This week's question was answered correctly by Ackbach, anemone, MarkFL and marthbalarka.
You can find Ackbach's solution here (using standard calculus):
[sp]Let $L$ be the illumination (luminosity), $I$ be the intensity (strength), $r$ the distance from the source to the point in question, and $k$ the proportionality constant. Then our general equation is
$$L= \frac{kI}{r^{2}}.$$
We assume the total luminosity adds linearly, so that $L_{ \text{tot}}=L_{1}+L_{2}$. Let $x$ be the distance along the line between the sources as measured from the stronger source. Then we seek to minimize the quantity
$$L_{ \text{tot}}= \frac{3kI}{x^{2}}+ \frac{kI}{(10-x)^{2}}.$$
The minimization of this quantity should occur at the same location as the minimization of
$$y:= \frac{L_{ \text{tot}}}{kI}= \frac{3}{x^{2}}+ \frac{1}{(10-x)^{2}}.$$
The bounds on $x$ are $x \in (0,10)$, since the luminosity technically blows up at each source. The minimum will obviously not occur at either source. Setting $y'=0$ is the same as setting
$$-6x^{-3}+2(10-x)^{-3}= \frac{2x^{3}-6(10-x)^{3}}{x^{3}(10-x)^{3}}=0,$$
which only happens when
$$2x^{3}=6(10-x)^{3} \implies x= \sqrt[3]{3}(10-x) \implies \boxed{x= \frac{10 \sqrt[3]{3}}{1+\sqrt[3]{3}} \, \text{feet} }.$$
Note that, since this value of $x$ is approximately $5.9$, it is in the interval. Also note that
$$y''=18x^{-4}+6(10-x)^{-4}>0;$$
it follows that we found a local minimum. Note also that the other two roots of the cubic are complex and therefore do not contribute critical points.[/sp]
You can find Mark's solution here (using Lagrange Multipliers):
[sp]Let's work this problem in general terms, and obtain a formula into which we can then plug our given data. Let:
$$0<I_1$$ be the illumination of first light source
$$0<I_2$$ be the illumination of second light source
$$0<D$$ be the distance between the light sources
$$x$$ be the object's distance from the first light source
$$y$$ be the object's distance from the second light source
Thus, our objective function, which is the total illumination received by the object is:
$$f(x,y)=I_1x^{-2}+I_2y^{-2}$$
subject to the constraint:
$$g(x,y)=x+y-D=0$$
Using Lagrange multipliers, we obtain the system:
$$-2I_1x^{-3}=\lambda$$
$$-2I_2y^{-3}=\lambda$$
From this we find:
$$\lambda=-2I_1x^{-3}=-2I_2y^{-3}\implies y=\sqrt[3]{\frac{I_2}{I_1}}x$$
Substituting for $y$ into the constraint, we obtain:
$$x+\sqrt[3]{\frac{I_2}{I_1}}x=D$$
$$x=\frac{D}{\sqrt[3]{\frac{I_2}{I_1}}+1}$$
Hence:
$$y=\sqrt[3]{\frac{I_2}{I_1}}\left(\frac{D}{\sqrt[3]{\frac{I_2}{I_1}}+1} \right)=\frac{D}{\sqrt[3]{\frac{I_1}{I_2}}+1}$$
Since the objective function is unbounded at the boundaries $(x,y)=(0,D),\,(D,0)$, we are assured these values for $x$ and $y$ are at the minimum.
Using the given data (where we define the second light source to be the stronger):
$$I_2=3I_1$$
$$D=10\text{ ft}$$
we find:
$$x=\frac{10}{\sqrt[3]{3}+1}\text{ ft}\approx4.09458563186124\text{ ft}$$
$$y=\frac{10}{\sqrt[3]{\frac{1}{3}}+1}\text{ ft}\approx5.90541436813876\text{ ft}$$[/sp]
