A boat leaves a dock at 3:00 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 4:00 PM. How many minutes after 3:00 PM were the two boats closest together? (Round your answer to the nearest minute.)

To find the time when the two boats were closest together, we can set up a problem where we find the distance between the two boats at different times.

Let's assume that the two boats were closest together x minutes after 3:00 PM.

Since the first boat is traveling due south at a speed of 20 km/h, after x minutes, it will be (20/60) * x = x/3 kilometers south of the dock.

Similarly, the second boat is traveling due east at a speed of 15 km/h. Since it reaches the dock at 4:00 PM, it travels for 60 minutes. Therefore, after x minutes, it will be (15/60) * x = x/4 kilometers east of the dock.

To find the distance between the two boats, we can use the Pythagorean theorem:

Distance^2 = (x/3)^2 + (x/4)^2
Distance^2 = x^2/9 + x^2/16
Distance^2 = (16x^2 + 9x^2)/(9*16)
Distance^2 = 25x^2/144

To minimize the distance between the two boats, we need to minimize the square of the distance. Therefore, we need to minimize 25x^2/144.

To find the minimum value, we can find the derivative of the function and set it equal to 0.

d(25x^2/144)/dx = 0
(50x/144) = 0
x = 0

However, x cannot be zero since the boats start moving after 3:00 PM. Therefore, we need to consider the closest time after 3:00 PM where the two boats were closest together, which will be the time that minimizes the distance.

To determine this time, we can evaluate the distance function at two consecutive values of x, such as x = 1 and x = 2.

When x = 1:
Distance^2 = 25(1^2)/144 = 25/144 = 0.1736

When x = 2:
Distance^2 = 25(2^2)/144 = 100/144 = 0.6944

From these calculations, we can see that the minimum square of the distance between the two boats is at x = 1, which is 1 minute after 3:00 PM.

Therefore, the two boats were closest together 1 minute after 3:00 PM.

Ahahahah