Hello, killasnake!

I *think* I have the first one . . .

The illumination at a point is inversely proportional to the square of the distance of the

point from the light source and directly proportional to the intensity of the light source.

If two light sources are s feet apart and their intensities are $\displaystyle I$ and $\displaystyle J$ respectively,

at what point between them will the sum of their illuminations be a minimum?

Instruction: Give your answer in terms of $\displaystyle \text{s, I, and J.}$

Let $\displaystyle x$ be the distance from $\displaystyle I$ at which the sum of the illuminations be minimum.

Then: .$\displaystyle x\:=\: \_\_\_ $

We are told that: .$\displaystyle L \:=\:\frac{ki}{d^2}$ ... where $\displaystyle L$ = illumination, $\displaystyle i$ = intensity, $\displaystyle d$ = distance Code:

P
I *-----------*---------* J
: x : s-x :

$\displaystyle P$ is the point we are seeking.

Let distance $\displaystyle PI = x$, then distance $\displaystyle PJ = s-x$

The illimunation from $\displaystyle I$ is: .$\displaystyle L_I\:=\:\frac{kI}{x^2}$

The illumination from $\displaystyle J$ is: .$\displaystyle L_J\:=\:\frac{kJ}{(s-x)^2}$

Their sum is: .$\displaystyle S \;= \;\frac{kI}{x^2} + \frac{kJ}{(s-x)^2}$

And we will minimize: .$\displaystyle S \;= \;kIx^{-2} + kJ(s-x)^{-2}$

Differentiate and equate to zero:

. . $\displaystyle S' \;=\;-2kIx^{-3} + 2kJ(s-x)^{-3} \;= \;0$

This simplifies to: .$\displaystyle J(s - x)^{-3} \;= \;Ix^{-3}\quad\Rightarrow\quad Jx^3\;=\;I(s-x)^3$

Take cube roots: .$\displaystyle x\sqrt[3]{J} \;= \;(s-x)\sqrt[3]{I}$

Solve for $\displaystyle x:\;\;x\sqrt[3]{J}\;=\;s\sqrt[3]{I} - x\sqrt[3]{I}$

. . . . .$\displaystyle x\sqrt[3]{I} + x\sqrt[3]{J} \;= \;s\sqrt[3]{I}$

. . . .$\displaystyle x\left(\sqrt[3]{I} + \sqrt[3]{J}\right) \;= \;s\sqrt[3]{I}$

. . Therefore: . . $\displaystyle \boxed{x \;=\;\frac{s\sqrt[3]{I}}{\sqrt[3]{I} + \sqrt[3]{J}}} $