Let the distance from A to his landing place be x
then from there it would be another 5-x miles to his house.
let the distance he rows be d
d^2 = x^2 + 2^2
d = (x^2 + 4)^(1/2)
time for the rowing part = (x^2 + 4)^(1/2) /3
time for the walking part = (5-x)/5
Time = (x^2 + 4)^(1/2) /3 + (5-x)/5
d(Time)/dx = (1/6)(x^2 + 4)^(-1/2) (2x) - 1/5
x/(3√(x^2+4) ) - 1/5
= 0 for a minimum of Time.
x/(3√(x^2+4) ) - 1/5
5x = 3√(x^2 + 4)
25x^2 = 9(x^2+4)
25x^2 = 9x^2 + 36
16x^2 = 36
4x = 6
x = 6/4
sub that into Time = .. and you got it
( I got 1 hour and 32 minutes)
check my arithmetic
(minimum commuting time) A lighthouse lies 2 miles offshore directly across from point A of a straight coastline. The lighthouse keeper lives 5 miles down the coast from point A.What is the minimum time it will take the lighthouse keeper to commute to work, rowing his boat at 3 miles per hour, and walking at 5 miles per hour?
1 answer