In 1998, 2000 runners entered a marathon in Joppetown U.S.A. In 2004 22,500 runners entered the race because the limited number of facilities the carrying pompo

Question

In 1998, 2000 runners entered a marathon in Joppetown U.S.A. In 2004 22,500 runners entered the race because the limited number of facilities the carrying pomposity for the number of racers is 60,00. the number of racers t in years can be modeled by the logistic function P(t)=C/1+Be^kt 

What is the value of C?

Find B?

Find K?

Answer 1

You have not fully specified the problem, so the following answer makes some assumptions about it.

 

The logistic function is given as

 

P(t)=C/(1+Be^kt)

 

With this form and for this problem

 

C is the maximum number of runners that can be handled.  Here, C=60,000.

 

k is a parameter that determines the speed at which P approaches its upper limit.  Here, k<0.

 

B is a scaling factor.  Here, B>0.

 

To simplify, assume t=0 in 1998 and t=6 in 2004.

 

Substituting the data in the logistic function, we get:

 

2000=C/(1+Be^k(0))=C/(1+B)

 

22500=C/(1+Be^k(6))

 

60000=C/(1+Be^k(∞))=C    (remember, k<0 and e raised to the negative infinity power is 0)

 

So, C=60000.

 

Substitute for C in the first equation.

 

2000=60000/(1+B)

 

(1+B)=60000/2000=30

 

B=29

 

Now substitute for B and C in the remaining equation.

 

22500=60000/(1+29ek(6))

 

1+29e^k6=60000/22500=8/3

 

 

29e^k6=8/3 - 1 = 5/3

 

e^6k=5/(3*29)=0.0575

 

6k=LN(0.0575)=-2.856

 

k=-2.856/6=-0.476