Đáp án:
Max `A = 2017` khi `y=1/4;x=5/4`
Giải thích các bước giải:
Ta có:
`A=-2x^2-10y^2+4xy +4x+4y+2013`
`=-(2x^2+10y^2-4xy-4x-4y-2013)`
`=-[(2x^2+2y^2-4xy)-(4x-4y)+2-2015+8y^2-8y]`
`=-[2(x-y)^2-4(x-y)+2+(8y^2-8y+2)-2017]`
`=-[2(x-y-1)^2+8(y-1/2)^2]+2017 <=2017`
Dấu "=" xảy ra khi:
\(\Leftrightarrow\left\{{}\begin{matrix}y-\dfrac{1}{2}=0\\x-y-1=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}y=\dfrac{1}{2}\\x-\dfrac{3}{2}=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=\dfrac{1}{2}\\x=\dfrac{3}{2}\end{matrix}\right.\)
Vậy Max `A = 2017` khi `y=1/2;x=3/2`
