Answer:
approximately ≅ 22minutes that is 3:22:00pm
Step-by-step explanation:
The two boats are at the dock at 3pm and 4pm respectively; let’s write everything in  terms of a common start time. Let 3pm be the time t = 0. At time t = 0, the first boat  is at the dock and the second boat is at 15 km west of the dock.
At time t, the first boat has moved distance 20t (recall that distance is speed × time),  so it is distance 20t from the dock. The second boat has moved distance 15t toward the  dock, but it was originally distance 15 away from the dock, so it is now 15 − 15t away  from the dock.
At time t, the distance between the two boats is then (by the Pythagorean Theorem)
[tex]d =\sqrt{(15- 15t)^{2} + (20t)^{2} }[/tex]
We want to minimize this quantity.we therefore need to minimize;
[tex]d^{2} = f(t) = (15 - 15t)^{2} + (20t)^{2} = 625t^{2} - 450t + 225.[/tex]
To minimize, we note that f ¹ (t) = 1250t − 450, so f ¹(t) = 0 occurs only at the time  t = [tex]\frac{450}{1250}[/tex] = [tex]\frac{9}{25}[/tex] = 0.36. This is the critical number, and since
f ¹¹(t) = 1250 > 0, it is indeed  a minimum.
So the boats are closest together at time t = 0.36, which is 0.36 hours (or 21.6 minutes)  approximately  22minutes
after 3pm.
that is 3:22:00pm
Diagram attached to the question for interpretation
By knowing how the boats move, we want to find when the boats will be closest together, we will find that it happens 22 minutes after 3:00 PM
Let's define 3:00 PM as our zero in time.
Boat 1 leaves the dock at t = 0, and it moves due south at 20km/h. If we define the positive y-axis as north and the positive x-axis as east, the position equation for this boat will be:
P(t) = (0km, -20km/h*t)
Now we know that boat 2 moves with a speed of 15km due east and it reaches the dock at t = 1h. Then the initial position is at 15km west of the dock, thus the position equation is:
P'(t) = (-15km + 15km/h*t, 0)
Now that we have the two position equations, we want to find the value of t such that the distance between them is minimized, the distance will be:
[tex]D = |P(t) - P'(t)| = \sqrt{(-15km + 15km/h*t)^2 + (20km/h*t)^2} \\\\D^2 = (-15km + 15km/h*t)^2 + (20km/h*t)^2\\\\D^2 = 625km^2/h^2*t^2 - 450km^2/h*t + 225km^2[/tex]
Notice that minimizing the distance squared is equivalent to minimizing the distance.
The distance squared is a quadratic equation of positive leading coefficient, thus the minimum is at the vertex, which is:
[tex]t = -\frac{-450km^2/h}{2*625km^2/h^2} = 0.36h[/tex]
So the time when the boats are closest together is at t = 0.36h. Knowing that 1 h = 60min, we have: 0.36h = 0.36*60 min = 21.6 min.
Then, rounding to the nearest minute, 22 minutes after 3:00 PM the boats are closest together.
If you want to learn more, you can read:
https://brainly.com/question/13169621
