Beer drinking: inspired by Martin Gardner

Lets introduce a few constants: $H$ is the height of the can, $\rho$ is the density of the beer, $A$ is the cross-sectional area of the can, and $m$ is the mass of the (empty) can. We also introduce the variable $h=$ the hight of the beer inside the can.

Now, the mass of the empty can is $m$ and the centre of of mass is at height $H/2$ . The mass of the beer (without the can) is $hA\rho$ and the centre of mass is at height $h/2$ . The centre of mass of the can with the beer in is thus $C=\frac{m\cdot H/2+hA\rho\cdot h/2}{m+hA\rho}$ Since we want the point where this is a minimum we should differentiate wrt $h$ and equate to zero: $\frac{d}{dh}C = \frac{(m+hA\rho)hA\rho - (mH/2+h^2A\rho/2)A\rho}{(m+hA\rho)^2} = 0$ Which gives $h^2A\rho/2+hm-Hm/2=0$ Solving for $h$ gives $h = \frac{-m\pm \sqrt{m^2+HmA\rho}}{A\rho}$

Now we note that the initial volume in the can is $V_0=HA$ and the volume left in the can when the height of the beer is $h$ is $V_1=hA$ . Thus our equation becomes $V_1 \rho = -m \pm \sqrt{m^2+V_0 \rho m}$ But we assume $\rho=1$ and we know $V_0=495$ and $m=12$ . Therefore $V_1 = -12 + \sqrt{144+495\times 12} = 66$ To be left with 66ml we must drink $495-66 = 429$ ml.

Assuming that beer is homogenous, the can is perfectly cylindrical, and other things that physicists like to assume .

Clearly we're only concerned about the height of the center of mass. By appropriate scaling and translating, assume that an empty (or full) can has center of mass at $0$ units of height, and $1$ ml of beer to occupy a height of $2$ units of height. Thus, the can has height $990$ units, and the base of the can is at height $-495$ units.

Now, if $x$ ml of the beer has been drunk, the center of mass of the beer alone is at height $-x$ units, and the mass is $495-x$ g. The center of mass of the overall system is thus:

$\begin{aligned} \text{Center of mass} &= \frac{\text{Contribution by can} + \text{Contribution by beer}}{\text{Total mass}} \\ &= \frac{(0 \cdot 12) + (-x \cdot (495-x))}{12 + (495-x)} \\ &= \frac{-x(495-x)}{507-x} \end{aligned}$

Our task is to minimize the above expression for $x \in [0, 495]$ . This should be easily solvable by calculus, but since I'm lazy, WolframAlpha says that the minimum is $-363$ , reached with $x= \boxed{429}$ . That is the amount of beer that must be drunk, in milliliters.