<p>Probably is easy i just forgot how to do it</p>
<p>The number of moose in a national park is modeled by the function M that satisfies the logistic differential equation dM/dt = 0.6M(1 - M/200), where t is the time in years and M(0) = 50. What is lim t-> infinity M(t)?</p>
<p>a) 50 B) 200 c) 500 d) 1000 e) 2000</p>
<p>Its off an old AP BC test</p>
<p>the answer is (b)</p>
<p>I think it's 200 because that's the carrying capacity, but I'm not positive..</p>
<p>you could always do partial fractions and separating variables though, I suppose</p>
<p>projekt howd you figure it out?</p>
<p>bummpp anyone?</p>
<p>i dont think logistics are on the bc test anymore....do you know what year it was from?</p>
<p>When M equals 200, dM/dt equals zero. The function approaches this value and stays at it, since as it gets closer, it changes at an ever lesser rate.</p>
<p>wow i just did this problem a few minutes ago.. in a logistic diff eq, dM/dt (where M is population and t is time) is equal to kM(1- M/C), where k is a constant and C is the carrying capacity (aka the highest population that will ever happen), so, you can use separation of variables to solve the diff eq and understand that you will get M(t) = 200/(1+3e^-.6t) ... as t approaches infinity, the that part of the denominator goes to zero, so you get 200/1.</p>
<p>wow i just did this problem a few minutes ago.. in a logistic diff eq, dM/dt (where M is population and t is time) is equal to kM(1- M/C), where k is a constant and C is the carrying capacity (aka the highest population that will ever happen), so, you can use separation of variables to solve the diff eq and understand that you will get M(t) = 200/(1+3e^-.6t) ... as t approaches infinity, the that part of the denominator goes to zero, so you get 200/1.</p>
<p>here's the formula and solution to the problem
<a href="http://putfile.com/pic.php?pic=4/11315132351.jpg&s=x402%5B/url%5D">http://putfile.com/pic.php?pic=4/11315132351.jpg&s=x402</a></p>
