Let $x,y \in R$ such that $|x+y| + |x-y| = 2$

Question

Let $x,y \in R$ such that $|x+y| + |x-y| = 2$, then find maximum values of $x^2 - 6x + y^2$ and $x^2 + y^2 + 10y$.

How do I go about solving this question? Is it possible to find all real values of $x$ and $y$ from the first equation? Please help.

Answer 1

Guide:

Notice that $|x+y|+|x-y|=2$ describes the boundary of a square.

$x^2-6x+y^2=(x-3)^2+y^2-9$

The largest value of $y^2$ is $1$. Given that $-1 \le x \le 1$. Try to solve the $-1 \le x \le 1$, how would you optimize $(x-3)^2$.

Answer 2

Hint:

If $f(x,y)=|x+y|+|x-y|-2,$

observe that $f(x,y)=f(-x,y)=f(x,-y)=f(-x,-y)$

WLOG $x+y, x-y\ge0;$

$$x=1, -x\le y\le x, y^2\le x^2=1$$

Now, $$x^2+y^2+10y=1+(y+5)^2-25$$

$-1\le y\le1\iff5-1\le5+y\le5+1\implies4^2\le(y+5)^2\le6^2$

Answer 3

Well, in the worst case when we are stuck we can do cases:

Case 1: $x+y\ge 0$ and $x-y \ge 0$ then

$x\ge -y$ and $x \ge y$ so $x \ge \max(y,-y) = |y|$.

And $|x+y| + |x-y| = (x+y)+(x-y) = 2x = 2$ so $x = 1$ and $|y| \le x=1$ so $-1 \le y \le 1$.

$x^2 - 6x +y^2 = -5+y^2$ and the max that $y^2$ can be is $1$ so the max of $x^2 -6x +y^2$ is $-4$ when $y = \pm 1$ if was assume $x+y\ge 0$ and $x-y \ge 0$.

(Likewise the max of $x^2 + y^2 + 10y = 1+y^2 + 10y$ is $12$ if $y=1$.)

Case 2:

$x +y < 0$ and $x-y \ge 0$ then

$x < -y$ and $x \ge y$ so $y \le x < -y$ so $y < 0$.

And $|x+y| + |x-y| = -x -y + x-y = -2y = 2$ so $y = -1$ and $-1\le x < 1$.

$x^2 -6x + y^2 = x^2 -6x +1$ and $x^2 \le 1$ and $x \ge -1$ so $-6x \le 6$ so the very max that $x^2-6x +y^2$ is $1+6 +1 = 8$ if $x=-1$.

(Likewise $x^2 +y^2 + 10y = x^2 + 11$ an the max that can be is $12$ if $x = -1$.)

Case 3: $x +y \ge 0$ and $x -y < 0$ then

$x \ge -y$ and $x < y$ so $-y \le x < y$ so $y > 0$.

$|x+y| + |x-y| = x+y+y-x = 2y =2$ so $y =1$ and $-1 \le x < 1$ and

the max value of $x^2-6x +y^2 = x^2-6x+1$ and $x^2 \le 1$ and $-6x \le 6$ so $x^2 -6x+1 \le 8$ and the max value is $8$ when $x=-1$.

case 4: $x+y < 0$ and $x-y < 0$ so $x< -y$ and $x < y$ so $x< \min (y,-y) = -|y|$

And $|x+y| + |x-y| = -x -y +y -2 =-2x = 2$ so $x = -1$ and $-|y| > x=-1$ so $|y|< 1$.

so $x^2 -6x +y^2 = 7 + y^2< 8$. but $x^2-6x +y^2 = 8$ never occurs; it's an unacheived least upper bound.

Of those four cases the max value of $x^2 -6x +y^2 = 8$.

....

that's probably not an efficient or clever way but it is a complete hit the ground running systematic way.