`A = 2020 - 2x^2 - y^2 - x + y`

`A = -(2x^2 + x) - (y^2 - y) + 2020`

`A = -2(x^2 + 1/2 x) - (y^2 - y) + 2020`

`A = -2[(x + 1/4)^2 - 1/16] - [(y - 1/2)^2 - 1/4] + 2020`

`A = -2(x + 1/4)^2 + 1/8 - (y - 1/2)^2 + 1/4 + 2020`

`A = -2(x + 1/4)^2 - (y - 1/2)^2 + (1/8 + 2/8 + 16160/8)`

`A = -2(x + 1/4)^2 - (y - 1/2)^2 + 16163/8`

Vì `-2(x + 1/4)^2 <= 0 AA x; -(y - 1/2)^2 <= 0 AA y`

Nên `A = -2(x + 1/4)^2 - (y - 1/2)^2 + 16163/8 <= 16163/8`

Dấu "`=`" xảy ra khi:

`{(x + 1/4 = 0),(y - 1/2 = 0):} <=> {(x = -1/4),(y = 1/2):}`

Vậy `A_{"max"} = 16163/8` khi `x = -1/4` và `y = 1/2`