Aboat leaves a dock at 5: 00 pm and travels due south at a speed of 20 km/h. another boat has been heading due east at 15 km/h and reaches the same dock at 6: 00 pm. how many minutes after 5: 00 pm were the two boats closest together? (round your answer to the nearest minute.

Place the dock at (0, 0) in the xy-plane. At 5:00 P.M. boat A is at (0, 0). It's position after 5:00 P.M. is given by (0, -20t) where t is in hours. At 6:00 P.M. boat B is at (0, 0). That's 1 hour after boat A left the point (0, 0) so 1 h x 15 Km/h=15 Km which means at 5:00 P.M. boat B was 15 Km west of the dock at (0, 0) which means it was at (-15, 0) at 5:00 P.M. Boat B's position after 5:00 P.M. is therefore (-15+15t, 0). Use the distance formula to find the distance between the two boats. 
d=√((x2-x1)²+(y2-y1)²) 
=√((-15+15t-0)²+(0+20t)²) 
=√(225-450t+225t²+400t²) 
=√(225-450t+625t²) 
Find the derivative 
d'= (1/2)(225-450t+625t²)^(-1/2)(-450+1250t) 
Set equal to zero and you get 
-450+1250t=0 
t=450/1250 
=0.36 h=22 minutes