Question

Is this right???? Can someone please help?

Answer 1

This is a logistic growth model, so, we can set the function's differential equation equal to 0 and solve for p, then for y. So, let's take the derivative of this function and set it to 0:
$\frac{dp}{dt} = 0.005p(300-p) = 1.5p-0.005p^2$
$\frac{d^2p}{dt^2} = 1.5 - 0.01p =0$
$-0.01p = -1.5$
$p = \frac{-1.5}{-0.01} = 150$

Now, for a logistic model, the general solution is in the form of:
$p = \frac{L}{1+Ce^{-ky}}$
for
$\frac{dp}{dt} = kp(L-p)$

You haven't been given the 'C' constant, so I don't have a clue as to where to go after this. Sorry, we are yet to go over logistic growth in BC class. I could take you this far by just researching. If you have further questions, I'd be happy to answer them.