Question

The maximum possible value of $x^{2}+y^{2}-4x-6y, x,y$ real, subject to the condition $|x+y|+|x-y|=4$ is

• A. 12
• B. 28
• C. 72
• D. does not exist

Answer 1

Answer: (B)

We have,
$\left|x+y\right|+\left|x-y\right|=4 $
$\Rightarrow x+y +x-y=4$
$\left[\because x+y \ge\,0 , x -y \ge\,0\right] $
$x=2 \Rightarrow x+y -x+y=4 $
$\left[\because x+y \ge\,0, x -y <\,0\right] $
$y=2 \Rightarrow -x-y+x-y=4 $
$\left[\because x+y <\,0, x-y \ge\,0\right]$
$y=-2 \Rightarrow -x-y-x+y=4 $
$\left[\because x+y<\,0, x-y <\,0\right]$
$x=-2$
The maximum possible value of
$x^{2}+y^{2}-4x-6y$ at $\left(-2, 2\right)$
$\therefore \left(-2\right)^{2}+\left(-2\right)^{2}-4\left(-2\right)-6\left(-2\right)$
$\Rightarrow 4+4+8+12=28$