Determine the region bounded by the inequalities: $$ 0 \leq x + y \leq 1 \\ 0 \leq x - y \leq x + y $$

I don't know what to solve for first, so I just added them:

$$ 0 \leq x \leq 1 + x + y \\ $$

I guess I can subtract $x$:

$$ -x \leq 0 \leq 1 + y \\ $$


$$ -y - 1 \leq 0 \leq x \\ $$

So from this inequality, it looks like some area in the 4th quadrant because $x \geq 0$ means everything to the right of the $y$-axis, and $-y - 1 \leq 0$ means $- 1 \leq y$ which is above the line $y = -1$. However, it looks like I'm analyzing incorrectly as the answer says that it is some area above $y = 0$. I'm not sure what I'm doing wrong.

  • 1
    $\begingroup$ There are four linear inequalities in two variables. These correspond to half-planes in the plane, and your region is their intersection. Draw each of these half-planes to see what your region looks like. $\endgroup$ – Servaes May 20 '14 at 18:50

The inequalities are: $$y\le 1-x$$ $$y\ge -x$$ $$y\ge 0$$ $$y\le x$$ You should graph these and determine the region where all the inequalities hold. Here is a picture of what it should look like

enter image description here


Note that you can also write your pair of inequalities as a single linear chain of inequalities:

$$0 \leq x-y \leq x+y \leq 1.$$

So all the information you need is contained in the three inequalities of the chain:

$$\begin{cases}0 \leq x-y,\\x-y \leq x+y,\\x+y \leq 1.\end{cases} \iff \begin{cases}y \leq x,\\0 \leq y,\\y \leq 1-x.\end{cases}$$

The region bounded by these three lines is found to be the interior of the right isosceles triangle with vertices $(0,0),(0,1),(\frac12,\frac12)$.


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