Question

The solution for minimizing the function $z = x + y$ under an L.P.P. with constraints $x + y \ge 2$, $x + 2y \le 8$, $y \le 3$, $x, y \ge 0$ is ______.

• A. at the point (0, 3)
• B. at the point (8, 0)
• C. at infinite number of points but bounded set
• D. at unbounded set

Hint:

If the objective function is exactly parallel to (a scalar multiple of) one of the boundary constraint lines, the optimal solution will ALWAYS lie along the entirety of that line segment, yielding infinite solutions!

Answer 1

The Correct Option is C

Step 1: Understanding the Question:
We must minimize an objective function $z = x + y$ bounded by several linear inequalities. The key lies in noticing the relationship between the objective function and the constraints.

Step 2: Detailed Explanation:
The objective function to minimize is $z = x + y$.
Look closely at the very first constraint:
$x + y \ge 2$.
Because $x$ and $y$ must physically exist in the feasible region, the lowest possible value that $x + y$ can ever geometrically take within this defined region is exactly 2.
Therefore, the minimum possible value of the objective function $z$ is 2, and this minimum occurs exclusively anywhere on the line segment defined by $x + y = 2$.
However, we must verify if this entire line segment (within the first quadrant $x, y \ge 0$) is actually valid and not blocked by the other constraints.
The segment $x + y = 2$ in the first quadrant runs perfectly from point $(2, 0)$ to point $(0, 2)$.
Let's test both boundary endpoints against the remaining constraints ($x + 2y \le 8$ and $y \le 3$):
1. Test point $(2, 0)$:
$2 + 2(0) = 2 \le 8$ (True)
$0 \le 3$ (True)
2. Test point $(0, 2)$:
$0 + 2(2) = 4 \le 8$ (True)
$2 \le 3$ (True)
Since both endpoints satisfy all constraints, the entire continuous line segment connecting $(2, 0)$ and $(0, 2)$ lies entirely inside the valid feasible region.
Because there are infinitely many points on any given line segment, the minimum value $z = 2$ is attained at an infinite number of points. Furthermore, because a line segment between two coordinates is geometrically finite, it constitutes a bounded set.

Step 3: Final Answer:
The solution is at infinite number of points but bounded set, matching option (c).