# Thread: Optimisation Problem - Ships

1. ## Optimisation Problem - Ships

Hello everyone,

I am having trouble with this problem, and I have made my problem and work available at ImageShack - Image Hosting :: shipoptimisationprobbx7.jpg (The diagram is my own, not given by the question)

Thank you very much for the help!

2. Hello, scherz0!

I hope I got it right . . .

Ship $\displaystyle A$ looks northeast and sees ship $\displaystyle B$ 5 km away.
Ship $\displaystyle A$ is moving east at 5 km/hr; ship $\displaystyle B$ is moving south at 4 km/hr.
When will they be at their closest?

At the very beginning, the diagram looks like this:
Code:
                      * Q
* |
*   |
*     |
5  *       | 5√2/2
*         |
*           |
* 45°         |
P * - - - - - - - * R
5√2/2
Ship $\displaystyle A$ is at $\displaystyle P$; ship $\displaystyle B$ is at $\displaystyle Q.\quad PQ \:=\: 5,\;PR \:=\: \tfrac{5\sqrt{2}}{2},\;QR \:=\: \tfrac{5\sqrt{2}}{2}$

Code:
                      * Q
* |
*   | 4t
*     |
5  *       * B
*       * |
*       *   | 5√2/2-4t
*       *     |
P * - - - * - - - * R
5t   A 5√2/2-5t
In $\displaystyle t$ hours, ship $\displaystyle A$ has moved $\displaystyle 5t$ km to point $\displaystyle A.\quad AR \:=\:\tfrac{5\sqrt{2}}{2} - 5t$

In the same $\displaystyle t$ hours, ship $\displaystyle B$ has move $\displaystyle 4t$ km to point $\displaystyle B.\quad BR \:=\:\tfrac{5\sqrt{2}}{2} - 4t$

Their distance is given by: .$\displaystyle D \;=\;AB^2 \;=\;\left(\tfrac{5\sqrt{2}}{2} - 5t\right)^2 + \left(\tfrac{5\sqrt{2}}{2} - 4t\right)^2$

. . which simplifies to: .$\displaystyle D \;=\;41t^2 - 45\sqrt{2}t + 25$

Differentiate and equate to zero: .$\displaystyle 82t - 45\sqrt{2} \:=\:0 \quad\Rightarrow\quad t \:=\:\frac{45\sqrt{2}}{82}$

Therefore, they are closest at: .$\displaystyle 0.776092809\text{ hours} \;\approx\;46\text{ mimutes, }34\text{ seconds}$