To find P(10), we substitute t=10 into the function:
P(10) = 3000 + 200(10) + 0.1(10)^2
P(10) = 3000 + 2000 + 1
P(10) = 5001
Therefore, P(10) represents the population of the community 10 years after it was incorporated, which is 5001.
Similarly, to find P(20):
P(20) = 3000 + 200(20) + 0.1(20)^2
P(20) = 3000 + 4000 + 4
P(20) = 7004
Therefore, P(20) represents the population of the community 20 years after it was incorporated, which is 7004.
Now, to find the average rate of change of P between t=10 and t=20, we use the formula:
Average rate of change = (P(20) - P(10)) / (20 - 10)
Average rate of change = (7004 - 5001) / 10
Average rate of change = 200.3
The average rate of change of P between t=10 and t=20 represents the average population growth per year during that time period.
To find when the population will be 6000, we set P(t) = 6000 and solve for t:
3000 + 200t + 0.1t^2 = 6000
0.1t^2 + 200t - 3000 = 0
Solving this quadratic equation, we get t ≈ 8.26 or t ≈ 71.74. Since t represents the number of years since 1985, we discard the negative value and take t ≈ 71.74 as the approximate time when the population will be 6000, which is approximately in the year 2056.
The population of a planned seaside community in Florida is given by the function P(t)=3000+200t+0.1t^2, where t represents the number of years since the community was incorporated in 1985.
find P(10) and P(20). What do these values represents?
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find the average rate of change of P between t=10 and t=20. What does the number represent?
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Approximately when will the population be 6000?
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