Linear Programming Class 12: Graph Accuracy Tips from a Home Tutor
Linear programming questions look long, but most marks are lost in two places: the graph and the corner points.
This post shows the common errors, a worked example, and a weekly routine you can follow at home.
Where marks are lost in Class 12 linear programming
Linear programming means finding the largest or smallest value of a quantity, such as profit or cost, when there are limits on what you can use. The quantity is the objective function. The limits are constraints, written as inequalities such as x + y ≤ 6.
The area that satisfies all constraints together is the feasible region. The answer sits at a corner of this region, so one wrong line or one wrong shaded side changes everything.
In my years of teaching, the same mistakes come back every year:
- A scale that is too small, so the lines cross in a cramped space.
- A line plotted with one correct point and one guessed point.
- The wrong side shaded.
- x ≥ 0 and y ≥ 0 forgotten.
- Corner points read off the graph instead of solved.
- Some corners left untested, or a corner put into the wrong function.
Set up the graph paper before you draw
A good graph starts before the first line.
- Find the intercepts of every line. An intercept is the point where a line meets an axis. Put x = 0 for the y-intercept and y = 0 for the x-intercept.
- Look at the largest intercept on each axis and choose a scale so it fits on the page.
- Use the same scale on both axes unless the numbers are very different. Write the scale in a corner.
- Draw the axes with a ruler, mark the origin and label x and y.
Why this matters: on a tiny scale, two different corners can look like one point, and you will read the wrong coordinates.
Draw the lines and shade the correct side
Treat every inequality as an equation first. Draw the line through its two intercepts. Then decide which side to shade.
Use the origin test. Put (0, 0) in the inequality. If it is true, shade the side that has the origin. If it is false, shade the other side. If the line passes through the origin, test another easy point such as (1, 0) or (0, 1).
- Label each line with its equation as soon as you draw it.
- Use light pencil shading, with small arrows pointing to the side you chose.
- For ≤ and ≥, draw a solid line. For strict inequalities (< or >), a dashed line shows the boundary is not included.
- Mark the final common region only after all lines are drawn.
Tip: if the common region comes out empty, do not rub everything out. Check the arrows line by line. Usually only one side is wrong.
Find corner points by solving, not by reading
A corner point is a vertex of the feasible region, where two boundary lines meet. The graph tells you which two lines meet at each corner. The coordinates must come from solving those two equations together.
A hand-drawn graph is never perfect, so a value read from it can be slightly off. Marks are given for correct coordinates, so solve them.
Also check that each corner satisfies every other constraint. Two lines may meet outside the region. That point is not a corner of the feasible region.
Worked example with a corner-point table
Problem: Maximise Z = 3x + 2y subject to x + y ≤ 6, x ≤ 4, x ≥ 0, y ≥ 0.
Step 1: Intercepts. The line x + y = 6 passes through (6, 0) and (0, 6). The line x = 4 is vertical, through 4 on the x-axis.
Step 2: Shade. Put (0, 0) in x + y ≤ 6. We get 0 ≤ 6, which is true, so shade towards the origin. The same holds for x ≤ 4. With x ≥ 0 and y ≥ 0, the region stays in the first quadrant.
Step 3: Corners. Solve each pair of boundary lines:
- x = 0 and y = 0 give (0, 0).
- y = 0 and x = 4 give (4, 0).
- x = 4 and x + y = 6 give y = 2, so (4, 2).
- x = 0 and x + y = 6 give (0, 6).
Step 4: Table.
| Corner point | Z = 3x + 2y |
|---|---|
| (0, 0) | 0 |
| (4, 0) | 12 |
| (4, 2) | 16 |
| (0, 6) | 12 |
Answer: the maximum value of Z is 16, at (4, 2).
Every corner is tested, including (0, 0). Skipping a corner is a common way to lose a mark. Write the answer as a sentence: the maximum value and the point where it occurs.
Two checks before you write the final answer
First, put your best corner back into each constraint. It must satisfy all of them. In the example, (4, 2) gives 4 + 2 = 6, which fits x + y ≤ 6, and x = 4 fits x ≤ 4.
Second, if the region is unbounded (it extends without limit), do not simply pick the largest value from the table. In the textbook method, you draw the half-plane of the objective function against that value and see whether it has a point in common with the feasible region. Only then can you say whether a maximum or minimum exists. Practise this separately, because many students stop short here.
A weekly routine for Gurugram and Dwarka students
This chapter rewards steady practice. Here is a simple plan for a Class 12 CBSE student:
- Day 1: Draw only the graphs for a few problems. Focus on scale, intercepts and shading.
- Day 2: Take the same graphs and solve all corner points.
- Day 3: Do full problems, one with a maximum and one with a minimum.
- Day 4: Read word problems and write only the constraints and the objective function. Many students find this translation hardest.
- Day 5: Do one unbounded-region problem with the half-plane check.
Keep one page of your own mistakes, such as "wrong side shaded", and read it before each session.
How a home tutor helps with this chapter
A printed solution shows only the final neat graph. A tutor sitting beside the student can watch the pencil move and catch the exact moment a slip happens: a wrong intercept, a skipped arrow, a corner copied from the picture.
The tutor can also ask the student to explain why one side is shaded, and fix the habit on the spot. Great Home Tutors offers home tuition only, with no group batch. A person calls you back to understand the student, and a first class lets you see the fit before anything is fixed.
Key takeaways
- Find intercepts first, then choose a scale that fits the largest value.
- Use the origin test to shade, and label every line with its equation.
- Solve for corner points; do not read them from the graph.
- Test every corner in the objective function and write the answer as a sentence.
- Practise graphs, corners and word-problem translation on separate days.
Frequently asked questions
Why is the maximum or minimum always at a corner point?
The objective function is a straight-line expression. Across the feasible region its value changes steadily in one direction, so the extreme value is reached at a corner. This is the corner point theorem in the NCERT chapter.
What if the origin lies on the line?
Then the origin test cannot decide the side. Pick another simple point not on the line, such as (1, 0) or (0, 1), and put it in the inequality.
Do I need to test points that are not corners?
No. For a bounded region, testing all corner points is enough. For an unbounded region, use the half-plane check described above.
Can a student in Gurugram take home tuition for Class 12 maths?
Great Home Tutors lists Class 12 Maths for Dwarka and a home tutor in Gurgaon. Tell us the class, subject, board and your locality, and we match a tutor.
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