Sample 2025-26 Class 12 Maths Q16 Solution: Missing constraint 2x + y ≥ 2 from the graph of an LPP

The question

A student of class XII studying Mathematics comes across an incomplete question in a book.

Maximise $Z = 3x + 2y + 1$

Subject to the constraints $x \ge 0, y \ge 0, 3x + 4y \le 12$ ⁠,

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 given graph, redrawn.

The missing constraint is

(A) $x + 2y \le 2$

(B) $2x + y \ge 2$

(C) $2x + y \le 2$

(D) $x + 2y \ge 2$

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) $3a + c = 4b$

Exam answer

The other boundary line passes through $(1, 0)$ and $(0, 2)$ ⁠:

Why?

In the graph, the line $3x + 4y = 12$ of the given constraint meets the axes at $(4, 0)$ and $(0, 3)$ ⁠. The shaded region has one more edge, on a line that meets the axes at $(1, 0)$ and $(0, 2)$ ⁠.

\frac{x}{1} + \frac{y}{2} = 1 \Rightarrow 2x + y = 2

Why?

A line with intercepts $a$ on the x-axis and $b$ on the y-axis is

\frac{x}{a} + \frac{y}{b} = 1

Here $a = 1$ and $b = 2$ ⁠. Multiplying by 2 clears the fractions:

\begin{aligned} 2 \cdot \frac{x}{1} + 2 \cdot \frac{y}{2} &= 2 \cdot 1 \\ 2x + y &= 2 \end{aligned}

Check: $(1, 0)$ gives $2 + 0 = 2$ ⁠, and $(0, 2)$ gives $0 + 2 = 2$ ⁠.

The origin is not in the shaded region, and at $(0, 0)$ ⁠, $2x + y = 0 < 2$ ⁠. So the region is $2x + y \ge 2$ ⁠.

Why?

The line splits the plane into two sides. The origin, on one side, gives $2x + y < 2$ ⁠. The shaded region is on the other side, so there $2x + y \ge 2$ ⁠, with the line included.

Options (A) and (D) use $x + 2y = 2$ ⁠, the line through $(2, 0)$ and $(0, 1)$ ⁠, which is not in the graph. Option (C) is the origin's side.

Answer: (B) $2x + y \ge 2$

For Visually Impaired

$Z$ is maximum at both corners, so it has the same value there:

Why?

The maximum value is a single number. If it is attained at $(4, 0)$ and at $(0, 3)$ ⁠, then $Z$ is equal at these two points (and at every point of the edge joining them).

\begin{aligned} &Z(4, 0) = Z(0, 3) \\ \Rightarrow\ &4a + c = 3b + c \\ \Rightarrow\ &4a = 3b \end{aligned}

Why?

Put the points into $Z = ax + by + c$ ⁠:

\begin{aligned} Z(4, 0) &= 4a + 0 + c \\ Z(0, 3) &= 0 + 3b + c \end{aligned}

Watch out: $c$ appears on both sides and cancels, so (C) and (D), which keep $c$ ⁠, are wrong.

Answer: (A) $4a = 3b$

Where this comes from

Chapter 12, Linear Programming: LPP: a constraint read from the feasible region.

Question type: (main question) Read the rules from a graph; (visually impaired version) Both maximum and minimum, and a tie along an edge

How to spot it: click to see

(main question) Type 7. Read the rules from a graph

Spot it: the graph of the region is given, and the question asks for its constraints, or for the rule that is missing.

Here: The missing edge meets the axes at $(1, 0)$ and $(0, 2)$ ⁠: $\frac{x}{1} + \frac{y}{2} = 1$ ⁠, so $2x + y = 2$ ⁠. The origin is not in the region and gives $0 \lt 2$ ⁠, so the rule is $2x + y \ge 2$ ⁠.

(visually impaired version) Type 3. Both maximum and minimum, and a tie along an edge

Spot it: "Minimise and maximise $Z$ ". Often one rule, like $x - 2y \ge 0$ ⁠, is a line through the origin.

Here: A maximum at both $(4, 0)$ and $(0, 3)$ means the two corner values tie: $4a + c = 3b + c$ ⁠, so $4a = 3b$ ⁠.

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