Question

Max $ z\, =\, x + y $
subject to $ y \le |x| - 1 $
$ y \ge - |x| $
$ x,y \ge 0 $
The Max $z$ is equal to

• A. 0
• B. 1
• C. 2
• D. -1

Answer 1

Answer: (B)

Given, $\max z=x+y$
subject to $y \leq|x|-1$
$y \geq 1-|x| $
$x, y \geq 0$
Consider above inequalities as equations, we get
$y=|x|-1=\begin{cases}-x-1, & x < 0 \\x-1, & x>0\end{cases} $
$y=1-|x|=\begin{cases}1-x, & x>0 \\1+x, & x < 0\end{cases}$
On plotting these lines on the graph paper, we get

Here the feasible region is $B O A$ and its corner points are $O(0,0), A(1,0)$ and $B(0,1)$
Now, by corner point method

Point	Max $z = x + y$
O(0, 0)	z = 0
A (1, 0)	z = 1
B (0, 1)	z = 1

Hence, the maximum value of $z$ is $1$ .