Answer:
Step-by-step explanation:
Given the population of yeast modelled by the function f(t)=a/1+be^-0.7t where t is measured in hours.
If at time t = 0 the population is 20 cells, then the equation becomes:
20 = a/1+be^-0.7t
20 = a/1+be^-0.7(0)
20 = a/1+be^-0
20 = a/1+b(1)
a/1+b = 20
a = 20(1+b) ........ equation 1
Also if the population is increasing at a rate of 12 cells/hour, then d(f(t)/dt = 12
Differentiate the expression with respect to time
f(t)=a(1+be^-0.7t)^-1
d(f(t)/dt =
a{-(1+be^-0.7t)^-2×(-0.7be^-0.7t)
a{-(1+be^-0.7t)^-2×(-0.7be^-0.7t) = 12
0.7abe^-0.7t/(1+be^-0.7t)² = 12 ..... equation 2
at t = 0, the equation becomes
0.7abe^-0/(1+be^-0)² = 12
0.7ab/(1+b)²= 12
0.7ab = 12(1+b)²
Substitute 1 into 2
0.7×20(1+b)b = 12(1+b)²
14(1+b)b = 12(1+b)²
Divide both sides by 1+b
14b = 12(1+b)
14b = 12+12b
14b-12b = 12
2b = 12
b = 6
Substitute b= 6 into the equation a = 20(1+b) to get a.
a = 20(1+6)
a = 20×7
a = 140
For us to be able to determine what happens to the yeast population in the long run. we will take the limit of f(t) as t approaches infinity.
Lim t -->/infty a/1+be^-0.7t
= a/1+be^-(infty)
= a/1+b(0)
= a
Since a = 140, hence the population of the yeast tends to 140 on the long run
