Problem 31 - Applied Mathematics: For the Managerial, Life, and Social Sciences

Question

Kane Manufacturing has a division that produces two models of fireplace grates, model A and model B. To produce each model A grate requires \(3 \mathrm{lb}\) of cast iron and \(6 \mathrm{~min}\) of labor. To produce each model B grate requires \(4 \mathrm{lb}\) of cast iron and 3 min of labor. The profit for each model A grate is $$\$ 2.00$$, and the profit for each model B grate is $$\$ 1.50$$. If \(1000 \mathrm{lb}\) of cast iron and 20 labor-hours are available for the production of fireplace grates per day, how many grates of each model should the division produce in order to maximize Kane's profit? What is the optimal profit?

Step-by-Step Solution

Answer

To maximize Kane's profit, the division should produce 400 Model A grates and no Model B grates per day, resulting in an optimal profit of $800 per day.

Step 1: Understand the problem and define variables

Let x be the number of Model A grates and y be the number of Model B grates produced per day. Given, the profit for each Model A grate is \(2.00, and the profit for each Model B grate is \)1.50. We want to maximize the profit, which can be represented as: Profit = 2x + 1.5y We also have constraints on cast iron and labor which are: - 3x + 4y ≤ 1000 (cast iron constraint) - 6x + 3y ≤ 2400 (labor constraint) We will now graph the feasible region and identify the corner points.

Step 2: Plot the constraints and find the feasible region

First, we need to convert both inequalities into equalities and plot the straight lines for each constraint: - 3x + 4y = 1000 (cast iron constraint) - 6x + 3y = 2400 (labor constraint) Now, we will plot both lines on a graph where x-axis represents the number of Model A grates (x) and y-axis represents the number of Model B grates (y). Both x and y must be non-negative, as it doesn't make sense to produce a negative number of grates. Don't forget to shade the feasible region, which includes all points that satisfy both constraints.

Step 3: Identify the corner points

We now need to find the corner points of the feasible region since the maximum profit will likely occur at one of these points. To find these points, we can use the intersection points of the constraint lines and the axes: 1. (0,0) - Origin 2. (0,250) - Intersection of the cast iron constraint and the y-axis 3. (400,0) - Intersection of the labor constraint and the x-axis 4. A point where the cast iron constraint intersects the labor constraint. For the last corner point, we can solve the system of linear equations: - 3x + 4y = 1000 - 6x + 3y = 2400

Step 4: Solve the system of linear equations

Now, we need to solve the system of linear equations to find the intersection point of the cast iron constraint and the labor constraint: 1. Multiply the first equation by 2 to make the coefficients of x the same in both equations: 6x + 8y = 2000 2. Subtract the second equation from the first: 5y = -400 3. Divide both sides by 5: y = -80 Since y must be non-negative, we can discard this solution as not feasible. Therefore, the feasible corner points are: 1. (0,0) 2. (0,250) 3. (400,0)

Step 5: Calculate the profit at each corner point

Now, we need to calculate the profit (2x + 1.5y) at each corner point: 1. (0,0) - Profit = $$(2\times0) + (1.5\times0) = \$0$$ 2. (0,250) - Profit = $$(2\times0) + (1.5\times250) = \$375$$ 3. (400,0) - Profit = $$(2\times400) + (1.5\times0) = \$800$$

Step 6: Determine the optimal solution

The maximum profit occurs at the corner point (400,0), with a profit of $800. Therefore, the division should produce 400 Model A grates and no Model B grates to maximize the profit, with the optimal profit being $800 per day.

Key Concepts

Profit Maximization

In linear programming, profit maximization is the primary goal when dealing with resource allocation in manufacturing businesses. The aim is to determine the best output levels of different products that bring in the most profit while still adhering to given constraints. Given the profitability of manufacturing models, profit can be expressed as a linear function:

• For Model A grates, with a profit of \(2 each, and Model B grates, with a profit of \)1.50 each, the company's total profit is calculated as \( 2x + 1.5y \).

The trick is to identify the combination of x (the number of Model A grates) and y (the number of Model B grates) that maximizes this profit equation. This involves analyzing the feasible region constrained by available resources such as materials and labor.
This process often requires calculating at the intersection points within the feasible region, as maximum profit tends to occur at these points, known as corner points.

Constraints

Constraints in a linear programming problem refer to the limitations or restrictions placed on the problem's decision variables. They carve out the conditions under which the objectives (like profit maximization) must be achieved. Here, the two main constraints involve resources available:

• **Cast Iron Constraint**: Given by \( 3x + 4y \leq 1000 \), indicating the total weight of cast iron available per day for producing both grate models.


• **Labor Constraint**: Represented by \( 6x + 3y \leq 2400 \), implying the available labor hours in minutes for daily production.

These constraints define how much of each resource can be used, ensuring the production plan remains feasible. They restrict the number of products manufactured to adhere to resource availability, forming the boundary lines on a graph for the feasible region.

Feasible Region

In linear programming, the feasible region is the area on a graph that represents all possible combinations of decision variables that satisfy the problem's constraints. It's typically a polygon in which all points comply with every constraint of the linear problem. The problem's solution lies somewhere within this shaded area.
To find the feasible region in this case, graph the equations for each constraint (changed from inequalities to equalities for graphing):

• \( 3x + 4y = 1000 \) (Cast Iron)
• \( 6x + 3y = 2400 \) (Labor)

The feasible region is bounded by these lines and the non-negative conditions \( x \geq 0 \) and \( y \geq 0 \), ensuring no negative production occurs. Remember, the feasible region is vital because it encompasses all potential solutions, from which the optimal point is selected.

Corner Points

Corner points are the vertices of the feasible region in a linear programming problem. These are crucial because the optimal solution for maximizing (or minimizing) the objective function often lies at one of these points. Finding these points involves determining where the lines of your constraints intersect each other and the axes.For the fireplace grates example, calculate the corner points by identifying where each pair of constraint lines intersects using algebraic reduction. In this problem's simplified scenario, the primary corner points are:

• \((0,0)\)
• \((0,250)\)
• \((400,0)\)

These intersections mark the endpoints of possible solutions within the feasible region. After identifying these points, calculate the potential profit at each one to determine where the maximum profit lies, thus guiding the optimal production strategy.