\(B=12x-8y-4x^2-y^2+1\)
\(=-\left(4x^2-12x+y^2+8y-1\right)\)
\(=-\left[\left(4x^2-12x+9\right)+\left(y^2+8y+16\right)-24\right]\)
\(=\left[\left(2x-3\right)^2+\left(y+4\right)^2-24\right]\)
\(=-\left(2x-3\right)^2-\left(y+4\right)^2+24\)
\(\Rightarrow B_{max}=24\Leftrightarrow-\left(2x-3\right)^2-\left(y+4\right)^2=0\)
\(\Rightarrow\hept{\begin{cases}2x-3=0\\y+4=0\end{cases}\Rightarrow\hept{\begin{cases}x=\frac{3}{2}\\y=-4\end{cases}}}\)
Ta có: B = 12x - 8y - 4x2 - y2 + 1 = (-4x2 + 12x - 9) - (y2 + 8y + 16) + 26 = -4(x2 - 3x + 9/4) - (y + 4)2 + 26 = -4(x - 3/2)2 - (y + 4)2 + 26
Ta luôn có: -4(x - 3/2)2 \(\le\) 0 \(\forall\) x (vì 4(x - 3/2)2 \(\ge\)0 \(\forall\)x)
-(y + 4)2 \(\le\) 0 \(\forall\)y (vì (y + 4)2 \(\ge\)0 \(\forall\) y)
=> -4(x - 3/2)2 - (y + 4)2 + 26 \(\le\) 26 \(\forall\)x,y
hay B \(\le\) 26 \(\forall\)x, y
Dấu "=" xảy ra khi : \(\hept{\begin{cases}\left(x-\frac{3}{2}\right)^2=0\\\left(y+4\right)^2=0\end{cases}}\) <=> \(\hept{\begin{cases}x-\frac{3}{2}=0\\y+4=0\end{cases}}\) <=> \(\hept{\begin{cases}x=\frac{3}{2}\\y=-4\end{cases}}\)
Vậy Bmax = 26 tại x = 3/2 và y = -4
