Three siblings are at a playground: Alice, Bob, and Carey. Alice, the oldest, gets a call from their dad \emdash their pancake dinner is ready! But they won’t get to eat until all three kids are home.
They each walk home at a different constant speed. Alice can walk home in 10 minutes, Bob can do it in 20, and Carey in 30. Fortunately, any of the kids can carry any of the others on their back without reducing their own walking speed. (However, none of them can carry a kid who is, in turn, carrying another kid.) Assume that they can pick someone up, set someone down, and change direction instantaneously.
What is the fastest they can all get home?
Clearly Alice, the oldest sister will have to do some amount of carrying, else everyone will be waiting for Carey to get home in 30 minutes. If Bob is never carried, then the best possible dinner time is 20 minutes after their dad's call. To do better, let's envision a scheme where indefatiguable Alice first carries one sibling for $t$ minutes, then drops that sibling and reverses course and travels away from home to meet up with the other sibling, then carries that second sibling the rest of the way home.
The time, let's say $t_1 = t_1(t, v_2),$ it will take Alice to travel away from home until she meets up with the other sibiling will be a function of the chosen first time period $t$ and the speed of the second sibling, $v_2$. In particular, if Alice and the second sibling were both traveling for $t$ minutes, then given the differential in their speeds, they would be a distance of $\frac{t}{10} - tv_2$ apart (assuming distance to home is the unit length here). In this case, in order for both of them to make up that distance at their relative combined speed of $v_2 + \frac{1}{10},$ we have $$t_1 = t\frac{\frac{1}{10} - v_2}{\frac{1}{10}+ v_2} = t \frac{1-10v_2}{1+10v_2}.$$ Additionally of note, Alice and her second siblings meeting occurs $$d = d(t, v_2) = (t + t_1) v_2 = t v_2 \left( 1 + \frac{1 - 10v_2}{1 + 10v_2} \right) = \frac{ 2t v_2 }{1 + 10v_2}$$ home lengths away from the playground. So in order for Alice and the second sibling to arrive home, it will take another, let's say $$t_2 = t_2(t, v_2) = \frac{1 - d(t,v_2)}{0.1} = 10 - 10d = 10 - \frac{20tv_2}{1 + 10v_2}$$ minutes. So Alice and one sibling will arrive at \begin{align*}T_A = T_A(t,v_2) &= t + t_1(t, v_2) + t_2(t, v_2) \\ &= t + t \frac{1 - 10v_2}{1 + 10v_2} + 10 - \frac{20tv_2}{1+10v_2} \\ &= 10 + \frac{ t(1 + 10v_2) + t(1-10v_2) - 20 t v_2) }{1 + 10v_2} \\ &= 10 + \frac{ 2t - 20t v_2}{1 + 10v_2} = 10 + 2t \frac{1-10v_2}{1+10v_2}\end{align*} minutes.
Meanwhile, the first sibling that Alice dropped off after $t$ minutes has been trudging home under their own power at a uniform rate of $v_1$ and will arrive home at $$T_1 = T_1(t, v_1) = t + \frac{1 - \frac{t}{10}}{v_1} = \frac{1}{v_1} - t\frac{1 - 10v_1}{10 v_1}.$$ So the time when all of the siblings are safely home and devouring pancakes for dinner is \begin{align*}T = T(t, v_1, v_2) &= \max \{ T_A(t, v_2), T_1(t, v_1) \} \\ &= \max \left\{ 10 + 2t \frac{1 - 10v_2}{1 + 10v_2}, \frac{1}{v_1} - t \frac{1-10v_1}{10v_1} \right\} \end{align*} So in particular, if Alice first picks up Bob and then Carey, we have $$T_{B,C} = T_{B,C} (t) = T\left(t, \frac{1}{20}, \frac{1}{30}\right) = \max \{ 10 + t, 20 - t \},$$ while if Alice first picks up Carey and then Bob, we have $$T_{C,B} = T_{C,B} (t) = T\left(t, \frac{1}{30}, \frac{1}{20}\right) = \max \left\{10 + \frac{2}{3}t, 30 -2t \right\}.$$ The optimal value of $T_{B,C}$ is $$T_{B,C}^* = \min_{t \in [0,10]} T_{B,C} (t) = 15 \,\text{minutes},$$ which occurs when $t^* = 5$ minutes; whereas, the optimal value of $T_{C,B}$ is $$T_{C,B}^* = \min_{t \in [0,10]} T_{C,B} (t) = 17.5 \, \text{minutes},$$ which occurs when $t^* = 6.5$ minutes. So between the two cases of carrying Bob first or carrying Carey first, the children will be eating pancakes sooner if Alice carries Bob first.
We can safely exclude the cases of either Bob carrying Carey at first, or Alice having more than one session of carrying any individual sibling, since each of these cases would lead to Carey, the slowest of the bunch, having a longer average speed home and thus delaying pancake festivities. Therefore, the fastest that Alice, Bob and Carey can all get home is in 15 minutes, where Alice carries Bob for 5 minutes, then sends him home, goes back to meet Carey and upon meeting Alice carries Carey the rest of the way home.
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