Question

Consider the following statements
I. The objective function
maximise $Z=-x+2 y$
subject to the constraints
$x \geq 3, x+y \geq 5, x+2 y \geq 6, x, y \geq 0 q$
has no maximum value.
II. The objective function
maximise $Z=x+y$
subject to $x-y \leq-1,-x+y \leq 0, x, y \geq 0$
has no maximum value.
Choose the correct option.

• A. Only $\mid$ is true
• B. Only II is true
• C. Both I and II are true
• D. Neither I nor II is true

Answer 1

Answer: (C)

I. Our problem is to maximise
$Z=-x+2 y$...(i)
Subject to the constraints are
$x \geq 3$...(ii)
$x+y \geq 5$...(iii)
$x+2 y \geq 6$...(iv)
$x \geq 0, y \geq 0$...(v)
Draw the graph of the line $x+y=5$

x	0	5
y	5	0

Putting $(0,0)$ in the inequality $x+y \geq 5$ we have
$0+0 \geq 5$
$\rightarrow 0 \geq 5$ (which is false)
So, the half plane is away from the origin.
Dwaw the graph of line $x+2 y=6$.

x	0	6
y	3	0

Putting $(0,0)$ in the inequality $x+2 y \geq 6$, we have
$0+2 \times 0 \geq 6$
$\rightarrow 0 \geq 6$ (which is false)
So, the half plane is away from the origin.

Since, $ x \geq 3 y \geq 0$
So, the feasible region lies in the first quadrant.
The points of intersection of lines $x=3$
and $-x+2 y=1 \text { is } C(3,2)$
$\text { and lines } x+2 y=6 $
$ \text { and } x+y=5 \text { is } B(4,1)$
It can be seen that the feasible region is unbounded.
The corner points of the feasible region are $A(6,0), B(4$, 1) and $C(3,2)$.
The values of $Z$ at these points are as follows

Corner point	$Z=-x+2 y$
$A(6,0)$	-6
$B(4,1)$	-2
$C(3,2)$	$1 \leftarrow$ Maximum

As the feasible region is unbounded. Therefore, $Z=1$ may or may not be the maximum value. For this, we graph the inequality $-x+2 y>1$ and check whether the resulting half plane has points in common with the feasible region or not.
The resulting feasible region has points in common with the feasible region. Therefore, $Z=1$ is not the maximum value.
Hence, $Z$ has no maximum value.
II. Our problem is to maximise
$Z=x+y$ ...(i)
Subject to the constraints are
$x-y \leq-1$ ...(ii)
$-x+y \leq 0 $...(iii)
$x \geq 0, y \geq 0$...(iv)
Draw the graph of the line $x-y=-1$.

x	0	-1
y	1	0

Putting $(0,0)$ in the inequality $x-y \leq-1$, we have
$0-0 \leq-1$
$\rightarrow 0 \leq-1$(which is false)
So, the half plane is away from the origin.
Draw the graph of the line $-x+y=0$.

x	0	-1
y	0	1

Putting $(2,0)$ in the inequality $-x+y \leq 0$, we have
$-2+0 \leq 0$
$\rightarrow -2 \leq 0$ (whict, is true)

So, the half plane is towards the origin.
Since, $ x, y \geq 0$
So, the feasible region lies in the first quadrant.
From the above graph, it is clearly shown that there is no common region. Hence, there is no feasible region and thus $Z$ has no maximum value.