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

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:

$a$ 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:

\(\ds x\)

\(=\)

\(\ds \dfrac m {3 a}\)

\(\ds y\)

\(=\)

\(\ds \dfrac m {3 b}\)

\(\ds z\)

\(=\)

\(\ds \dfrac m {3 c}\)

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 $a x = b y = c z = \dfrac m 3$.

$\blacksquare$