Math Problem Statement

Guadalupe runs a factory that makes DVD players. Each T50 takes 4 ounces of plastic and 4 ounces of metal. Each G150 requires 2 ounces of plastic and 8 ounces of metal. The factory has 164 ounces of plastic, 464 ounces of metal available, with a maximum of 20 T50 that can be built each week. If each T50 generates $10 in profit, and each G150 generates $18, how many of each of the DVD players should Guadalupe have the factory make each week to make the most profit?

Solution

The question asks about optimizing production to maximize profit, and this can be framed as a linear programming problem. Let's break down the setup.

Problem Summary:

  • T50 DVD Player:

    • Uses 4 ounces of plastic
    • Uses 4 ounces of metal
    • Profit: $10 per unit
    • Max units: 20 units per week
  • G150 DVD Player:

    • Uses 2 ounces of plastic
    • Uses 8 ounces of metal
    • Profit: $18 per unit
  • Available materials:

    • 164 ounces of plastic
    • 464 ounces of metal

Goal:

Maximize profit by determining how many units of T50 and G150 should be produced, subject to material and production constraints.


Variables:

  • Let xx = number of T50 units
  • Let yy = number of G150 units

Constraints:

  1. Plastic Constraint:
    4x+2y1644x + 2y \leq 164 (Total plastic used)

  2. Metal Constraint:
    4x+8y4644x + 8y \leq 464 (Total metal used)

  3. Production Limit for T50:
    x20x \leq 20

  4. Non-negativity:
    x0x \geq 0, y0y \geq 0


Objective Function:

Maximize profit:
Profit=10x+18y\text{Profit} = 10x + 18y


Solution Approach:

We will solve this using linear programming to find the values of xx and yy that maximize the profit while satisfying the constraints.


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Math Problem Analysis

Mathematical Concepts

Linear Programming
Optimization
Algebra

Formulas

Plastic constraint: 4x + 2y ≤ 164
Metal constraint: 4x + 8y ≤ 464
Production limit for T50: x ≤ 20
Profit function: 10x + 18y

Theorems

Linear Programming Theorem

Suitable Grade Level

Grades 11-12, College Level