Calculus Optimization Problem. Help?

Question

A light is suspended at a height h above the floor. The illumination at the point P is inversely proportional to the square of the distance from the point P to the light and directly proportional to the cosine of the angle theta. How far from the floor should the light be to maximize the illumination at the point P?

Additional information:
Point O is the point directly below the hanging light.
Point P is 10m from point O.
r is the distance from the light to point p.

Answer 1

Assuming theta is the angle the ray from the light to P makes with the vertical. Call the height x, then distance to light is sqrt(x^2+100), intensity = 1/(x^2+100) * x/sqrt(x^2+100), or x/(x^2+100)^1.5.
From the ref., the derivative is
1 / (x^2 + 100) ^ 1.5 - 3 * x^2 / (x^2 + 100)^2.5 (sparing myself the tedium of the chain rule.) Set the derivative = 0 and multiply by (x^2 + 100)^2.5, yielding x^2 + 100 - 3 * x^2 = 100 - 2 * x^2 = 0, so the derivative is 0 at x = sqrt(50) = sqrt(100*0.5) =10*sqrt(0.5). Thus the light should be at 7.07m height.