A swimmer wishes to cross a 500m wide riverflowing at 5Km/h.. His speeed with respect to water is 3km/h.The man wishes to reach directly opposite to his starting point but he reach some wherere else on the opposite shore and he has to walk to that point.. Find the minimum distance that he has to walk

This is more complicated than I first thought!

Let us suppose it takes time t, and the swimmer has to come back x km

You can now draw a right angled triangle whose sides are (kms):

3t (hypotenuse), 5 (width of river), 5t - x

(The first is the line along which the swimmer swims, while 5t is the distance the river flows in time t)

Hence (5t -x)^2 + .5^2 = (3t)^2 (Equation A)

Considering x as a function of t, and differentiating, (calling x` the derivative of x)

2 (5t -x) (5 - x`) = 18 t

(x` = 0 in case of maximum/minimum)

2 (5t -x) 5 = 18 t

x = 3.2 t

Solving this with equation A

I get t = 5/24 and x = 2/3 km

So I find that the minimum value of x is 2/3 km

Why is this "minimum" and not maximum?

Common sense tells us that there is no maximum! (He can keep swimming along with the river for ever without getting across!)