Suppose that a population develops according to the logistic equation \[ \frac{d P}{d t}=0.05 P-0.00006667 P^{2} \] where \( t \) is measured in weeks. (a) What is the carrying capacity? (b) Is the solution increasing or decreasing when \( P \) is between 0 and the carriying capacity? ? (c) Is the solution increasing or decreasing when \( P \) is greater than the carrying capacity? ?

The carrying capacity of a population is defined as the maximum population size that the environment can sustain indefinitely without being degraded. To find it mathematically from the logistic equation \(\frac{d P}{d t} = 0.05 P - 0.00006667 P^{2}\), we set the equation to zero: \[0 = 0.05 P - 0.00006667 P^{2}\] Factoring gives us \(P(0.05 - 0.00006667 P) = 0\). This implies two solutions: \(P = 0\) and \(P = \frac{0.05}{0.00006667} \approx 750\). Thus, the carrying capacity is approximately 750. When the population \(P\) is between 0 and the carrying capacity (0 < \(P\) < 750), the growth rate \(\frac{d P}{d t}\) is positive. This is because substituting \(P\) into the equation yields a positive value: as long as \(P\) is less than 750, the term \(0.05 P\) outweighs the negative effect of \(-0.00006667 P^{2}\). Therefore, in this range, the population is increasing. Conversely, if \(P\) is greater than the carrying capacity (when \(P > 750\)), the growth rate \(\frac{d P}{d t}\) becomes negative. In this scenario, the quadratic term \(-0.00006667 P^{2}\) dominates, leading to a decrease in the population. Thus, when \(P\) exceeds 750, the size of the population will decline towards the carrying capacity.