2 Consider a bioreactor used by a yogurt factory to grow the

2. Consider a bioreactor used by a yogurt factory to grow the bacteria needed to make yogurt. The growth of the bacteria is governed by the logistic equation dP k P(M P) where P is the population in millions and t is the time in days. Recall that M is the carrying capacity of the reactor and k is a constant that depends on the growth rate. (a) Through obeservation it is found that after a long time the population is the reactor stabilizes at 30 million bacteria, and that when the population of the reactor is 10 million the population increases at a rate of 2 million bacteria per day. From this, find k and M in the governing equation. (b) If the colony starts with a population of 500,000 bacteria, how long will it take for the population to reach 80% of the carrying capacity.

Solution

Answer:

(a) The population that the reactor will stabilize is the carrying capacity.

Here, M = 30 million.

The point at which the population is at 10 million can be calculated as follows:

dP/dt = 2 = k * 10 * (30 10)

2 = 200k

Therefore, k = 0.01

So, the differential equation stands as:

dP/dt = 0.01 * P * (30 P)

(b) Given, P(0) = 0.5 million.

It is a logistic equation and is separable.

So,

P(t) = M*P₀ / {P₀ + (M P₀) * e ^{kM t}

= 15 / {0.5 + (29.5) * e ^0.315t }

Thus, P(t) = 0.8M = 24

Solving for t, we get:

24 = 15 / {0.5 + (29.5) * e ^0.315t}

12 + 752.25 * e ^0.315t = 15

752.25 * e ^0.315t = 3

e ^0.315t = 0.003988

0.315 t = ln (0.003988) = 5.5244

t = 17.5377 days

(c) For solving this, we require to find the the derivative of dP/dt with respect to P and set it to 0.

So, 0 = d/dP ( dP/dt ) = d/dP (0.0105 * P * (30 P))

0 = 0.315 0.021 * P

Therefore, P = 15

So for P = 15 after 3 days, we set P0 = 15.

t = 0 will occur 3 days into the 6-day period (t = 3 to t = 3).

P(t) = M * P₀ / {P₀ + (M P₀)e ^{kM t}}

= 450 / {15 + (15) e ^0.315t }

At the start of the 6-day interval, the population is

P(3) = 455 / {15 + 15e ^0.315(3) }

= 8.397 million

Therefore, an initial population of 8.397 million will yield the greatest growth over the 6-day period.

So,

P(3) P(3) = [455 / {15 + 15e ^0.315(3)}] [455 / {15 + 15e ^0.315(3)}] = 13.207 million

(d) The differential equation can be given by

dP/dt = k * P * (M P) h

dP/dt = 0.0105 * P * (30 P) h

Rearranging, we get

dP/dt = 0.0105 * (P² 30P + 95.24h)

Equilibrium points occur when dP/dt = 0.

If h = 0, there will be 2 stable equilibrium points.

If h is a value so that (P² 30P + 95.24h) can be factored as (P C)², then there will only be one equilibrium point.

Completing for the square, we get

95.24h = C² = 152 = 225

or h = 2.362

For any higher value of h, the equation

0 = dP dt = 0.0105P(30 P) h has no solutions.

Thus, there are no stable equilibrium points if h > 2.362.

So, h = 2.362 million/day is the maximum allowable harvesting rate.