Answer
The object should be placed a distance of $~~\frac{10~\sqrt[3] 3}{1+\sqrt[3] 3}~ft~~$ from the stronger light source.
Work Step by Step
Let the strength of the light sources be $P$ and $3P$
Let $x$ be the distance of the object from the $3P$ light source.
We can write an expression for the intensity at position $x$:
$I = \frac{3P}{x^2}+\frac{P}{(10-x)^2}$
We can find $\frac{dI}{dx}$:
$\frac{dI}{dx} = (-2)\frac{3P}{x^3}+(-2)\frac{P}{(10-x)^3}(-1)$
$\frac{dI}{dx} = -\frac{6P}{x^3}+\frac{2P}{(10-x)^3} = 0$
$\frac{6P}{x^3} = \frac{2P}{(10-x)^3}$
$\frac{3}{x^3} = \frac{1}{(10-x)^3}$
$3(10-x)^3 = x^3$
$\sqrt[3] 3~(10-x) = x$
$10~\sqrt[3] 3 = x~(1+\sqrt[3] 3)$
$x = \frac{10~\sqrt[3] 3}{1+\sqrt[3] 3}$
The object should be placed a distance of $~~\frac{10~\sqrt[3] 3}{1+\sqrt[3] 3}~ft~~$ from the stronger light source.
