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The least value of $2 x^{2}+y^{2}+2 x y+2 x-3 y+8$ for real
numbers $x$ and $y,$ is
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numbers $x$ and $y,$ is
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Verified Answer
The correct answer is:
$-1 / 2$
$2 x^{2}+y^{2}+2 x y+2 x-3 y+8$
$=\frac{1}{2}\left|4 x^{2}+2 y^{2}+4 x y+4 x-6 y+16\right|$
$=\frac{1}{2}\left[\left(y^{2}-8 y\right)+\left(4 x^{2}+y^{2}+4 x y+4 x+2 y\right)+16\right].$
$=\frac{1}{2}\left|(y-4)^{2}+(2 x+y+1)^{2}-1\right|$
So, least value will be $-\frac{1}{2}$ at $y=4$ and $x=\frac{-5}{2}$
$=\frac{1}{2}\left|4 x^{2}+2 y^{2}+4 x y+4 x-6 y+16\right|$
$=\frac{1}{2}\left[\left(y^{2}-8 y\right)+\left(4 x^{2}+y^{2}+4 x y+4 x+2 y\right)+16\right].$
$=\frac{1}{2}\left|(y-4)^{2}+(2 x+y+1)^{2}-1\right|$
So, least value will be $-\frac{1}{2}$ at $y=4$ and $x=\frac{-5}{2}$
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