Is it correct to say: Taking the smallest corner value as the minimum in an unbounded region without checking?

For an unbounded region, draw the open half-plane ax + by < m. If it meets the region, Z has no minimum. Write the test and its conclusion.

Is it correct to say: Shading the wrong side of a constraint line?

Substitute (0, 0). If the inequality is satisfied, the origin side is feasible. Otherwise, shade the other side.

Is it correct to say: Including a point where two lines meet outside the feasible region?

Not every intersection of two lines is a corner. Check each candidate point in every constraint before adding it to the table.

Is it correct to say: Forgetting the corner points on the axes?

The non-negativity constraints x ≥ 0 and y ≥ 0 are boundary lines too. Intercepts on the axes are often corners.

Is it correct to say: Stating only the two corners when they give equal optimal values?

Every point on the segment joining them is also optimal. Mention the whole segment.

Is it correct to say: Using the closed half-plane (≥ or ≤) in the unbounded test?

The test uses the OPEN half-plane ax + by > M (or < m). The line ax + by = M itself always touches the region at the corner.

📄 Formula sheet (PDF)

Linear Programming

Q: Is it correct to say: Shading the wrong side of a constraint line?

Substitute (0, 0). If the inequality is satisfied, the origin side is feasible. Otherwise, shade the other side.

Q: Is it correct to say: Including a point where two lines meet outside the feasible region?

Not every intersection of two lines is a corner. Check each candidate point in every constraint before adding it to the table.

Q: Is it correct to say: Forgetting the corner points on the axes?

The non-negativity constraints x ≥ 0 and y ≥ 0 are boundary lines too. Intercepts on the axes are often corners.

Q: Is it correct to say: Stating only the two corners when they give equal optimal values?

Every point on the segment joining them is also optimal. Mention the whole segment.

Q: Is it correct to say: Using the closed half-plane (≥ or ≤) in the unbounded test?

The test uses the OPEN half-plane ax + by > M (or < m). The line ax + by = M itself always touches the region at the corner.

Objective function, constraints, feasible and infeasible regions, bounded and unbounded regions, and the corner point method for two-variable problems — NCERT Class 12 Maths Ch 12

📐 13 formulas✏️ 4 examples🎯 5 practice⚖️ Unit V: 5 marks🏫 CBSE📚 Class 12✓ 2025–26 syllabus

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Board Exam Tips

→Draw each constraint line accurately using its two intercepts. Shade the correct side by testing the origin, then mark the feasible region clearly.
→Find every corner point by solving the two boundary lines that meet there. Check each corner in ALL the constraints.
→Lay out a table of corner points and their Z values. It is the standard layout and makes the maximum and minimum easy to spot.
→If the feasible region is unbounded, the largest or smallest corner value is not automatically the answer. Do the open half-plane test and write the conclusion.
→If two corners give the same optimal value, every point on the segment joining them is optimal. Say so explicitly.
→Always include x ≥ 0, y ≥ 0. They are constraints too, and they usually give corner points on the axes.

📐 Formulas(13)

Objective Function

Z = a x + b y

Symbol	Meaning
$Z$	Objective function (profit, cost, ...)
$x, y$	Decision variables (non-negative)
$a, b$	Constants (for example profit per unit)

Constraints

a_{1} x + b_{1} y \leq c_{1} (or \geq c_{1}), \dots, x \geq 0, y \geq 0

Feasible Region

R = {(x, y) : all constraints, including x \geq 0, y \geq 0, hold}

Plotting a Constraint Line

a x + b y = c meets the axes at (\frac{c}{a}, 0) and (0, \frac{c}{b})

Choosing the Side to Shade

Put (0, 0) in a x + b y \leq c : 0 \leq c true \Rightarrow origin side

Corner Point (Intersection of Two Lines)

x = \frac{c _{1} b _{2} - c _{2} b _{1}}{a _{1} b _{2} - a _{2} b _{1}}, y = \frac{a _{1} c _{2} - a _{2} c _{1}}{a _{1} b _{2} - a _{2} b _{1}}

Corner Point Theorem

If Z has an optimal value on R, it occurs at a corner point of R

Bounded Feasible Region★ Board fav

R bounded \Rightarrow Z_{m a x} = M, Z_{m i n} = m

Symbol	Meaning
$M$	Largest value of Z among the corner points
$m$	Smallest value of Z among the corner points

Unbounded Region: Maximum Test★ Board fav

M is the maximum \Leftrightarrow the open half-plane a x + b y > M has no point in common with R

Unbounded Region: Minimum Test★ Board fav

m is the minimum \Leftrightarrow the open half-plane a x + b y < m has no point in common with R

Multiple Optimal Solutions

Z (P) = Z (Q) = Z^{*} \Rightarrow Z = Z^{*} at every point of segment P Q

No Feasible Region

R = ϕ \Rightarrow no feasible solution, so no optimal value

Corner Point Method (Steps)★ Board fav

Sketch R \to find corners \to evaluate Z \to compare (bounded) or test half-plane (unbounded)

✏️ Solved Examples

1Solved Exampleeasy3 steps

Maximise Z = 3x + 2y subject to x + y ≤ 6, x ≤ 4, x ≥ 0, y ≥ 0.

Draw x + y = 6 (through (6, 0) and (0, 6)) and x = 4. Testing (0, 0) shows the feasible region is on the origin side of both lines. It is bounded.

Corners: O (0, 0), A (4, 0), B (4, 2), C (0, 6)

2Solved Exampleboard4 steps

Find the minimum and maximum of Z = 5x + 3y subject to x + y ≥ 2, x + 2y ≤ 8, 2x + y ≤ 10, x ≥ 0, y ≥ 0.

The region lies on or above x + y = 2 and on or below both x + 2y = 8 and 2x + y = 10, in the first quadrant. It is bounded.

Corners: (2, 0), (5, 0), (4, 2), (0, 4), (0, 2)

3Solved Exampleboard4 steps

Minimise Z = 3x + 2y subject to x + 2y ≥ 6, 2x + y ≥ 6, x ≥ 0, y ≥ 0.

Both constraints are of the ≥ type, so the feasible region lies away from the origin and is UNBOUNDED.

Corners: A (0, 6), B (2, 2), C (6, 0)

4Solved ExampleHOTS3 steps

Maximise Z = 2x + 4y subject to x + 2y ≤ 10, x + y ≤ 7, x ≥ 0, y ≥ 0.

Draw x + 2y = 10 (through (10, 0) and (0, 5)) and x + y = 7 (through (7, 0) and (0, 7)). The region on the origin side of both lines is bounded.

x + 2 y = 10, x + y = 7 \Rightarrow y = 3, x = 4

⚠️ Traps & Common Mistakes

⚠️Common Mistakes6

1
Taking the smallest corner value as the minimum in an unbounded region without checking
✓For an unbounded region, draw the open half-plane ax + by < m. If it meets the region, Z has no minimum. Write the test and its conclusion.
2
Shading the wrong side of a constraint line
✓Substitute (0, 0). If the inequality is satisfied, the origin side is feasible. Otherwise, shade the other side.
3
Including a point where two lines meet outside the feasible region
✓Not every intersection of two lines is a corner. Check each candidate point in every constraint before adding it to the table.
4
Forgetting the corner points on the axes
✓The non-negativity constraints x ≥ 0 and y ≥ 0 are boundary lines too. Intercepts on the axes are often corners.
5
Stating only the two corners when they give equal optimal values
✓Every point on the segment joining them is also optimal. Mention the whole segment.
6
Using the closed half-plane (≥ or ≤) in the unbounded test
✓The test uses the OPEN half-plane ax + by > M (or < m). The line ax + by = M itself always touches the region at the corner.

🎯 Practice Yourself

🎯Practice Yourself5 questions

Q1
Maximise Z = 5x + 4y subject to x + y ≤ 5, x ≤ 3, x ≥ 0, y ≥ 0.
Q2
Minimise Z = 4x + 6y subject to x + y ≥ 5, x ≤ 4, y ≤ 6, x ≥ 0, y ≥ 0.
Q3
Minimise Z = x + 3y subject to x + y ≥ 4, x + 3y ≥ 6, x ≥ 0, y ≥ 0.
Q4
Find the minimum and maximum (if they exist) of Z = 3x + 2y subject to x + 2y ≥ 4, x ≥ 0, y ≥ 0.
Q5
Maximise Z = x + y subject to x + 2y ≤ 3, 2x + y ≥ 8, x ≥ 0, y ≥ 0.

📝 Notes

Linear Programming

Chapter 12 optimises a linear objective Z = ax + by over a region cut out by linear inequalities. In two variables the whole problem is solved on a graph, and whenever an optimal value exists it occurs at a corner point of the region.

Setting up the graph

Turn each inequality into an equation and draw the line using its intercepts.
Test the origin to decide which side to shade.
The feasible region is the part common to all the shaded half-planes in the first quadrant (x ≥ 0, y ≥ 0).

If no common region exists, the problem has no feasible solution. Stop there and say so.

Corner point method

List every corner. Each one is the intersection of two boundary lines and must satisfy all the constraints.
Make a table of the corner points and their values of Z.
Bounded region: the largest value in the table is the maximum and the smallest is the minimum.
Unbounded region: the extreme value in the table is only a candidate. For a maximum M, check that the open half-plane ax + by > M has no point in common with the region. For a minimum m, check ax + by < m. If the half-plane does meet the region, that optimum does not exist.

Special outcomes

Multiple optima: two corners give the same best value, so the whole edge between them is optimal. This happens when the objective line is parallel to that edge, i.e. Z is a multiple of the left-hand side of the edge's equation.
No optimum: an unbounded region can fail to have a maximum (or a minimum), even when it has the other.
Infeasible: the constraints contradict each other, so there is no region and no solution.

Writing the answer

End with a clear sentence, for example "Maximum value of Z is 26 at x = 4, y = 2." For unbounded regions, include a line describing the half-plane test.