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October 31, 2014

October 31, 2014

Posted by **robert** on Tuesday, June 30, 2009 at 4:28pm.

a. 6.57 ft

b. 6.5 ft

c. 4.8 ft

d. 6.7 ft

e. 6.55 ft

f. 5.4 ft

- calculus -
**Reiny**, Tuesday, June 30, 2009 at 7:01pmlet x ft be the distance from the stronger light, so 11-x is the distance from the weaker light

let TI be the total illumination, then

TL = 3I/x^2 + I/(11-x)^2

d(TL)/dx = -6I(x)^-3 - I(11-x)^-3(-1)

= 0 for a min of TL

divide both sides by I, cross-multiplying and reducing I go

x^3 = 3(11-x)^3

take the cube root of both sides:

x = 3^(1/3)(11-x)

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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