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

Use the origin test for each inequality separately. If the line passes through the origin, test (1, 0) or (0, 1).

Is it correct to say: Leaving out x ≥ 0, y ≥ 0?

Write the non-negativity constraints in the formulation and use the axes as boundaries of the region.

Is it correct to say: Writing 'at least' as ≤ (or 'at most' as ≥)?

'At least 6' means ≥ 6; 'at most 10' or 'not more than 10' means ≤ 10.

Is it correct to say: Estimating the intersection corner from the graph?

Solve the two boundary equations algebraically; Z must be evaluated at the exact corner.

Is it correct to say: Declaring a maximum on an unbounded region just because one corner has the largest value?

On an unbounded region check the half-plane Z > M. If it meets the region, Z has no maximum.

Is it correct to say: Reporting only one optimal point when two corners tie?

If Z is equal at two adjacent corners, every point on the segment joining them is optimal.

📄 Formula sheet (PDF)

Linear Programming

Linear inequations in two variables, formulation of a linear programming problem, feasible region and the graphical corner-point method — Maharashtra HSC Maths Part I Ch 7

📐 12 formulas✏️ 4 examples🎯 5 practice🏫 Maharashtra State Board📚 Class 12✓ 2025–26 syllabus

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Board Exam Tips

→Start every word problem with 'Let x = … and y = …' in words, then write the objective function and one constraint per resource, with its name beside it.
→Plot each boundary line from its two intercepts, decide the side with the origin test, and mark the feasible region clearly.
→Find corner points that come from two lines by solving the equations — do not read them off the graph paper.
→Make a table of corner points and the value of Z at each; the answer comes straight from that table.
→Never forget the non-negativity constraints x ≥ 0, y ≥ 0 — they keep the region in the first quadrant.
→If two corner points give the same optimum value, say that every point on the segment joining them is optimal.

📐 Formulas(12)

Linear Inequation as a Half-plane

a x + b y \leq c or a x + b y \geq c

Plotting the Boundary Line by Intercepts

a x + b y = c passes through (\frac{c}{a}, 0) and (0, \frac{c}{b})

Origin Test

a (0) + b (0) \leq c ⟺ 0 \leq c

Translating Word Conditions

at most k : \leq k, at least k : \geq k, does not exceed k : \leq k

Mathematical Form of an LPP★ Board fav

Maximise / Minimise Z = c_{1} x + c_{2} y subject to a_{i 1} x + a_{i 2} y \leq b_{i} (or \geq b_{i}), x \geq 0, y \geq 0

Symbol	Meaning
$x, y$	Decision variables (quantities to be decided)
$Z$	Objective function (profit, cost, …)
$c_{1}, c_{2}$	Profit or cost per unit of x and y

Non-negativity Constraints

x \geq 0, y \geq 0

Feasible Region

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

Corner Point from Two Boundary Lines

a_{1} x + b_{1} y = c_{1}, a_{2} x + b_{2} y = c_{2} \Rightarrow 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★ Board fav

Z_{m a x} and Z_{m i n} (when they exist) occur at corner points of R

Multiple Optimal Solutions★ Board fav

Z (P) = Z (Q) = Z_{opt} \Rightarrow Z = Z_{opt} at tP + (1 - t) Q, 0 \leq t \leq 1

Unbounded Feasible Region

M = largest corner value; {(x, y) : c_{1} x + c_{2} y > M} \cap R = \emptyset \Rightarrow Z_{m a x} = M

Infeasible Problem

R = \emptyset \Rightarrow no feasible solution

✏️ Solved Examples

1Solved Exampleeasy3 steps

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

Boundary lines: x + y = 4 through (4, 0) and (0, 4); x = 3 is vertical. The origin satisfies both inequalities, so the region is towards the origin.

x + y = 4, x = 3

2Solved Exampleboard4 steps

A workshop makes two products A and B. Each unit of A needs 2 hours on machine M₁ and 1 hour on machine M₂; each unit of B needs 1 hour on M₁ and 2 hours on M₂. M₁ is available for 10 hours and M₂ for 8 hours a day. The profit is ₹40 per unit of A and ₹30 per unit of B. How many units of each should be made daily to maximise profit?

Let x units of A and y units of B be made. Formulate the LPP.

Maximise Z = 40 x + 30 y subject to 2 x + y \leq 10 (M_{1}), x + 2 y \leq 8 (M_{2}), x, y \geq 0

3Solved Exampleboard4 steps

Food F₁ costs ₹3 per unit and food F₂ costs ₹2 per unit. One unit of F₁ gives 1 unit of vitamin A and 2 units of vitamin B; one unit of F₂ gives 1 unit of each vitamin. At least 4 units of vitamin A and 6 units of vitamin B are needed. Find the minimum cost.

Let x units of F₁ and y units of F₂ be used. Requirements give '≥' constraints.

Minimise Z = 3 x + 2 y subject to x + y \geq 4, 2 x + y \geq 6, x, y \geq 0

4Solved ExampleHOTS3 steps

Maximise Z = 2x + 4y subject to x + 2y ≤ 8, 3x + y ≤ 9, x ≥ 0, y ≥ 0.

Intercepts: x + 2y = 8 through (8, 0), (0, 4); 3x + y = 9 through (3, 0), (0, 9). Inner corner:

x + 2 y = 8, 3 x + y = 9 \Rightarrow x = 2, y = 3

⚠️ Traps & Common Mistakes

⚠️Common Mistakes6

1
Shading the wrong side of a boundary line
✓Use the origin test for each inequality separately. If the line passes through the origin, test (1, 0) or (0, 1).
2
Leaving out x ≥ 0, y ≥ 0
✓Write the non-negativity constraints in the formulation and use the axes as boundaries of the region.
3
Writing 'at least' as ≤ (or 'at most' as ≥)
✓'At least 6' means ≥ 6; 'at most 10' or 'not more than 10' means ≤ 10.
4
Estimating the intersection corner from the graph
✓Solve the two boundary equations algebraically; Z must be evaluated at the exact corner.
5
Declaring a maximum on an unbounded region just because one corner has the largest value
✓On an unbounded region check the half-plane Z > M. If it meets the region, Z has no maximum.
6
Reporting only one optimal point when two corners tie
✓If Z is equal at two adjacent corners, every point on the segment joining them is optimal.

🎯 Practice Yourself

🎯Practice Yourself5 questions

Q1
Maximise Z = 4x + 3y subject to x + y ≤ 6, 2x + y ≤ 8, x ≥ 0, y ≥ 0.
Q2
Minimise Z = 2x + 5y subject to x + 2y ≥ 4, 3x + y ≥ 6, x ≥ 0, y ≥ 0.
Q3
The corner points of a bounded feasible region are (0, 3), (2, 0), (5, 0), (5, 4) and (0, 6). Find the maximum and minimum of Z = 3x + 2y.
Q4
A shopkeeper buys pressure cookers at ₹1000 each and mixers at ₹2000 each. He can invest at most ₹50,000 and has space for at most 40 items. Profit is ₹200 per cooker and ₹300 per mixer. Formulate and solve the LPP.
Q5
Show that the LPP 'Maximise Z = x + y subject to x + y ≤ 2, x + y ≥ 5, x ≥ 0, y ≥ 0' has no feasible solution.

📝 Notes

Linear Programming

Linear programming chooses the best value of a linear quantity — profit, cost, time — when the choices are limited by linear inequalities. In the HSC course everything is done with two variables, so every problem is solved on a graph by the corner-point method.

Formulation

In a word problem, setting up the LPP correctly is the key step:

Define x and y in words ("Let x be the number of …").
Write the objective function $Z = c_1x + c_2y$ and say whether it is to be maximised or minimised.
Write one inequality per resource or requirement — available hours, money or space give $\le$ ; minimum requirements give $\ge$ .
Add $x \ge 0$ , $y \ge 0$ .

A table with the products in rows and the resources in columns makes the coefficients easy to read off.

Solving graphically

Draw each boundary line from its intercepts, use the origin test to choose the side, and identify the common region. Find every corner point exactly (intersections by elimination or Cramer's rule), then evaluate Z at each corner in a table. By the corner point theorem the optimum, if it exists, is the largest or smallest entry in that table.

Special cases

Unbounded region: the corner value is the optimum only if the open half-plane beyond it ( $Z > M$ for a maximum, $Z < m$ for a minimum) has no point in common with the region.
Multiple optima: two adjacent corners with the same optimal value mean every point on the edge joining them is optimal.
Infeasible problem: contradictory constraints leave no feasible region, so there is no solution.