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Two years ago the population of a town was 40000. The population of the town at present has reached 44100. Calculate the population growth rate of the city.​

Sagot :

Answer:

5% annual population growth rate

Step-by-step explanation:

Let the percent the population grows by be [tex]X\%[/tex]. The total population, [tex]f(x)[/tex], after [tex]t[/tex] years can be modeled by the function:

[tex]f(x)=40,000\cdot (\frac{X}{100}+1)^t[/tex]

Why?

Let's take a look at a simple example. If we said a number [tex]n[/tex] grew by 10%, we could represent the number after it grew by multiplying [tex]n[/tex] by [tex]1.10[/tex]. This is because growing by 10% is equivalent to taking [tex]100\%+10\%=110\%[/tex] of that number and we convert a percentage to a decimal by dividing by 100.

Therefore, if the population grew [tex]X\%[/tex], we would divide it by 100 to convert it to a decimal, then add 1 (100%) and raise to the power of [tex]t[/tex] (number of years) to multiply by the initial population of 40,000 to get the total population after [tex]t[/tex] years.

Since the population of the town after two years is 44,100, substitute [tex]f(x)=44,100[/tex] and [tex]t=2[/tex] into [tex]f(x)=40,000\cdot (\frac{X}{100}+1)^t[/tex]:

[tex]44,100=40,000\cdot (\frac{X}{100}+1)^2,\\\\(\frac{X}{100}+1)^2=\frac{44,100}{40,000},\\\\(\frac{X}{100}+1)^2=1.1025,\\\\(\frac{X}{100}+1)^2=\pm \sqrt{1.1025},\\\\\begin{cases}\frac{X}{100}+1=1.05,\frac{X}{100}=0.05, X=\boxed{5\%},\\*\text{negative case is extraneous since X must be positive}\end{cases}[/tex]

Therefore, the city has an annual population growth rate of 5%.