The population of the planet Magrathea doubles every 40 years. In the year zero the population was 20 million. The total land area of Magrathea is 200 million square kilometers. One square kilometer of land can house 200 Magratheans. Land that is not used for housing is used for agriculture. Write down an equation whose solution is the year in which there are 30 m2m2 of agricultural land available for each inhabitant. Solve this equation.

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].

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:

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:

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]

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