Length Optimization

[racuna]

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

I don't know how to set the problem up this way.
I think I had done a problem like this before, but it was a simple differentiation problem.
How do I figure this out?

 

[Gene]

The first boat is at
y = -20(t-2)
The second boat is at
x = 25(t-3)
The distance is
s = sqrt(x²+y²)
with the dock at P(0,0)

 

[soroban]

Hello, racuna!

I had to baby-talk through it . . . Is anyone surprised?

A boat leaves a dock at 2:00p.m. and travels due south at a speed of 20kph.
Another boat has been heading due east at 15 kph and reaches the same dock at 3:00p.m.
At what time were the two boats closest together?

Here's my reasoning . . .

At 2:00 the first boat \(\displaystyle \#1\) is at the dock \(\displaystyle P\).
. . It moves south at 20 kph.
. . In \(\displaystyle t\) hours, \(\displaystyle \#1\) will be \(\displaystyle 20t\) km south of the dock at point \(\displaystyle A\).

The second boat \(\displaystyle \#2\) had been moving east at 15 kph since 2:00 and reaches the dock at 3:00.
. . Back at 2:00, \(\displaystyle \#2\) must have been 15 km west of the dock at point \(\displaystyle Q\).
. . In \(\displaystyle t\) hours, \(\displaystyle \#2\) will be \(\displaystyle 15t\) km closer to the dock at point \(\displaystyle B\).

. . .| - - -  15  - - - |

     Q  15t   B  15-15t P
     *- - - - * - - - - *
                \       |
                  \     |20t
                  d \   |
                      \ |
                        *A

The distance between the boats is: \(\displaystyle d\,=\,AB\)

From Pythagorus, we get: \(\displaystyle \;d^2\:=\20t)^2\,+\,(15-15t)^2\)

. . which simplifies to: \(\displaystyle \;d\:=\625t^2\,-\,450t\,+\,225)^{\frac{1}{2}}\)

And that is the function we must minimize.