MathsISC Class 12 · Linear Programming

Graphical Method of Solving a Linear Programming Problem

Q: Why does the optimum occur at a corner?

Z is constant along the parallel lines ax + by = c. Sliding the line in the direction that increases Z, the last point of a bounded region it touches is a vertex (or a whole edge containing two vertices).

Q: Do I need to draw the graph to scale?

Yes, reasonably. A neat graph with labelled lines and a shaded region earns marks on its own, and helps you avoid choosing a false corner.

Q: What if two corners give the same best value?

Then every point on the edge joining them is optimal, so the problem has infinitely many optimal solutions. Say so in your answer.

Q: Is the simplex method in the ISC syllabus?

No. The 2027 syllabus asks for the graphical method for problems in two variables with up to three non-trivial constraints.

Q: How many marks does this carry?

Linear programming carries 5 marks in 2027. The 2024 and 2025 papers each set a 4-mark word problem solved graphically.

Updated 15 October 2026 · Fact-checked

To solve an LPP graphically, draw each constraint as a line, shade the region satisfying all of them (the feasible region), find its corner points, and evaluate the objective function Z = ax + by at each. For a bounded feasible region, the largest and smallest of these values are the maximum and minimum of Z.

Evaluate Z at every corner of the shaded feasible region; the largest value, 42 at (6, 3), is the maximum.

Illustration drawn for QuestionBuddy.

Understand Graphical Method of Solving an LPP

The idea. A linear programming problem (LPP) in two variables asks for the largest or smallest value of a linear objective function $Z = a x + b y$ over all points $(x, y)$ that satisfy a set of linear constraints and the non-negativity restrictions $x \geq 0$ , $y \geq 0$ . The graphical method draws those constraints and lets geometry find the answer.

Step 1: draw the boundary lines. For each constraint, replace the inequality sign by "=" and plot the line through its two intercepts. For $3 x + y \leq 21$ : intercepts $(7, 0)$ and $(0, 21)$ .

Step 2: choose the side. Test a point not on the line, usually the origin. If $(0, 0)$ satisfies the inequality, shade the side containing the origin; if not, the other side. For " $\leq$ " constraints with a positive right-hand side, the origin side is shaded; for " $\geq$ ", the far side. The ISC 2024 and 2025 examiners both reported candidates who could not graph the inequalities or identify the feasible region, so do this test explicitly every time.

Step 3: the feasible region. The common region of all the constraints, including $x, y \geq 0$ , is the feasible region; every point in it is a feasible solution. Points outside are infeasible. In ISC problems there are at most three non-trivial constraints.

Step 4: corner points. The vertices of the feasible region are the intersections of pairs of boundary lines that lie in the region. Find them by solving the pair of equations; check each against the remaining constraints, because an intersection outside the region is not a corner.

Step 5: the corner point theorem. If the feasible region is bounded, Z has both a maximum and a minimum, and each occurs at a corner point. Evaluate Z at every corner, tabulate, and pick.

Why corners? The lines $a x + b y = c$ for different c are parallel (iso-profit or iso-cost lines). As c increases the line moves steadily in one direction. The last point of the region the line touches as it leaves is a vertex (or a whole edge). That is the geometric reason behind the theorem, and it gives a second method: slide a ruler parallel to $a x + b y = 0$ .

Ties. If Z takes the same optimal value at two adjacent corners, it takes that value at every point on the edge joining them: there are infinitely many optimal solutions. This happens when the objective line is parallel to that edge.

Integer answers. When the variables count objects (bags, books), the optimum should be checked to be a whole number; ISC problems are set so that the optimal corner has whole-number coordinates.

Presenting the answer. Graph with labelled lines and shaded region; table of corner points and Z values; a final sentence in context ("maximum profit ₹1300 when 20 loaves and 30 cakes are made").

Graphical Method of Solving an LPP: formulas to remember

Standard LPP: $Max/Min Z = a x + b y subject to a_{i} x + b_{i} y \leq c_{i}, x, y \geq 0$; Up to three non-trivial constraints in ISC.
Intercepts of a boundary: $p x + q y = r : (\frac{r}{p}, 0), (0, \frac{r}{q})$; Two points to draw the line.
Origin test: $p (0) + q (0) \leq r ⟺ r \geq 0$; True: shade towards the origin.
Corner point theorem: $Z_{m a x}, Z_{m i n} \in {Z (corners)}$; For a bounded feasible region.
Iso-profit lines: $a x + b y = c, slope - \frac{a}{b}$; Parallel family; slide outward to maximise.
Multiple optima: $Z (P) = Z (Q) = Z_{opt} \Rightarrow every point of P Q$; Objective line parallel to edge PQ.

How to solve Graphical Method of Solving an LPP questions

A full-marks graphical solution has the same five pieces every time.

1Write the LPP clearly: variables, objective, constraints, $x, y \geq 0$ .
2Find the two intercepts of each boundary line and plot the lines on one graph with a sensible scale.
3Test the origin in each inequality and shade; mark the common feasible region.
4Find every corner by solving pairs of equations; reject any intersection that breaks another constraint.
5Tabulate Z at each corner and choose the maximum or minimum.
6State the answer in words, with units, and mention multiple optima if two corners tie.

Graphical Method of Solving an LPP 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.

Corner point method

Suppose a phone shop can stock at most 30 phones. A basic phone costs ₹5000 and a smartphone ₹15 000, and the owner can spend at most ₹3 00 000. The profit is ₹500 on a basic phone and ₹1200 on a smartphone.

How many of each should be stocked for maximum profit?

Try it first, then show the solutionSolution

Maximise $Z = 500 x + 1200 y$ subject to $x + y \leq 30$ , $5000 x + 15000 y \leq 300000$ (that is $x + 3 y \leq 60$ ), $x, y \geq 0$ .
Corners: (0, 0), (30, 0), (15, 15) from $x + y = 30$ and $x + 3 y = 60$ , and (0, 20).
Z: 0, 15 000, 25 500, 24 000.

Answer: 15 basic phones and 15 smartphones, for a maximum profit of ₹25 500.

Memory hook: Shade, corner, table, choose.

Why the optimum is at a corner

Imagine drawing the profit lines 500x + 1200y = c for the phone shop above, for larger and larger c.

Explain why the last feasible point touched is (15, 15).

Try it first, then show the solutionSolution

The profit lines have slope $- \frac{500}{1200} = - \frac{5}{12} \approx - 0.42$ .
The edges meeting at (15, 15) have slopes −1 (stock limit) and $- \frac{1}{3} \approx - 0.33$ (budget). The profit slope lies between them.
So as the line moves outward it leaves the region exactly at the vertex where these two edges meet.

Answer: Because the profit line's slope lies between the slopes of the two edges meeting at (15, 15), that vertex is the last point it touches.

Memory hook: Slide the ruler outward; the last corner it touches wins.

Tricks to remember Graphical Method of Solving an LPP

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

S-C-T-C
Shade, Corners, Table, Conclude.
Use it when: The order of a graphical solution.
Origin test
"Is zero allowed?" If yes, shade towards the origin.
Use it when: Choosing the side of each line.
Check every crossing
A crossing of two lines is a corner only if it obeys the third constraint.
Use it when: Avoiding false corners.
Tie means a whole edge
Two equal best corners: the whole segment between them is optimal.
Use it when: Multiple optimal solutions.

Graphical Method of Solving an LPP: solved examples

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

Example 1: Reading a corner table · 1 mark · MCQBoard level

The corner points of a bounded feasible region are (0, 0), (5, 0), (3, 4) and (0, 6). The maximum of $Z = 3 x + 4 y$ is:
(a) 15
(b) 24
(c) 25
(d) 21

Try it first, then show the solutionSolution

Z = 0, 15, 25, 24 at the four corners.

Answer: (c) 25, at (3, 4).

Example 2: Two constraints · 3 marks · Short answerBoard level

Solve graphically: maximise $Z = 4 x + 3 y$ subject to $x + y \leq 8$ , $2 x + y \leq 12$ , $x, y \geq 0$ .

Try it first, then show the solutionSolution

Lines: $x + y = 8$ through (8, 0), (0, 8); $2 x + y = 12$ through (6, 0), (0, 12). The origin satisfies both, so shade towards it. (1 mark)
Corners: (0, 0), (6, 0), (0, 8), and $x + y = 8$ , $2 x + y = 12$ give $x = 4$ , $y = 4$ . (1 mark)
Z: 0, 24, 24, 28. Maximum 28 at (4, 4). (1 mark)

Answer: $Z_{m a x} = 28$ at $x = 4$ , $y = 4$ .

Example 3: Three constraints · 5 marks · Long answerHarder board level

Solve graphically: maximise $Z = 2 x + 5 y$ subject to $x + 4 y \leq 24$ , $3 x + y \leq 21$ , $x + y \leq 9$ , $x, y \geq 0$ .

Try it first, then show the solutionSolution

Intercepts: $x + 4 y = 24$ : (24, 0), (0, 6); $3 x + y = 21$ : (7, 0), (0, 21); $x + y = 9$ : (9, 0), (0, 9). Shade towards the origin for all three. (1 mark)
On the x-axis the binding limit is (7, 0); on the y-axis it is (0, 6). (1 mark)
$3 x + y = 21$ and $x + y = 9$ : $x = 6$ , $y = 3$ ; check $6 + 12 = 18 \leq 24$ . $x + y = 9$ and $x + 4 y = 24$ : $y = 5$ , $x = 4$ ; check $12 + 5 = 17 \leq 21$ . (1 mark)
Corners (0, 0), (7, 0), (6, 3), (4, 5), (0, 6); Z = 0, 14, 27, 33, 30. (1 mark)
Maximum 33 at (4, 5). (1 mark)

Answer: $Z_{m a x} = 33$ at $x = 4$ , $y = 5$ .

Example 4: Minimum on a bounded region · 3 marks · Short answerHarder board level

Find the minimum and maximum of $Z = 5 x + 7 y$ subject to $x + y \geq 4$ , $x \leq 6$ , $y \leq 5$ , $x, y \geq 0$ .

Try it first, then show the solutionSolution

The region lies above $x + y = 4$ , left of $x = 6$ and below $y = 5$ : corners (4, 0), (6, 0), (6, 5), (0, 5), (0, 4). (1 mark)
Z: 20, 30, 65, 35, 28. (1 mark)
Minimum 20 at (4, 0); maximum 65 at (6, 5). (1 mark)

Answer: $Z_{m i n} = 20$ at (4, 0); $Z_{m a x} = 65$ at (6, 5).

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

For the feasible region of the three-constraint example above (corners (0, 0), (7, 0), (6, 3), (4, 5), (0, 6)), show that $Z = x + y$ has infinitely many optimal solutions.

Try it first, then show the solutionSolution

Z = 0, 7, 9, 9, 6 at the corners. (1 mark)
The maximum 9 occurs at both (6, 3) and (4, 5). (1 mark)
Both lie on the edge $x + y = 9$ , which is parallel to the objective line, so every point of the segment joining them gives Z = 9. (1 mark)

Answer: Maximum 9 at every point of the segment from (6, 3) to (4, 5).

Example 6: Bakery planning · 4 marks · Case-basedHOTS / competency

Suppose a bakery makes loaves (x) and cakes (y) each day. The oven has room for at most 50 items, and decorating time allows $x + 2 y \leq 80$ . The profit is ₹20 per loaf and ₹30 per cake.
(i) Formulate the LPP. (1 mark)
(ii) Find the corner points of the feasible region. (2 marks)
(iii) Find the maximum profit and the plan that gives it. (1 mark)

Try it first, then show the solutionSolution

(i) Maximise $Z = 20 x + 30 y$ subject to $x + y \leq 50$ , $x + 2 y \leq 80$ , $x, y \geq 0$ .
(ii) $x + y = 50$ and $x + 2 y = 80$ give $y = 30$ , $x = 20$ . Corners: (0, 0), (50, 0), (20, 30), (0, 40).
(iii) Z: 0, 1000, 1300, 1200. Maximum ₹1300 at 20 loaves and 30 cakes.

Answer: ₹1300 from 20 loaves and 30 cakes.

Where students lose marks in Graphical Method of Solving an LPP

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

Mistake: Feasible region not identified
Both the 2024 (rice and wheat) and 2025 (library books) examiners reported candidates who could not graph the inequalities and shade the region.
Do this: Test the origin for each inequality, shade lightly, and mark the common region boldly.
Usually costs: 2 to 3 marks.
Mistake: Wrong corner points
Reported in 2024: candidates used an intersection that lies outside the region.
Wrong working
Using (0, 12.5) from $3 x + 2 y = 25$ although $x + y \leq 10$ cuts it off.
Right working
Use (0, 10), and the intersection (5, 5) of the two lines.
Do this: Check each candidate corner against every constraint.
Usually costs: 1 to 2 marks.
Mistake: Wrong objective function
In 2025 some candidates wrote the objective wrongly from the words.
Do this: Write what x and y count, then what is being maximised in the same units.
Usually costs: 1 mark plus follow-through.
Mistake: No conclusion
A table without a sentence leaves the question unanswered.
Do this: End with the optimal values of x, y and Z in context.
Usually costs: 1 mark.

Exam tips

Use graph paper scale so that all intercepts fit; label each line with its equation.
Show the working for every intersection, not just the answer.
Present corner values in a table; it is quick for the examiner to mark.
For a minimisation on a bounded region, the same corner method works.

Practice questions on Graphical Method of Solving an LPP

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 feasible region are (0, 2), (3, 0), (6, 0), (6, 8) and (0, 5). For $Z = 4 x + 6 y$ :
(a) min 12 at (0, 2) only
(b) min 12 at every point of the segment from (0, 2) to (3, 0)
(c) max 30 at (0, 5)
(d) min 0
Hint
Evaluate at all five corners.
Answer
(b): Z = 12, 12, 24, 72, 30.
Q2 · 3 marks · Short answerBoard level
Solve graphically: maximise $Z = 3 x + 2 y$ subject to $x + 2 y \leq 10$ , $3 x + y \leq 15$ , $x, y \geq 0$ .
Hint
The lines meet at (4, 3).
Answer
Corners (0, 0), (5, 0), (4, 3), (0, 5); Z = 0, 15, 18, 10; maximum 18 at (4, 3).
Q3 · 3 marks · Short answerHarder board level
Maximise $Z = 2 x + 3 y$ subject to $x + y \leq 6$ , $x \geq 2$ , $y \leq 3$ , $y \geq 0$ .
Hint
Corners (2, 0), (6, 0), (3, 3), (2, 3).
Answer
Z = 4, 12, 15, 13; maximum 15 at (3, 3).
Q4 · 1 mark · Assertion-reasonHarder board level
Assertion (A): If Z is maximum at two corner points of a bounded feasible region, it is maximum at every point of the segment joining them.
Reason (R): A linear function takes, on a segment, every value between its values at the ends; equal end values make it constant on the segment.
(a) Both true, R explains A. (b) Both true, R does not explain A. (c) A true, R false. (d) A false, R true.
Hint
Z is linear along any segment.
Answer
(a).
Q5 · 2 marks · Short answerHarder board level
On the region with corners (0, 0), (7, 0), (6, 3), (4, 5), (0, 6), find the minimum of $Z = - x + 2 y$ .
Hint
Negative coefficients are allowed.
Answer
Z = 0, −7, 0, 6, 12; minimum −7 at (7, 0).
Q6 · 3 marks · Short answerHOTS / competency
For the region with corners (0, 0), (6, 0), (4, 4), (0, 8), find all values of k for which $Z = k x + y$ has its maximum at (4, 4) only.
Hint
Compare 4k + 4 with 6k and with 8.
Answer
$4 k + 4 > 6 k$ gives $k < 2$ ; $4 k + 4 > 8$ gives $k > 1$ . So $1 < k < 2$ .
Q7 · 4 marks · Case-basedHOTS / competency
Case study: Suppose a student plans x hours of Maths and y hours of Physics revision on a holiday, with $x + y \leq 10$ , $x \leq 6$ and $y \leq 7$ . Her expected gain in marks is $Z = 5 x + 4 y$ .
(i) List the corner points of the feasible region. (2 marks)
(ii) Find the plan that maximises Z. (1 mark)
(iii) If the gain changed to $Z = 4 x + 4 y$ , how many optimal plans would there be? (1 mark)
Hint
Corners include (6, 4) and (3, 7).
Answer
(i) (0, 0), (6, 0), (6, 4), (3, 7), (0, 7). (ii) Z = 0, 30, 46, 43, 28: 6 h Maths, 4 h Physics, Z = 46. (iii) Infinitely many: Z = 40 along the whole edge from (6, 4) to (3, 7).

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.State the corner point theorem.Show answer
On a bounded feasible region, the maximum and minimum of a linear objective occur at corner points.
2.What is a feasible solution?Show answer
Any point satisfying all the constraints, including non-negativity.
3.Corners (0, 0), (4, 0), (2, 3), (0, 4): max of Z = x + 2y?Show answer
8, at (2, 3) and (0, 4), so along the whole edge between them.
4.What are iso-profit lines?Show answer
Lines $a x + b y = c$ on which Z is constant; they are all parallel.
5.How many non-trivial constraints appear in ISC LPPs?Show answer
At most three.

Graphical Method of Solving an LPP: frequently asked questions

Why does the optimum occur at a corner?

Z is constant along the parallel lines ax + by = c. Sliding the line in the direction that increases Z, the last point of a bounded region it touches is a vertex (or a whole edge containing two vertices).

Do I need to draw the graph to scale?

Yes, reasonably. A neat graph with labelled lines and a shaded region earns marks on its own, and helps you avoid choosing a false corner.

What if two corners give the same best value?

Then every point on the edge joining them is optimal, so the problem has infinitely many optimal solutions. Say so in your answer.

Is the simplex method in the ISC syllabus?

No. The 2027 syllabus asks for the graphical method for problems in two variables with up to three non-trivial constraints.

How many marks does this carry?

Linear programming carries 5 marks in 2027. The 2024 and 2025 papers each set a 4-mark word problem solved graphically.