Draw the diagram. Label the points. Now label DB as x, and CD as 2-x
So the distance AD is sqrt(3^2+(2-x)^2)
so the time for the trip is
time=DB/6 + x/8
= 1/6 (sqrt(above )) + x/8
take the deriviative of time with respect to x, set to zero, and solve for x.
I assume you made a diagram according to your description.
Let CD = x km, 0 ≤ x ≤ 2
then DB = 2-x km
let AD = y
then y= (x^2 + 9)^(1/2)
time rowing = (x^2 + 9)^(1/2)/6
time running = (2-x)/8
Total Time = (x^2 + 9)^(1/2)/6 + (2-x)/8
d(Total Time)/dx = x/[6((x^2 + 9)^(1/2)] - 1/8 = 0 for a max/min of TT
This simplified to
8x = 6√(x^2+9)
4x = 3√(x^2+9) , I then squared both sides
16x^2 = 9x^2 + 81
x = 3.4
but that is outside of our domain,
unless I made an arithmetic error.
Better check it.
So let's check our trivial routes:
all rowing: y = √13
time = √13/6 = .6 hrs
combination rowing + running
time = 3/6 + 2/8 = .75 hours
So he should just row directly to B
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