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

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

- calculus -
**Reiny**, Tuesday, June 30, 2009 at 7:01pm
let 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)

## Answer this Question

## Related Questions

Math - The illumination of an object by a light source is directly proportional ...

Math - The illumination of an object by a light source is directly proportional ...

MATH 100 - The illumination of an object by a light source is directly ...

math - The illumination of an object by a light source is directly proportional...

Calculus - "The illumination of an object by a light source is directly ...

Math - The illumination produced by a light source is inversely proportional to ...

College Algebra - The illumination produced by a light source is inversely ...

Math - The intensity of illumination at a given point is directly proportional ...

Algebra II - The intensity of illumination at a given point is directly ...

Algebra II - The intensity of illumination at a given point is directly ...