The feasible region for a LPP is shown in figure. Evaluate $Z=4x+y$ at each of the corner points of this region. Find the minimum value of Z, if it exists.
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → 3
As per the given figure, ABC is the feasible region which is open unbounded.
Here, we have
$x+y=3$ …(i)
and $x+2y= 4$ ...(ii)
$Z=4x+y$
Solving eq. (i) and (ii), we get
$x=2$ and $y = 1$
So, the corner points are
$A(4,0), B(2, 1)$ and $C(0,3)$
Let us evaluate the value of Z
|
Corner points |
$Z=4x+y$ |
|
A(4, 0) |
$Z=4(4)+(0)=16$ |
|
B(2, 1) |
$Z=4(2)+(1)=9$ |
|
C(0, 3) |
$Z=4(0)+(3)=3$ → Minimum |
Now, the minimum value of Z is 3 at (0, 3) but since, the feasible region is open bounded so it may or may not be the minimum value of Z.
Therefore, to face such situation, we draw a graph of $4x + y < 3$ and check whether the resulting open half plane has no point in common with feasible region. Otherwise Z will have no minimum value. From the graph, we conclude that there is no common point with the feasible region.
Hence, Z has the minimum value 3 at (0, 3).