Code:

` 5 miles`

|------------|

(x). | -

(boat) . 20mph | |

. | |

. | |

- |(a) | 10 miles

| | . |

d | | . 35mph |

| | . |

- |(y) (hotel) -

That was fun to draw :) .

From the boat (x), the power boat will take the passengers to the shore at (a), then the minivan will take the passengers to the hotel (y).

Let the distance between (a) and (y) be equal to d, then the distance from (x) to (a) is sqrt[5^2 + (10 - d)^2]

The total time traveled will be:

*T = 20*sqrt[25 + (10 - d)^2] + 35*d*
Now we have a function relating distance and time. Taking the derivative, we get:

T' = 20*1/2(25 + (10 - d)^2)^(-1/2)*(-2)(10 - d) + 35

= -20*(10 - d)/sqrt[25 + (10 - d)^2] + 35

Letting T' = 0:

0 = -20*(10 - d)/sqrt[25 + (10 - d)^2] + 35

35*sqrt[25 + (10 - d)^2] = 200 - 20d

Sqare both sides:

35^2*[25 + (10 - d)^2] = 40000 - 8000d + 400d^2

1225(25 + 100 - 20d + d^2) = 40000 - 8000d + 400d^2

153125 - 24500d + 1225d^2 = 40000 - 8000d + 400d^2

825d^2 - 16500d + 113125 = 0

25(33d^2 - 660d + 4525) = 0

d = (660 +/- sqrt[435600 - 597300])/66

I made an "oops" somewhere. There's too many digits! :eek: . Give me a few minutes to gain my compsure and maybe I'll figure this one out.