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

Global Maxima

For real valu...

Question

For real values of x and y, the minimum value of $f (x, y) = x^{2} + 2 x y + 2 y^{2} + 6 y + 10$ , is?

0

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

1

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

3

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

10

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

Open in App

Solution

The correct option is A $1$
$f_x^{'} = 2x+2y = 2(x+y) = 0$
$⟹ x = - y$
$f_{y}^{^{'}} = 2 x + 4 y + 6 = 0$
or, $2 y = - 6$
$y = - 3$ and $x = 3$
Putting in the equation, we get
$9 - 18 + 18 - 18 + 10$
$⟹ 1$

Suggest Corrections

Similar questions

x^{2} + 2 x y + 3 y^{2} - 6 y \geq \frac{- 9}{2}

f (x) = max (y^{2} - 2 x y + 1)

0 \leq y \leq 2,

f (x)

\forall x \in R

x + y = 3

x - y = 1

x - y + 5 = 0

x^{2} + 2 x y - y^{2} = 6,

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

x

y

x^{2} + y^{2} - 2 x y + 4 x + 4 y + 12 = 0

y

Join BYJU'S Learning Program