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Higher Order Equations

The least val...

Question

The least value of expression $x^{2} - 4 x y + 5 y^{2} - 2 y + 6,$ if $x, y \in R$ is

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Solution

$f (x, y) = x^{2} - 4 x y + 5 y^{2} - 2 y + 6$
$= x^{2} - 2 x (2 y) + (2 y)^{2} + y^{2} - 2 y + 1 + 5$
$= (x - 2 y)^{2} + (y - 1)^{2} + 5$
$\geq 5$

Hence, least value of $f (x, y)$ is $5.$

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x^{2} - 4 x y + 5 y^{2} - 2 y + 6,

x, y \in R

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2 x^{2} + 6 x y + 5 y^{2} = 1

A = {(x, y) \in R \times R | 2 x^{2} + 2 y^{2} - 2 x - 2 y = 1},

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| r |

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