math question help?

Question

The illumination at a point is inversely proportional to the square of the distance of the point from the light source and directly proportional to the intensity of the light source. If two light sources are 20 feet apart and their intensities are 30 and 20 respectively, at what point between them will the sum of their illuminations be a minimum?

Answer 1

Let S1 be the source of intensity 30, S2 be the source with intensity 20 and let x be the distance from the point to S1. According to the statement, the illumination of S1 at x is given by

I1 = 30K/Cx, where K and C are positive constants. So, the illumination of S2 at x is I2 = 20K/(20 -x), for 0 < x < 20. So, the total illumination at x is

I + I1 + I2 = 30K/Cx + 20K/(C(20 -x)). Taking the derivative of I with respect to x, we get

I' = K/C [ -30/(x^2) +20/(20 - x)^2]. Setting I' = 0, we get

-30/(x^2) +20/(20 - x)^2 = 0 => ((20 - x)^2)/20 = x^2/30

3 (20 - x)^2 = 2x^2 => 3(x^2 - 40 x + 400) = 2x^2
x^2 - 120 x + 1200 = 0. We have 2 solutions,

x = (120 + sqrt(14400 - 4800))/2 = 60 + 20 sqrt(6) and
x = 60 - 20 sqrt(6) =~= 11.01. The first solution is out of our range, so it shouldn't be considered. The other one is in (0, 20). We must check if it is really a point of minimum.

Actually, we readily see that I' is negative for x in (0, 11.01) and positive on (11.01, 20), which shows x = 11.01 is a point of minimum.

It's interesting to point out that this point of minimum depends on the intensities of the sources, but not on K and C.

Answer 2

Put the 30 lumen source at 0, the 20 at 20. At a point x between, intensity should be 30k/x² + 20m/(20-x)², where k and m are constants of proportionality. Minimizes when dI/dx = 0,

-60k/x^3 + 40m/(20-x)^3 = 0
40m/(20-x)^3 = 60k/x^3
2mx^3 = 3k(20-x)^3

probably k=m and should cancel out.
let cr be cube root.

x cr(2) = (20-x) cr(3)
x cr(2) + x cr(3) = 20 cr(3)
x = [20 cr(3)] / [cr(2) + cr(3)]
x = 10.6747

Steiner forgot to square his distances in the proportion.