Math Problem Statement

Kane Manufacturing has a division that produces two models of fireplace grates, x units of model A and y units of model B. To produce each model A grate requires 3 lb of cast iron and 6 min of labor. To produce each model B grate requires 4 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. Also, 1000 lb of cast iron and 20 labor-hours are available for the production of fireplace grates per day.

Because of a backlog of orders for model A grates, Kane's manager had decided to produce at least 150 of these grates a day. Operating under this additional constraint, how many grates of each model should Kane produce to maximize profit?

Solution

We need to formulate this problem as a linear programming (LP) model in order to maximize Kane Manufacturing's profit, given the resource constraints and requirements for each model of fireplace grate. Here's the step-by-step process.

1. Define the variables:

Let:

  • xx = the number of model A grates produced per day
  • yy = the number of model B grates produced per day

2. Objective function:

The profit for model A is $2 per grate, and for model B is $1.50 per grate. We want to maximize the total profit: Maximize P=2x+1.5y\text{Maximize } P = 2x + 1.5y where PP is the total profit.

3. Constraints:

There are several constraints based on the available resources and the requirement for producing at least 150 model A grates.

Cast iron constraint:

Each model A grate uses 3 lb of cast iron, and each model B grate uses 4 lb of cast iron. The total cast iron available is 1000 lb. So: 3x+4y10003x + 4y \leq 1000

Labor constraint:

Each model A grate requires 6 minutes of labor, and each model B grate requires 3 minutes of labor. There are 20 labor-hours available, which is equivalent to 1200 minutes. So: 6x+3y12006x + 3y \leq 1200

Production constraint for model A:

The company has decided to produce at least 150 model A grates per day: x150x \geq 150

Non-negativity constraints:

The number of grates produced must be non-negative: x0,y0x \geq 0, \quad y \geq 0

4. Formulation of the linear programming problem:

We now have the complete LP formulation:

Maximize P=2x+1.5y\text{Maximize } P = 2x + 1.5y subject to: 3x+4y1000(Cast iron constraint)3x + 4y \leq 1000 \quad (\text{Cast iron constraint}) 6x+3y1200(Labor constraint)6x + 3y \leq 1200 \quad (\text{Labor constraint}) x150(Minimum production of model A)x \geq 150 \quad (\text{Minimum production of model A}) x0,y0(Non-negativity constraints)x \geq 0, \quad y \geq 0 \quad (\text{Non-negativity constraints})

5. Solve the system:

To solve this linear programming problem, you can use methods like the graphical method (since there are only two variables) or use a simplex algorithm.

Graphical Solution Approach (Basic Outline):

  1. Plot the constraint inequalities on the xyxy-plane.
  2. Identify the feasible region where all the inequalities overlap.
  3. Evaluate the objective function at each corner (vertex) of the feasible region, because the maximum profit will occur at one of these points.

Finding the vertices:

  • The intersection of the constraints will give us the vertices (corner points) of the feasible region.
  • After determining the coordinates of these points, we evaluate the profit function P=2x+1.5yP = 2x + 1.5y at each vertex to find which combination maximizes profit.

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Math Problem Analysis

Mathematical Concepts

Linear Programming
Optimization
Inequalities

Formulas

Profit function: P = 2x + 1.5y
Cast iron constraint: 3x + 4y ≤ 1000
Labor constraint: 6x + 3y ≤ 1200
Minimum production constraint for model A: x ≥ 150

Theorems

Simplex Algorithm
Graphical Method for Linear Programming

Suitable Grade Level

Grades 11-12, College