# Optimisation Problem - Ships

• Dec 9th 2008, 08:48 AM
scherz0
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!
• Dec 9th 2008, 12:42 PM
Soroban
Hello, scherz0!

I hope I got it right . . .

Quote:

Ship $A$ looks northeast and sees ship $B$ 5 km away.
Ship $A$ is moving east at 5 km/hr; ship $B$ 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 $A$ is at $P$; ship $B$ is at $Q.\quad PQ \:=\: 5,\;PR \:=\: \tfrac{5\sqrt{2}}{2},\;QR \:=\: \tfrac{5\sqrt{2}}{2}$

Code:

* Q
* |
*  | 4t
*    |
5  *      * B
*      * |
*      *  | 5√2/2-4t
*      *    |
P * - - - * - - - * R
5t  A 5√2/2-5t

In $t$ hours, ship $A$ has moved $5t$ km to point $A.\quad AR \:=\:\tfrac{5\sqrt{2}}{2} - 5t$

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

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

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

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

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