Homework Help: Calculus

Posted by Dominick on Friday, May 1, 2009 at 7:24pm.

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 θ. How far from the floor should the light be to maximize the illumination at the point P? (Let d = 6 ft.)

Calculus - Damon, Saturday, May 2, 2009 at 7:00am
I guess d is from below the light to a point on the floor?
If so:

I = k cos T / (h^2+d^2)
where cos T = h/(h^2+d^2)^.5
so
I = k h/(h^2+d^2)^1.5
dI/dh = k [ (h^2+d^2)^1.5 -h(1.5)(h^2+d^2)^.5 (2h) ] / (h^2+d^2)^3
= 0 for extreme
(h^2+d^2)^1.5 = 3 h^2 (h^2+d^2)^.5
(h^2+d^2) = 3 h^2
d^2 = 2 h^2
d = h sqrt 2
h = 6/sqrt 2 = 4.24 ft

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