# Thread: Optimization(rowing across river)

1. ## Optimization(rowing across river)

A man can row at 8 km/h, and run at 15 km/h. He needs to get from a point A, on the south bank of a stretch of still water, to point B on the north bank of the water. The direct distance from A to B is 17 km, and the water is 8 km wide. He starts rowing with an angle X between North and the direction in which he rows. Find an expression for the time he will take to get from A to B, in terms of X.

Any helps would be great

2. Hello, squeeze101!

A man can row at 8 km/h, and run at 15 km/h.
He needs to get from a point $A$, on the south bank of a stretch of still water,
to point $B$ on the north bank of the water.
The direct distance from $A$ to $B$ is 17 km, and the water is 8 km wide.
He starts rowing with an angle $\theta$ between North and the direction in which he rows.
Find an expression for the time he will take to get from $A$ to $B$, in terms of $\theta$.
Code:
      : - - - - - - - 15  - - - - - - - - :
C      8tanθ       P    15-8tanθ    B
* - - - - - - - - * - - - - - - - - * - -
|              *              *
|           *           *
8 |        *        * 17
| θ   *     *
|  *  *
A * - - - - - - - - - - - - - - - - - - - -

The distance directly across the water is: $AC = 8$
We are told that: $AB = 17$
From Pythagorus, we find that: $CB = 15$

He will row toward point $P$ at an angle of $\theta = \angle CAP$

In right triangle $PCA\!:\;\;AP = 8\sec\theta$
. . and: . $CP = 8\tan\theta \quad\Rightarrow\quad PB \:=\:15-8\tan\theta$

He will row $AP \:=\: 8\sec\theta$ km at 8 km/hr

. . This will take him: . $\frac{8\sec\theta}{8} \:=\:\sec\theta\,\text{ hours.}$

He will run $PB \:=\:15-8\tan\theta$ km at 15 km/hr.

. . This will take him: . $\frac{15-8\tan\theta}{15} \:=\:1 - \tfrac{8}{15}\tan\theta\,\text{ hours.}$

Therefore, his total time is: . $T \;=\;\sec\theta + 1 - \tfrac{8}{15}\tan\theta\,\text{ hours}$