Length Optimization

racuna

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Sep 28, 2005
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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?
 
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)
 
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 #1\displaystyle \#1 is at the dock P\displaystyle P.
. . It moves south at 20 kph.
. . In t\displaystyle t hours, #1\displaystyle \#1 will be 20t\displaystyle 20t km south of the dock at point A\displaystyle A.

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

     Q  15t   B  15-15t P
     *- - - - * - - - - *
                \       |
                  \     |20t
                  d \   |
                      \ |
                        *A
The distance between the boats is: d=AB\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.
 
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