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

Asked on by shayaan

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embizze | High School Teacher | (Level 1) Educator Emeritus

Posted on

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)

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