A swimmer S is in the sea at a distance of 4km from the closest point A on a straight shore.His house B is on the shore at a distance 5km from A.He can swim at a speed of 3km/hr.The distance of the point on the shore from B so that he reaches his home in the minimum time is=?

Answer 1

d^2 = X^2 + Y^2 = 4^2 + 5^2 = 41
d = 6.4 km. = Shortest distance to his
home.

Answer 2

Being familiar with this type of optimization problem, I was expecting to be given the rate at which he can run or walk along the beach.

If the rate along the beach is faster than his swimming rate, then the hypotenuse that Henry found will not give him the minimum time.
he would aim for some point between A and B

I will solve it with some arbitrary speed along the beach.
Suppose the speed running along the beach is 6 km/h
Pick a point P somewher between A and B
Let AP = x , then BP = 1-x
ST^2 = x^2 + 4^2
ST = √(x^2 + 16) or (x^2+16)^(1/2)

time swimming = (x^2+16)^(1/2) /3
time running along beach = (5-x)/6 = 5/6 - x/6

Total time = T = (x^2+16)^(1/2) /3 + 5/6 - x/6
dT/dx = (1/2)(1/3)(x^2+16)^(-1/2) (2x) - 1/6
= 0 for a min of T

2x/(6√(x^2+16) = 1/6
2x/√(x^2+16) = 1
2x = √(x^2+16)
square both sides
4x^2 = x^2 + 16
3x^2 = 16
x^2 = 16/3
x = appr 2.3

so distance from point B is 5-x = 2.7 km

Adjust the above solution using your given rate along the beach.