<s>[INST] The maximum possible value of $x^2+y^2-4x-6y$ subject to the condition $|x+y|+|x-y|$ The maximum possible value of
$x^2+y^2-4x-6y$
subject to the condition $|x+y|+|x-y|$=4
My workout...
Now if we add 13 to the equation we get 
$x^2+y^2-4x-6y+13-13$
or,$x^2+y^2-4x-6y+4+9-13$
or,$(x-2)^2+(y-3)^2-13$
Are there any methods other than function.
 [/INST] 
Solution is x=-2, y=-2.
Let us call $S(x,y)=x^2+y^2-4x-6y$.
Let us analyze our condition $|x+y|+|x-y|=4$. If we now rise this to quadrat :
$$
(|x+y|+|x-y|)^2=4^2
$$
$$
x^2+y^2+2xy+x^2+y^2-2xy+2x^2-2y^2=4x^2=4^2
$$
$$
x=+2,-2
$$
2 Solutions, or
$$
x^2+y^2+2xy+x^2+y^2-2xy-2x^2+2y^2=4y^2=4^2
$$
$$
y=+2,-2
$$
For x=2 or x=-2, we have for y a range between -2 and 2. But because function is convex, a maximal value can be only at y=-2, or y=2 but not between. Same if happen fixing y : For y=2 or y=-2, we have for x a range between -2 and 2, but it need to be -2 or 2. So we only need to check all 4 possibilities : 
S(-2,2)=4
S(2,2)=-12
S(-2,-2)=28
S(2,-2)=12
Clearly a maximal value appear for S(x=-2,y=-2)=28.
greatings, Daniel
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