A fish population (in fish) grows in a lake

MATHEMATICS · Single correct · AP · Calculus BC · Differential Equations · Differential Equations

Problem

A fish population F(t)F(t) (in fish) grows in a lake. Biologists believe the growth rate is proportional to both the current population and the difference between the population and a carrying capacity of 80008000 fish, with constant of proportionality 0.050.05 per month. Which differential equation models F(t)F(t)? (a) dFdt=0.05F (8000−F)\displaystyle \frac{dF}{dt}=0.05F\,(8000-F) (b) dFdt=0.05F(1−F8000)\displaystyle \frac{dF}{dt}=0.05F\Bigl(1-\dfrac{F}{8000}\Bigr) (c) dFdt=0.05 (8000−F)\displaystyle \frac{dF}{dt}=0.05\,(8000-F) (d) dFdt=0.05F(1+F8000)\displaystyle \frac{dF}{dt}=0.05F\Bigl(1+\dfrac{F}{8000}\Bigr)

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