Geometry problem 5 by Dhaval Furia

Geometry Level pending

A man makes complete use of $405 \text{ cm}^3$ of iron, $783 \text{ cm}^3$ of aluminium, and $351\text{ cm}^3$ of copper to make a number of solid right circular cylinders of each type of metal. These cylinders have the same volume and each of these has radius $3\text{ cm}$ . If the total number of cylinders is to be kept at a minimum, find the total surface area of all these cylinders in $\text{ cm}^2$ .

8464 \pi

928 \pi

1026 (1 + \pi)

1044 (4 + \pi)

1 solution

A Former Brilliant Member
May 22, 2020

G. C. F. of $405, 783,351$ is $27$ . Hence, the volume of each cylinder is $27$ cm $^3$ , there will be $15$ iron cylinders, $29$ aluminium cylinders, and $13$ copper cylinders. So, in all, there will be $57$ cylinders.

We also have, volume of each cylinder $=π\times 3^2\times h=27\implies h=\dfrac{3}{π}$ . Here $h$ is the height of each cylinder.

Total surface area of each cylinder is

$2π\times 3\times (\frac{3}{π}+3)=18(1+π)$ cm $^2$ .

So the total surface area of all the cylinders is $57\times 18(1+π)=\boxed {1026(1+π)}$ .

Geometry problem 5 by Dhaval Furia

1 solution

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