Set the closest landing point from the boat as A(it will be directly perpendicular from the shore to the boat )

Set an assumed landing point you have chosen as B

Assign the distance between A and B as x

For whatever value of x, the distance you have to row is

sqr root of ( 2^2 + x^2) miles

(Pythagoras theorem. I hope you’ve studied)

Therefore time taken to row will be

sqr root of ( 2^2 + x^2) miles /2 mph

=sqr of ( 2^2 + x^2)/2 hrs

( time = distance / speed )

Next you find the time you use to run from where you have landed to the house

The remaining distance to run is

6-x miles

And the time used is

6-x miles ÷ 6mph = 6-x/6hours

( note that speed = distance /time

So time = distance / speed)

Now you’ve got the eqn:

time to row + time to run

= sqr of ( 2^2 + x^2)/2 hrs+ 6-x/6hours

=sqr of( 2^2 + x^2)/2 + 6-x/6 hours

THIS will be the function of your graph ofagainsttimedistance of landing from boathouse

Y = sqr of (2^2 + x^2)/2 + 6-x/6

In which Y is time and x is distance between point A and B ( refer drawing )

Now differentiate the graph to find its minimum point

I hope you've understood. too bad I can't draw anything in this stupid replying column. e mail me for a better description.

You should have studied this when you were in secondary school...