Coffee now, regret later

Let $C(t)$ denote the amount of caffeine, in milligrams, present in the engineer’s body $t$ hours after drinking the coffee.

1. Since the rate at which caffeine is eliminated is assumed to be proportional to the amount currently present in the body, we have
   $$C'(t)=-kC(t),$$where $k>0$. The negative sign indicates that the amount of caffeine decreases over time.

2. Separating variables gives
   $$\frac{dC}{C}=-k\,dt.$$Integrating,
   $$\ln C=-kt+c,$$

$$C(t)=C_0e^{-kt},$$where $C_0>0$ is a constant.

3. Immediately after drinking the coffee, the engineer has $200$ mg of caffeine in their body. Hence
   $$C(0)=200.$$Since
   $$C(0)=C_0,$$we obtain
   $$C_0=200.$$Thus
   $$C(t)=200e^{-kt}.$$

4. The half-life is $5$ hours, so after $5$ hours the amount of caffeine has been divided by two:
   $$C(5)=100.$$Therefore
   $$100=200e^{-5k},$$so
   $$e^{-5k}=\frac12.$$Taking logarithms,
   $$-5k=\ln\left(\frac12\right)=-\ln 2,$$and hence
   $$\boxed{k=\frac{\ln 2}{5}}.$$

The model is therefore
$$\boxed{C(t)=200e^{-\frac{\ln 2}{5}t}}.$$

5. We want to determine when the amount of caffeine reaches $50$ mg:
   $$200e^{-\frac{\ln 2}{5}t}=50.$$Thus
   $$e^{-\frac{\ln 2}{5}t}=\frac14.$$Since
   $$\frac14=2^{-2},$$we obtain
   $$-\frac{\ln 2}{5}t=-2\ln 2,$$and therefore
   $$\boxed{t=10\text{ hours}}.$$

6. The coffee was consumed at noon. Ten hours later, the caffeine level reaches $50$ mg at
   $$\boxed{22{:}00}.$$

7. To have at most $50$ mg of caffeine remaining by $23{:}00$, the engineer must allow at least $10$ hours between drinking the coffee and going to bed.

Therefore, the latest possible time to drink the coffee is
$$23{:}00-10\text{ hours}=13{:}00.$$

Hence,
$$\boxed{13{:}00}.$$

8. Suppose the engineer drinks one coffee at $12{:}00$ and another identical coffee at $16{:}00$.

At $23{:}00$, the caffeine remaining from the first coffee is
$$C_1=200e^{-\frac{\ln 2}{5}\cdot 11} =200\cdot 2^{-11/5}.$$

The second coffee has been in the body for $7$ hours, so its contribution is
$$C_2=200e^{-\frac{\ln 2}{5}\cdot 7} =200\cdot 2^{-7/5}.$$

Thus the total amount of caffeine at $23{:}00$ is
$$C_{\mathrm{total}} = 200\left(2^{-11/5}+2^{-7/5}\right).$$

Numerically,
$$C_{\mathrm{total}}\approx 43.5+75.8 \approx119.3\text{ mg}.$$

Therefore,
$$\boxed{C_{\mathrm{total}}>50\text{ mg}}.$$

According to our model, the engineer will not reach the $50$ mg threshold before $23{:}00$. So perhaps that second coffee was not such a good idea.