Tìm giá trị lớn nhất của biểu thức: $B=-x^{2}-y^{2}-xy+2x+3y$

$4B=-4x^2-4y^2-4xy+8x+12y$

$= - (4x^2+4y^2+4xy -8x-12y)$

$= - [(4x^2 +4xy + y^2) - (8x + 4y) + (3y^2-8y+\dfrac{16}{3})-\dfrac{16}{3}]$

$= - [(2x+y)^2-4(2x+y)+4 + 3 (y-\dfrac{4}{3})^2-\dfrac{28}{3}]$

$= - [(2x+y-2)^2 +3(y-\dfrac{4}{3})^2 - \dfrac{28}{3}]$

$= - (2x+y-2)^2-3(y-\dfrac{4}{3})^2 +\dfrac{28}{3}$

$=> B=\dfrac{-1}{4}(2x+y-2)^2 -\dfrac{3}{4}(y-\dfrac{4}{3})^2 +\dfrac{7}{3}=< \dfrac{7}{3}$

Dấu "$=$" xảy ra khi :

$2x+y=2, y=\dfrac{4}{3}$

$<=>x=\dfrac{1}{3},y=\dfrac{4}{3}$

Vậy $B_{max}=\dfrac{7}{3}<=>x=\dfrac{1}{3},y=\dfrac{4}{3}$