CBSE Sample Paper 2026-27 · Section C · 3 marks

Q31: Maximise $Z = 40x + 30y$ graphically

Sample 2026-27Ch 12 · Linear Programming

The question

Solve the linear programming problem graphically:

Maximize $Z = 40x + 30y$

Subject to constraints: $x + y \le 6$ ⁠, $2x \ge y$ ⁠, $y \ge 2$

For Visually Impaired:

For a linear programming problem, vertices of the feasible region are $(20, 0)$ ⁠, $(40, 0)$ ⁠, $(30, 10)$ ⁠, $(0, 20)$ and $(0, 10)$ ⁠. Minimize and maximize the objective function $Z = 50x - 20y$

Exam answer

Draw the lines (graph below):

$x + y = 6$ through $(6, 0)$ and $(0, 6)$ ⁠;

$y = 2x$ through $(0, 0)$ and $(2, 4)$ ⁠;

$y = 2$ ⁠, parallel to the x-axis.

Why?

Replace each " $\le$ " or " $\ge$ " by "=" to get the boundary line, and plot two points of it. For $x + y = 6$ ⁠, the intercepts are easiest: $x = 0$ gives $y = 6$ ⁠, and $y = 0$ gives $x = 6$ ⁠.

$y = 2x$ passes through O, so a second point is needed: $x = 2$ gives $y = 4$ ⁠.

$x + y \le 6$ is the side of $x + y = 6$ containing O; $2x \ge y$ is the side of $y = 2x$ containing $(1, 0)$ ⁠; $y \ge 2$ is on or above $y = 2$ ⁠. The feasible region is the triangle ABC (shaded). It is bounded.

Why?

Test-point method. At O: $0 + 0 \le 6$ is true, so $x + y \le 6$ is the side containing O.

O lies on $y = 2x$ ⁠, so it cannot be the test point for that line. Try $(1, 0)$ ⁠: $2 \ge 0$ is true, so $2x \ge y$ is the side containing $(1, 0)$ ⁠, below the line.

The feasible region is the part where all three sides overlap.

The feasible region ABC (shaded), with corners A(1, 2), B(4, 2) and C(2, 4).

The corner points are found by solving the boundary lines in pairs:

\begin{aligned} y = 2,\ y = 2x &\Rightarrow A(1, 2) \\ y = 2,\ x + y = 6 &\Rightarrow B(4, 2) \\ y = 2x,\ x + y = 6 &\Rightarrow C(2, 4) \end{aligned}

Why?

A: $2 = 2x$ gives $x = 1$ ⁠.

B: $x + 2 = 6$ gives $x = 4$ ⁠.

C: put $y = 2x$ in $x + y = 6$ ⁠: $3x = 6$ ⁠, so $x = 2$ and $y = 4$ ⁠.

\begin{aligned} \text{A}&: x = \frac{2}{2} = 1 \\ \text{B}&: x = 6 - 2 = 4 \\ \text{C}&: x + 2x = 6 \\ &\ \ 3x = 6,\ x = \frac{6}{3} = 2 \\ &\ \ y = 2(2) = 4 \end{aligned}

\begin{array}{c|c} \text{Corner} & Z = 40x + 30y \\ \hline A(1, 2) & 100 \\ B(4, 2) & 220 \leftarrow \text{max} \\ C(2, 4) & 200 \end{array}

Why?

Corner point method: when the feasible region is bounded, Z has a maximum, and it occurs at a corner point. So we only need Z at A, B and C.

\begin{aligned} Z_A &= 40(1) + 30(2) \\ &= 40 + 60 = 100 \\[4pt] Z_B &= 40(4) + 30(2) \\ &= 160 + 60 = 220 \\[4pt] Z_C &= 40(2) + 30(4) \\ &= 80 + 120 = 200 \end{aligned}

$\therefore$ Maximum $Z = 220$ at $x = 4$ ⁠, $y = 2$ ⁠.

For Visually Impaired

Let the vertices be $A(20, 0)$ ⁠, $B(40, 0)$ ⁠, $C(30, 10)$ ⁠, $D(0, 20)$ and $E(0, 10)$ ⁠.

Why?

The feasible region is the polygon with these five vertices, so it is bounded. By the corner point method, the minimum and the maximum of Z are both found among the values of Z at the vertices.

\begin{aligned} Z_A &= 50 \times 20 - 20 \times 0 = 1000 \\ Z_B &= 50 \times 40 - 20 \times 0 = 2000 \\ Z_C &= 50 \times 30 - 20 \times 10 = 1300 \\ Z_D &= 50 \times 0 - 20 \times 20 = -400 \\ Z_E &= 50 \times 0 - 20 \times 10 = -200 \end{aligned}

Why?

Each value is two products, then a subtraction:

\begin{aligned} Z_A &= 1000 - 0 = 1000 \\ Z_B &= 2000 - 0 = 2000 \\ Z_C &= 1500 - 200 = 1300 \\ Z_D &= 0 - 400 = -400 \\ Z_E &= 0 - 200 = -200 \end{aligned}

The largest value in the list is 2000, at B, and the smallest is $-400$ ⁠, at D.

Watch out: Z has a minus sign, so a corner with a large $y$ and $x = 0$ gives a negative value. The minimum is $-400$ ⁠, not 0 or 1000.

$\therefore$ Minimum $Z = -400$ at $D(0, 20)$ and maximum $Z = 2000$ at $B(40, 0)$ ⁠.

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: $y \ge 2$ keeps the region above $y = 2$ ⁠, away from the origin, and $y = 2x$ goes through the origin, so test $(1, 0)$ ⁠. The triangle ABC is bounded, so the largest value, 220 at $(4, 2)$ ⁠, 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 five vertices are given, so make one table of $Z = 50x - 20y$ ⁠: it gives the maximum 2000 at $(40, 0)$ and the minimum $-400$ at $(0, 20)$ ⁠.

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Q31: Maximise Z=40x+30yZ = 40x + 30yZ=40x+30y graphically

Q31: Maximise $Z = 40x + 30y$ graphically