Calculus
posted by Stuck .
"The illumination of an object by a light source is directly proportional to the strength of the source and inversely proportional to the square of the distance from the source. Two streetlights are 40m apart and one is twice as strong as the other. Where is the darkest spot between the two lights?"
I can draw the diagram, and can organize all the information, but can't seem to bring it all together into an equation I can work with.

Distance between street lights, L = 40m
Distance of darkest point from left light = x
Distance of darkest point from right light (twice as strong) = Lx
Height of street light from ground in metres
= h
Assuming
1. The right light is twice as strong as the left, and
2. the "darkest" point is located along a straight line joining the two lights.
Illumination by left light, I1
= K/(h²+x²) K=light constant
Illumination by right light, I2
= 2K/(h²+(Lx)²)
Total illumination, I
= I1+I2
=K/(h²+x²)+2K/(h²+(Lx)²)
Differentiate with respect to x and equate to zero and solve for x (in terms of K and h) to find the minimum luminosity.
I get 17.7 m from the weaker light, assuming the height above ground h=0. 
Oh I see! Thank you!

A street light is at the top of a 19 ft tall pole. A woman 6 ft tall walks away from the pole with a speed of 5 ft/sec along a straight path. How fast is the tip of her shadow moving when she is 40 ft from the base of the pole?
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