Question

A fisherman is in a row boat on a lake 2 miles form shore when he catches a huge fish. He wants to show the fish to his buddies in a tavern 3 miles down the straight shore. He can row 5 miles an hour and run 13 miles per hour. To what point on the shore should he row to get to the tavern as quickly as possible?

Answer 1

Given value are: distance between boat and shore = 2 miles

distance between shore and tavern = 3 miles

Suppose fisherman rows the boat for 'x' distance, then

In right angle triangle we know that:

x^2 = 2^2 + y^2

y = sqrt (x^2 - 4)

Now after that remaining distance for which he needs to run then

remaining distance = d = 3 - y = 3 - sqrt (x^2 - 4)

Now given that his speed during rowing = 5 miles/hr

his speed during running = 13 miles/hr

So total time taken by fisherman will be:

t = t1 + t2

t1 = time taken during rowing = distance/speed = x/5

t2 = time taken during running = (3 - sqrt (x^2 - 4))/13

So,

t = t1 + t2 = x/5 + (3 - sqrt (x^2 - 4))/13

t = x/5 + 3/13 - (sqrt (x^2 - 4))/13

Now for him to reach as quickly as possible,

dt/dx = 0, So

dt/dx = 1/5 + 0 - [1/(2*13*sqrt (x^2 - 4))]*(2x - 0)

dt/dx = 1/5 - x/(13*sqrt (x^2 - 4)) = 0

1/5 = x/(13*sqrt (x^2 - 4))

13*sqrt (x^2 - 4) = 5x

x^2 - 4 = 25x^2/169

4 = x^2 - 25x^2/169

4 = 144x^2/169

x = sqrt (169*4/144) = 13*2/12 = 13/6 miles

x = 2.167 miles = distance for which he should be rowing

y = sqrt (x^2 - 4) = sqrt (169/36 - 4) = sqrt (25/36) = 5/6 miles

y = 0.833 miles = point on the shore where he should row

Let me know if you've any query.