Q.The feasible region of a linear programming problem is bounded but the objective function attains its minimum value at more than one point. One of the points is (5,0). Then one of the other possible points at which the objective function attains its minimum value is
(A) (2,9)
(B) (6,6)
(C) (4,7)
(D) (0,0) For Visually Impaired: The graph of the inequality 3𝑥 + 5𝑦 < 10 is the (A) Entire 𝑋𝑌 −plane (B) Open Half plane that doesn’t contain origin (C) Open Half plane that contains origin, but not the points of the line 3𝑥 + 5𝑦 = 10 (D) Half plane that contains origin and the points of the line 3𝑥 + 5𝑦 = 10

This item has two parts.

Main MCQ. If a bounded feasible region gives the same minimum at more than one point, the objective line must be parallel to one edge, and both corners of that edge are optimal. So the 'other' optimal point is the corner joined to $(5,0)$ by that edge. Which corner this is can only be read from the feasible-region graph, and that graph is not supplied here — so the specific option cannot be fixed from the printed text without inventing constraints. …

Main MCQ — the principle

On a bounded feasible region the optimum of a linear objective sits at a corner. When the minimum is attained at more than one point, the objective line $Z=ax+by=\text{const}$ is parallel to one edge of the region. Sliding it inward, its last contact is that whole edge, so both end-corners of the edge (and every point between) share the minimum value.

Given one optimal corner $(5,0)$, the 'other' optimal point is the corner at the far end of that edge. Deciding which of $(2,9),(6,6),(4,7),(0,0)$ that is requires knowing the edge — i.e. the actual constraint lines in the figure.

Visually-Impaired part — a strict inequality

The VI sub-question is fully self-contained. Consider $3x+5y<10$. …

Method: Classifying the Half-Plane of a Linear Inequality

Use this to decide which region a single linear inequality $px+qy \;(\lessgtr)\; r$ describes — whether the boundary is included, and which side is kept.

Steps

Step 1: Decide open vs. closed from the sign.

• A strict inequality ($<$ or $>$) excludes the boundary line — the region is an open half-plane; points on $px+qy=r$ do not satisfy it.
• A weak inequality ($\le$ or $\ge$) includes the boundary — a closed half-plane.

Step 2: Pick the correct side with a test point.

Substitute a point off the line (the origin is simplest) into the inequality. If it is satisfied, the wanted half-plane is the one containing that point; if not, it is the other side.

Step 3 (related principle — multiple optima): …

Common Mistakes

Mistake 1: Treating a strict inequality as if it included the boundary.

Why it's wrong: $3x+5y<10$ uses "$<$", so points on the line $3x+5y=10$ do not satisfy it — the region is an open half-plane. Choosing option (D), which includes the line, is the error. Correct approach: strict $<$/$>$ means an open half-plane (boundary excluded); only $\le$/$\ge$ include it.

Mistake 2: Picking the wrong side of the line.

Why it's wrong: without a test point students may shade away from the origin. Correct approach: check $(0,0)$ — $3(0)+5(0)=0<10$ is true, so the region is the side containing the origin, giving option (C). …