Linear Programming — Complete Guide with Solved Examples

What Linear Programming Actually Is

Linear programming (LP) is one of those topics where the name sounds intimidating but the core idea is beautifully simple: we want to maximize or minimize something (profit, cost, area) subject to certain constraints — and every constraint and the objective itself must be linear.

Think of it this way: a factory makes two products. More of product A means more profit, but it uses more raw material. Product B is less profitable but lighter on resources. How do we decide the mix? That’s exactly LP.

Class 12 LP is restricted to two variables, so we can solve everything graphically. No matrices, no simplex method — just drawing lines and finding a corner point. Once you understand the geometry behind it, this becomes one of the most predictable scoring topics in CBSE boards.

The examiners love LP because it tests multiple skills together: inequalities, graphing, and optimization — all in one 5-mark question. CBSE 2023, 2024 both had LP as a compulsory question in Section D.

Key Terms and Definitions

Decision variables are the unknowns we’re solving for — typically $x$ and $y$ in Class 12 problems. They represent quantities like units of product A and B, or kilograms of two food types.

Objective function is what we want to maximize or minimize. It’s always linear: $Z = ax + by$ , where $a$ and $b$ are constants.

Constraints are the limitations — inequalities like $2x + 3y \leq 120$ (raw material limit) or $x \geq 0$ , $y \geq 0$ (non-negativity conditions). Every constraint is a linear inequality.

Feasible region is the set of all points $(x, y)$ that satisfy every constraint simultaneously. This is the shaded region in the graph.

Feasible solution is any point inside the feasible region. There are infinitely many of these.

Optimal solution is the feasible solution that gives the best (maximum or minimum) value of $Z$ . This is what we’re hunting for.

Corner points (or vertices) are the intersection points at the boundary of the feasible region. The Corner Point Theorem guarantees the optimal solution always occurs at one of these.

CBSE always asks you to state and use the Corner Point Theorem explicitly. Write: “Since the feasible region is bounded, the maximum/minimum value of Z occurs at one of the corner points.” You get 1 mark just for this statement.

The Graphical Method — Step by Step

The entire Class 12 LP syllabus reduces to this one method. Master these 5 steps and every question becomes the same question.

Step 1: Formulate the Problem

Read the word problem carefully. Identify:

What are $x$ and $y$ ? (Define them clearly — examiners deduct marks for missing definitions)
What is $Z$ ? (The objective function)
What are the constraints?

Step 2: Convert Inequalities to Equations and Graph

Replace each inequality with = and find two points on each line. Plot all constraint lines on the coordinate axes.

To find the two points quickly: set $x = 0$ to get the y-intercept, and set $y = 0$ to get the x-intercept. Connect them.

Step 3: Identify the Feasible Region

For each constraint inequality, test the origin $(0, 0)$ (unless the line passes through origin). If $(0,0)$ satisfies the inequality, shade the side containing the origin. Otherwise, shade the other side.

The feasible region is the intersection of all shaded regions including $x \geq 0$ and $y \geq 0$ (so we stay in the first quadrant).

Students often forget to include non-negativity constraints $x \geq 0, y \geq 0$ when identifying the feasible region. This shrinks the region to the first quadrant — always double-check you’re not using points with negative coordinates.

Step 4: Find All Corner Points

The corner points are where constraint lines intersect each other, or where they meet the axes. Solve pairs of simultaneous equations to find exact coordinates.

Don’t approximate — use exact fractions. A corner point like $(20/3, 14/3)$ is valid and must be calculated precisely.

Step 5: Evaluate Z at Every Corner Point

Substitute each corner point into $Z = ax + by$ . The point giving the largest value is your maximum; the smallest gives the minimum.

If the feasible region is bounded, then both the maximum and minimum values of $Z$ exist and are attained at corner points.

If the feasible region is unbounded:

For minimum: check if the open half-plane $ax + by <$ minimum value has points in the feasible region. If yes, no minimum exists.
For maximum: check if $ax + by >$ maximum value has common points with feasible region. If yes, no maximum exists.

Solved Examples

Example 1 (CBSE Level — Easy)

Maximise $Z = 3x + 2y$

Subject to:

x + y \leq 4, \quad x \geq 0, \quad y \geq 0

Solution:

The constraint $x + y = 4$ gives us intercepts $(4, 0)$ and $(0, 4)$ .

Corner points of the feasible region (triangle OAB):

$O = (0, 0)$ : $Z = 0$
$A = (4, 0)$ : $Z = 3(4) + 2(0) = 12$
$B = (0, 4)$ : $Z = 3(0) + 2(4) = 8$

Maximum $Z = 12$ at $(4, 0)$ .

Example 2 (CBSE Level — Medium, 5-marker pattern)

Minimise $Z = 3x + 5y$

Subject to:

x + 3y \geq 3, \quad x + y \geq 2, \quad x \geq 0, \quad y \geq 0

Solution:

Line 1: $x + 3y = 3$ → points $(3, 0)$ and $(0, 1)$

Line 2: $x + y = 2$ → points $(2, 0)$ and $(0, 2)$

Test $(0,0)$ for constraint 1: $0 + 0 = 0 \geq 3$ ? No. So shade away from origin.

Test $(0,0)$ for constraint 2: $0 + 0 = 0 \geq 2$ ? No. So shade away from origin.

Intersection of the two lines: subtract equation 1 from equation 2:

$(x + y) - (x + 3y) = 2 - 3$

$-2y = -1 \Rightarrow y = 1/2$ , so $x = 3/2$

Corner points:

$A = (3, 0)$ : $Z = 9 + 0 = 9$
$B = (3/2, 1/2)$ : $Z = 9/2 + 5/2 = 7$
$C = (0, 2)$ : $Z = 0 + 10 = 10$

The feasible region is unbounded. We check: does $3x + 5y < 7$ have common points with the feasible region?

The line $3x + 5y = 7$ — test a point in the open half-plane $3x + 5y < 7$ , say $(1, 0.5)$ : $3 + 2.5 = 5.5 < 7$ . But is $(1, 0.5)$ in the feasible region? Check $x + 3y = 1 + 1.5 = 2.5 < 3$ . No, it’s outside. So no common points exist.

Minimum $Z = 7$ at $(3/2, 1/2)$ .

Example 3 (CBSE Board Exam Style — Hard Word Problem)

A manufacturer makes two types of furniture: chairs and tables. A chair requires 3 hours of carpentry and 1 hour of painting. A table requires 3 hours of carpentry and 2 hours of painting. Available: 36 hours of carpentry, 20 hours of painting per week. Profit on a chair is ₹80 and on a table is ₹120. How many of each should be produced to maximize profit?

Solution:

Let $x$ = number of chairs, $y$ = number of tables.

Objective function: Maximise $Z = 80x + 120y$

Constraints:

Carpentry: $3x + 3y \leq 36 \Rightarrow x + y \leq 12$
Painting: $x + 2y \leq 20$
Non-negativity: $x \geq 0, y \geq 0$

Corner points:

From $x + y = 12$ and $x = 0$ : point $(0, 12)$

From $x + 2y = 20$ and $y = 0$ : point $(20, 0)$

Intersection of $x + y = 12$ and $x + 2y = 20$ :

Subtract: $y = 8$ , so $x = 4$ . Point: $(4, 8)$

Intersection of $x + y = 12$ and $x = 0$ : $(0, 12)$ — but check $x + 2y \leq 20$ : $0 + 24 = 24 > 20$ . Not feasible!

So the corner points are: $(0, 0)$ , $(12, 0)$ , $(4, 8)$ , $(0, 10)$

Corner Point	$Z = 80x + 120y$
$(0, 0)$	0
$(12, 0)$	960
$(4, 8)$	320 + 960 = 1280
$(0, 10)$	1200

Maximum profit = ₹1280, by producing 4 chairs and 8 tables.

In word problems, always write the definition of variables, objective function, and constraints as separate labelled steps. CBSE awards 1 mark each for the formulation — don’t skip these.

Exam-Specific Tips

CBSE Class 12

LP appears in Section D (5 marks). The marking scheme is typically:

1 mark: defining variables and stating objective function
1 mark: writing all constraints correctly
1 mark: correct graph with feasible region shaded
1 mark: finding all corner points correctly
1 mark: correct optimal value with conclusion

The word problem format is standard — almost always a manufacturing or diet problem. CBSE 2024 asked about maximising profit from producing two types of items with constraints on machine hours and raw material.

CBSE 2023 Term 1 included an LP MCQ. CBSE 2024 Board had a full 5-mark LP word problem. The pattern is consistent — expect one LP question every year, always 5 marks, always solvable graphically in the first quadrant.

ICSE / ISC

ISC Class 12 LP problems are similar in style but sometimes involve two objective functions or ask you to verify optimality more rigorously. The feasible region shading must be shown clearly.

JEE / NEET

LP is not in the JEE or NEET syllabus as of 2024 (it was removed when the curriculum was revised). If you’re a CBSE student also targeting JEE, this topic gives you board marks but needs zero JEE preparation time.

Common Mistakes to Avoid

Mistake 1: Not defining variables explicitly. Writing “Let x and y be the number of items” loses a mark if the items aren’t named. Write: “Let x = number of chairs and y = number of tables.”

Mistake 2: Plotting lines correctly but shading the wrong region. Always test the origin $(0,0)$ for each constraint (when the line doesn’t pass through origin). Half the class shades the infeasible region by habit.

Mistake 3: Missing a corner point. Students often find 2-3 corner points and stop. You must find ALL vertices of the feasible polygon. An unbounded region has infinitely many boundary points — focus on intersections of constraint lines with each other and with axes.

Mistake 4: Assuming maximum always exists for unbounded regions. For unbounded feasible regions, you must verify using the open half-plane check. Just picking the largest $Z$ value among corner points and calling it maximum is incomplete — the examiner wants to see the verification step.

Mistake 5: Solving the intersection equations incorrectly. When two constraint lines intersect, students make arithmetic errors solving simultaneous equations. Always substitute back to verify. Getting a corner point wrong invalidates the entire optimization.

Practice Questions

Q1. Maximise $Z = x + y$ subject to $x - y \leq -1$ , $-x + y \leq 0$ , $x, y \geq 0$ .

The constraint $x - y \leq -1$ means $y \geq x + 1$ and $-x + y \leq 0$ means $y \leq x$ . So we need $y \geq x + 1$ and $y \leq x$ simultaneously — this is impossible (no feasible region exists). There is no solution; the feasible region is empty.

Q2. Minimise $Z = -3x + 4y$ subject to $x + 2y \leq 8$ , $3x + 2y \leq 12$ , $x \geq 0$ , $y \geq 0$ .

Corner points: $(0,0)$ , $(4,0)$ , $(2,3)$ , $(0,4)$ .

$Z$ at $(0,0) = 0$ ; at $(4,0) = -12$ ; at $(2,3) = -6+12=6$ ; at $(0,4) = 16$ .

Minimum $Z = -12$ at $(4,0)$ .

Q3. A diet must contain at least 80 units of Vitamin A and 100 units of minerals. Two foods F1 and F2 are available. F1 costs ₹4 per kg and contains 3 units of Vitamin A and 4 units of minerals per kg. F2 costs ₹6 per kg and contains 6 units of Vitamin A and 3 units of minerals per kg. Find the minimum cost diet.

Let $x$ = kg of F1, $y$ = kg of F2.

Minimise $Z = 4x + 6y$

Constraints: $3x + 6y \geq 80$ , $4x + 3y \geq 100$ , $x \geq 0$ , $y \geq 0$

Corner points: $(0, 100/3)$ , $(24, 4/3)$ , $(100/4, 0) = (25, 0)$

$Z$ at $(0, 100/3)$ : $Z = 200$ ; at $(24, 4/3)$ : $Z = 96 + 8 = 104$ ; at $(25, 0)$ : $Z = 100$ .

Since region is unbounded, check $4x + 6y < 100$ — no common point with feasible region.

Minimum cost = ₹104 at 24 kg of F1 and 4/3 kg of F2.

Q4. Maximise $Z = 6x + 4y$ subject to $x \leq 2$ , $x + y \leq 3$ , $x \geq 0$ , $y \geq 0$ .

Corner points: $(0,0)$ , $(2,0)$ , $(2,1)$ , $(0,3)$ .

$Z$ : $0$ , $12$ , $16$ , $12$ .

Maximum $Z = 16$ at $(2,1)$ .

Q5. A factory makes products P and Q. Each unit of P requires 2 hours on machine M1 and 1 hour on M2. Each unit of Q requires 1 hour on M1 and 2 hours on M2. M1 is available 10 hours/day, M2 is available 8 hours/day. Profit: ₹5 per unit P, ₹4 per unit Q. Maximise daily profit.

Let $x$ = units of P, $y$ = units of Q.

Maximise $Z = 5x + 4y$ , subject to $2x + y \leq 10$ , $x + 2y \leq 8$ , $x,y \geq 0$ .

Intersection: $2x + y = 10$ and $x + 2y = 8$ . Solve: $3x = 12 \Rightarrow x = 4, y = 2$ .

Corner points: $(0,0)$ , $(5,0)$ , $(4,2)$ , $(0,4)$ .

$Z$ : $0$ , $25$ , $28$ , $16$ .

Maximum profit = ₹28 by producing 4 units of P and 2 units of Q.

FAQs

What is the difference between a feasible solution and an optimal solution?

Every point satisfying all constraints is a feasible solution — there are infinitely many. The optimal solution is the specific feasible point that gives the best value of the objective function. We find it using the Corner Point Theorem.

Can there be more than one optimal solution?

Yes. If the objective function line $Z = ax + by$ is parallel to one side of the feasible region polygon, then every point on that side gives the same optimal $Z$ value. This is called multiple optimal solutions. CBSE sometimes asks about this — state it as “optimal solution exists on the line segment between [corner point 1] and [corner point 2].”

What if the feasible region is empty?

Some problems have no feasible region — no point satisfies all constraints simultaneously. In that case, there is no solution to the LP problem. Always check that the feasible region is non-empty before proceeding.

What is an unbounded feasible region?

When the feasible region extends to infinity in some direction (usually when all constraints are $\geq$ type), it’s unbounded. Minimum problems with $\geq$ constraints typically give unbounded regions on one side. The optimal may or may not exist — use the half-plane verification test.

Do we always get integer corner points in CBSE?

No. CBSE questions often have fractional corner points like $(3/2, 1/2)$ . Don’t round off — always use the exact fraction. The objective function value must also be computed exactly.

What’s the quickest way to identify the feasible region in an exam?

After plotting all constraint lines, shade each region using the origin test. The feasible region is where all shadings overlap. Then mark all corner points — these are where lines intersect each other or meet the axes. This takes about 4-5 minutes once you’re practiced.

Can the objective function equal zero at the optimal point?

Yes, $(0,0)$ is often a corner point and gives $Z = 0$ . For a maximization problem, this would only be optimal if all other corner points also give $Z = 0$ — very unlikely in real problems. For minimization, $Z = 0$ at the origin is often the minimum when $a, b > 0$ , but check whether $(0,0)$ is actually in the feasible region.

How many marks is LP in the CBSE Class 12 board exam?

5 marks, always in Section D (long answer). It’s a guaranteed question — not optional — and follows a predictable format. Given that the entire method is learnable in a day, LP has among the highest marks-per-study-hour ratios in the Class 12 math paper.