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 $\left(-2,2\right)$
$\therefore {\left(-2\right)}^2+{\left(-2\right)}^2-4\left(-2\right)-6\left(-2\right)$
$\Rightarrow 4+4+8+12=28$