Solving Constant Product for ratio

Hi Maxbry,

You can follow these steps:

First, express $Y'$ in terms of $R'$ and $X'$:

\[ Y' = R' \cdot X' \]

Substitute $X' = X + z$  and  $Y' = Y - a$:

\[ R' \cdot (X + z) = Y - a \]

Since $a = Y - \frac{k}{X + z}$, substitute $a $:

\[ R' \cdot (X + z) = Y - \left(Y - \frac{k}{X + z}\right) \]

Simplify the equation:

\[ R' \cdot (X + z) = \frac{k}{X + z} \]

We have $k = XY$, so substitute $k$:

\[ R' \cdot (X + z) = \frac{XY}{X + z} \]

Multiply both sides by $X + z$ to clear the fraction:

\[ R' (X + z)^2 = XY \]

Rearrange to form a quadratic equation in terms of $z$:

\[ R' z^2 + 2R' X z + (R' X^2 - XY) = 0 \]

This quadratic equation can be solved using the quadratic formula:

$z = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$, where $a = R'$, $b = 2R' X$, and $c = R' X^2 - XY$.

\[ z = \frac{-2R' X \pm \sqrt{(2R' X)^2 - 4(R')(R' X^2 - XY)}}{2R'} \]

Simplify the terms inside the square root:

\[ z = \frac{-2R' X \pm \sqrt{4R'^2 X^2 - 4R'^2 X^2 + 4R' XY}}{2R'} \]
\[ z = \frac{-2R' X \pm \sqrt{4R' XY}}{2R'} \]
\[ z = \frac{-2R' X \pm 2\sqrt{R' XY}}{2R'} \]
\[ z = -X \pm \sqrt{\frac{XY}{R'}} \]

So, we get two potential solutions for $z$:

\[ z = -X + \sqrt{\frac{XY}{R'}} \]
\[ z = -X - \sqrt{\frac{XY}{R'}} \]

Since $z$ represents the amount of funds of currency $X$ given into the pool, it must be a positive value.

Thus:

\[ z = -X + \sqrt{\frac{XY}{R'}} \]

Thus, the solution for $z$ given $X$, $Y$, and $R'$ is:

\[ z = \sqrt{\frac{XY}{R'}} - X \]

Example (using your numbers):
- $X = 20$
- $Y = 10$
- $R' = 0.4535$

Compute $\frac{XY}{R'}$:

\[ \frac{XY}{R'} = \frac{20 \times 10}{0.4535} = \frac{200}{0.4535} \approx 440.87 \]

Then: $\sqrt{\frac{XY}{R’} }$:

\[ \sqrt{440.87} \approx 21 \]

Finally, $z$:

\[ z = 21 - 20 = 1 \]

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Kav10