Well done to Nithiya and JY who have sent in the solution of 843 cubic centimetres, which is very nearly correct. Alex sent in his work on this problem, which shows how he gets the right answer.

I thought of four ways to arrange things.

In the blue diagram, the size could be chosen in two ways.

If the length of the paper fixes the circumference of the cylinder (so the height is across the page), then we know that

2 π r = 29.6

, so

r = 14.8 / π

. Also,

2 r + h = 21

(looking at the width), so

h = 21 - (29.6 / π)

. So the volume is

π r^{2} h = π \times \frac{14 . 8^{2}}{π^{2}} \times (21 - \frac{29.6}{π}) \approx 807 {cm}^{3}

(to the nearest whole number).

If the length of the paper fixes the height of the cylinder (think of the label as being the other way round on the page), then we know that $h = 29.6$ , so $2 r + 2 π r = 21$ , so $r = 10.5 / (1 + π)$ . So the volume is

$π r^{2} h = π \times {(\frac{10.5}{1 + π})}^{2} \times 29.6 \approx 598 {cm}^{3}$

(to the nearest whole number).

Also, in the red diagram the size could be chosen in two ways.

If the width of the paper fixes the circumference of the cylinder, then $2 π r = 21$ , so $r = 10.5 / π$ . Also, $2 r + h = 29.6$ , so $h = 29.6 - (21 / π)$ . So the volume is

$π r^{2} h = π \times \frac{10 . 5^{2}}{π^{2}} \times (29.6 - \frac{21}{π}) \approx 804 {cm}^{3}$

(to the nearest whole number).

Finally, if the width of the paper fixes the height of the cylinder, then $h = 21$ and $2 r + 2 π r = 29.6$ so $r = 14.8 / (1 + π)$ . So the volume is

$π r^{2} h = π \times {(\frac{14.8}{1 + π})}^{2} \times 21 \approx 842 {cm}^{3}$

(to the nearest whole number). This last option is clearly the best.