Question 890509
A light house, a shop, and a housekeeper.  Where does the housekeeper begin?  From the light house?   I could analyze and solve the problem if you make the description understandably precise.


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I believe I have an understanding of the description after thinking through the description.


Draw a right triangle.  Vertical segment OB, showing B below O, and point A to the right of point O.   Segment OB is 4, and segment OA is 2.  The angle AOB is a right angle.


The housekeeper will row from the light house (?)  to point P between O and B.  


Let y = Distance for walking which is segment PB.
That means 4-y is length of OP.


Distance Rowing, {{{highlight_green(sqrt(2^2+(4-y)^2))}}}
Distance Walking, {{{highlight_green(y)}}}.


Uniform Rates usually for travel can be represented RT=D, or T=D/R.
Using a function for total time, T you have total time for this housekeepers row & walk trip as {{{highlight_green(T=sqrt(4+(4-y)^2)/4+y/6)}}}.


Find the derivative, {{{dT/dy}}}.
In one of its raw forms, this starts as:


{{{dT/dy=(1/4)(1/2)(4-(4-y)^2)^(-1/2)*2(4-y)(-1)+(1/6)}}}


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Equating that to 0, and then omitting the many algebra steps here,  I am finding {{{highlight(13y^2+40y+192=0)}}};  if this has a real solution, then I believe I made no mistakes...


I MADE A MISTAKE IN MY WORK, SOMEWHERE.  MAYBE YOU WILL FIND IT AND GET THE RIGHT ANSWER.



RETRIED SOLUTION-------------------------------------------------------

The same formula for T was found.
Derivative {{{dT/dy=-((4-y)/(sqrt(4+(4-y)^2)))+1/6}}}.
Simplifying and then focusing on the numerator being 0 gave {{{highlight(35y^2-280y+4=0)}}}.
Two solutions for y were found, and the choice which makes sense was {{{highlight(y=0.18)}}}.
The calculations and computation to find that are omitted here, taking nearly a whole page on paper.


The housekeeper would row to a point between O and B, 0.18 km from the point B (the shop).