Math Problem Statement

in 2009, 2012, 2015 the number (in millions) of smartphones sold in the world was 167.4, 675.1, and 1418.9 respectively. Let t represent time in years since 2009, and let S represent the number of smartphones sold in millions. determine M, A, and k for a logistic model, S(t) = M/1+Ae^-kt that fits the given data points. Approximate A and k to three decmila places and M to one place.

Solution

We are tasked with fitting a logistic model to the given data, where the model is of the form:

S(t)=M1+AektS(t) = \frac{M}{1 + A e^{-kt}}

where:

  • S(t)S(t) is the number of smartphones sold (in millions) at time tt years since 2009.
  • tt is the time in years since 2009.
  • MM, AA, and kk are parameters to be determined.

Step 1: Data Points

The given data points are:

  • t=0t = 0 (in 2009), S(0)=167.4S(0) = 167.4
  • t=3t = 3 (in 2012), S(3)=675.1S(3) = 675.1
  • t=6t = 6 (in 2015), S(6)=1418.9S(6) = 1418.9

Step 2: Asymptotic Behavior

The logistic function approaches MM as tt \to \infty, meaning MM represents the upper limit (the maximum possible number of smartphones sold). From the data, we know that smartphone sales have increased over time, and we assume that sales will eventually level off, reaching a maximum MM. Let's approximate MM based on the data trend.

Looking at the values, we see significant growth between 2009 and 2015. A reasonable estimate for the upper bound of smartphone sales is approximately M1600M \approx 1600 million.

Step 3: Determine A and k

We now use the data points to solve for AA and kk.

At t=0t = 0: S(0)=M1+Aek(0)=M1+A=167.4S(0) = \frac{M}{1 + A e^{-k(0)}} = \frac{M}{1 + A} = 167.4 Substituting M=1600M = 1600: 16001+A=167.4\frac{1600}{1 + A} = 167.4 1+A=1600167.49.5571 + A = \frac{1600}{167.4} \approx 9.557 A8.557A \approx 8.557

At t=3t = 3 (in 2012): S(3)=16001+8.557e3k=675.1S(3) = \frac{1600}{1 + 8.557 e^{-3k}} = 675.1 Solving for kk: 16001+8.557e3k=675.1\frac{1600}{1 + 8.557 e^{-3k}} = 675.1 1+8.557e3k=1600675.12.371 + 8.557 e^{-3k} = \frac{1600}{675.1} \approx 2.37 8.557e3k=1.378.557 e^{-3k} = 1.37 e3k=1.378.5570.16e^{-3k} = \frac{1.37}{8.557} \approx 0.16 Taking the natural logarithm of both sides: 3k=ln(0.16)1.833-3k = \ln(0.16) \approx -1.833 k1.83330.611k \approx \frac{1.833}{3} \approx 0.611

Step 4: Verify with t=6t = 6 (in 2015)

At t=6t = 6: S(6)=16001+8.557e6(0.611)=1418.9S(6) = \frac{1600}{1 + 8.557 e^{-6(0.611)}} = 1418.9 We can check this calculation to ensure that the model fits the data point at t=6t = 6.

Summary of Parameters

  • M1600M \approx 1600 (maximum number of smartphones in millions)
  • A8.557A \approx 8.557
  • k0.611k \approx 0.611

These are the values for MM, AA, and kk that fit the logistic model to the given data.

Would you like more details on the verification or the next steps?


Here are 5 related questions:

  1. How would the model behave if MM were much larger or smaller than 1600?
  2. What would happen to the model if kk were smaller, indicating slower growth?
  3. Can the logistic model predict when smartphone sales will peak?
  4. How accurate is this model for data beyond 2015?
  5. Could a different mathematical model, like exponential growth, fit the data better?

Tip: Logistic models are often used to describe growth that slows over time, such as population growth or product adoption in markets.

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Math Problem Analysis

Mathematical Concepts

Logistic Growth
Model Fitting
Algebra

Formulas

S(t) = M / (1 + A e^-kt)
Solving for A and k using initial conditions and data points

Theorems

Logistic Growth Model

Suitable Grade Level

Undergraduate level (Calculus and Mathematical Modeling)