Expert: Paul Klarreich - 5/13/2007

Question
the number of bacteria in a culture at time t is given approximated by y=1000(25+te^(-t/20)for 0¡Üt¡Ü100.
a)find the largest and smallest number of bacteria in the culture during the interval.
b). at what time during the interval is the rate of change in the number of bacteria in minimum?

Answer
Questioner:   bhaivka
Category:  Advanced Math
 
Subject:  bhavika ???????????
Question:  the number of bacteria in a culture at time t is given approximated by  y = 1000(25 + te^(-t/20))    for 0 <= t <= 100.
a)find the largest and smallest number of bacteria in the culture during the interval.
b). at what time during the interval is the rate of change in the number of bacteria in minimum?

I assume you meant to write   0 <= t <= 100.  Don't try to include special characters in your questions -- they just don't get through the site correctly.  [Hmm -- does that sound like something you have heard before?]

a) Finding absolute max and min involves:
  1. Find the critical points.  That includes:
       Stationary points:  dy/dx = 0
       Singular points:    dy/dx is undefined.
       Endpoint of the interval.
  2. Test the values at these, if possible.
  3. Pick biggest and smallest.

y = f(t) = 25 + te^(-t/20)   [Ignore the 1000]

dy/dt = e^-t/20 + t(-1/20)e^-t/20
dy/dt = e^-t/20 - t/20 e^-t/20
dy/dt = e^-t/20(1 - t/20)

Stationary point:  1 - t/20 = 0,  t = 20
Endpoints:  t = 0, t = 100

Test points:

t = 0;  f(0) = 25
t = 20: f(20) = 25 + 20 e^(-20/20) = 25 + 20 = 45
t = 100: f(100) = 25 + 100 e^(-5) = 25.67 by calculator.

Min at t = 0,  max at t = 20.
.........................................
b). at what time during the interval is the rate of change in the number of

bacteria a minimum?

You want  f''(t) and set that equal to zero.

dy/dt = e^-t/20(1 - t/20)

f''(t) = e^-t/20(- 1/20) + (-1/20)e^-t/20(1 - t/20)
= e^-t/20[ - 1/20  - 1/20(1 - t/20) ]

= e^-t/20[ - 1/20  - 1/20 +  t/400 ]

= e^-t/20[ - 1/10 +  t/400 ]

- 1/10 +  t/400 = 0
t/400 = 1/10
t = 40

Make the same tests as before.