# Suppose that the number of bacteria in a culture at time t is given by `N=5000(25+te^(-t/20))` a) Find the largest and smallest number of bacteria in the culture during the time...

Suppose that the number of bacteria in a culture at time t is given by `N=5000(25+te^(-t/20))`

a) Find the largest and smallest number of bacteria in the culture during the time interval`0<=t<=100`

b) At what time during the time interval in part a) is the number of bacteria decreasing the most rapidly?

I have a feeling that you have to graph this or at least find the critical points and find the max and min, but I am not sure.
Also, part b just throws me for a loop. Am I supposed to find the possibly inflection points or something?``

ishpiro | College Teacher | (Level 1) Educator

Posted on

Sorry, part of the answer has gone missing:

From comparison of the values it can be seen that N(0) is the smallest and N(20) is the greatest.

So the largest number of bacteria is `` , which happens at t = 20, and the smallest number of bacteria is

N(0) = 125000, which happens at t = 0.

b) You are also right that the second part involves looking for an inflection point. You are supposed to be looking for the point at which the number of bacteria is "decreasing the most rapidly". This means the rate of change (that is, derivative) of N(t) is negative and has maximum absolute value. This happens when the second derivative of N(t) is zero, that is, at the inflection point.

To find this point, first, calculate the second derivative of N(t) by taking the derivative of N'(t):

``

Simplifying this results in N''(t) = (-1/10)e^(-t/20) + (t/400)e^(-t/20)

Setting N''(t) = 0 yields

-1/10 + (1/400)t = 0

From here t = 400/10 = 40.

At t = 40, the first derivative of N(t) is negative and the second derivative is 0. So, at t = 40 the number of bacteria is decreasing most rapidly.

ishpiro | College Teacher | (Level 1) Educator

Posted on

Yes, you need to find the critical points of N(t) and then determine which points correspond to maximum and minimum of the function on the given interval.

To find critical points, first find the derivative N'(t). The critical points are the points where derivative is 0 or does not exist.

We can find the derivative using the product rule:

`N'(t) = 5000(e^(-t/20) + t(-1/20)e^(-t/20))`

The derivative exists everywhere, so critical points can be found by solving the equation

N'(t) = 0

`5000(e^(-t/20) -t/20e^(-t/20)) = 0`

The exponential function is always positive, so we can divide both sides of equation by 5000 and `e^(-t/20)` :

`1- t/20 = 0`

From here, t = 20. This is the only critical point of this function.

To find maximum and minimum values of N(t) on the interval [0, 100], compare the values of N(t) at the edges of this interval and at the critical point: N(0), N(20) and N(100).

`N(0) = 5000(25 + 0*e^(-0/20)) = 125000`

`N(20) =5000(25 + 20*e^(-20/20)) = 125000 + 20/e`

`N(100) = 5000(25 + 100e^(-100/20))= 125000 + 100/e^5`