Answer is: option2

Solution:

The growth rate is given by the derivative of N(t):

N'(t) = d/dt [200 ln(t² + 36)] = 200 · (2t / (t² + 36))

Simplifying:

N'(t) = 400t / (t² + 36)

The change in growth is given by the derivative of N'(t):

N''(t) = d/dt [400t / (t² + 36)]

Using the quotient rule:

N''(t) = [(400)(t² + 36) − (400t)(2t)] / (t² + 36)²
         = [400(t² + 36 − 2t²)] / (t² + 36)²
         = [400(36 − t²)] / (t² + 36)²

Simplifying:

N''(t) = [400(36 − t²)] / (t² + 36)²

To find where the change in growth is maximized, set N''(t) = 0:

400(36 − t²) = 0      ⟹      36 − t² = 0      ⟹      t² = 36      ⟹      t = 6

(We discard t = −6 since time cannot be negative.)

The second derivative N''(t) changes from positive to negative at t = 6, indicating a maximum in the rate of change of growth.

Answer: B