Henry Ernest Dudeney/Modern Puzzles/142 - Economy in String/Solution
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.
- 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.
General Solution
Let the string pass:
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:
\(\ds x\) | \(=\) | \(\ds \dfrac m {3 a}\) | ||||||||||||
\(\ds y\) | \(=\) | \(\ds \dfrac m {3 b}\) | ||||||||||||
\(\ds z\) | \(=\) | \(\ds \dfrac m {3 c}\) |
Proof 1
From the general solution:
Let the string pass:
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:
\(\ds x\) | \(=\) | \(\ds \dfrac m {3 a}\) | ||||||||||||
\(\ds y\) | \(=\) | \(\ds \dfrac m {3 b}\) | ||||||||||||
\(\ds z\) | \(=\) | \(\ds \dfrac m {3 c}\) |
$\Box$
Thus it follows that the total length of string that passes along the length, breadth or depth must in every case be the same to allow of the maximum dimensions, that is: $4$ feet.
That is:
- $a x = b y = c z = \dfrac m 3$
The string passes $2$ times along length, $4$ times along breadth, and $6$ times along depth.
Therefore $4$ feet divided by $2$, $4$ and $6$ feet will give us $2$ feet, $1$ foot, and $\tfrac 2 3$ feet respectively for the length, breadth or depth of the largest possible parcel.
Proof 2
From the general solution:
Let the string pass:
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
- 1926: Henry Ernest Dudeney: Modern Puzzles ... (previous) ... (next): Solutions: $142$. -- Economy in String
- 1968: Henry Ernest Dudeney: 536 Puzzles & Curious Problems ... (previous) ... (next): Answers: $313$. Economy in String