Hello, singh1030!

I get an ugly cubic to solve . . .

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 4 times as strong as the other, are placed 20 feet apart,

where should an object be placed on the line between the sources

so as to receive the least illumination?

Let $\displaystyle S$ = strength of the light source

and $\displaystyle d$ = distance fronm the light source.

We are told that the illumination is: .$\displaystyle I \:=\:\frac{kS}{d^2}$ Code:

(S) (4S)
* - - - - * - - - - - - - - *
A x P 20-x B

Let $\displaystyle A$ be the source with strength $\displaystyle S$

and $\displaystyle B$ be the source with strength $\displaystyle 4S.$

The object is placed at $\displaystyle P$, which is $\displaystyle x$ feet from $\displaystyle A$ and $\displaystyle (20-x)$ feet from $\displaystyle B.$

The object's illumination from $\displaystyle A$ is: .$\displaystyle I_A \:=\:\frac{kS}{x^2}$

The object's illumination from $\displaystyle B$ is: .$\displaystyle I_B \:=\:\frac{k(4S)}{(20-x)^2} $

The total illumination is: .$\displaystyle I \;=\;\frac{kS}{x^2} + \frac{4kS}{(20-x)^2} \;=\;kS\left[x^{-2} + 4(20-x)^{-2}\right] $

. . and **that** is the function we must minimize . . .