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.
 A: 
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
A: Hint: your objective function $x^2+y^2-4x-6y$ is convex, so its maximum is attained at an extreme point of your convex feasible region. 
A: $$x^2-4x+y^2-6x= (x-2)^2 + (y-3)^2-13$$
now this means you have to maximize distance from $(2,3)$
now lets analyze $|x-y| + |x+y|$
in quadrant 1: $x>y$
$$x-y+x+y=4 \rightarrow x=2$$
similiarly you have for $y>x$
$$y=2$$
now if you take negative of both $x$ and $y$ you will get same result, there will be symmetry around origin and figure will recreate as $x=-2$ and $y=-2$ in 3rd quadrant for $x<y$ and $y<x$ respectively
for 2nd quadrant we analyze $|x|>y$
$$ y-x -x-y=4 \rightarrow x=-2$$
after doing similar analysis for $y$ and using symmetry arguments you observe that you get a square of side $4$ centered around origin. Now you just need to pick a point from this square which gives you the maximum distance from $(2,3)$

