You visited us 1 times! Enjoying our articles? Unlock Full Access!

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

Open in App

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 $5.$

Suggest Corrections

Similar questions

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

x, y \in R

x, y \in R

x^{2} + y^{2} - 4 x - 2 y + 5 = 0

\frac{[(\sqrt{x} - \sqrt{y})^{2} + 4 \sqrt{x y})]}{(x + \sqrt{x y})}

x, y \in R

x^{2} + y^{2} - 4 x - 2 y + 5 = 0

\frac{(\sqrt{x} - \sqrt{y})^{2} + 4 \sqrt{x y}}{x + \sqrt{x y}}

2 x^{2} - 2 y + 4 x y - x

Join BYJU'S Learning Program

Explore more

Join BYJU'S Learning Program