Minimize $Z = x + 2y$ subject to $2x + y \ge 3,x + 2y \ge 6,x,y \ge 0.$
Linear Programming — Class 12 Maths Solution
Step-by-step Solution
.: The system of constraints is :
$2x + y \ge 3$ ...(1)
$x + 2y \ge 6$ ...(2)
and $x \ge 0,y \ge 0$ ...(3)
${l_1}:2x + y = 3$
${l_2}:x + 2y = 6$
The shaded region in the adjoining figure is
the feasible region determined by the system of constraints (1) to (3).
It is observed that the feasible region is unbounded.
The co-ordinates of B and C are (0, 3) and (6, 0).
Applying Corner Point Method, we have
Since the region is unbounded, we need to check whether 6 is the minimum value or not.
To decide this, we graph the inequality $x + 2y < 6$.
Now, in the graph we observe 6 does not have points in common with the feasible region.
So, 6 is the minimum value.
Hence, ${Z_{\min }} = 6$
at all points on the line segment joining the points (6, 0) and (0, 3).
Show that the maximum of Z occurs at more than two points.
NCERT & Exemplar solution for CBSE Class 12 Mathematics, Linear Programming. Curated by Sachin Sharma. Free for all students.