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 $y\le |x|-1$
$y\ge 1-|x|$
$x,y\ge 0$
Consider above inequalities as equations, we get
$y=|x|-1=\{\begin{array}{ll}-x-1,& x<0\\ x-1,& x>0\end{array}$
$y=1-|x|=\{\begin{array}{ll}1-x,& x>0\\ 1+x,& x<0\end{array}$
On plotting these lines on the graph paper, we get

Here the feasible region is $BOA$ 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 $1$ .