Question by Kaya Renger - Discuss with MathYouLike

Kaya Renger Coordinator

28/09/2017 at 17:21

Given x,y satisfy: x² + 2xy + 6x + 6y + 2y² + 8 = 0

Find minimum and maximum value of B = x + y + 2016

List of answers

Phan Thanh Tinh Coordinator 28/09/2017 at 18:48

\(x^2+2xy+6x+6y+2y^2+8=0\)

\(\Leftrightarrow x^2+y^2+9+2xy+6y+6x+y^2-1=0\)

\(\Leftrightarrow\left(x+y+3\right)^2+y^2=1\)

\(y^2\ge0\Rightarrow\left(x+y+3\right)^2\le1\Rightarrow-1\le x+y+3\le1\)

\(\Leftrightarrow2012\le x+y+2016\le2014\)

Hence :

B = 2012 only when : y = 0 ; x = 2012 - 2016 = -4

B = 2014 only when : y = 0 ; x = 2014 - 2016 = -2
Kaya Renger selected this answer.
Vũ Trung Dũng 30/09/2017 at 18:07

$x^{2} + 2 x y + 6 x + 6 y + 2 y^{2} + 8 = 0$

$\Leftrightarrow x^{2} + y^{2} + 9 + 2 x y + 6 y + 6 x + y^{2} - 1 = 0$

$\Leftrightarrow {(x + y + 3)}^{2} + y^{2} = 1$

$y^{2} \geq 0 \Rightarrow {(x + y + 3)}^{2} \leq 1 \Rightarrow - 1 \leq x + y + 3 \leq 1$

$\Leftrightarrow 2012 \leq x + y + 2016 \leq 2014$

Hence :

B = 2012 only when : y = 0 ; x = 2012 - 2016 = -4

B = 2014 only when : y = 0 ; x = 2014 - 2016 = -2

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