ilovepdf_merged (6).pdf). Content covers sections 37.1–37.x.Linear programming optimises a linear objective under linear inequality constraints. In two variables, solve by graphing the feasible region and evaluating the objective at corner points.
Unique optimum at a vertex; alternate optima along an edge; unbounded feasible region may yield unbounded objective; infeasible if constraints contradict.
Most exam-important points from this chapter:
Start every answer with the key definition or standard form from Linear Programming.
Copy the formula, then substitute values; never jump to the number alone.
Check domain, quadrant, non-zero denominators, and applicability of the theorem.
Memorise and apply: type ‘B’ toys subject to the limitation on the investment.
Memorise and apply: The above problem was easy to handle because the choice was limited to two types, and the
English questions from NIOS Mathematics (311) public / sample papers (Sample 2024, Apr 2024, Oct 2024, Apr 2025; board-style where scanned papers had no extractable text). Mapped exclusively to this chapter — no cross-chapter duplicates. Use Model Answer for the marking key and Explanation for working.
4 question(s) · Sources: Board-style corner method, Board-style feasible point, Board-style min, Board-style model
PYQ1. A dealer sells pens at ₹3 profit and pencils at ₹2 profit. Constraints: pens ≤ 100, pencils ≤ 200, pens+pencils ≤ 250, non-negative. Formulate the LPP to maximise profit (x=pens, y=pencils).
Model Answer
1) Max z=3x+2y s.t. x≤100, y≤200, x+y≤250, x≥0, y≥0
Explanation
1) Objective uses profit coefficients
2) stock limits become linear inequalities with non-negativity.
PYQ2. Feasible vertices are (0,0), (4,0), (3,3), (0,2). Maximise z=2x+5y.
Model Answer
1) Maximum z=21 at (3,3)
Explanation
1) z values: 0, 8, 21,
2) 1
0. Largest is 21 at (3,3).
PYQ3. Is (2,2) feasible for x≥0, y≥0, x+y≤5, x≤3?
Model Answer
1) Yes
Explanation
1) All inequalities hold: 2+2=4≤5 and 2≤3.
PYQ4. Using the same vertices as Q2, minimise z=2x+5y.
Model Answer
1) Minimum z=0 at (0,0)
Explanation
1) Smallest among 0,8,21,10 is 0.
10 English exam-style problems for this chapter (NIOS Mathematics 311). Each question states the formula, gives full step-by-step working with the final answer, plus formal/easy explanations. English only.
A shop sells pens at ₹3 profit and pencils at ₹2. Constraints: pens≤100, pencils≤200, pens+pencils≤250, non-negative. Write the LPP to maximise profit (x=pens, y=pencils).
1) Max z=3x+2y.
2) Subject to x≤100, y≤200, x+y≤250, x≥0, y≥0.
Answer: Max z=3x+2y; x≤100, y≤200, x+y≤250, x,y≥0
Standard modelling of a two-variable LPP.
Profit coefficients become objective; stock limits become inequalities.
Non-negativity always for counts.
Do not maximise x+y unless that is profit.
Feasible vertices of a region are (0,0),(4,0),(3,3),(0,2). Maximise z=2x+5y.
1) z(0,0)=0; z(4,0)=8; z(3,3)=6+15=21; z(0,2)=10.
2) Maximum 21 at (3,3).
Answer: Maximum z=21 at (3,3)
Corner-point theorem for linear objectives over polygons.
Compute z at each corner; pick largest.
Minimum would be 0 here.
List all vertices first.
Is (2,2) feasible for x≥0,y≥0,x+y≤5, x≤3?
1) 2≥0,2≥0,2+2=4≤5,2≤3 all true ⇒ feasible.
Answer: Yes, feasible
A point is feasible if it satisfies every constraint.
Check each inequality.
Interior points can be feasible even if not optimal.
One failed inequality is enough to reject.
The feasible vertices include (0,2) with constraint y≤2. Is y≤2 binding at (0,2)?
1) At (0,2), y=2, so y≤2 holds as equality.
2) Therefore the constraint is binding at (0,2).
Answer: Yes — binding (y=2)
A inequality constraint is binding at a point when it holds with equality.
The point sits on the boundary line y=2.
If y were 1, the constraint would be non-binding (slack).
Binding ⇔ on the boundary of that inequality.
If feasible region is x≥0,y≥0,x+y≥1 (unbounded) and z=x+y, is max z finite?
1) Along the ray x=t,y=0 for t≥1, z=t→∞.
2) Maximum is unbounded (no finite max).
Answer: No finite maximum (unbounded)
Linear objective may be unbounded on unbounded regions.
You can go to infinity with z growing.
Minimising z here has minimum 1 on x+y=1.
Graphical sense: open direction of increase.
Using vertices (0,0),(4,0),(3,3),(0,2), minimise z=2x+5y.
1) Values 0,8,21,10 ⇒ minimum 0 at (0,0).
Answer: Minimum z=0 at (0,0)
Same evaluation as maximisation; pick smallest.
Zero at origin.
If origin not feasible, min is at another vertex.
Do not pick max by mistake.
A factory makes product X for ₹5 profit and Y for ₹4. Constraints: x≥0, y≥0, x+y≤10. Write the LPP to maximise profit.
1) Maximise z = 5x + 4y.
2) Subject to x + y ≤ 10, x ≥ 0, y ≥ 0.
Answer: Max z=5x+4y s.t. x+y≤10, x≥0, y≥0
Linear objective with linear constraints.
Non-negativity for quantities.
Feasible region is a right triangle.
Check each algebraic step carefully.
Maximise z=x+2y on vertices (0,0), (4,0), (0,3). Which point is optimal?
1) z(0,0)=0; z(4,0)=4; z(0,3)=6.
2) Maximum is 6 at (0,3).
Answer: Maximum at (0, 3); z=6
Corner point theorem for linear programmes.
Compare three values only.
Optimum at a vertex of the feasible polygon.
Check each algebraic step carefully.
Is (2, 2) feasible for x≥0, y≥0, x+y≤5, x≤3?
1) 2≥0, 2≥0, 2+2=4≤5, 2≤3 — all hold.
2) Yes, feasible.
Answer: Yes
A point is feasible if it satisfies every constraint.
Check each inequality.
On the boundary is still feasible for ≤.
Check each algebraic step carefully.
For max z=x on x≥0, y≥0 (no other constraints), does a finite maximum exist?
1) x can increase without bound.
2) No finite maximum (unbounded feasible region in the objective direction).
Answer: No finite maximum
Unbounded region can make linear objectives unbounded.
Need enough constraints to close the region.
Minimum would be 0 at x=0.
Check each algebraic step carefully.