Math Problem Statement

In the year 2008, a company made $3.5 million in profit. For each consecutive year after that, their profit increased by 13%. How much would the company's profit be in the year 2010, to the nearest tenth of a million dollars?

Solution

The company's profit follows an exponential growth pattern, given that it increases by 13% each year.

Step 1: Define the formula

The formula for exponential growth is:

Pn=P0(1+r)nP_n = P_0 (1 + r)^n

where:

  • PnP_n is the profit after nn years
  • P0=3.5P_0 = 3.5 million (initial profit in 2008)
  • r=0.13r = 0.13 (13% growth rate)
  • nn is the number of years after 2008

Step 2: Find P2P_2 for the year 2010

Since 2010 is 2 years after 2008 (n=2n = 2), we calculate:

P2=3.5(1.13)2P_2 = 3.5 (1.13)^2

P2=3.5×1.2769P_2 = 3.5 \times 1.2769

P24.4692P_2 \approx 4.4692

Step 3: Round to the nearest tenth

P24.5P_2 \approx 4.5

Final Answer:

The company's profit in 2010 would be 4.5 million dollars.

Would you like a breakdown of a different approach or a verification of the calculation?

Here are five related questions you might find interesting:

  1. What would the profit be in 2015 using the same formula?
  2. How long would it take for the profit to double?
  3. What is the total profit earned from 2008 to 2012?
  4. If the growth rate changed to 10% per year, how much would the profit be in 2012?
  5. What is the average annual profit from 2008 to 2015?

Tip: Exponential growth problems often require logarithms when solving for time!

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Math Problem Analysis

Mathematical Concepts

Exponential Growth
Percentage Increase
Algebra

Formulas

P_n = P_0 (1 + r)^n

Theorems

Exponential Growth Formula

Suitable Grade Level

Grades 9-12