Maximum volume of water one can drink without being burnt

Question

A friend asked me the following interesting question:

We have a bottle with 1L water of 20°C and an unlimited supply of boiling water (100°C). A person can only drink water of temperature $\leq 60$°C. What is the maximum amount of water one can drink if we assume waiting doesn't cool the water and we can only drink directly from the bottle?

Note: the temperature of the result of mixing $m_1$ L water of $t_1$°C and $m_2$ L water of $t_2$°C is $\frac{m_1t_1+m_2t_2}{m_1+m_2}$°C

A strategy is to always drink a fixed portion of water from the bottle, add boiling water until the bottle is full again, drink, add, ..., until when adding would make the water exceed 60°C, at which time we add water until we reach 60°C and drink all of it.

Let the portion of the remaining water in the bottle after drinking be $0<r<1$ (i.e. we drink $1-r$ L from the bottle). I could prove that after drinking and adding the $n$-th time, the temperature in the bottle is (°C) $$t_n=100-80r^n.$$ Therefore we would have drunk $$n=\lfloor-\log_r 2\rfloor$$ times until we reach 60°C, $(1-r)$ L each time.

Since $\lim_{r \to 1}n(1-r)=\ln 2$, we can drink up to $(1+\ln 2)$ L water together with the remaining whole bottle 60°C water.

My question: Is this strategy optimal? How can I prove it if so? How can we do better otherwise?

Any help is appreciated, thanks!

Answer 1

Assume one has separated the $1L$ bottle to components each of $m_1$,$m_2$,$m_3$,... $20^\circ$ water and decides to mix each of those components with $100^\circ$ water and drink. Obviously the optimal strategy is to mix $m_i$ at each step to an amount of water to obtain an exactly $2m_i$ of $60^\circ$ water. This is because $$\dfrac{20m_i+100x}{m_i+x}=60$$ which yields to $$x=m_i$$ therefore the most amount of water one can drink is $2m_1+2m_2+2m_3+...$. From the other side, we have $$m_1+m_2+m_3+...=1$$ therefore the maximum amount of water he can drink is $2L$. Since we have no extra constraint on $m_i$'s we easily assume $m_1=1$ and $m_2=m_3=...=0$ so our optimal strategy is:

He mixes whole the 1L $20^\circ$ water with 1L $100^\circ$ water to obtain 2L $60^\circ$ water then drinks it.

Answer 2

If you drink at each step $m_i$ of the bottle and fill up with hot water, you will get, $$t_n=20\prod_{i=1}^n (1-m_i) + 100\sum_{k\geq 1}^nm_k\prod_{i\geq k+1}(1-m_i) $$ I think, to prove your method, you have to show that this quantity is minimum, for a fixed $n$ and a fixed $0<k<1$ such that $m_1+\cdots+m_n=k$, when $m_1=\dots = m_n$, which seems to me a rather classical optimisation problem.

For exemple for $n=2$, $$t_1=20+ m_1 80$$ and $$t_2= 20+(m_1+m_2-m_1m_2)80 $$ With $m_1+m_2=k$, $$t_2= 20+(k-m_1(k-m_1))80 =20+k80-m_1(k-m_1)80,$$ so you need $m_1(k-m_1) $ being maximal (with $0<m_1<k$) so $$m_1 =\frac{k}{2} $$ Since $k$ could be any value, it is proven that, for two steps, the minimal temperature is obtain (for a subtracted fixed volume) by two equal portion.

Answer 3

If the temperature of the water in the bottle is at least $100-80e^{-x}$ degrees and we add one $y$-liter portion of boiling water (after making room for it), the temperature will rise to at least $$\frac{100y+(100-80e^{-x})(1-y)}{y+(1-y)}=100-80e^{-x}(1-y)>100-80e^{-x-y}$$ degrees. By induction on the number of portions, we can show that after adding a total of $x>0$ liters of boiling water, the temperature in the bottle is greater than $100-80e^{-x}$ degrees. In particular, after adding a total of $\ln 2$ liters of water, no matter how you divide it into portions, the water will be too hot to drink. In other words, the limit you found is the best possible.