Question 179110: The fish population in a certain lake rises and falls accourding to the formula:
F=1000(30+15t-t^2)
Here, F is the number of fish at t, time, where t is measured in years since January 1,2002.
When will the fish population be the same as it was on January 1,2002?
All of the fish will die after approximately t=?
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! The fish population in a certain lake rises and falls accourding to the formula:
F=1000(30+15t-t^2)
Here, F is the number of fish at t, time, where t is measured in years since January 1,2002.
When will the fish population be the same as it was on January 1,2002?
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In 2002, t = 0
Then F = 1000(30 + 15*0-0^2) = 30,000
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Solve F=30,000 for another value of "t":
1000(30+15t-t^2) = 30000
Rearrange:-t^2 + 15t + 30 = 30
-t^2+15t = 0
-t(t-15) = 0
t = 0 or t=15
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Year? 2002 + 15 = 2017
There will be 30,000 fish in 2017
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All of the fish will die after approximately t=?
F=1000(30+15t-t^2)
Let F=0 and solve for "t"
1000(30+15t-t^2) = 0
30 + 15t -t^2 = 0
t^2 - 15t -30 = 0
t = [15 +- sqrt(225 - 4*-30)]/2
t = [15 +- sqrt(355)]/2
Positive solution:
t = 16.92
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No fish in 2002 + 16.92 = 2019
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Cheers,
Stan H.
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