# Optimization Problem

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.

## Answers and Replies

Staff Emeritus
Gold Member
What are you intial thoughts?

suspenc3
I cant picture, or draw a diagram to illustrate the problem. I dont understand what the setup would look like.

Staff Emeritus
Gold Member
$$I = \frac{S}{r^2} + \frac{3S}{(10-r)^2}$$

Perhaps something like that?

suspenc3
Yes, that makes sence, but what can I use to eliminate one of the variables?

Staff Emeritus
Gold Member
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 dont get a real number

Staff Emeritus
Gold Member
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?

Last edited:
Staff Emeritus
Gold Member
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 #### Attachments

• illumination.jpg
25 KB · Views: 314
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.