An island is 1 mile due north of its closest point along a straight shoreline. A visitor is staying at a cabin on the shore that is 15 miles west of that point.

Question

The visitor is traveling from the cabin to the island. The visitor swims at the rate of 2.5 mph and runs at a rate of 6 mph. How far should the visitor run before swimming to minimize the time it takes to reach the island?

Answer 1

Time (T)= x/6 + sqrt((15-x)^2+1^2)/2.5

dT/dx = 1/6 -(15-x)/sqrt(15-x)^2+1)/2.5

1/6 = (15-x)/sqrt(15-x)^2+1)/2.5

6(15-x) = 2.5sqrt(15-x)^2+1)

Let 15 - x = y

6y = 2.5sqrt(y^2+1)

36y^2 = 6.25y^2 + 6.25

29.75y^2 = 6.25

y = sqrt(25/119)

x = 15 - sqrt(25/119)