minimizing this expression: $\frac{(x+y)(c+d)-(z+w)(a+b)}{(a+b)(c+d)}$

Question

Jody and Shelli each receive a box of buttons on Saturday and on Sunday. Each button is either red or white, and the number of buttons in each of the four boxes is from $1$ to $100$, inclusive.

On both days, the percentage of red buttons in Jody's box is greater than the percentage of red buttons in Shelli's box. If $J \%$ of Jody's total number of buttons are red and $S \%$ of Shelli's total number of buttons are red, what is the least possible value of $J-S$? Express your answer to the nearest integer.

Ans. -96 (Source: 2021 MathCounts Target Round, calculators allowed)

I had Jody receiving on Saturday and Sunday respectively $a$ and $b$ buttons of which $x$ and $y$ are red. Shelli on Saturday and Sunday receive respectively $c$ and $d$ buttons, of which $z$ and $w$ are red.

We have $\frac{x}{a} > \frac{z}{c}$ and $\frac{y}{b} > \frac{w}{d}$ and

$J-S=100 \cdot (\frac{x+y}{a+b} - \frac{z+w}{c+d})$. So,

$J-S=100 \cdot \frac{(x+y)(c+d)-(z+w)(a+b)}{(a+b)(c+d)}$

I'm not sure how to minimize this expression.

Answer 1

Heuristic: use a graphical reasoning on the percentages transformed into slopes of the diagonals in these parallelograms:

(in fact, take a grid $200 \times 200$).

Answer 2

("guessing from extreme scenarios" Approach for competitions that only needs final answer. It's most likely true, but isn't proven. )
For the first day, Jody has 1 red button (100%), Sheeli has 99 red buttons and 1 white button (99%).
For the second day, Jody has 1 red button and 99 white buttons (1%), Sheeli has 1 white button (0%).
Then, $ J = \frac{ 1 + 1 } { 1 + 100 } = 2/ 100$ and $ s = \frac{99+0 } { 100 + 1 } = 99 / 101.$
This gives us $2 / 101 - 99/101 \approx -96\%$.

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Now, prove it.