Q.A student of class XII studying Mathematics comes across an incomplete question in a book. Maximise 𝑍 = 3𝑥 + 2𝑦 + 1 Subject to the constraints 𝑥 ≥ 0, 𝑦 ≥ 0, 3𝑥 + 4𝑦 ≤ 12, He/ She notices the below shown graph for the said LPP problem, and finds that a constraint is missing in it: Help him/her choose the required constraint from the graph. The missing constraint is
(A) 𝑥 + 2𝑦 ≤ 2
(B) 2𝑥 + 𝑦 ≥ 2
(C) 2𝑥 + 𝑦 ≤ 2
(D) 𝑥 + 2𝑦 ≥ 2

The feasible region is fixed by $x\ge 0$, $y\ge 0$, $3x+4y\le 12$ plus one more line read off the graph. That fourth boundary passes through $(2,0)$ and $(0,1)$.

A line with $x$-intercept $2$ and $y$-intercept $1$ is

$\frac{x}{2}+\frac{y}{1}=1 \;\Rightarrow\; x+2y=2.$ …

The idea

Every constraint of an LPP is a straight line together with a choice of which side to keep. The stem already lists $x\ge 0$, $y\ge 0$ and $3x+4y\le 12$. The graph shows one more boundary that is not in this list — that hidden line is the missing constraint. To name it we (1) find its equation from its intercepts and (2) fix the inequality sign from the side on which the shaded region lies.

Step 1 — Equation of the fourth line

The fourth boundary meets the axes at $(2,0)$ and $(0,1)$, so $x$-intercept $a=2$ and $y$-intercept $b=1$. Using the intercept form $\frac{x}{a}+\frac{y}{b}=1$:

Step 2 — Which side?

Test the origin $(0,0)$, which lies inside the shown feasible region:

$0+2(0)=0\le 2 \;\;\text{(true)}.$ …

Method: Reading a Constraint's Inequality off its Graph

Use this when a boundary line is drawn on a feasible-region graph and you must name the inequality it represents (equation and direction).

Steps

Step 1: Get the line's equation from its intercepts.

Read where the line crosses the axes, say the $x$-axis at $(a,0)$ and the $y$-axis at $(0,b)$. Then use the intercept form:

Clearing denominators gives the constraint line in the form $px+qy=r$.

Step 2: Fix the $\le$ or $\ge$ by a test point.

A line splits the plane into two half-planes; the constraint keeps only one. Pick any point not on the line — the origin $(0,0)$ is easiest when the line misses it — and substitute into $px+qy$.

• If the test point lies inside the shaded feasible region and makes $px+qy \le r$ true, the constraint is $\le$. …

Common Mistakes

Mistake 1: Using the wrong line's intercepts.

Why it's wrong: the graph's fourth boundary passes through $(2,0)$ and $(0,1)$, giving $x+2y=2$; students who misread the intercepts as $(1,0),(0,2)$ get $2x+y=2$ (options B/C) — a different line entirely. Correct approach: read the exact axis crossings and build the equation with the intercept form $\frac{x}{a}+\frac{y}{b}=1$.

Mistake 2: Guessing the inequality sign. …