Minimum value of the function $z = 2 x^{2} + 2 x y + y^{2} - 2 x + 2 y + 2$ where $x$ and $y$ belongs to $R$ , is

3

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

- 3

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

7

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

2

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

Open in App

Solution

The correct option is A $- 3$
$z = x^{2} + y^{2} + 1 + 2 x y + 2 x + 2 y + x^{2} - 4 x + 4 - 4 + 1$
$= {(x + y + 1)}^{2} + {(x - 2)}^{2} - 3$
Minimum value of $z$ is $- 3$ .

Suggest Corrections

Similar questions

Z = 50 x + 20 y . . . (1)

2 x y \geq 5 . . . (2)

3 x + y \geq 3 . . . (3)

2 x - 3 y \leq 12 . . . (4)

x \geq 0, y \geq 0 . . . (5)

x, y > 0

2 x^{2} + \frac{2}{x} - 2 x + 2 y^{2} + \frac{2}{y} - 2 y + 2

Minimum value of $x^{2} + 2 x y + 3 y^{2} - 6 x - 2 y$ $x, y, ϵ R$ is

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}}

| r |

A \cup B \subseteq C

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

x

y

Join BYJU'S Learning Program