Optimization Problem: A man is at a point on the bank of....

[Mitch885885]

Hello everyone

Q: A man is at a point on one bank of a straight river, 3 km wide, and wants to reach a point 8 kn downstream on the opposite bank as quickly as possible. If the man can row at 6 km/h and run at 8 kn/h, how far downstream should he land across the river in order to get to the other point as quickly as possible?

So far, I have constructed a diagram, and have come up with the formulas: C^2=A^2 + B^2, and Time = distance/speed. But I have no clue how to set up the equation or find factors to substitute in. Any help with this would be greatly appreciated! Thank you.

Mitch

 

[galactus]

t=d/r.

Let $\displaystyle R_{row}=rowing\;\ rate=6\;\ km/hr$

Let $\displaystyle R_{run}=running\;\ rate=8\;\ km/hr$

Rowing is from A to B; running is from B to C.

His time would then be given by:

\(\displaystyle \L\\t=\frac{\sqrt{x^{2}+9}}{R_{row}}+\frac{8-x}{R_{run}}\)

You know the given rates. Enter them in the function and optimize.

 

[soroban]

Re: Optimization Problem ~

Hello, Mitch!

Your preliminary work is great . . . you were that close.

A man is at a point on one bank of a straight river, 3 km wide, and wants to
reach a point 8 km downstream on the opposite bank as quickly as possible.
If the man can row at 6 km/h and run at 8 kn/h, how far downstream should
he land across the river in order to get to the other point as quickly as possible?

      A
      *
      | \
      |   \    ____
    3 |     \ √x²+9
      |       \
      |         \
      * - - - - - * - - - - - - - *
      B     x     P     8 - x     C

The man is at $\displaystyle A$, his destination is at $\displaystyle C.\;AB\,=\,3,\;BC\,=\,8$
He will row to point $\displaystyle P$ at 6 km/hr, then run to point $\displaystyle C$ at 8 km/hr.
Let $\displaystyle BP\,=\,x$

He will row the distance $\displaystyle AP\,=\,\sqrt{x^2\,+\,9}$
$\displaystyle \;\;$It will take him: $\displaystyle \,\frac{\sqrt{x^2\,+\,9}}{6}$ hours.

He will run the distance $\displaystyle PC\:=\:8\,-\,x$
$\displaystyle \;\;$It will take him: $\displaystyle \,\frac{8\,-\,x}{8}$ hours.

And there is the time-function: \(\displaystyle \L\,T\;=\;\frac{\sqrt{x^2\,+\,9}}{6}\,+\,\frac{8\,-\,x}{8}\)

Edit: Fixed the typos . . . Thanks for the heads-up. Cody!