Question

A man makes complete use of 405 cc of iron, 783 cc of aluminium, and 351 cc 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 cm. If the total number of cylinders is to be kept at a minimum, then the total surface area of all these cylinders, in sq cm, is

• A. $8464\pi$
• B. $928\pi$
• C. $1044(4+\pi)$
• D. $1026(1+\pi)$

Answer 1

Answer: (D)

$1026(1+\pi)$

Answer 2

To determine the minimum number of cylinders, the volume of each cylinder must be the Highest Common Factor (HCF) of 405, 783, and 351.
$HCF(405,783,351)=27$
As a result, the number of cylinders for iron is $\frac{405}{27}=15$, for aluminum is $\frac{783}{27}=29$, and for copper is $\frac{351}{27}=13$.
Therefore, the total number of cylinders is $15+29+13=57.$
Additionally, the volume of each cylinder is $27 cc.$
$\pi r^2h=27$
$\pi \times3^2\times h=27$
$h=3\pi$

The total surface area of each cylinder is $2πr(r+h)=2π×3(3+3π)=18(π+1)$.

Hence, the total surface area of 57 cylinders is $57×18(π+1)=1026(π+1).$