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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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Similar questions

Q. The least value of expression

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

x, y \in R

x, y \in R

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},

B = {(x, y) \in R \times R | 4 x^{2} + 4 y^{2} - 16 y + 7 = 0}

C = {(x, y) \in R \times R | x^{2} + y^{2} - 4 x - 2 y + 5 \leq r^{2}}

.
Then the minimum value of

| r |

A \cup B \subseteq C

Q. Subtract x²− 2xy + 5y² − 4 from 4xy − 5x² − y² + 6.

x + y = 3

x y = 6

, what is the value of the expression

(x^{2} + y^{2})^{2} + 4 x^{2} y^{2} + 4 x y^{2} + 4 x^{2} y

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Higher Order Equations

Standard XIII Mathematics

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