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February 1, 2015

February 1, 2015

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

- Calculus -
**Damon**, Saturday, May 2, 2009 at 7:00amI 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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