1. ## Maximum value

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.

2. ## Re: Maximum value

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.

3. ## Re: Maximum value

Sorry mister, I don't get it at all, or perhaps what you have said is not on my syllabus, can you please explain it ?

4. ## Re: Maximum value

I am speaking of methods you should have been taught in algebra/pre-calculus. Given the parabola:

$y=ax^2+bx+c$ where $a\ne0$

the axis of symmetry is the line:

$x=-\frac{b}{2a}$

It is on this line that the vertex will be. If $0 then the vertex is the global minimum, and if $a<0$ then the vertex is the global maximum.

5. ## Re: Maximum value

My bad, sir, my math's foundation was bad, I will like to know how did you get the equation for the axis of symmetry?

6. ## Re: Maximum value

Suppose we equate the quadratic to zero, and require it to have a repeated root $r$. We know then that the axis of symmetry is the line $x=r$.

$ax^2+bx+c=a(x-r)^2=ax^2-2arx+ar^2$

Equating the linear coefficients, we find:

$b=-2ar\,\therefore\,r=-\frac{b}{2a}$

And so, the axis of symmetry is the line:

$x=r=-\frac{b}{2a}$

7. ## Re: Maximum value

Thank you very much for the explanation, but still, the method I have listed out at the beginning can solve this problem also? Is that something wrong with that?

8. ## Re: Maximum value

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:

$\frac{dy}{dx}=2ax+b=0\,\therefore\,x=-\frac{b}{2a}$

9. ## Re: Maximum value

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.

11. ## Re: Maximum value

Thank you for provide me an alternative way, sir, now I have more choice of answers.