Sample 2026-27 Class 12 Maths Q25 Solution: Write the constraints of a shaded feasible region

The question

The feasible region of a linear programming problem is shown in the given figure:

Write all the constraints on the decision variables involved

For Visually Impaired:

For a linear programming problem, the maximum value of an objective function $Z = 3x + 4y$ exists at the points $(p, q)$ and $\left(2p, \dfrac{q}{2}\right)$ ⁠. Find the relation between $p$ and $q$ ⁠.

Exam answer

The shaded region ACB lies on or above the x-axis, between the lines $3x + 2y = 10$ and $x + 2y = 6$ ⁠.

Why?

Every edge of a feasible region lies on the line of one constraint. Here the edges are AC on $3x + 2y = 10$ ⁠, CB on $x + 2y = 6$ and AB on the x-axis, $y = 0$ ⁠. So there are three constraints, one for each line.

Each constraint is the equation of its line with "=" replaced by " $\le$ " or " $\ge$ ". The side of the line where the region lies decides which.

At $O(0, 0)$ ⁠: $0 \ge 10$ is false and $0 \le 6$ is true. So the region is on the side of $3x + 2y = 10$ away from O, and on the side of $x + 2y = 6$ that contains O.

Why?

Test-point method: put a point that is not on the line into the inequality. If it is true, the inequality describes the side containing that point. O is the easiest point to try.

At O, $3x + 2y = 0$ ⁠, and $0 \ge 10$ is false. The shaded region is on the far side of that line from O, so it needs $3x + 2y \ge 10$ ⁠.

At O, $x + 2y = 0$ ⁠, and $0 \le 6$ is true. The region is on the same side as O, so it needs $x + 2y \le 6$ ⁠.

On or above the x-axis means $y \ge 0$ ⁠.

Why?

Check with a point inside the triangle, say $\left(4, \frac12\right)$ ⁠: $3(4) + 2\left(\frac12\right) = 13 \ge 10$ ⁠, $4 + 2\left(\frac12\right) = 5 \le 6$ and $\frac12 \ge 0$ ⁠. All three hold.

\begin{aligned} &3(4) + 2\left(\tfrac12\right) = 12 + 1 = 13 \\ &13 \ge 10\ \checkmark \\[4pt] &4 + 2\left(\tfrac12\right) = 4 + 1 = 5 \\ &5 \le 6\ \checkmark \end{aligned}

Constraints: $3x + 2y \ge 10$ ⁠, $x + 2y \le 6$ ⁠, $y \ge 0$

For Visually Impaired

The maximum value exists at $(p, q)$ and $\left(2p, \frac{q}{2}\right)$ ⁠, so Z has the same value at both points:

Why?

Both points give the maximum, and there is only one maximum value. So Z must be equal at the two points.

\begin{aligned} 3p + 4q &= 3 \times 2p + 4 \times \frac{q}{2} \\ \Rightarrow 3p + 4q &= 6p + 2q \\ \Rightarrow 6p - 3p &= 4q - 2q \end{aligned}

Why?

On the left, put $x = p$ ⁠, $y = q$ in $Z = 3x + 4y$ ⁠. On the right, put $x = 2p$ ⁠, $y = \frac{q}{2}$ ⁠.

\begin{aligned} 3 \times 2p &= 6p \\ 4 \times \frac{q}{2} &= \frac{4q}{2} = 2q \end{aligned}

Take $3p$ from both sides, then take $2q$ from both sides:

\begin{aligned} 3p + 4q &= 6p + 2q \\ 4q &= 6p - 3p + 2q \\ 4q - 2q &= 6p - 3p \\ 2q &= 3p \end{aligned}

Then collect the $p$ terms on one side and the $q$ terms on the other: $6p - 3p = 3p$ and $4q - 2q = 2q$ ⁠.

$\therefore$ The relation is $3p = 2q$ ⁠.

Where this comes from

Chapter 12, Linear Programming: LPP: constraints from a 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 region ACB is on the side of $3x + 2y = 10$ away from O and on O's side of $x + 2y = 6$ ⁠, so the rules are $3x + 2y \ge 10$ and $x + 2y \le 6$ ⁠. It sits on or above the $x$ -axis, so add $y \ge 0$ ⁠.

(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: Both points give the maximum, so their values tie: $3p + 4q = 6p + 2q$ ⁠, which gives $3p = 2q$ ⁠.

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Q25: Write the constraints of a shaded feasible region