# Math Help - Difficult Relative Rates Problem

1. ## Difficult Relative Rates Problem

A hunter is at a point on a straight river bank running east-west. He wants to get to his cabin located 3 miles north and 8 miles west. He can travel 5 mph on the river but only 2 mph on the rocky land. How far upriver should he go in order to reach the cabin in minimum time?

I assume the question means downriver and not upriver.

I've set two rates for the different speeds. dR/dt= -5(speed on river) and dL/dt = 2(speed over land) I didn't use x and y because the speed over land is not necessarily in the y direction. I'm stuck now though. I drew out a picture and labeled the distances. I'm not sure what equation I'm supposed to be minimizing. I think it's supposed to be time correct?

Any help would be appreciated.
Thanks.

2. Hello, ShawnC!

A hunter is at a point on a straight river bank running east-west.
He wants to get to his cabin located 3 miles north and 8 miles west.
He can travel 5 mph on the river but only 2 mph on the rocky land.
How far upriver should he go in order to reach the cabin in minimum time?
The set-up is easier than you think . . .
Code:
    : - - - -  8  - - - - :

C    8-x    P    x    A
o - - - - - o - - - - o
*       |
_____ *     | 3
√x²+3²   *   |
* |
o
H

The hunter is at $H.$
He wants to get to his cabin $C.$

He crosses the river to point $P$
then hikes the remaining distance to $C.$

The distance across the river is: , $HP \:=\:\sqrt{x^2+9}$ miles.
. . At 5 mph, this will take him: . $\frac{\sqrt{x^2+9}}{5}$ hours.

The distance he hikes on rocky land is: . $PC \;=\;8-x$ miles.
. . At 2 mph, this will take him: . $\frac{8-x}{2}$ hours.

Hence, his total time is: . $T \;=\;\tfrac{1}{5}\left(x^2+9\right)^{\frac{1}{2}} + 4 - \tfrac{1}{2}x$ hours.

And that is the function we must minimize.

3. I think you reversed the distances on it though. The distance he travels over water should be 8-x i believe, And since I'm trying to determine the distance he should travel down river. Dont you think I should set that distance to x instead? and set the the distance on the triangle to 8-x?

4. I figured it out.

The correct function to minimize is:

T = (x/5) + sqrt((8-x)^2+9)/2

when you take the derivative to minimize it you get

(1/5) + (x-8)/(2(sqrt(x^2-16x+73)))

And when you solve the derivative for 0 to find the critical points you get, I believe, 9.31 and 6.69. However 9.31 is not in the domain so it is omitted leaving 6.69.

So the hunter must travel 6.69 miles downriver before getting off the river to hike to the cabin.

Could anyone verify that this is correct? It seems right in my mind conceptually. I know that he should travel more than 5 miles and less than 8 just based on intuition.