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

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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?


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 1

We have that:

$a x + b y + c z = m$

The maximum area of $x y$ is found as follows:

Put:

\(\ds a x + b y\) \(=\) \(\ds n\)
\(\ds \leadsto \ \ \) \(\ds x\) \(=\) \(\ds \dfrac {n - b y} a\)
\(\ds \leadsto \ \ \) \(\ds x y\) \(=\) \(\ds \dfrac {n y} a - \dfrac {b y^2} a\)
\(\ds \leadsto \ \ \) \(\ds \dfrac \d {\d y} x y\) \(=\) \(\ds \dfrac n a - \dfrac {2 b y} a\)
\(\ds \) \(=\) \(\ds 0\)
\(\ds \leadsto \ \ \) \(\ds y\) \(=\) \(\ds \dfrac n {2 b}\)
\(\ds \leadsto \ \ \) \(\ds b y\) \(=\) \(\ds \dfrac n 2\)

Similarly also by differentiation with respect to $x$:

$a x = \dfrac n 2$

and so:

$a x = b y$


Similarly:

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

and so:

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

Hence:

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

$\blacksquare$


Proof 2

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

$\blacksquare$


Sources

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