Probem44+James+Sam+O

very good!

Problem 44, A boat leaves a dock a 2:00 PM and travels due south at a speed of 20 km/h. Another boat has been heading east at 15 km/h and reaches the same dock at 3:00 PM. At what time were the two boats closest together?

This problem creates a right triangle so I started by using the pythagorean theorem. Another equation I used was the d=rt that represents the sides of the triangle.

__Given:__ r1= 20 km/h --> 1/3 km/min (you need to convert to km/min, because later you will add the minutes to 2:00 PM to find the time when the boats are closest)r2= 15 km/h ---> 1/4 km/min d2= 15 km (it takes one hour for the second boat to reach the dock, and since the boat travels at 15 km/h then the boat starts 15 kms away) d1= 0 (the first boat starts at the dock)

**A = 15km - ((1/4 km/min) * t) ** = = NOW lets make B the distance travel by the first boat... **B**  **= (r * t)** =**B** = (1/3 km/min) * t= = = NOW lets substitute the equations of A and B into the original function...
 * C^2 = A^2 + B^2** (I am looking for a minimum value for C)NOW lets make A the distance traveled by the second boat...
 * A =  ****d - (r * t)** ** (the d is there because the boat starts 15km away from the dock. As the boat moves that side of the triangle decreases) **

C^2= (15km - ((1/4 km/min) * t))^2 + ((1/3 km/min) * t)^2C^2= (15 - (1/4)t)^2 + ((1/3)t)^2 C^2= 225 - (30/4)t + t^2 + (1/9)t^2 C^2= 225 - 7.5t + .1736t^2 (combining like terms) NOW take the derivative....

2C * C'= -7.5 + .3472t 0= -7.5 + .3472t (set equal to zero to find critical point) 7.5 = .3472t t= 21.6 min, therefore the two boats are closest at **2:21 PM (and 36 seconds)**