You really don't need calculus at all, the given derivative is parabolic in P, and so the critical value is the axis of symmetry. Since the coefficient of the squared term is negative, we know the vertex is the maximum.
Determine the maximum value of dP/dt where dP/dt = rP(1-(P/k)), r and k are positive constant, and interpret the meaning of this maximum value.
I do it like this,
let y = dP/dt
y=rP(1-(P/k))
dy/dP = r-2P(r/k)
(d^2)y/dP^2 = -2(r/k)
(d^2)y/dP^2 < 0 , so it is a maximum value.
When dy/dP = 0,
0 = r-2P(r/k)
P = k/2
Substitute P = k/2 into dP/dt = rP(1-(P/k)),
and I get dP/dt = rk/4
So the maximum value is (k/2 , rk/4)
Is that correct? I didn't use the differentiation for (d^2)P/dt^2 and (d^3)P/dt^3 because it is too complicated, can I really do like this? Please correct me if I am wrong and provide me a way to do it. Thanks in advance.
I am speaking of methods you should have been taught in algebra/pre-calculus. Given the parabola:
where
the axis of symmetry is the line:
It is on this line that the vertex will be. If then the vertex is the global minimum, and if then the vertex is the global maximum.
No, there's nothing wrong with what you did, I was merely pointing out that when optimizing a quadratic, we don't need the calculus, but there's nothing wrong with using it. If you take the derivative of the general quadratic, and equate it to zero and solve for x, you will find:
Your help will greatly appreciated, sir. You really save my life, I gonna pass up this assignment next Monday, because what I learn this term was calculus, that's why I need to use that method to solve the problem. Really, thank you very much. Since both two method will bears the same result.