Question

[hải linh]

tìm GTNN của A=2x^2+y^2-2xy-2x+y-12

tìm GTLN của A=-2x^2-y^2-2xy-2x+y-12

Answer 1

[_Guiltykamikk_]

\(A=2x^2+y^2-2xy-2x+y-12\)

\(A=\left(x^2-2xy+y^2\right)+x^2-2x+y-12\)

\(A=\left[\left(x-y\right)^2-2\left(x-y\right).\frac{1}{2}+\frac{1}{4}\right]+\left(x^2-x+\frac{1}{4}\right)-\frac{25}{2}\)

\(A=\left(x-y-\frac{1}{2}\right)^2+\left(x-\frac{1}{2}\right)^2-\frac{25}{2}\)

Do \(\left(x-y-\frac{1}{2}\right)^2\ge0\forall x;y\)

     \(\left(x-\frac{1}{2}\right)^2\ge0\forall x\)

\(\Rightarrow A\ge-\frac{25}{2}\)

Dấu "=" xảy ra khi :  \(\hept{\begin{cases}x-y-\frac{1}{2}=0\\x-\frac{1}{2}=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=\frac{1}{2}\\y=0\end{cases}}\)

Vậy  \(A_{Min}=-\frac{25}{2}\Leftrightarrow\left(x;y\right)=\left(\frac{1}{2};0\right)\)

Answer 2

[_Guiltykamikk_]

\(A=-2x^2-y^2-2xy-2x+y-12\)

\(-A=2x^2+y^2+2xy+2x-y+12\)

\(-A=\left(x^2+2xy+y^2\right)+x^2+2x-y+12\)

\(-A=\left[\left(x+y\right)^2-2\left(x+y\right).\frac{1}{2}+\frac{1}{4}\right]+\left(x^2+3x+\frac{9}{4}\right)+\frac{19}{2}\)

\(-A=\left(x+y-\frac{1}{2}\right)^2+\left(x+\frac{3}{2}\right)^2+\frac{19}{2}\)

Do  \(\left(x+y-\frac{1}{2}\right)^2\ge0\forall x;y\)

      \(\left(x+\frac{3}{2}\right)^2\ge0\forall x\)

\(\Rightarrow-A\ge\frac{19}{2}\Leftrightarrow A\le-\frac{19}{2}\)

Dấu "=" xảy ra khi :  \(\hept{\begin{cases}x+y-\frac{1}{2}=0\\x+\frac{3}{2}=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=-\frac{3}{2}\\y=2\end{cases}}\)

Vậy \(A_{Max}=-\frac{19}{2}\Leftrightarrow\left(x;y\right)=\left(-\frac{3}{2};2\right)\)