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:
[tex] \frac{dp}{dt} = 0.005p(300-p) = 1.5p-0.005p^2 [/tex]
[tex] \frac{d^2p}{dt^2} = 1.5 - 0.01p =0 [/tex]
[tex]-0.01p = -1.5[/tex]
[tex]p = \frac{-1.5}{-0.01} = 150 [/tex]
Now, for a logistic model, the general solution is in the form of:
[tex]p = \frac{L}{1+Ce^{-ky}} [/tex]
for
[tex] \frac{dp}{dt} = kp(L-p) [/tex]
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.
