1_ Tìm max:
A= -3x^2-y^2-3y-2
B=-4x^2-3y^2+6xy-6x+12y-23
[TEX]A=-3x^2-(y^2+3y+\frac{9}{4})+\frac{9}{4}-2[/TEX]
[TEX]A=-3x^2-(y+\frac{3}{2})^2+\frac{1}{4}[/TEX]
Ta thấy:
[TEX] -(y+\frac{3}{2})^2 \leq 0 ;-3x^2\leq 0[/TEX]
[TEX]\Rightarrow A\leq \frac{1}{4}[/TEX]
Vậy Max A=[TEX]\frac{1}{4}\Leftrightarrow x=0;y=\frac{-3}{2}[/TEX]
[TEX]B=-[4x^2+6x(1-y)+\frac{9(1-y)^2}{4}]+\frac{9-18y+9y^2}{4}-3y^2+12y-23[/TEX]
[TEX]B=-(2x+\frac{3(1-y)}{2})^2+\frac{9-18y+9y^2-12y^2+48y}{4}-23[/TEX]
[TEX]B=-(2x+\frac{3(1-y)}{2})^2+\frac{-3y^2+30y-75+84}{4}-23[/TEX]
[TEX]B=-(2x+\frac{3(1-y)}{2})^2+\frac{-3(y-5)^2}{4}+21-23[/TEX]
Vậy [TEX]max B=-2\Leftrightarrow x=3;y=5[/TEX]
P/s: Mỏi tay quá,thanks giùm cái nha!