1. ## least distance

1.) A ship lies 6 mi. from shore, and opposite a point 10 mi. farther along the shore another ship lies 18 mi. offshore. A boat from the first ship is to land a passenger and then proceed to the other ship. What is the least distance the boat can travel?

Help pls...can anyone show me the diagram of this problem.

-tnx

2. Hello, ihmth!

There is a back-door approach to this problem,
. . involving optics and striaght-line paths.
But I'll use the expected Calculus set-up.

1) A ship lies 6 mi. from shore.
Opposite a point 10 mi. farther along the shore another ship lies 18 mi. offshore.
A boat from the first ship is to land a passenger and then proceed to the other ship.
What is the least distance the boat can travel?
Code:
* C
* |
*   |
*     |
A *                       *       |
| *                   *         | 16
|   *               *           |
6 |     *           *             |
|       *       *               |
|         *   *                 |
B * - - - - - * - - - - - - - - - * D
x     P       10 - x

The first ship is $A\!:\;AB = 6$

The second ship is at $C\!:\;CD = 16$
And: . $BD = 10$

Let $P$ be the landing point.
Then: . $BP \:=\: x\:\text{ and }\:PD \:=\:10-x$

In right triangle $ABP\!:\;\;AP \:=\:\sqrt{x^2 + 6^2}$

In right triangle $CDP\!:\;\;CP \:=\:\sqrt{(10-x)^2 + 18^2}$

The distance travelled is: . $D \;=\;\left(x^2+36\right)^{\frac{1}{2}} + \left(x^2-20x+424\right)^{\frac{1}{2}}$

. . and that is the function we must minimize.