Posted by **Avilene** on Friday, December 4, 2009 at 9:30pm.

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**, Friday, December 4, 2009 at 9:37pm
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.

- calculus -
**Reiny**, Friday, December 4, 2009 at 9:52pm
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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