Question

In 1999, 500 runners entered a marathon in Joppetown U.S.A. In 2004, 19,500 runners entered the race. Because of the limited number of facilities, the carrying C capacity for the number of racers is 58,000. The number of racers at any time, t, in years, can be modeled by the logistic function P(t) In what year should 1+Be kt at least 48,000 runners be expected to run in the race?

Answer 1

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

Since, the carrying capacity for the number of racers = 58000

This implies , C = 58000

Given, Initial number of racers = 500

Substituting t = 0 in P(t)

P(0) = C / (1 + Be^k(0)) = C / (1 + B) = 58000 / (1 + B)

This implies,

58000 / (1 + B) = 500

=> 1 + B = 116

=> B = 115

Therefore,

P(t) = 58000 / (1 + 115e^kt)

Number of runners in 2004 i.e. 5 years after 1999 = 19500

Therefore,

P(5) = 19500

=> 58000 / (1 + 115e^k(5)) = 19500

=> (1 + 115e^5k) = (58000 / 19500)

=> 115e^5k = (58000 / 19500) - 1

=> e^5k = (38500) / 2242500

=> ln e^5k = ln (385 / 22425)

=> 5k =  ln (385 / 22425)

=> k = -0.8129

Therefore,

P(t) = 58000 / (1 + 115e^-0.8129t)

Let us assume n years after 1999, the number of runners will be 48,000

Therefore,

  58000 / (1 + 115e^-0.8129n) = 48000

=> 1 + 115e^-0.8129n = 58000 / 48000 = 58 / 48

=> 115e^-0.8129n = (58 / 48) - 1

=> 115e^-0.8129n = 10 / 48

=> e^-0.8129n = 1 / 552

=> ln e^-0.8129n = ln (1 / 552)

=> - 0.8129n =  ln (1 / 552)

=> n = 7.77 $\approx$ 8 years

Therefore,

At least 48000 runners to be expected in the race 8 years after 1999 i.e. in the year 2007