Question

The fitness \( f(n) \) of an individual in a group of size \( n \) is given by \[ f(n) = n(10 - n). \] At evolutionary equilibrium, groups are found in two different sizes. If one group size is 6, the other group size must be ________ (Answer in integer).

Hint:

At evolutionary equilibrium, the fitness values for different group sizes must be equal. Use this condition to solve for the unknown group size.

Answer 1

Correct Answer: 4

At evolutionary equilibrium, the fitness for the two group sizes must be equal. Therefore, we have: \[ f(6) = f(n), \] where \( n \) is the other group size. First, calculate \( f(6) \): \[ f(6) = 6(10 - 6) = 6 \times 4 = 24. \] Now, set \( f(n) = 24 \), and solve for \( n \): \[ f(n) = n(10 - n) = 24. \] This simplifies to: \[ n(10 - n) = 24, \] \[ 10n - n^2 = 24, \] \[ n^2 - 10n + 24 = 0. \] Solving this quadratic equation: \[ n = \frac{10 \pm \sqrt{10^2 - 4 \times 1 \times 24}}{2 \times 1} = \frac{10 \pm \sqrt{100 - 96}}{2} = \frac{10 \pm \sqrt{4}}{2} = \frac{10 \pm 2}{2}. \] Thus, \[ n = \frac{10 + 2}{2} = 6 \quad \text{or} \quad n = \frac{10 - 2}{2} = 4. \] Since one group size is already 6, the other group size must be \( 4 \). Thus, the other group size is \( 4 \).