The least value of $2 x^{2} + y^{2} + 2 x y + 2 x - 3 y + 8$ for real number $x$ and $y$ is

2

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8

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3

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\frac{- 1}{2}

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Solution

The correct option is D $\frac{- 1}{2}$
$2 x^{2} + y^{2} + 2 x y + 2 x - 3 y + 8$
$= \frac{1}{2} [4 x^{2} + 2 y^{2} + 4 x y + 4 x - 6 y + 16]$

$= \frac{1}{2} [(y^{2} - 8 y) + (4 x^{2} + y^{2} + 4 x y + 4 x + 2 y) + 16]$

$= \frac{1}{2} [(y^{2} - 8 y + 16) + (4 x^{2} + y^{2} + 4 x y) + (4 x + 2 y)]$

$= \frac{1}{2} [(y - 4)^{2} + (2 x + y)^{2} + 2 (2 x + y) + 1 - 1]$

$= \frac{1}{2} [(y - 4)^{2} + (2 x + y + 1)^{2} - 1]$
So, least value will be $- \frac{1}{2}$ at $y = 4$ and $x = \frac{- 5}{2}$ .

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