Answer:
a) d(t) = -3t³ + 3t² + 18t
= -3t(t² - t - 6)
= -3t(t - 3)(t + 2)
b) The ship returns to the harbor at
t = 3 hours.
c) t = -2 (extraneous solution)
d) Here are some points:
t d(t)
0 0
.5 9.375
1 18
1.5 23.625
2 24
2.5 16.875
3 0
e) d'(t) = -9t² + 6t + 18
3t² - 2t - 6 = 0
t = (2 ± √((-2)² - 4(3)(-6)))/6
= (2 + √76)/6
= (2 + 2√19)/6
= (1 + √19)/3
= about 1.79 hours (1 hour, 47 minutes)