Thread: Man on a boat - Optimization

1. Man on a boat - Optimization

A man with a boat is located at point on the shore of a circular lake of radius 5 miles. He wants to reach the point on the shore diametrically opposed to as quickly as possible. He plans to paddle his boat at an angle () to to some point on the shore, then walk along the shore to his . If he can paddle 2.9 miles per hour and walk at 4 miles per hour, what is the shortest possible time it will take him to reach ?

To solve this problem, we need to minimize the following function of the angle :

A stationary point for is

(Write DNE if there is none.)

We conclude that the minimal possible time for the trip is .
The maximal possible time for the trip is .

2. Hello, rawkstar!

This one takes a bit of work.
Hope you're sitting down . . .

$\text{A man with a boat is located at point }P\text{ on the shore}$
. . $\text{of a circular lake of radius 5 miles.}$
$\text{He wants to reach the point }Q\text{ diametrically opposite }P\text{ as soon as possible.}$

$\text{He plans to paddle his boat at an angle }\theta\text{ to }PQ\:(0 \le \theta \le \frac{\pi}{2})$
. . $\text{to some point }R\text{ on the shore, then walk along the shore to }Q .$

$\text{If he can paddle at 2.9 mph and walk at 4 mph,}$
. . $\text{ what is the shortest possible time it will take him to reach }Q?$
Code:
              * * *
*           *   R
*               o
*     x    *   *  *
*     *5     y
*   * @     *       *
P o - - - - o - - - - o Q
*    5    O         *

*                 *
*               *
*           *
* * *

The center of the circle is $\,O.$
. . And: . $OP = OQ = OR = 5$

Let $\theta = \angle RPQ$
. . Note that: . $\angle ROQ = 2\theta$

Let $\,x \,=\,\overline{PR}$, the distance he rows.
Let $y \,=\,\text{arc}(RQ)$, the distance he walks.

Find the distance $\,x.$

In isosceles triangle $POQ,\:\angle POQ \,=\,\pi - 2\theta$

Use the Law of Cosines:

. . $x^2 \:=\:5^2 + 5^2 - 2(5)(5)\cos(\pi - 2\theta) \;=\;50 + 50\cos2\theta$

. . $x^2 \;=\;50(1 + \cos2\theta) \;=\;100\left(\dfrac{1+\cos2\theta}{2}\right) \;=\;100\cos^2\!\theta$

. . $x \;=\;10\cos\theta$

He rowed $10\cos\theta$ miles at 2.9 mph.
. . This took: . $T_1 \:=\:\dfrac{10\cos\theta}{2.9}$ hours.

Find the distance $\,y.$

The arc length $RQ$ is: . $\text{radius} \times \text{angle}$

Hence: . $y \:=\:5(2\theta) \;=\:10\theta$

He walked $10\theta$ miles at 4 mph.
. . This took: . $T_2 \:=\:\dfrac{10\theta}{4} \:=\:\dfrac{5}{2}\theta$ hours.

The total time is: . $T \;=\;\dfrac{10}{2.9}\cos\theta + \dfrac{5}{2}\theta$

And that is the function you must minimize.

Go for it!

3. ok so f(t)=10/2.9 cos(t)+5/2t
f'(t) thus equals -10/2.9sin(t)+5/2
you set f'(t) equal to 0 to find the stationary point
i did that using my calculator and found the stationary point to be at t=.81103
i also know that the minimal time will be when t=0 so it will be approx. 3.4483
i am now confused at how to get the maximum time possible for the trip
my guess would be to put the greatest value of t possible but I don't know how to find that for this case

4. No, the time optimum time is at t = 0.81103.

Use the equation that Soroban gave you, that is

$x = 10 \cos\theta$

To find the distance of x then find the time it takes to cover x.

The find the distance y and the time to cover this.

Lastly, add those two time taken to get the optimum time.

5. thank you
i got it

6. From the graph of

$T \;=\;\dfrac{10}{2.9}\cos\theta + \dfrac{5}{2}\theta$

The point that you got seems to be a maximum, meaning that the value you've got is the value when the time t is a maximum.

The minimum time occurs either at t = 0 degrees or somewhere near pi/2.

EDIT:

At 0 degrees, the time is 3.4483 hours as you got.