A fish population grows according to

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

Problem

A fish population grows according to dydt=0.05 y (1000−y)\dfrac{dy}{dt}=0.05\,y\,(1000-y), where yy is the number of fish. Which statement correctly compares the instantaneous growth rates when y=200y=200 and when y=800y=800? (a) The instantaneous growth rate at y=200y=200 is less than that at y=800y=800, and both rates are positive. (b) The instantaneous growth rates at y=200y=200 and y=800y=800 are equal. (c) The instantaneous growth rate at y=200y=200 is greater than that at y=800y=800. (d) The instantaneous growth rate at y=200y=200 is negative, whereas at y=800y=800 it is positive.

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