Analysis of Models Algebraically

Introduction

So far, we have illustrated how a model is turned into a mathematical equation how the variables in a business situation are related. Given the value of one of the variables, we able to calculate the other by substitution. In this case, there is no need for graphical analysis to find answers for our questions.

Examples

Example 1: The monthly income of a salesmen is calculated from the equation: y = 0.03x + 4000
Where x is the monthly sales, in dollars, and y is the monthly income, in dollars

Questions:
  1. If the salesmen makes sales of $10,000, find the corresponding income
  2. If the salesmen wants to earn $7,000 in a month, find the sales that should be generated
Solution:
  1. y=0.03 × 10,000 + 4,000 = 300 + 4,000 = $4,300
  2. $7,000 = 0.03x + 4,000; after substitution, x = 3,000 / 0.03 = $100,000
 Example 2: A production manager studies the relationship between the length of time an employee has been working in a day and the number of items made in the day. He uses the following model: N = 20t − 10
Where t is the time an employee has been working in a day, in hours and N is the number of items made in that day.

Questions:
  1. How many items will an employee make in 5 hours? in 10 hours
  2. How long would an employee have to work to make 250 items?
Solution:
  1. In 5 hours, N = 20 × 5 − 10 = 90 items
  2. In 10 hours, N = 20 × 10 − 10 = 190 items
  3. To make 250 items, 250 = 20t − 10; 20t = 240; and after substitution, t = 12 hours
    4.

Exercises with Solutions

Please access the following link to do additional exercises on how to analyze data algebraically.
Exercises with solution on analyzing data algebraically


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Date of last modification: 2022