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qaws
2 months ago
8

The number of students in a cafeteria is modeled by the function p that satisfies the logistic differential equation dp/dt = 1/2

000 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?
Physics
2 answers:
Softa [3K]2 months ago
7 0
The function provided describes how the number of students changes over time. With p representing the number of students, taking the derivative with respect to time gives the rate of change. We aim to find the highest value (maximum) of this rate. Normally, this involves finding where the derivative of the rate equals zero, but since we don't have p(t) explicitly, solving the differential equation entirely would be complex.
Instead, because the rate function is a parabola, its maximum is at the vertex, which we can calculate directly.
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}
Given the parabola parameters:
a=-\frac{1}{2000}\\ b=\frac{1}{10}\\ c=0
The vertex (p-coordinate) is found using:
p_v=\frac{-b}{2a}\\ p_v=100
Plugging this into the original function provides the maximum rate:
p'_{max}=-\frac{100^2}{2000}+\frac{100}{10}=5$students/s
For a graph of this derivative, see: https://www.desmos.com/calculator/tgmnxqb7fd

serg [3.5K]2 months ago
4 0

Answer:

The maximum rate of change occurs at

(\frac{dp}{dt})_{max} = 5 when p equals 100.

Explanation:

The rate of change in student numbers per second is expressed as

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

We need to find the peak rate of change.

To find this maximum, we set the derivative of the rate function to zero:

So we have

\frac{dr}{dp} = 0

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

Solving yields

p = 100

Therefore, the greatest rate of change is when p = 100.

The value of this maximum rate is

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

\frac{dp}{dt} = 5

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