Target Exam

CUET

Subject

Section B1

Chapter

Linear Programming

Question:

Consider the following Linear Programming Problem: Minimise $Z = x + 2y$, Subject to $2x + y \ge 3, x + 2y \ge 6, x, y \ge 0$. Show graphically that the minimum of $Z$ occurs at more than two points.

Options:

The minimum value is $Z = 3$ and occurs at the point $(1.5, 0)$.

The minimum value is $Z = 6$ and occurs only at the vertex $(6, 0)$.

The minimum value is $Z = 6$ and occurs at every point on the line segment joining $(0, 3)$ and $(6, 0)$.

The minimum value is $Z = 0$ at the origin $(0, 0)$.

Correct Answer:

The minimum value is $Z = 6$ and occurs at every point on the line segment joining $(0, 3)$ and $(6, 0)$.

Explanation:

The correct answer is Option (3) → The minimum value is $Z = 6$ and occurs at every point on the line segment joining $(0, 3)$ and $(6, 0)$. ##

The feasible region determined by the constraints, $2x + y \ge 3, x + 2y \ge 6, x \ge 0, y \ge 0$ is as shown.

The corner points of the unbounded feasible region are $A(6, 0)$ and $B(0, 3)$.

The values of $Z$ at these corner points are as follows:

Corner point

Value of the objective function Z=x+2y

$A(6, 0)$

$6$

$B(0, 3)$

$6$

We observe the region $x + 2y < 6$ has no points in common with the unbounded feasible region. Hence the minimum value of $Z = 6$.

It can be seen that the value of $Z$ at points $A$ and $B$ is same. If we take any other point on the line $x + 2y = 6$ such as $(2, 2)$ on line $x + 2y = 6$, then $Z = 6$.

Thus, the minimum value of $Z$ occurs for more than 2 points and is equal to 6.