Answer:
1. (0, 1/40000) and (1/900, +∞)
2. (1/40000, 1/900).
Step-by-step explanation:
The population is increasing when dP/dt is positive and the population is decreasing when dP/dt is negative.
First, replacing the values of K, k and m, we get that dP/dt are equal to:
dP/dt=0.1P(1-40000P)(1-900P)
To find the intervals when dP/dt is positive, we need to make dP/dt greater than zero as:
dP/dt=0.1P(1-40000P)(1-900P) > 0
So, we have 3 terms in the equation, so:
0.1P > 0 if P > 0
1 - 40000 P > 0 if P < 1/40000
1 - 900 P > 0 if P < 1/900
We can said that 0.1P is positive if P is positive, (1 - 40000P) is positive is P is lower than 1/40000 and (1 -900P) is positive is P is lower than 1/900
Therefore, we have the following intervals: (0, 1/40000), (1/40000, 1/900) and (1/900, +∞)
For the interval (0, 1/40000): 0.1P is positive, (1 - 40000P) is positive and (1 -900P) is positive. So we can said that dP/dt is positive.
For the interval (1/40000, 1/900): 0.1P is positive, (1 - 40000P) is negative and (1 -900P) is positive. So we can said that dP/dt is negative.
For the interval (1/900, +∞): 0.1P is positive, (1 - 40000P) is negative and (1 -900P) is negative. So we can said that dP/dt is positive.
Finally, the population increase at values of P between: (0, 1/40000) and (1/900, +∞) and the population decrease at values of P between: (1/40000, 1/900).