Percentage Changes and Optimization

Given $P_{2020} = 100000$. $P_{2021} = 100000\left(1 - \frac{y}{100}\right)$. $P_{2022} = 100000\left(1 - \frac{y}{100}\right)\left(1 + \frac{x}{100}\right) > 100000$.

So $\left(1 - \frac{y}{100}\right)\left(1 + \frac{x}{100}\right) > 1 \implies (100 - y)(100 + x) > 10000$. $100(x - y) - xy > 0$

Since $x, y \in \mathbb{N}$ and $P_{2022} > P_{2020}$, we must have $x > y$, so $x - y = 10 \implies x = y + 10$. $100(10) > y(y + 10) \implies y^2 + 10y - 1000 < 0$ $(y + 5)^2 < 1025$

Since $\sqrt{1025} \approx 32.015$, we have $y + 5 \le 32 \implies y \le 27$.

To minimize $P_{2021} = 100000(1 - y/100)$, we maximize $y$. The maximum natural number value of $y$ is $27$.

Lowest $P_{2021} = 100000 \left(1 - \frac{27}{100}\right) = 73000$.