The intensity of light varies inversely as the square of the distance from its source. If two searchlights are 600 meters apart and one light is 8 times as strong as the other, where should a man cross the line between them in order to be illuminated as little as possible

Answer 1

As Tom showed you,

I = A/s^2 + 8A/(600-s^2)

Actually, that is not quite right. It should be

1/s^2 + 8/(600-s)^2

You should have caught that. In any case,

The A is just a constant, so it does not affect the solution, and we can say

I = 1/s^2 + 8/(600-s)^2
dI/ds = -2/s^3 + 16/(600-s)^3
= 18(200-s)(s^2+120000) / s^3(600-s)^3

clearly, dI/ds=0 when s=200 meters from the weaker source.