Posted by Avilene on .
A man launches his boat from point A on a bank of a straight river, 3 km wide, and wants to reach point B, 2 km downstream on the opposite bank, as quickly as possible. He could row his boat directly across the river to point C and then run to B, or he could row directly to B, or he could row to some point D between B and C and then run to B. If he can row 6 km/h and run 8 km/h, where should he land to reach B as soon as possible? (We assume that the speed of the water is negligible compared to the speed at which the man rows.)

calculus 
bobpursley,
Draw the diagram. Label the points. Now label DB as x, and CD as 2x
So the distance AD is sqrt(3^2+(2x)^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. 
calculus 
Reiny,
I assume you made a diagram according to your description.
Let CD = x km, 0 ≤ x ≤ 2
then DB = 2x km
let AD = y
then y= (x^2 + 9)^(1/2)
time rowing = (x^2 + 9)^(1/2)/6
time running = (2x)/8
Total Time = (x^2 + 9)^(1/2)/6 + (2x)/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