Question

The number of students in a cafeteria is modeled by the function p that satisfies the logistic differential equation dp/dt = 1/2000 p(200-p), where t is the time in seconds and p(0) = 25. what is the greatest rate of change, in students per second, of the number of students in the cafeteria?

Answer 1

The function that we are given represents the rate of change. If p is the number of students in the cafeteria and you take the first derivative with respect to time you get the rate of change. Now we have to find the maxima of this function. This is usually done by finding the first derivate and then finding its roots. In this case, we are not explicitly given the function p(t) so we can't do that. You could solve the given differential equation and then find the second derivative. However, there is an easier way. We know that parabola has its maximum in the vertex. So all we have to do is find the vertex of the parabola we are given.
$p'(t)=\frac{1}{2000}p(200-p)\\ p'(t)=\frac{1}{2000}(200p-p^2)\\ p'(t)=-\frac{p^2}{2000}+\frac{p}{10}$
So we are given parabola with the following parameters:
$a=-\frac{1}{2000}\\ b=\frac{1}{10}\\ c=0$
The x (or in our case p coordinate) coordinate of the vertex is given with this formula:
$p_v=\frac{-b}{2a}\\ p_v=100$
We plug this back into the original equation to obtain the maximum rate of change:
$p'_{max}=-\frac{100^2}{2000}+\frac{100}{10}=5 students/s$
You can check out the graph of the first derivative on this link: https://www.desmos.com/calculator/tgmnxqb7fd

Answer 2

Answer:

Maximum rate of change will be

$(\frac{dp}{dt})_{max} = 5$ at p = 100

Explanation:

As we know that rate of change in students per second is given by

$r = \frac{dp}{dt} = \frac{1}{2000} p(200 - p)$

here we need to find the greatest rate of change in the number of students

So in order to find the greatest value of the rate of students we have to put the differentiation of of the above function to be zero

so we have

$\frac{dr}{dp} = 0$

$0 = \frac{1}{2000} (200 - 2p)$

by solving above equation we have

$p = 100$

so maximum rate of students will be at the condition when p = 100

so the value of maximum rate will be

$\frac{dp}{dt} = \frac{1}{2000}(100)(200 - 100)$

$\frac{dp}{dt} = 5$