CBSE Sample Paper 2025-26 · Section C · 3 marks

Q30: Maximise $Z = 2x + y$ graphically

Sample 2025-26Ch 12 · Linear Programming

The question

Solve graphically:

Maximise $Z = 2x + y$ subject to

$x + y \le 1200$

$x + y \ge 600$

$y \le \dfrac{x}{2}$

$x \ge 0, y \ge 0$ ⁠.

For Visually Impaired:

The objective function $Z = 3x + 2y$ of a linear programming problem under some constraints is to be maximized and minimized. The corner points of the feasible region are $A(600, 0)$ ⁠, $B(1200, 0)$ ⁠, $C(800, 400)$ and $D(400, 200)$ ⁠. Find the point at which $Z$ is maximum and the point at which $Z$ is minimum. Also, find the corresponding maximum and minimum values of $Z$ ⁠.)

Exam answer

Draw the lines (graph below):

$x + y = 1200$ through $(1200, 0)$ and $(0, 1200)$ ⁠;

$x + y = 600$ through $(600, 0)$ and $(0, 600)$ ⁠;

$y = \frac{x}{2}$ through $(0, 0)$ and $(800, 400)$ ⁠.

Why?

Each inequality is first drawn as its boundary line, by replacing the sign with "=". Two points fix a line.

The easiest points are where the line cuts the axes: put $y = 0$ ⁠, then $x = 0$ ⁠. For $x + y = 1200$ that gives $(1200, 0)$ and $(0, 1200)$ ⁠.

$y = \frac{x}{2}$ passes through the origin, so we need one more point: $x = 800$ gives $y = 400$ ⁠.

$x + y \le 1200$ is the side of $x + y = 1200$ containing O; $x + y \ge 600$ is the side of $x + y = 600$ away from O; $y \le \frac{x}{2}$ is the side of $y = \frac{x}{2}$ containing $(1, 0)$ ⁠. With $x, y \ge 0$ ⁠, the feasible region is ABCD (shaded). It is bounded.

Why?

To find the correct side of a line, test a point not on it. The origin is easiest:

\begin{aligned} 0 + 0 \le 1200 &\text{ is true} \\ 0 + 0 \ge 600 &\text{ is false} \end{aligned}

So we keep the side of the first line with O, and the side of the second line without O.

The origin lies on $y = \frac{x}{2}$ ⁠, so test $(1, 0)$ instead: $0 \le \frac12$ is true.

$x, y \ge 0$ keeps us in the first quadrant. The region is bounded because it fits inside a finite triangle.

C: $x + y = 1200$ ⁠, $y = \frac{x}{2}$ give $x = 800$ ⁠, $y = 400$ ⁠. D: $x + y = 600$ ⁠, $y = \frac{x}{2}$ give $x = 400$ ⁠, $y = 200$ ⁠.

Why?

A corner where two lines cross satisfies both equations. Put $y = \frac{x}{2}$ into $x + y = 1200$ ⁠:

\frac{3x}{2} = 1200 \Rightarrow x = 800

and $y = \frac{800}{2} = 400$ ⁠. In the same way $\frac{3x}{2} = 600$ gives $x = 400$ ⁠, $y = 200$ ⁠.

In full, for C:

\begin{aligned} x + \frac{x}{2} &= 1200 \\ \frac{2x}{2} + \frac{x}{2} &= 1200 \\ \frac{3x}{2} &= 1200 \\ 3x &= 1200 \times 2 = 2400 \\ x &= \frac{2400}{3} = 800 \\ y &= \frac{800}{2} = 400 \end{aligned}

For D:

\begin{aligned} x + \frac{x}{2} &= 600 \\ \frac{3x}{2} &= 600 \\ 3x &= 600 \times 2 = 1200 \\ x &= \frac{1200}{3} = 400 \\ y &= \frac{400}{2} = 200 \end{aligned}

A and B are where $x + y = 600$ and $x + y = 1200$ meet the x-axis.

The feasible region ABCD (shaded), drawn with equal scales. Its corners are A(600, 0), B(1200, 0), C(800, 400) and D(400, 200).

\begin{array}{c|c} \text{Corner} & Z = 2x + y \\ \hline A(600, 0) & 1200 \\ B(1200, 0) & 2400 \leftarrow \text{max} \\ C(800, 400) & 2000 \\ D(400, 200) & 1000 \end{array}

Why?

Corner point method: when the feasible region is bounded, the maximum (and minimum) of $Z$ is reached at a corner point. So we only need $Z$ at A, B, C and D:

\begin{aligned} A&: 2(600) + 0 = 1200 \\ B&: 2(1200) + 0 = 2400 \\ C&: 2(800) + 400 \\ &= 1600 + 400 = 2000 \\ D&: 2(400) + 200 \\ &= 800 + 200 = 1000 \end{aligned}

The largest value, 2400, is at B.

$\therefore$ Maximum $Z = 2400$ when $x = 1200$ and $y = 0$ ⁠.

For Visually Impaired

The value of $Z = 3x + 2y$ at each corner point:

Why?

The corner points are given, and the feasible region they enclose is bounded. So, by the corner point method, both the maximum and the minimum of $Z$ occur at corner points.

\begin{aligned} Z_A &= 3 \times 600 + 2 \times 0 = 1800 \\ Z_B &= 3 \times 1200 + 2 \times 0 = 3600 \\ Z_C &= 3 \times 800 + 2 \times 400 = 3200 \\ Z_D &= 3 \times 400 + 2 \times 200 = 1600 \end{aligned}

Why?

Substitute each corner's $x$ and $y$ into $Z = 3x + 2y$ ⁠. For C, $2400 + 800 = 3200$ ⁠; for D, $1200 + 400 = 1600$ ⁠.

\begin{aligned} Z_A &= 1800 + 0 = 1800 \\ Z_B &= 3600 + 0 = 3600 \\ Z_C &= 2400 + 800 = 3200 \\ Z_D &= 1200 + 400 = 1600 \end{aligned}

Compare the four values: the largest is 3600 (at B) and the smallest is 1600 (at D).

$\therefore$ Maximum $Z = 3600$ at $B(1200, 0)$ ⁠, and minimum $Z = 1600$ at $D(400, 200)$ ⁠.

Where this comes from

Chapter 12, Linear Programming: LPP: maximise graphically, bounded region.

Question type: (main question) A " $\ge$ " rule; (visually impaired version) Both maximum and minimum, and a tie along an edge

How to spot it: click to see

(main question) Type 2. A " $\ge$ " rule: shade away from the origin

Spot it: a rule like $x + 2y \ge 10$ ⁠, often with "Minimise $Z$ ". The region no longer touches the origin.

Here: $x + y \ge 600$ fails at $(0, 0)$ ⁠, so shade away from the origin for it; $y = \frac{x}{2}$ goes through the origin, so test $(1, 0)$ ⁠. The region ABCD is bounded, and the largest value, 2400 at $(1200, 0)$ ⁠, is the maximum.

(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: The four corners are given, so make one table of $Z = 3x + 2y$ ⁠: the largest, 3600 at $B(1200, 0)$ ⁠, is the maximum and the smallest, 1600 at $D(400, 200)$ ⁠, the minimum.

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