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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!

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    $\begingroup$ Are you sure you don't have an empty glass on hand? Then you could drink 2L water one glass at a time. $\endgroup$ – user856 Feb 18 '18 at 12:40
  • $\begingroup$ @Rahul How can we drink 2L? After drinking 1L of 20°C water, we would have to drink 100°C water, which is impossible. Edit: Ah.. interesting idea, but no. $\endgroup$ – ftfish Feb 18 '18 at 12:41
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    $\begingroup$ So to clarify, you are only allowed to drink the water straight out of the bottle, and have no access to any other empty containers, is that right? $\endgroup$ – user856 Feb 18 '18 at 12:47
  • $\begingroup$ @Rahul, yes, that's how I understand it. That makes it more a math question than a brain teaser.. $\endgroup$ – ftfish Feb 18 '18 at 12:49
  • $\begingroup$ I doubt there's better strategy than yours. Doesn't make sense to drink more colder water than the absolute minimum at each step $\endgroup$ – Yuriy S Feb 18 '18 at 12:58
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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.

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  • $\begingroup$ You can use compactness to show that the minimum is achieved for a fixed number of portions. Then your calculation for the case of two portions shows that any two successive portions have to be equal in the optimal division, and consequently all portions have to be equal. $\endgroup$ – Taneli Huuskonen Feb 19 '18 at 17:08
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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.

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    $\begingroup$ Thanks. What if we don't have extra containers and we can only drink directly from the bottle? $\endgroup$ – ftfish Feb 18 '18 at 12:53
  • $\begingroup$ Then I think your solution is correct... $\endgroup$ – Mostafa Ayaz Feb 18 '18 at 12:56
  • $\begingroup$ A proof would be appreciated. I don't see directly why using different ratios for different steps wouldn't make it better... $\endgroup$ – ftfish Feb 18 '18 at 13:00
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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.

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