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

No worries! We‘ve got your back. Try BYJU‘S free classes today!

8

No worries! We‘ve got your back. Try BYJU‘S free classes today!

3

No worries! We‘ve got your back. Try BYJU‘S free classes today!

1

No worries! We‘ve got your back. Try BYJU‘S free classes today!

- \frac{1}{2}

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

Open in App

Solution

The correct option is E $- \frac{1}{2}$
The given equation is,
$2 x^{2} + y^{2} + 2 x y + 2 x - 3 y + 8 = \frac{1}{2} (4 x^{2} + 2 y^{2} + 4 x y + 4 x - 6 y + 16) = \frac{1}{2} [{(y - 4)}^{2} + {(2 x + y + 1)}^{2} - 1] \geq - \frac{1}{2}$

Therefore, the minimum value of the given expression is $- \frac{1}{2}$ , when $(y - 4)^{2} = 0 \Rightarrow y = 4$
and $(2 x + y + 1)^{2} \Rightarrow x = \frac{- 5}{2}$

Note: No option is correct.

Suggest Corrections

Similar questions

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

x

y

y - x = 1

x = y^{2}

y = 3^{x - 1} + 3^{- x - 1}

y^{2} = 8 x a n d x y = - 1

\frac{1}{x + y} - \frac{1}{2 x} = \frac{1}{30}, \frac{5}{x + y} + \frac{1}{x} = \frac{4}{3}

2 x^{2} - y^{2}

Join BYJU'S Learning Program