Jane is 2 miles offshore in a boat and wishes to reach a coastal village 6 miles down a straight shoreline from the point nearest the boat. She can row 2 mph and can walk 5 mph. Where should she land her boat to reach the village in the least amount of time

x  = 0,87 miles

Step-by-step explanation:

Let´s look at the stuation.

Jane is two miles away (in point B) from point A at the beach  and the horizontal distance between Jane and the village( Point C ) is 6 miles, she will leave the boat somewhere (Point D,  at x distance from A) between A and C.

Then distances  for the travel of Jane are:

L₁  distance to row      L₁²  =  2² + x²     L₁  = √ 4 + x²

And the distance to walk    L₂  =  6 - x

The relation between distances, time, and speed is

d = v*t   then  t =  d/v

According to that the time for rowing will be  

t₁  = ( √4  +  x² ) / 2      and for walking   t₂  = (  6  -  x  ) /5

T = t₁  +   t₂

T(x)  = ( √4  +  x² ) / 2   + (  6  -  x  ) /5

Tacking derivatives on both sides of the equation

T´(x)  =  1/2  [ ( 2*x / 2√ ( 4 + x² ) ]  -1/5

T´(x)  = ( 1/2 )*  x / √ ( 4 + x² )   - 1/5

T´(x) = 0

( 1/2 )*  x / √ ( 4 + x² )   - 1/5   = 0

5*x  - 2* √ ( 4 + x² )  =  0

5*x  =  2 *√ ( 4 + x² )

Squaring

25*x²  =  4* ( 4 + x² )

25*x²  =  16  +  4*x²

21*x²  =  16

x  =  √ 16/21

x  = 0,87 miles

do you mean add both of the equations together

because if you do the answer is 10x-5

step-by-step explanation:

false

step-by-step explanation: