Maxi-Mini-ma

Let the trashcan's surface be expressed as $A = 2\pi r^2 + \pi r^2 + 2\pi rh = 3\pi r^2 + 2\pi rh$ , which solving for $h$ in terms of $r$ yields $h =\frac{A - 3\pi r^2}{2\pi r}.$

Let the trashcan's volume be expressed as a function of the radius $r$ , or:

$V(r) = \pi r^{2}h + \frac{2}{3} \pi r^3 = \pi r^2 \cdot (\frac{A - 3\pi r^2}{2\pi r}) + \frac{2}{3} \pi r^3 = \frac{Ar}{2} - \frac{5\pi r^3}{6}.$ (i)

Differentiating (i) with respect to $r$ now produces $\frac{dV}{dr} = \frac{A}{2} - \frac{5\pi r^2}{2} = 0 \Rightarrow r = \sqrt{\frac{A}{5\pi}}.$ The second derivative at this critical radius shows that:

$\frac{d^{2}V}{dr^{2}} = -5\pi r \Rightarrow -\sqrt{5\pi A} < 0$

we have our maximal radius. Thus, the desired ratio $\frac{h}{r} = \frac{A - 3\pi r^2}{2\pi r^2} = \frac{A}{2\pi r^2} - \frac{3}{2} = \frac{A}{2\pi (\frac{A}{5\pi})} - \frac{3}{2} = \frac{5-3}{2} = \boxed{1}.$