Optimizing Illumination: Placing an Object to Receive Least Illumination

[suspenc3]

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 the two light sources, one three times as strong as the other, are placed 10ft apart, where should an object be placed on the line between the sources so as to receive the least illumination?

Im having trouble setting up the problem..any help would be appreciated.

 

[Hootenanny]

What are you intial thoughts?

 

[suspenc3]

I can't picture, or draw a diagram to illustrate the problem. I don't understand what the setup would look like.

 

[Hootenanny]

I = \frac{S}{r^2} + \frac{3S}{(10-r)^2}

Perhaps something like that?

 

[suspenc3]

Yes, that makes sence, but what can I use to eliminate one of the variables?

 

[Hootenanny]

Before we proceed, it is probably more appropriate if we say that;

I\;\; \alpha \;\; \frac{S}{r^2} + \frac{3S}{(10-r)^2}

As, S is constant, the only two variables are I and r. We want to know how I varies with respect to r, if I understant the problem correctly?

So how about finding I'(r)\;\;dr?

 

[suspenc3]

ok well, I found I' but I think it is wrong. I get

I' = \frac{-10r^2}{r^4} - \frac{30(r^2-2r+100)}{(r^2-2^r+100)^2}

When I try to solve for r I don't get a real number

 

[Hootenanny]

I get

I'(r) = \frac{6S}{(10-r)^3} - \frac{2S}{r^3}

You don't want to solve for I. What do you know about the gradient at a minimum point?

 

[Hootenanny]

If you are having trouble visulaising the functions I have attached plot of them. The blue curve is that of the original function, the red is that of the derrivative. As S is constant(strength of source) I assigned an arbitray value of S=1 for this plot.

Hope this helps

 

[BeccaWoodard]

Hey Suspenc, did you ever figure out this problem. I am having trouble figuring it out myself and was wondering if you could assist.