Minimizing illumination intensity between two lamps

Differentiate $I(x) = 8x^{-2} + 2(10-x)^{-2}$ to get $I'(x) = -16x^{-3} + 4(10-x)^{-3}$. Setting $I'(x) = 0$ gives $\frac{4}{(10-x)^3} = \frac{16}{x^3}$, which simplifies to $x^3 = 4(10-x)^3$. Taking cube roots yields $x = 4^{1/3}(10-x)$, so $x(1 + 4^{1/3}) = 10\cdot 4^{1/3}$.

Solving gives $x = \dfrac{10\cdot 4^{1/3}}{1+4^{1/3}} \approx 6.13$ feet. Since $I(x) \to \infty$ as $x \to 0^+$ or $x \to 10^-$, this unique critical point in $(0,10)$ must give the minimum. Substituting back into $I(x) = \dfrac{8}{x^2} + \dfrac{2}{(10-x)^2}$ gives the minimum intensity, approximately $I(6.13) \approx 0.347$.