Question

Brandon is on one side of a river that is 80 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 2 m/s and run at 6 m/s.

(Use decimal notation. Give your answer to two decimal places.)

The minimum amount of time is

Answer 1

Brandon has other options besides swimming straight across and swimming straight to his destination. Let's assume he swims "x" downstream. That means he swims a distance "d" such that d = √(80² + x²), and he runs "D" where D = 300 - x.

The time it takes to do so is

t = √(6400 + x²) / 2 + (300 - x)/6

Differentiate with respect to x we get

dt/dx =x/2√(6400+x^2)-(1/6)

For extremum of t, we have dx/dt=0,which gives

x/(2√(6400+x^2)=(1/6)

i.e..9x^2-x^2=6400

i.e..x^2 =800

i.e..x=28.28

i.e..x= 28.28 m downstream.

At x=28.28 the second derivative os greater than zero

Then The minimum amount of time is

min t = √(6400 +28.28²) /2 + (300-28.28) /6= 87. 71 seconds