My sketch shows the following:
M is the man's position 16 m off shore
A is the point on the shore perpendicular to M
B is the point where he wants to go, AB = 30
Let P be the optimum point between A and B, let
AP = x, then PB = 30-x
distance in water = MP
MP = (x^2 + 256)^(1/2)
time spent rowing = (x^2 + 256)^(1/2) /3
time spent jogging = (30-x)/5
T = total time
= (1/3)(x^2 + 256)^(1/2) + (1/5)(30-x)
dT/dx = (1/6)(x^2 + 256)^(-1/2) (2x) - 1/5
= 0 for a min of T
x/(3√(x^2 + 256) = 1/5
3√(x^2+256) = 5x
square both sides
9x^2 + 2304 = 25x^2
16x^2 = 2304
x^2 = 144
x = 12
so he should land at 30-12 or 18 miles west of his destination
Optimization: A man on an island 16 miles north of a straight shoreline must reach a point 30 miles east of the closet point on the shore to the island. If he can row at a speed of 3 mph and jog at a speed of 5 mph, where should he land on the shore in order to reach his destination as soon as possible?
1 answer