document.write( "Question 1175054: Assume there is a certain population of fish in a pond whose growth is described by the logistic equation. It is estimated that the carrying capacity for the pond is 1500 fish. Absent constraints, the population would grow by 200% a year.
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document.write( "If the population is given by po=600, then after one breeding season the population of the pond is given by
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document.write( "p1=
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document.write( "After two breeding seasons the population of the pond is given by
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document.write( "p2= \n" );
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Algebra.Com's Answer #800576 by Theo(13342)![]() ![]() You can put this solution on YOUR website! if the growth is 200% per year, then the growth rate is 2 per year. \n" ); document.write( "the formula for future value is f = p * (1 + r) ^ n \n" ); document.write( "f is the future value \n" ); document.write( "p is the present value \n" ); document.write( "r is the rat4e of growth per year. \n" ); document.write( "n is the number of years.\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "in your problem, the formula becomes f = 600 * (1 + 2) ^ n\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "p1 is the number at the end of 1 year. \n" ); document.write( "p2 is the number at the end of 2 years.\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "p1 = 600 * 3^1 = 1800 \n" ); document.write( "p2 = 600 * 3^2 = 5400\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "how does this work?\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "the population is 600 in p0. \n" ); document.write( "the growth is 200% per year, meaning the growth is 1200 from p0 to p1. \n" ); document.write( "the population is 600 + 1200 = 1800 in p1. \n" ); document.write( "the growth is 200% per year, meaning the growth is 3600 from p1 to p2. \n" ); document.write( "the population is 1800 + 3600 = 5400 in p2.\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "since the carrying capacity of the pond is 1500 fish, it's pretty clear that there's not a lot of room for growth. \n" ); document.write( "at the given growth rate of 200% per year, the pond is over capacity at the end of 1 year.\r \n" ); document.write( " \n" ); document.write( " \n" ); document.write( " \n" ); document.write( " \n" ); document.write( " \n" ); document.write( " \n" ); document.write( " \n" ); document.write( " \n" ); document.write( " \n" ); document.write( "\n" ); document.write( " \n" ); document.write( " |