Đáp án: $B_{max}=4$ khi $x=y=2$
Giải thích các bước giải:
$B=-x^2-y^2+xy+2x+2y$
$=-(x^2+y^2-xy-2x-2y)$
`=-(x^2-xy+\frac{1}{4}y^2-2x+y+1+\frac{3}{4}y^2-3y+3-4)`
`=4-[(x^2-xy+\frac{1}{4}y^2)-(2x-y)+1]-(\frac{3}{4}y^2-3y+3)`
`=4-[(x-\frac{1}{2}y)^2-2.(x-\frac{1}{2}y)+1-\frac{3}{4}(y^2-4y+4)`
`=4-(x-\frac{1}{2}y-1)^2-\frac{3}{4}(y-2)^2`
Do `(x-\frac{1}{2}y-1)^2≥0∀x;y⇒-(x-\frac{1}{2}y-1)^2≤0`
`(y-2)^2≥0∀y⇒-\frac{3}{4}(y-2)^2≤0`
`⇒-(x-\frac{1}{2}y-1)^2-\frac{3}{4}(y-2)^2≤0`
`⇒B=4-(x-\frac{1}{2}y-1)^2-\frac{3}{4}(y-2)^2≤4`
Dấu bằng xảy ra
`⇔(x-\frac{1}{2}y-1)^2=(y-2)^2=0`
Từ $(y-2)^2=0⇔y-2=0⇔y=2$
`(x-\frac{1}{2}y-1)^2=0⇔x-\frac{1}{2}y-1=0⇔x=\frac{1}{2}y+1=\frac{1}{2}.2+1=2`
