I need help setting this up.

jsucal · **Joined:** Mon, 16 Apr 2012 04:40:35 UTC **Posts:** 4

A man in a rowboat at point P 6 miles from shore desires to reach point Q on the shore at a straight-line distance of 10 miles from his present position. If he can walk 4 mph and row 3 mph, at what point L should he land in order to reach Q in the shortest time?

Shadow · **Posted:** Mon, 16 Apr 2012 06:03:30 UTC

jsucal wrote:

A man in a rowboat at point P 6 miles from shore desires to reach point Q on the shore at a straight-line distance of 10 miles from his present position. If he can walk 4 mph and row 3 mph, at what point L should he land in order to reach Q in the shortest time?

What have you tried?

jsucal · **Joined:** Mon, 16 Apr 2012 04:40:35 UTC **Posts:** 4

I know that the distance of the shore is 11.66 miles and that i will divide the rowing time by 3 and the walking time by 4 and i know that i have to set up an equation but don't know what to use.

Shadow · **Posted:** Mon, 16 Apr 2012 22:42:38 UTC

jsucal wrote:

I know that the distance of the shore is 11.66 miles and that i will divide the rowing time by 3 and the walking time by 4 and i know that i have to set up an equation but don't know what to use.

Whoa, I thought the shore distance was 6, why is it now 11.66?

Soroban · **Posted:** Tue, 17 Apr 2012 01:24:44 UTC

Hello, jsucal!

You've twisted some of the facts . . .

Quote:

A man in a rowboat at point P 6 miles from shore desires to reach point Q on the shore
at a straight-line distance of 10 miles from his present position.
If he can walk 4 mph and row 3 mph, at what point L should he land in order to reach Q in the shortest time?

Code:

    P o
      | * *
      |   *   *    10
    6 |     *     *
      |       *       *
      |         *         *
      o-----------o-----------o
      R     x     L    8-x    Q
      : - - - - - 8 - - - - - :

The man is at $P\!:\;PR = 6.$

is 10 miles from $P\!:\;PQ = 10.$
Using Pythagorus: RQ = 8.

He will row to point

. . Let $RL \,=\, x.$ .Then $PL \,=\,\sqrt{x^2+36}$
Then he will walk to point

. . $LQ \,=\,8-x.$

He will row $\sqrt{x^2+36}$ miles at 3 mph. . This will take $\dfrac{\sqrt{x^2+36}}{3}$ hours.
He will walk 8-x

miles at 4 mph. . This will take $\dfrac{8-x}{4}$ hours.

His total time is: . $T \;=\;\frac{1}{3}(x^2+36)^{\frac{1}{2}} + \dfrac{8-x}{4}\:\text{ hours.}$

And that is the function you must minimize.

S.O.S. Mathematics CyberBoard

I need help setting this up.

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