In some cases, a population can exceed its carrying capacity for a brief period of time. In a population of dragonflies that approaches its carrying capacity, many females use the last of their energy reserves to reproduce. As a result, the population size reaches 675 dragonflies, but the estimated carrying capacity ($K$) of the area is 550 dragonflies. The r_max value for this population is 0.5. Calculate the rate of population growth (dN/dt) & the per capita rate of growth, r(1 – N/k).

• A. dN/dt = –97, r(1 – N/k) = –0.14
• B. dN/dt = –77, r(1 – N/k) = –0.11
• C. dN/dt = –153, r(1 – N/k) = –0.23
• D. dN/dt = 828, r(1 – N/k) = 0.23

Answer: (B)

Verified step by step guidance

1. Understand the concept of carrying capacity (K), which is the maximum population size that an environment can sustain indefinitely. In this problem, K is given as 550 dragonflies.
2. Identify the current population size (N), which is 675 dragonflies, and note that it exceeds the carrying capacity.
3. Use the formula for the rate of population growth: \( \frac{dN}{dt} = r_{max} \times N \times \left(1 - \frac{N}{K}\right) \). Here, \( r_{max} \) is the maximum per capita rate of increase, given as 0.5.
4. Substitute the values into the formula: \( \frac{dN}{dt} = 0.5 \times 675 \times \left(1 - \frac{675}{550}\right) \). Calculate the term \( \left(1 - \frac{675}{550}\right) \) to find the per capita rate of growth.
5. Evaluate the expression to find \( \frac{dN}{dt} \) and \( r(1 - N/K) \). This will give you the rate of population growth and the per capita rate of growth.