# 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.)

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1. 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.

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bobpursley
Let CD = x km, 0 ≤ x ≤ 2
then DB = 2-x km
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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