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

2

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8

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3

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1

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- \frac{1}{2}

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Solution

The correct option is A $- \frac{1}{2}$
$E = 2 x^{2} + y^{2} + 2 x y + 2 x - 3 y + 8$
$\Rightarrow E = \frac{1}{2} (4 x^{2} + 2 y^{2} + 4 x y + 4 x - 6 y + 16)$
$\Rightarrow E = \frac{1}{2} {(y - 4)^{2} + (2 x + y + 1)^{2} - 1} \geq - \frac{1}{2}$
This happen when the ${(y - 4)^{2} + (2 x + y + 1)^{2} - 1} = - 1$
Which i possible when $(x, y) \equiv (- \frac{5}{2}, 4)$

So, least value is $- \frac{1}{2}$

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2 x^{2} + y^{2} + 2 x y + 2 x - 3 y + 8

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1. {(3 x - y)}^{2} - ({5 x}^{2} - 2 x y)

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