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'Tubular Stand' printed from https://nrich.maths.org/
There were a number of partial solutions but
this well explained one is almost entirely the work ofAndrei of
Tudor Vianu National College and shows how "obvious" the answer is
with a little visualising and careful reasoning.
Let $l$ be the exterior length of the cylinders and $r$ - the
radius of the cylinders.
In the first solution, I have $4$ cylinders (the $4$ squares length
$l-4r$ and $2r$) and $8$ half-cylinders - cylinders cut through the
diagonal (the 8 small right-angled triangles from the
figure).
Each of the four cylinders has the volume: $ V_1 = \pi r^2 (l -
4r)$
Each of the half-cylinders has a volume of half a cylinder: $ V_2 =
\frac{1}{2} \times \pi r^2 \times 2r = \pi r^3$ Now, the total
volume of the tubular stand is: $ V = 4 \times V_1 + 8 \times V_2 =
4\pi r^2 (l - 2r) $
With the second method, I arrange the $8$ half-cylinders to make
$4$ bigger cylinders, and these $4$ to the bigger ones. The height
of one 'big' cylinder is $(l - 2r)$, and the total volume is: $$ V
= 4 \pi r^2 (l - 2r) $$ which is exactly the result obtained above.
Substituting the numerical values: $l = 10 \; \text{cm}$ and $r
=0.5 \; \text{cm}$, I obtain:
$$ V = 4 \pi \times 0.25 \times(10 - 1) = 9 \pi = 28.27 \;
{\text{cm}}^3 $$
If the volume wood would be double $(18 \pi)$, then the outside
dimension of the dowel would be: \begin{eqnarray} 18 \pi
&=& 4\pi \times 0.25 (l -1) \\ l - 1 & =& 18 \;
\text{cm}\\ l &=& 19 \; \text{cm}\end{eqnarray}
If the volume of wood would be the same but the radius would be $1
\; \text{cm}$, then the outside dimension would be:
\begin{eqnarray} 9 \pi &=& 4\pi (l - 2) \\ l - 2
&=& \frac {9}{4} \\ l &=& 4.25 \;
\text{cm}\end{eqnarray} The general formula for the
volume of the dowel (proved above) is: $$ V = 4 \pi r^2 (l - 2r) $$
One could see that the volume is proportional to the square of the
radius, and in the limit of long outside dimensions, proportional
to the outside dimension.