Asked by Bob
francois is stokcing a new fish pond on his farm. The pond is filling slowly and the volume of water in it can support a certain number of fish,(variable-F). F=42m+250, where m is the number of months the pond has been filling.
the pond was stocked initially with 110 fish, a population that will grow at a rate of 8 percent per month. P=110(1.08) to the power of m. this is the equation that models the population of fish after m number of months.
when will p equal f?
what will happen after this point has been reached? explain your answer.
the pond was stocked initially with 110 fish, a population that will grow at a rate of 8 percent per month. P=110(1.08) to the power of m. this is the equation that models the population of fish after m number of months.
when will p equal f?
what will happen after this point has been reached? explain your answer.
Answers
Answered by
drwls
P = F when
42 m + 250 = 110*1.08^m
That equation can be solved by iteration. If m = 30,
42 m + 250 = 1510
110*1.08^m = 1106
If m = 35
F = 42 m + 250 = 1720
P = 110*1.08^m = 1626
If m = 36
42 m + 250 = 1762
110*1.08^m = 1756
m is slightly more than 36 months
At longer times, P will exceed F, and fish will start to die because the pond is too small to support the population.
42 m + 250 = 110*1.08^m
That equation can be solved by iteration. If m = 30,
42 m + 250 = 1510
110*1.08^m = 1106
If m = 35
F = 42 m + 250 = 1720
P = 110*1.08^m = 1626
If m = 36
42 m + 250 = 1762
110*1.08^m = 1756
m is slightly more than 36 months
At longer times, P will exceed F, and fish will start to die because the pond is too small to support the population.
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