Brandon is on one side of a river that is 60m wide and wants to reach a point 300m downstream on the opposite side as quickly.

Question

Brandon is on one side of a river that is 60 m wide and wants to reach a point 300 m downstream on the opposite side as quickly as possible by swimming diagonally across the river and then running the rest of the way. Find the minimum amount of time if Brandon can swim at 2m/s and run at 4 m/s.

The minimum amount of time is =

Answer 1

Hey Raymond B.-- Cute, but wrong?

Must set up t as a function of angle to the perpendicular swam at.

So,

(300-60tanθ)/4 + (60secθ)/2 = t

dt/dθ = 0 for a minimum (or maximum) = -15sec^2(θ) + 30secθtanθ ;

solves through to x = 20(3)^0.5 ft "downstream" is his optimal aim-point.

You should be able to take it from there.

-- Cheers, --Mr. d. (a Bi(den) supporter)

Answer 2

60 meters wide river

300 meters downstream

run 4 m/s

swim 2 m/s

time = distance divided by speed

t = d/r = distance on the shoreline/4 + distance on water/2

265.36/4 + 69.28/2 = 66.34 + 34.64 = 100.98 seconds minimum

he runs downstream along the shore line at

4 m/s until he stops and swims 2 m/s across the river

let 300-d = the distance he runs before swimming

then time t = (300-d)/4 + sqr(d^2 +60^2)/2

take the derivative and set equal to zero, then solve for d

t' = 1/4 -(1/4)(2d)/sqr(d^2 +3600) = 0

1 = 2d/sqr(d^2 +3600)

sqr(d^2+3600) = 2d

d^2 +3600 = 4d^2

3d^2 = 3600

d^2 = 1200

d= sqr1200 34.64

He runs 300-34.64 = 265.36 meters

then swims 69.28 meters

total time = 265.36/4 + 69.28/2 = 100.98 seconds minimum time