Minimise the metal in a fixed-volume cylindrical can

Both copies print the complete working. Let r be radius, h height and A surface area. The source model includes both circular ends: $V=\pi r^2h$ and $A=2\pi r^2+2\pi rh$. Assume fixed $V>0$ and $r,h>0$.

From the volume, $h=V/(\pi r^2)$. Thus $A=2\pi r^2+2V/r$, with $dA/dr=4\pi r-2V/r^2$. Setting this to zero gives $r^3=V/(2\pi)$, hence $r=(V/(2\pi))^{1/3}$.

$d^2A/dr^2=4\pi+4V/r^3$. At the stationary radius this is $4\pi+8\pi=12\pi>0$, exactly as printed. Independently, it is positive for every feasible $r>0$, while A grows without bound as r tends to 0 or infinity. This is the unique global minimum.

The printed conclusion gives radius $r=(V/(2\pi))^{1/3}$. Independently substituting $V=2\pi r^3$ into $h=V/(\pi r^2)$ gives $h=2r$: the optimal height equals the diameter. This answers the can-proportion question posed at the chapter opening.