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What is the volume of the solid formed by rotating this right angled triangle about the hypotenuse?

Efficient Cutting

Stage: 4 Challenge Level: Challenge Level:2 Challenge Level:2

Ysanne and Joanna from Central Newcastle High School sent us their thoughts:

Roughly half of the paper should be used to make the circles and roughly half to make the rectangle as the height of the cylinder and the area of the circle are equally important in creating the volume.

We decided to use the side of length $29.6$ as the circumference of the cylinder and the length of the rectangle. As this was the circumference of the cylinder, and therefore the circles, we divided $29.6$ by $\pi$ to find the appropriate diameter for the circles. This came to $9.42$, so the remaining paper left for the rectangle meant that the height of the cylinder would be 11.58. To find the volume, we found the area of the circle and multiplied by the height of the cylinder: $\pi \times 4.71^2 \times 11.58 \approx 807.5$ cubic centimetres as our volume.

Lyman from Nanjing International School sent us two different arrangements for the circles and rectangles to try to maximise the area - click here to see them.

It is possible to improve on this volume:

Let the width of the paper fix the height of the cylinder, then h=21
If $r$ is the radius, then the length of the rectangle is $2 \pi r$ so if we fit the rectangle plus the circle along the $29.6$cm length we have $$2r + 2 \pi r = 29.6$$
$$r(2+2\pi) = 29.6$$
$$r = \frac{29.6}{2+2\pi}$$

So the volume of the cylinder is:
$$\pi r^2 h = \pi \times \frac{29.6^2}{(2+2\pi)^2} \times 21 \approx 842$$