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The maximum possible value of $x^{2}+y^{2}-4 x-6 y, x, y$ real, subject to the condition $|x+y|+|x-y|=4$
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The correct answer is:
is 28
$|x+y|+|x-y|=4$ represent a square
$x^{2}+y^{2}-4 x-6 y=(x-2)^{2}+(y-3)^{2}-13$
$=$ (distance point on square from $(2,3))^{2}-13$
Maximum $=(-2-2)^{2}+(-2-3)^{2}-13=28$
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