Calculus BC Question

Question

After pollution-abatement efforts, conservation researchers introduce 100 trout into a small lake. The researchers predict that after m months the rate of growth F of the trout population will be modeled by the differential equation dF/dm = 0.0002F(600-F)

(a) How large is the trout population when it is growing the fastest?

My solution: set the derivative of growth rate equal to 0: dF/dm = 0.0002F(600-F) = 0

The textbook suggests setting the second derivative equivalent to 0. If such, doesn’t that mean dF/dm is rate of growth, and F is the population? I don’t understand why the second derivative is necessary at this point?

Answer 1

I think you generally have the right idea but you may have applied the test for a maximum point to the wrong function.

Consider a function (say F(x)) and the tangent of that function at some point (lets all this point p). The derivative of F at point p is the slope of the tangent. For the function to be a maximum or minimum at point p, its tangent must be horizontal and so its slope and hence gradient is 0. Hence to find the maximum or minimum, we need to solve for dF/dx=0.

Now, the question asks "How large is the trout population when it is growing the fastest?". To answer this question, you first need to know "when it is growing fastest".

Note that growing means rate of change or how quickly something is changing which means you need to know the slope of the graph. In other words, the growth is derivative, dF/dm.

But remember that we need to know "when it is growing fastest". This is means we need to find the maximum of the function dF/dm. To do this, we need to solve for d(dF/dm)/dm=0. (i.e. Instead of applying the maximum or minimum criteria to F, we apply it to dF/dm since we are looking for the maximum of dF/dm, not F). And d(dF/dm)/dm is just the second derivative of F.

Once you solve for m by setting the second derivate of F to zero, you need to plug it back in to F to answer the original question.