1. ## optimization problem

A boat leaves a dock at 3:00 PM and travels due south at a speed of 21 km/h. Another boat has been heading due east at 13 km/h and reaches the same dock at 4:00 PM. At some time t hours after 3:00 PM the two boats were closest together. Find time t.

2. Originally Posted by singh1030
A boat leaves a dock at 3:00 PM and travels due south at a speed of 21 km/h. Another boat has been heading due east at 13 km/h and reaches the same dock at 4:00 PM. At some time t hours after 3:00 PM the two boats were closest together. Find time t.
note that if we draw a diagram of the situation, we will have a right-triangle. the hypotenuse of this triangle is the distance between the two boats, and it is the function for this distance that we want to minimize for the period 3pm to 4pm.

Let $t = 0$ at 3pm
so, $t = 1$ at 4pm

Let the boat moving at 21 km/h be boat $A$
Let the boat moving at 13 km/h be boat $B$
Let $D$ be the distance between the two boats

thus, for $0 \le t \le 1$,

the distance boat $A$ is from the dock is given by: $21 t$

the distance boat $B$ is from the dock is given by: $13 - 13t$

By Pythagoras' theorem:

$D^2 = (21t)^2 + (13 - 13t)^2$

the distance is minimized if the square of the distance is minimized, so let $x = D^2$, we have:

$x = (21t)^2 + (13 - 13t)^2$

now we want to find $t$ such that $x$ is minimized.