# A man in a boat is 3 km off shore and wishes to go directly to a point on the shore that is 5 km from the point directly opposite his present positionThe man can walk at 4 km/h and row at 2 km/h....

A man in a boat is 3 km off shore and wishes to go directly to a point on the shore that is 5 km from the point directly opposite his present position

The man can walk at 4 km/h and row at 2 km/h. At what point on the shore should he land so that he can reach his destination in the shortest possible time?

*The answer at the back of the textbook is 3.27 km but I don't know how to get that. Please help! Thanks*

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### 1 Answer

Let A be the current position,B the point on the shore 3km from A, C the point to land at, and D the destination.

Let BC=d.

Then AB=3,BC=d,AC=`sqrt(d^2+9)` , CD=5-d.(All distances in km)

We want to minimize the time. The total time is the sum of the time rowing plus the time walking. Using unit analysis, we see that `hr=(hr)/(km)*km=(km)/((km)/(hr))` . Thus:

`t=(sqrt(d^2+9)km)/(2 (km)/(hr))+((5-d)km)/(4(km)/(hr))`

`t=(sqrt(d^2+9))/2+(5-d)/4`

`t=1/2(d^2+9)^(1/2)+1/4(5-d)`

In order to minimize this function, we take the derivative and set it equal to zero.

`(dt)/(dd)=1/4(d^2+9)^(-1/2)(2d)-1/4`

`=(2d)/(4sqrt(d^2+9))-1/4`

Setting this equal to zero we get:

`(2d)/(4sqrt(d^2+9))=1/4`

`2d=sqrt(d^2+9)`

`4d^2=d^2+9`

`3d^2=9`

`d^2=3`

`d=sqrt(3)~~1.73`

**Thus you want to land 1.73km from B or 3.27km from D(the destination)**