calculus
posted by robert .
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. If two light sources, one three times as strong as the other, are placed 11 ft apart, how far from the stronger source should an object be placed on the line between the sources so as to receive the least illumination? Round the result to the nearest hundredth.
a. 6.57 ft
b. 6.5 ft
c. 4.8 ft
d. 6.7 ft
e. 6.55 ft
f. 5.4 ft

let x ft be the distance from the stronger light, so 11x is the distance from the weaker light
let TI be the total illumination, then
TL = 3I/x^2 + I/(11x)^2
d(TL)/dx = 6I(x)^3  I(11x)^3(1)
= 0 for a min of TL
divide both sides by I, crossmultiplying and reducing I go
x^3 = 3(11x)^3
take the cube root of both sides:
x = 3^(1/3)(11x)
expand the right side, bring the x term to the left, take out a common factor of x, ...
I got
x = 11(3)^(1/3) / (1+(3)^(1/3))
= 6.4959..
looks like b)
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