Henry Ernest Dudeney/Modern Puzzles/142 - Economy in String/Solution/Proof 2

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Modern Puzzles by Henry Ernest Dudeney: $142$

Economy in String
Owing to the scarcity of string a lady found herself in this dilemma.
In making up a parcel for her son, she was limited to using $12$ feet of string, exclusive of knots,
which passed round the parcel once lengthways and twice round its girth, as shown in the illustration.
Dudeney-Modern-Puzzles-142.png
What was the largest rectangular parcel that she could make up, subject to these conditions?


Solution

$2$ feet by $1$ foot by $\tfrac 2 3$ feet.


Proof

From the general solution:

Let the string pass:

$a$ times along length $x$
$b$ times along breadth $y$
$c$ times along depth $z$.

Let the string be length $m$.

Then the maximum volume $xyz$ of the parcel is given by:

$x y z = \dfrac {m^2} {27 a b c}$

where:

\(\ds x\) \(=\) \(\ds \dfrac m {3 a}\)
\(\ds y\) \(=\) \(\ds \dfrac m {3 b}\)
\(\ds z\) \(=\) \(\ds \dfrac m {3 c}\)

$\Box$


Setting:

\(\ds a\) \(=\) \(\ds 2\)
\(\ds b\) \(=\) \(\ds 4\)
\(\ds c\) \(=\) \(\ds 6\)
\(\ds m\) \(=\) \(\ds 12\)

we obtain:

\(\ds x\) \(=\) \(\ds 2\)
\(\ds y\) \(=\) \(\ds 1\)
\(\ds z\) \(=\) \(\ds \dfrac 2 3\)
\(\ds x y z\) \(=\) \(\ds 1 \tfrac 1 3\)

Hence the result.

$\blacksquare$


Sources

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