We have AB + BC + CD + DE = AB + BC' + C'D' + D'E' \(\ge AE'\)

=> min of (AB + BC + CD + DE) = AE' \(=\sqrt{\left(2\right)^2+\left(\dfrac{1}{3}+1+\dfrac{1}{6}\right)^2}=\dfrac{5}{2}\)

When \(B\equiv B'',C'\equiv C'',D'\equiv D''\)

We can infer positions of C, D.

In once hour, the first candle consumes 1/4 of its height, the second candle sonsumes 1/3 of its height. (note that the height of two candle is the same.)

Assume after x hours, the first candle twice the height of the second, we have:

\(1-\dfrac{1}{4}x=2\left(1-\dfrac{1}{3}x\right)\)

\(\Rightarrow x=\dfrac{12}{5}=2.4\) (hours)