Question

Given the LPP :
Minimize $f=2x_1-x_2$
$x_1\ge 0,x_2\ge 0$
$x_1+x_2\ge 5$
$-x_1+x_2\le 1$
$5x_1+4x_2\le 40$
The solution is

• A. 1
• B. -1
• C. 2
• D. -2

Answer 1

Answer: (A)

Given, minimize $f=2x_1-x_2$
$x_1\ge 0,x_2\ge 0$
$x_1+x_2\ge 5$
$-x_1+x_2\le 1$
and $x_1+x_2\le 1$
Graph of $x_1+x_2=5$

$x_1$	0	5
$x_2$	5	0

Putting $(0,0)$ in the inequality $x_1+x_2\ge 5$
$\Rightarrow 0+0\ge 5$
$\Rightarrow 0\ge 5$
(which is false)
$\Rightarrow$ Half plane away from origin.

Graph of $-x_1+x_2=1$
$\Rightarrow x_1-x_2=1$

$x_1$	0	-1
$x_2$	1	0

Putting $(0,0)$ in the inequality $-x_1+x_2\le 1$
$\Rightarrow 0+0\le 1\Rightarrow 0\le 1$
(which is true)
Half plane towards the origin. and graph of $5x_1+4x_2=40$

$x_1$	0	8
$x_2$	10	0

Putting $(0,0)$ in the inequality $5x_1+4x_2\le 40$
$\Rightarrow 0+0\le 40$
$\Rightarrow 0\le 40$ (which is true)
$\Rightarrow$ Half plane towards the origin.
Here, we obtain feasible region $ABPQ$ s Coordinates of $Q$.
Solving $x_1-x_2=-1$
$5x_1+4x_2=40$
We get $x_1=4,x_2=5,Q\equiv (4,5)$,
Coordinates of $P$.
Solving $x_1-x_2=-1$
and $x_1+x_2=5$
We get $x_1=2,x_2=3$
$P\equiv (2,3)$

Corner point	$Z=2x_1-x_2$
A (5, 0)	10
B(8, 0)	16
P(2, 3)	1 (minimum)
Q(4, 5)	3