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(A) Let the height of the closed cylinder be $h$ and the base radius be $r$. The volume $V$ of the cylinder is given by $V = \pi r^2 h$,which implies

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$2 : 1$

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(A) Let the height of the closed cylinder be $h$ and the base radius be $r$. The volume $V$ of the cylinder is given by $V = \pi r^2 h$,which implies $h = \frac{V}{\pi r^2}$. The total surface area $S$ of the closed cylinder is $S = 2 \pi r h + 2 \pi r^2$. Substituting the value of $h$ in the surface area formula: $S = 2 \pi r (\frac{V}{\pi r^2}) + 2 \pi r^2 = \frac{2V}{r} + 2 \pi r^2$. To find the minimum surface area,we differentiate $S$ with respect to $r$ and set it to zero: $\frac{dS}{dr} = -\frac{2V}{r^2} + 4 \pi r = 0$. This gives $4 \pi r = \frac{2V}{r^2}$,so $V = 2 \pi r^3$. Substituting $V = \pi r^2 h$ into this equation: $\pi r^2 h = 2 \pi r^3$. Dividing both sides by $\pi r^2$,we get $h = 2r$,which means the ratio $h : r = 2 : 1$.

$A$ closed cylinder of given volume will have the least surface area when the ratio of its height to its base radius is:

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