Respuesta :
Answer:
The year in which there are 30 m² of agricultural land available for each inhabitant is approximately 438.286 years
Step-by-step explanation:
The growth rate of the planet Magrathea = Double every 40 years
The start population of Magrathea = 20 million
The area of the Magrathea = 200 million km²
The area that can hose 200 Magratheans = 1 km²
The land used for agriculture = The land not used for housing
The equation for a population that doubles every 20 years is given as follows;
[tex]y = a \cdot b^t[/tex]
[tex]y = a \cdot (1 + r)^{t}[/tex]
We have;
[tex]2 = (1 + r)^{40}[/tex]
㏑2 = 40·㏑(1 + r)
㏑(1 + r) = ㏑(2)/40
1 + r = ㏑(2)/40
1 + r = e^(㏑(2)/40)
r = 1 - e^(㏑(2)/40) = 0.0174796921
30 m²
200·x
The area per individual = (1/200) km²
The total number
We have;
(x/200) km²) + (x × 0.00003 km²) = 200,000,000 km²
x × ((1/200) km² + 0.00003 km²) = 200,000,000 km²
x = 200,000,000 km²/(((1/200) km² + 0.00003 km²)) = 3.97614314 × 10¹⁰
The population at the time = 3.97614314 × 10¹⁰
We have;
3.97614314 × 10¹⁰/(2×10⁷) = [tex]e^{(ln(2)/40)}^{t}[/tex]
t = ㏑((3.97614314 × 10¹⁰/(2×10⁷)))/(㏑[tex]e^{(ln(2)/40)}[/tex])) ≈ 438.286 years
The year in which there are 30 m² of agricultural land available for each inhabitant ≈ 438.286 years
The equation for [tex]30\,\frac{m^{2}}{hab}[/tex] is [tex]30\times 10^{-6} = \frac{10}{2^{\frac{t}{40} }}-\frac{1}{200}[/tex], whose solution is [tex]t \approx 438.28\,yr[/tex].
How to calculate the time associate to a given available area per capita for agricultural use
Dimensionally speaking, the given available area per capita for agricultural use ([tex]r[/tex]), in square kilometers per habitant, is defined by the following formula:
[tex]r = \frac{A_{T}-A_{p}}{n}[/tex] (1)
Where:
- [tex]A_{T}[/tex] - Total area of the planet, in square kilometers.
- [tex]A_{p}[/tex] - Populated area of the planet, in square kilometers.
- [tex]n[/tex] - Total population of the planet, in habitants.
The populated area is defined by this formula:
[tex]A_{p} = m\cdot n[/tex] (2)
Where [tex]m[/tex] is the populated area per capita, in square kilometers per habitant.
By (2) in (1):
[tex]r = \frac{A_{T}}{n}-m[/tex] (1b)
And the population as function of time can be modeled after this geometric progression:
[tex]n = p_{o}\cdot 2^{\frac{t}{\tau} }[/tex] (3)
Where:
- [tex]t[/tex] - Time, in years.
- [tex]\tau[/tex] - Doubling time, in years.
- [tex]p_{o}[/tex] - Initial population, in habitants.
By (3) in (1b):
[tex]r = \frac{A_{T}}{p_{o}\cdot 2^{\frac{t}{\tau} }} -m[/tex] (1c)
If we know that [tex]p_{o} = 20\times 10^{6}[/tex], [tex]\tau = 20\,yr[/tex], [tex]A_{T} = 200\times 10^{6}[/tex], [tex]m = \frac{1}{200}\,km^{2}[/tex] and [tex]r = 30\times 10^{-6}\,km^{2}[/tex], then the time is:
[tex]30\times 10^{-6}\,km^{2} = \frac{200\times 10^{6}\,km^{2}}{(20\times 10^{6})\cdot 2^{\frac{t}{40\,yr} }} - \frac{1}{200}\,km^{2}[/tex]
[tex]30\times 10^{-6} = \frac{10}{2^{\frac{t}{40} }}-\frac{1}{200}[/tex]
[tex](5.03\times 10^{-3})\cdot 2^{\frac{t}{40} } = 10[/tex]
[tex]2^{\frac{t}{40} } = 1988.072[/tex]
[tex]\frac{t}{40} \cdot \log_{2} 2 = \log_{2} 1988.072[/tex]
[tex]\frac{t}{40} \approx 10.957[/tex]
[tex]t \approx 438.28\,yr[/tex]
The equation for [tex]30\,\frac{m^{2}}{hab}[/tex] is [tex]30\times 10^{-6} = \frac{10}{2^{\frac{t}{40} }}-\frac{1}{200}[/tex], whose solution is [tex]t \approx 438.28\,yr[/tex]. [tex]\blacksquare[/tex]
To learn more on populations, we kindly invite to check this verified question: https://brainly.com/question/905400
