Henry Ernest Dudeney/Modern Puzzles/142 - Economy in String/General Solution/Proof 2
Jump to navigation
Jump to search
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?
General Result
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
We have that:
- $a x + b y + c z = m$
Then:
\(\ds x y z\) | \(=\) | \(\ds \frac 1 {a b c} \sqrt[3] {\paren {a x} \paren {b y} \paren {c z} }^3\) | ||||||||||||
\(\ds \) | \(\le\) | \(\ds \frac 1 {a b c} \paren {\frac {a x + b y + c z} 3}^3\) | Cauchy's Mean Theorem | |||||||||||
\(\ds \) | \(=\) | \(\ds \frac 1 {a b c} \paren {\frac m 3}^3\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \frac {m^3} {27 a b c}\) |
Equality holds when $a x = b y = c z = \dfrac m 3$.
$\blacksquare$