Applesellers' Problem

: Chapter $2$: Ancient and Curious Problems: $41$
Two market-women were selling apples, one at $2$ for $1$ cent and the other at $3$ for $2$ cents.

They had $30$ apples apiece.

In order to end their competition they formed a trust, pooling their stocks and selling the apples at $5$ for $3$ cents.

This was to their advantage because under the new arrangement they took in total $36$ cents.

Under the old system they would have taken a total of only $35$ cents.

Their example was contagious.

Two other women, who also had $30$ apples apiece and who were selling them at $2$ for $1$ cent and $3$ for $1$ cent, formed a trust to sell their apples at $5$ for $2$ cents.

But instead of the $25$ cents which they would have taken in operating separate enterprises, their trust grossed only $24$ cents.

Why?

Solution
Consider a general trust such as these.

Let one participant sell $a$ apples for $b$ cents, and the other sell $c$ apples for $d$ cents.

Then the first sells apples at $\dfrac b a$ cents each, and the other at $\dfrac d c$ cents each.

, let $\dfrac b a > \dfrac d c$.

The average price per apple is $\dfrac 1 2 \paren {\dfrac b a + \dfrac d c} = \dfrac {b c + a d} {2 a c}$.

The trust sets the price at $\dfrac {b + d} {a + c}$.

This trust price will be advantageous :
 * $\dfrac {b + d} {a + c} > \dfrac {b c + a d} {2 a c}$

After algebra, this works out at:
 * $\paren {b c - a d} \paren {a - c} > 0$

For this to hold, it is important for both factors to be positive.

We have that:
 * $b c - a d > 0 \implies \dfrac b a > \dfrac d c$

This, combined with $a - c > 0$, means that the trust will be advantageous only if:
 * the original prices are unequal

and:
 * the denominator of the higher price is greater than that of the lower price.

For the first trust, we configure our variables such that:
 * $a = 3$, $b = 2$
 * $c = 2$, $d = 1$

in order to make $a - c > 0$.

Then we see:
 * $b c - a d = 2 \times 2 - 3 \times 1 = 1$

and so the trust is favourable.

For the second trust, we configure our variables such that:
 * $a = 3$, $b = 1$
 * $c = 2$, $d = 1$

in order to make $a - c > 0$.

Then we see:
 * $b c - a d = 2 \times 1 - 3 \times 1 = -1$

revealing that the trust is not favourable.

It is interesting to note that the trust price is arbitrarily dependent upon the form in which the original prices are presented.

Suppose the first participant in the second trust sold apples at $4$ for $2$ cents each.

Then the trust price of $\dfrac {2 + 1} {4 + 3} = \dfrac 3 7$ per apple is advantageous, as we see:
 * $a = 4$, $b = 2$
 * $c = 3$, $d = 1$

and:
 * $b c - a d = 3 \times 2 - 4 \times 1 = 2$

As puts it:
 * ... the formula which by chance or accident fits both the presently considered trust prices is a wholly unsound one.