MathsCBSE Class 12 · Linear Programming

Corner Point Method: Solving LPPs Graphically (Class 12)

Updated 15 October 2026 · Fact-checked

Draw the feasible region, find every corner point by solving the boundary equations, and evaluate Z = ax + by at each corner. On a bounded region the largest value is the maximum and the smallest is the minimum. If two adjacent corners tie, every point on the edge joining them is optimal.

Understand Corner Point Method

The corner point method is the graphical method of solving an LPP in two variables. It rests on two facts from NCERT Section 12.2.2:

Theorem 1. If the objective function $Z = a x + b y$ has an optimal (maximum or minimum) value on the feasible region, that value occurs at a corner point of the region.
Theorem 2. If the feasible region is bounded, $Z$ has both a maximum and a minimum, each at a corner point.

Why corners? The lines $a x + b y = k$ for different $k$ are parallel. Sliding such a line across the region, $k$ increases in one direction. The last point of the region the line touches is a corner (or a whole edge, if the line is parallel to that edge).

The method for a bounded region.

1. Draw the feasible region and shade it.
2. Find every corner point exactly by solving pairs of boundary equations.
3. Make a table of $Z$ at each corner.
4. The largest value in the table is the maximum and the smallest is the minimum.

Ties (multiple optimal solutions). If two adjacent corners give the same optimal value, the objective line is parallel to the edge joining them, and every point on that edge is optimal. Write: 'Z is maximum at every point of the line segment joining $P$ and $Q$ '. Boards test this as 'show that the optimum occurs at more than two points'.

Condition questions. For $Z = p x + q y$ , asking 'Z is optimal at both $P$ and $Q$ ' just means $Z (P) = Z (Q)$ : equate and simplify to get a relation between $p$ and $q$ .

Word problems. Choose the decision variables, write the objective (profit, cost, people served), write each 'at most' or 'at least' as a constraint, add $x, y \geq 0$ , then use the corner method. Answer in words with units: 'make 3 chairs and 6 tables for a maximum profit of Rs 3,900'.

If the region is unbounded, the table alone is not enough; that case has its own extra test (see the unbounded feasible regions topic). The 2026-27 syllabus allows up to three non-trivial constraints.

Corner Point Method: formulas to remember

Objective function: $Z = a x + b y$; Linear, of degree 1; to be maximised or minimised.
Corner point theorem: $Z_{opt} \in {Z (P_{1}), Z (P_{2}), \dots, Z (P_{n})}$; $P_{i}$ are the corner points of the feasible region.
Bounded region: $Z_{m a x} = max_{i} Z (P_{i}), Z_{m i n} = min_{i} Z (P_{i})$; Both always exist when the region is bounded.
Iso-profit line: $a x + b y = k, slope - \frac{a}{b}$; Parallel lines; the optimum is where the last one touches the region.
Tie on an edge: $Z (P) = Z (Q) \Rightarrow Z optimal on all of P Q$; Only when $P Q$ is an edge of the region and the value is optimal.
Condition for a tie: $p x_{1} + q y_{1} = p x_{2} + q y_{2}$; Gives the relation between $p$ and $q$ .

How to solve Corner Point Method questions

The 3-mark routine marking schemes follow: graph, corners, table, conclusion.

1Write the LPP: objective $Z$ , the constraints and $x, y \geq 0$ (formulate first in a word problem).
2Draw each boundary line from its intercepts and shade the feasible region using a test point.
3Find every corner point by solving the two equations that meet there; check each satisfies all constraints.
4Tabulate $Z$ at every corner.
5Bounded region: read off the maximum and/or minimum. If two adjacent corners tie, state that every point on the edge between them is optimal.
6Write the conclusion: the optimal value and where it occurs (and, in a word problem, the plan in words with units).

Corner Point Method in real life

Each concept, met in an everyday situation. Read the situation, try the problem, then check the solution. The one-line hook is what to remember.

Maximising takings with two limits

Suppose a juice stall sells mango juice at Rs 30 a glass and lemon juice at Rs 20 a glass. It can serve at most 50 glasses in an evening, and a mango glass takes 2 minutes to make while a lemon glass takes 1 minute, with 80 minutes available.

How many glasses of each should it sell for the largest takings?

Try it first, then show the solutionSolution

Maximise $Z = 30 x + 20 y$ subject to $x + y \leq 50$ , $2 x + y \leq 80$ , $x, y \geq 0$ .
Corners: $(0, 0)$ , $(40, 0)$ , $(0, 50)$ , and $x + y = 50$ with $2 x + y = 80$ gives $(30, 20)$ .
$Z$ : 0, 1200, 1000, 1300.

Answer: 30 mango and 20 lemon glasses, for Rs 1,300.

Memory hook: The best plan sits where two limits bite at the same time.

Minimising cost

Imagine a family that needs at least 6 hours of tutoring a week. Tutor A charges Rs 400 an hour and can give at most 4 hours; tutor B charges Rs 250 an hour and can give at most 5 hours.

Find the cheapest plan.

Try it first, then show the solutionSolution

Minimise $Z = 400 x + 250 y$ subject to $x + y \geq 6$ , $x \leq 4$ , $y \leq 5$ , $x, y \geq 0$ .
Corners: $(1, 5)$ , $(4, 2)$ , $(4, 5)$ (the region is a bounded triangle).
$Z$ : 1650, 2100, 2850.

Answer: 1 hour with A and 5 hours with B, costing Rs 1,650.

Memory hook: Use the cheaper option as far as its limit allows, then top up.

Tricks to remember Corner Point Method

Short memory aids. Say each one aloud once, then use it on a question the same day.

Graph, Corners, Table, Tell
GCTT: draw the Graph, solve the Corners, make the Table, Tell the answer.
Use it when: Order of a 3-mark LPP answer.
Slide the ruler
Place a ruler along $a x + b y = 0$ and slide it parallel; the last corner it touches is the maximum, the first is the minimum.
Use it when: Checking your table quickly.
Tie means a whole edge
Two adjacent corners with the same best $Z$ : the whole edge between them is optimal, so infinitely many solutions.
Use it when: 'More than two points' questions.
Bounded = both exist
On a bounded region, both the maximum and minimum always exist; no extra test needed.
Use it when: Knowing when the table alone settles it.

Corner Point Method: solved examples

Ordered from board level to HOTS. Try each one before you open the solution.

Example 1: Reading the optimum from given corners · 1 mark · Very short answerBoard level

The corner points of a bounded feasible region are $(0, 0)$ , $(7, 0)$ , $(5, 4)$ and $(0, 6)$ . Find the maximum value of $Z = 3 x + 5 y$ and where it occurs.

Try it first, then show the solutionSolution

$Z (0, 0) = 0$ , $Z (7, 0) = 21$ , $Z (5, 4) = 15 + 20 = 35$ , $Z (0, 6) = 30$ . (½ mark)
The largest is 35. (½ mark)

Answer: Maximum 35 at $(5, 4)$ .

Example 2: A standard maximisation · 3 marks · Short answerBoard level

Solve graphically: maximise $Z = 5 x + 4 y$ subject to $x + 2 y \leq 10$ , $3 x + y \leq 15$ , $x \geq 0$ , $y \geq 0$ .

Try it first, then show the solutionSolution

$x + 2 y = 10$ through $(10, 0)$ , $(0, 5)$ ; $3 x + y = 15$ through $(5, 0)$ , $(0, 15)$ ; both shaded towards the origin. (1 mark)
Solving: $x = 10 - 2 y$ , so $30 - 6 y + y = 15$ , $y = 3$ , $x = 4$ . Corners $(0, 0)$ , $(5, 0)$ , $(4, 3)$ , $(0, 5)$ . (1 mark)
$Z$ : 0, 25, $20 + 12 = 32$ , 20. Maximum $Z = 32$ at $(4, 3)$ . (1 mark)

Answer: Maximum $Z = 32$ at $x = 4$ , $y = 3$ .

Example 3: Maximum and minimum on the same region · 3 marks · Short answerHarder board level

Find the maximum and minimum values of $Z = 3 x + 5 y$ subject to $x + y \geq 2$ , $x + 2 y \leq 10$ , $x \leq 6$ , $x \geq 0$ , $y \geq 0$ .

Try it first, then show the solutionSolution

The region lies above $x + y = 2$ , below $x + 2 y = 10$ and left of $x = 6$ ; it is bounded. (1 mark)
Corners: $(2, 0)$ , $(6, 0)$ , $x = 6$ with $x + 2 y = 10$ gives $(6, 2)$ , $(0, 5)$ , $(0, 2)$ . (1 mark)
$Z$ : 6, 18, $18 + 10 = 28$ , 25, 10. Maximum 28 at $(6, 2)$ ; minimum 6 at $(2, 0)$ . (1 mark)

Answer: Maximum 28 at $(6, 2)$ ; minimum 6 at $(2, 0)$ .

Example 4: A production word problem · 5 marks · Long answerHarder board level

A small furniture unit makes chairs and tables. A chair needs 2 hours of carpentry and a table 3 hours, with 24 carpentry hours a day. At most 10 items can be polished a day, and at most 6 tables can be sold. Profit is Rs 300 per chair and Rs 500 per table. Formulate and solve the LPP graphically for maximum profit.

Try it first, then show the solutionSolution

Let $x$ chairs and $y$ tables. Maximise $Z = 300 x + 500 y$ subject to $2 x + 3 y \leq 24$ , $x + y \leq 10$ , $y \leq 6$ , $x, y \geq 0$ . (1 mark)
Graph the three lines and shade towards the origin for each; the region is bounded. (1 mark)
Corners: $(0, 0)$ , $(10, 0)$ ; $x + y = 10$ with $2 x + 3 y = 24$ gives $y = 4$ , $x = 6$ : $(6, 4)$ ; $y = 6$ with $2 x + 3 y = 24$ gives $x = 3$ : $(3, 6)$ ; $(0, 6)$ . (1½ marks)
$Z$ : 0, 3000, $1800 + 2000 = 3800$ , $900 + 3000 = 3900$ , 3000. (1 mark)
Maximum profit Rs 3,900 from 3 chairs and 6 tables a day. (½ mark)

Answer: 3 chairs and 6 tables; maximum profit Rs 3,900.

Example 5: Infinitely many optimal solutions · 3 marks · Short answerHOTS / competency

Maximise $Z = 4 x + 6 y$ subject to $2 x + 3 y \leq 18$ , $x + y \leq 7$ , $x \geq 0$ , $y \geq 0$ . Show that the maximum occurs at more than two points.

Try it first, then show the solutionSolution

Solving $2 x + 3 y = 18$ and $x + y = 7$ : $y = 4$ , $x = 3$ . Corners $(0, 0)$ , $(7, 0)$ , $(3, 4)$ , $(0, 6)$ . (1 mark)
$Z$ : 0, 28, $12 + 24 = 36$ , 36. The maximum 36 occurs at both $(3, 4)$ and $(0, 6)$ . (1 mark)
These are adjacent corners on the edge $2 x + 3 y = 18$ , and $Z = 2 (2 x + 3 y)$ is constant (36) along it. So every point of the segment joining $(3, 4)$ and $(0, 6)$ , such as $(1.5, 5)$ , is optimal. (1 mark)

Answer: Maximum 36 at every point of the segment joining $(3, 4)$ and $(0, 6)$ .

Example 6: Case study: equipping a computer lab · 4 marks · Case-basedHOTS / competency

Suppose a school is buying $x$ desktops (Rs 40,000 each) and $y$ laptops (Rs 60,000 each) with a budget of Rs 6,00,000. The room has space for at most 12 machines, and at least 2 laptops are needed for teachers. A desktop serves 3 students a period and a laptop 4.
(i) Write the LPP to maximise students served, $Z = 3 x + 4 y$ . (1 mark)
(ii) Find the corner points of the feasible region. (1 mark)
(iii) Find the best purchase. (1 mark)
(iv) If a laptop could serve 5 students, would the best plan change? (1 mark)

Try it first, then show the solutionSolution

(i) Maximise $Z = 3 x + 4 y$ subject to $40000 x + 60000 y \leq 600000$ , that is $2 x + 3 y \leq 30$ ; $x + y \leq 12$ ; $y \geq 2$ ; $x \geq 0$ . (1 mark)
(ii) $y = 2$ with $x = 0$ : $(0, 2)$ ; with $x + y = 12$ : $(10, 2)$ ; $x + y = 12$ with $2 x + 3 y = 30$ : $(6, 6)$ ; $2 x + 3 y = 30$ with $x = 0$ : $(0, 10)$ . (1 mark)
(iii) $Z$ : 8, 38, 42, 40. Best: 6 desktops and 6 laptops, serving 42 students a period. (1 mark)
(iv) $Z = 3 x + 5 y$ : 10, 40, 48, 50. Now the best is 10 laptops and no desktops (50 students), so yes, the plan changes. (1 mark)

Answer: (ii) $(0, 2)$ , $(10, 2)$ , $(6, 6)$ , $(0, 10)$ (iii) 6 desktops, 6 laptops, 42 students (iv) yes: 10 laptops, 50 students.

Where students lose marks in Corner Point Method

The slips examiners see most often, and what to write instead.

Mistake: Missing a corner point
With three constraints the region can have five corners; skipping one can skip the optimum.
Wrong working
Furniture problem: corners $(0, 0)$ , $(10, 0)$ , $(6, 4)$ , $(0, 6)$ ; maximum Rs 3,800.
Right working
Also $(3, 6)$ where $y = 6$ meets $2 x + 3 y = 24$ ; maximum Rs 3,900 there.
Do this: Walk round the boundary of the shaded region and list a corner at every change of line.
Usually costs: The answer mark and the conclusion.
Mistake: Including intersection points outside the region
Two lines may meet at a point that breaks a third constraint.
Wrong working
$x + y = 10$ and $y = 6$ meet at $(4, 6)$ ; $Z = 4200$ , so the maximum is 4200.
Right working
$(4, 6)$ gives $2 (4) + 3 (6) = 26 > 24$ : not feasible, so it is not a corner.
Do this: Check every candidate corner in all constraints.
Usually costs: The final answer.
Mistake: Saying 'two optimal solutions' for a tie
A tie at adjacent corners means the whole edge is optimal.
Wrong working
Maximum 36 at $(3, 4)$ and $(0, 6)$ only.
Right working
Maximum 36 at every point of the segment joining $(3, 4)$ and $(0, 6)$ .
Do this: Write 'infinitely many optimal solutions on the segment ...'.
Usually costs: 1 mark.
Mistake: No graph or no shading
Marking schemes give a mark for the correct feasible region before the table.
Wrong working
Only a list of corner points and a table.
Right working
Labelled lines, shaded region, then the table.
Do this: Always draw and shade, even if the corners are easy to find.
Usually costs: 1 mark of 3.

Exam tips

A 3-mark graphical LPP appeared in every board paper from 2023 to 2026 (5 marks in 2023 Q34 and 2024 Q35); the scheme is roughly 1 mark graph, 1 mark corners, 1 mark table and conclusion.
Corner-value MCQs were asked in 2023 (Q17), 2025 (Q14) and 2026 (Q15): just tabulate Z, no graph needed.
The condition on $p$ and $q$ for a tie was asked in the 2023-24 sample paper (Q7) in a 'minimum at two corners' form; equate Z at the two corners.
Write the conclusion in words: 'Z is maximum at $x = 4$ , $y = 3$ , and the maximum value is 32.'
If a corner has fractions, keep exact fractions in the table; do not round before comparing.

Practice questions on Corner Point Method

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

Q1 · 1 mark · MCQBoard level
The corner points of a bounded feasible region are $(0, 0)$ , $(8, 0)$ , $(6, 3)$ and $(0, 5)$ . The maximum value of $Z = 2 x + 5 y$ is:
(a) 16
(b) 25
(c) 27
(d) 40
Hint
Evaluate Z at all four corners.
Answer
(c) 27, at $(6, 3)$ (the others give 0, 16 and 25).
Q2 · 2 marks · Very short answerBoard level
Maximise $Z = 3 x + y$ subject to $x + y \leq 6$ , $x \leq 4$ , $x \geq 0$ , $y \geq 0$ .
Hint
Corners $(0, 0)$ , $(4, 0)$ , $(4, 2)$ , $(0, 6)$ .
Answer
$Z$ : 0, 12, 14, 6. Maximum $Z = 14$ at $(4, 2)$ .
Q3 · 3 marks · Short answerHarder board level
Find the maximum and minimum values of $Z = x - 3 y$ subject to $x + y \leq 6$ , $- x + 2 y \leq 4$ , $x \geq 0$ , $y \geq 0$ .
Hint
One corner has fractional coordinates.
Answer
Corners $(0, 0)$ , $(6, 0)$ , $(0, 2)$ and, from $x + y = 6$ with $- x + 2 y = 4$ , $(\frac{8}{3}, \frac{10}{3})$ . $Z$ : 0, 6, $- 6$ , $\frac{8}{3} - 10 = - \frac{22}{3}$ . Maximum 6 at $(6, 0)$ ; minimum $- \frac{22}{3}$ at $(\frac{8}{3}, \frac{10}{3})$ .
Q4 · 3 marks · Short answerHarder board level
Solve graphically: maximise $Z = 10 x + 6 y$ subject to $3 x + y \leq 12$ , $2 x + 5 y \leq 34$ , $x \geq 0$ , $y \geq 0$ .
Hint
Intersection: $y = 12 - 3 x$ in the second constraint.
Answer
$2 x + 5 (12 - 3 x) = 34 \Rightarrow x = 2$ , $y = 6$ . Corners $(0, 0)$ , $(4, 0)$ , $(2, 6)$ , $(0, 6.8)$ . $Z$ : 0, 40, 56, 40.8. Maximum 56 at $(2, 6)$ .
Q5 · 1 mark · Assertion-reasonHarder board level
Assertion (A): For the bounded feasible region with corners $(0, 0)$ , $(3, 0)$ , $(2, 2)$ and $(0, 3)$ , $Z = x + y$ attains its maximum at infinitely many points.
Reason (R): On a bounded feasible region, the maximum of a linear objective function occurs at a corner point.
(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are true but R is not the correct explanation of A.
(c) A is true but R is false.
(d) A is false but R is true.
Hint
Tabulate Z.
Answer
(d). $Z$ : 0, 3, 4, 3, so the maximum 4 occurs only at $(2, 2)$ ; A is false. R is the corner point theorem, which is true.
Q6 · 5 marks · Long answerHarder board level
Suppose a tiffin service makes $x$ veg and $y$ non-veg thalis a day. A veg thali takes 10 minutes of kitchen time and a non-veg thali 15 minutes, with 300 minutes available. It can pack at most 25 thalis, and at most 12 non-veg thalis can be sold. Profits are Rs 40 and Rs 70 per thali. Formulate and solve for maximum profit.
Hint
Kitchen time: $10 x + 15 y \leq 300$ , that is $2 x + 3 y \leq 60$ .
Answer
Maximise $Z = 40 x + 70 y$ subject to $2 x + 3 y \leq 60$ , $x + y \leq 25$ , $y \leq 12$ , $x, y \geq 0$ . Corners: $(0, 0)$ , $(25, 0)$ , $(15, 10)$ from $2 x + 3 y = 60$ with $x + y = 25$ , $(12, 12)$ from $y = 12$ with $2 x + 3 y = 60$ , and $(0, 12)$ . $Z$ : 0, 1000, 1300, 1320, 840. Maximum profit Rs 1,320 from 12 veg and 12 non-veg thalis.
Q7 · 2 marks · Very short answerHOTS / competency
The corner points of a bounded feasible region are $(0, 0)$ , $(6, 0)$ , $(5, 2)$ and $(0, 4)$ . For $Z = p x + q y$ with $p, q > 0$ , find the condition on $p$ and $q$ so that the maximum of $Z$ occurs at both $(6, 0)$ and $(5, 2)$ .
Hint
Equate $Z$ at the two points, then check that this common value beats the other corners.
Answer
$6 p = 5 p + 2 q \Rightarrow p = 2 q$ . Then $Z (6, 0) = Z (5, 2) = 12 q$ , which exceeds $Z (0, 4) = 4 q$ and $Z (0, 0) = 0$ , so the condition is $p = 2 q$ .

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1.Where does the optimal value of $Z$ occur, if it exists?Show answer
At a corner point of the feasible region.
2.On a bounded region, does $Z$ always have a maximum and a minimum?Show answer
Yes, both exist and occur at corner points.
3.What does a tie between two adjacent corners mean?Show answer
Every point on the edge joining them is optimal: infinitely many solutions.
4.Corners $(0, 0)$ , $(4, 0)$ , $(4, 2)$ , $(0, 6)$ : the maximum of $Z = x + y$ ?Show answer
6, at $(4, 2)$ and $(0, 6)$ , so on the whole edge joining them.
5.What is the degree of an objective function?Show answer
1 (it is linear).
6.What are the four parts of a graphical LPP answer?Show answer
Graph with shaded region, corner points, table of Z, conclusion.

Corner Point Method: frequently asked questions

What is the corner point method?

It is the graphical way to solve an LPP: draw the feasible region, find its corner points, evaluate the objective function Z at each, and choose the largest (maximum) or smallest (minimum) value.

Why is the optimum always at a corner?

The objective function is linear, so its level lines are parallel straight lines. Sliding such a line across the region, the last point it touches is a corner, or a whole edge if the line is parallel to that edge.

What if two corner points give the same maximum?

If they are adjacent corners, every point on the edge between them gives the same maximum, so the problem has infinitely many optimal solutions. State this in your answer.

Which NCERT section covers the corner point method?

Section 12.2.2 (Graphical method of solving linear programming problems), within Section 12.2 of Chapter 12 in NCERT Mathematics Part II for Class 12.

Can I use the iso-profit line method instead?

Yes, sliding the line Z = k gives the same answer. The corner point table is the usual method in CBSE answers because it is easy to mark, so show the table even if you use the line to check.