours, we need to convert the available labor hours (20 hours) to minutes. Since there are 60 minutes in an hour, we have: \[6x + 3y \leq 20 * 60\] \[6x + 3y \leq 1200\] Also, note that the number of hibachis produced cannot be negative: \[x \geq 0\] \[y \geq 0\]

Step 3: Solve the linear programming problem

The linear programming problem can be written as: \[ \begin{aligned} \text{Maximize } & P(x, y) = 2x + 1.5y \\ \text{Subject to } & 3x + 4y \leq 1000 \\ & 6x + 3y \leq 1200 \\ & x \geq 0 \\ & y \geq 0 \\ \end{aligned} \] In order to solve this linear programming problem, we can graph the constraints and find the feasible region. Then, we will identify the vertices of this region and evaluate the objective function at each vertex to find the optimal solution.

Step 4: Find the optimal solution

After graphing the constraints and finding the feasible region, we identify the vertices of the feasible region as: \((0, 0)\), \((200, 0)\), \((0, 250)\), and \((100, 100)\). Now, we evaluate the objective function at each vertex: \[ \begin{aligned} P(0, 0) &= 2(0) + 1.5(0) &= 0 \\ P(200, 0) &= 2(200) + 1.5(0) &= 400 \\ P(0, 250) &= 2(0) + 1.5(250) &= 375 \\ P(100, 100) &= 2(100) + 1.5(100) &= 350 \\ \end{aligned} \] The largest profit occurs at the vertex \((200, 0)\), where \(x=200\) and \(y=0\). Therefore, the company should produce 200 hibachis of model A and no hibachis of model B in order to maximize the profit.

Step 5: Calculate the largest profit

Plugging the optimal values for \(x\) and \(y\) into the objective function, we can find the largest profit the company can realize: \[P(200, 0) = 2(200) + 1.5(0) = 400\] The largest profit the company can realize is \(\$400\).

Step 6: Determine if any raw material is left over

To find out if there is any cast iron left after producing the optimal number of hibachis, we plug the optimal values of \(x\) and \(y\) back into the cast iron constraint: \[3(200) + 4(0) = 600\] Since the company initially had \(1000 \mathrm{lb}\) of cast iron, the amount of cast iron left over after production is \(1000 - 600 = 400 \mathrm{lb}\).

Key Concepts

Optimization in Management

In the context of business mathematics, optimization plays a crucial role in management. It involves making the best possible decisions within the given constraints to achieve maximum efficiency, productivity, or profitability. In our example, Kane Manufacturing is aiming to optimize its profit by determining the ideal quantity of two hibachi models to produce while considering resource limitations.

Optimization problems, like the one presented, are usually addressed using linear programming—a mathematical technique for finding the most favorable outcome such as minimum cost or maximum revenue. It's essential for managers to understand how to formulate these problems with an objective function—here, it's the profit function, \(P(x, y) = 2x + 1.5y\), which needs to be maximized. By doing so, they can instruct production to focus on goods that yield higher profits, create more efficient manufacturing schedules, and better manage resources.

Linear Programming Constraints

When laying out a linear programming model, it is fundamental to recognize the constraints that restrict the feasible solutions. Constraints represent the limits within which the solution must fit, such as budget limitations, production capacities, or resource availability.

In the exercise, the constraints are defined by the availability of cast iron and labor hours, represented mathematically by \(3x + 4y \leq 1000\) for cast iron and \(6x + 3y \leq 1200\) for labor minutes. There's also the non-negativity constraint, indicating that negative production levels are not possible: \((x \geq 0\) and \((y \geq 0\)). These constraints are vital in any linear programming problem because they outline the feasible region and directly influence the solution.

Feasible Region in Linear Programming

The feasible region in linear programming is a graphical representation that shows all the potential solutions that satisfy the problem's constraints. It is where all constraints overlap and is bounded by the constraint lines on a graph.

In our case with Kane Manufacturing, the feasible region is determined by plotting the constraints onto a graph and finding where they intersect within the first quadrant, as the production quantities cannot be negative. The vertices of this region are calculated points where these lines intersect, and they can provide possible solutions to the linear programming problem. Importantly, the optimal solution lies at one of the vertices of the feasible region. After evaluating the profit function at each vertex, we can deduce that producing 200 model A hibachis maximizes the profit, clearly showing how the feasible region plays a key role in decision-making in linear programming scenarios.