Q.For Visually Impaired: If 𝑍 = 𝑎𝑥 + 𝑏𝑦 + 𝑐, where 𝑎, 𝑏, 𝑐 > 0, attains its maximum value at two of its corner points (4,0) and (0,3) of the feasible region determined by the system of linear inequalities, then
(A) 4𝑎 = 3𝑏
(B) 3𝑎 = 4𝑏
(C) 4𝑎 + 𝑐 = 3𝑏
(D) 3𝑎 + 𝑐 = 4𝑏

Concept: Linear Programming Optimality — when a linear objective attains the same maximum at two distinct corner points, the objective line is parallel to the edge joining them.

Reasoning:

1. The objective is $Z = ax + by + c$, with $a, b, c > 0$.

   Maximum occurs at both $(4,0)$ and $(0,3)$, so $Z$ is equal at these points.

2. Equate the values: …

When a linear programming problem has a maximum at two different corner points, it means the entire line segment between them is optimal. This happens because the objective function’s gradient is perpendicular to that edge — the objective lines are parallel to the edge itself. Let’s see why.

1. The geometry of optimality

   In linear programming, the feasible region is a convex polygon. The objective function $Z = ax + by + c$ is a family of parallel lines (one for each value of $Z$). As you increase $Z$, the line shifts in the direction of the gradient $(a, b)$. The maximum occurs at the last point (or edge) of the feasible region that the line touches.

   If the maximum occurs at two distinct corner points, say $P(4,0)$ and $Q(0,3)$, then the entire edge $PQ$ must be optimal. That means the objective line at the maximum value coincides with the line through $P$ and $Q$ — they are parallel.

2. Condition for parallelism

   Two lines are parallel if their direction vectors are proportional. The edge $PQ$ has direction vector $\vec{PQ} = (0-4, 3-0) = (-4, 3)$. The objective function’s level lines have normal vector $(a, b)$, so their direction vector is perpendicular to $(a, b)$, i.e., $(-b, a)$ or $(b, -a)$.

   For the edge $PQ$ to be parallel to the objective lines, the direction of $PQ$ must be proportional to the direction of the objective lines. So:

This gives:

Cross-multiplying:

3. What about the constant $c$? …

Method: Equating the Objective at Two Optimal Corners

Use this whenever a linear objective $Z=ax+by\,(+c)$ is said to attain the same optimum at two given corner points and you must find a relation between the coefficients.

Steps

Step 1: Write $Z$ at each of the two corners.

If the optimum is reached at both $(x_1,y_1)$ and $(x_2,y_2)$, then $Z$ has the same value at both:

This is the defining fact of a tie — the two corners are equally good, so their $Z$-values are equal.

Step 2: Cancel the constant and simplify.

Any constant term $c$ appears identically on both sides and cancels. What remains is a linear relation between $a$ and $b$ alone:

Step 3: Read the geometric meaning (why $c$ never appears). …

Common Mistakes

Mistake 1: Thinking the constant $c$ must appear in the answer.

Why it's wrong: equating $Z$ at $(4,0)$ and $(0,3)$ gives $4a+c=3b+c$, and the $c$ cancels — so the relation is $4a=3b$, with no $c$. Options (C) and (D), which keep $c$, are distractors. Correct approach: cancel the common constant; a shift $c$ never affects which points are optimal.

Mistake 2: Mixing up the coordinates. …