Q.Refer to Exercise 12. What will be the minimum cost?
🔒You're viewing a preview — the full solution, concept, methods & PYQ mapping are locked.
🔒 Start your 14-day free trial to unlock the full solution →Concept understanding — Linear Programming Graphical Method
The Graphical Method for Linear Programming
When a linear programming problem has just two decision variables, x and y, you can solve it by drawing a picture. This is the graphical method, and it is the technique the CBSE Class-12 course expects you to use.
The idea
Each constraint is a linear inequality such as 2x+3y≤100. On the xy-plane its boundary is a straight line, and the inequality picks one side of that line (a half-plane). The points that satisfy all the constraints at once form a single region — the feasible region. Your job is to find the point inside this region that makes the objective function Z=ax+by largest or smallest.
The step-by-step procedure
- Draw each constraint line. Replace every inequality by an equation and plot the line, usually by finding where it meets the axes.
- Shade the correct side. Test a simple point (often the origin (0,0)) in the inequality. If it holds, the origin's side is the wanted half-plane; if not, take the other side. Always include the non-negativity conditions x≥0, y≥0, which keep you in the first quadrant.
- Identify the feasible region. It is the overlap of all the shaded half-planes — the region satisfying every constraint together.
- Find the corner (vertex) points. These are the points where the boundary lines cross. Read them off the graph or solve the two relevant lines simultaneously.
- Evaluate Z at every corner and pick the largest value (for a maximum) or the smallest (for a minimum).
The whole method rests on the Corner-Point Theorem: if an optimum exists, it occurs at a vertex of the feasible region. So you never test interior points — only the corners.
Bounded vs unbounded …
Exercise 12 (vans). 1200 packages go by large vans (capacity 200, cost ₹400) and small vans (capacity 80, cost ₹200); at most ₹3000 may be spent and large vans ≤ small vans. With x = large, y = small:
Minimise Z=400x+200y,200x+80y≥1200, 400x+200y≤3000, x≤y, x,y≥0,
which reduce to 5x+2y≥30, 2x+y≤15, x≤y.
Corners and cost:
- (0,15):3000
- (5,5):3000 …
For the vans LPP of Exercise 12, the cost Z=400x+200y is least at x=y=730, giving 718000≈₹2571.43.
The referenced problem (Exercise 12)
1200 packages must be transported. A large van carries 200 (cost ₹400), a small van carries 80 (cost ₹200). At most ₹3000 may be spent, and the number of large vans cannot exceed the number of small vans. Let x = large vans, y = small vans.
Minimise Z=400x+200y
200x+80y≥1200,400x+200y≤3000,x≤y,x,y≥0,
which simplify to
5x+2y≥30,2x+y≤15,x≤y.
Step 1 — Corner points
- 5x+2y=30 and 2x+y=15: from the second y=15−2x, so 5x+2(15−2x)=30⇒x=0,y=15⇒(0,15).
- 2x+y=15 and x=y: 3x=15⇒(5,5).
- 5x+2y=30 and x=y: 7x=30⇒x=y=730⇒(730,730). …
Method: Graphical (Corner-Point) Method for a Two-Variable LPP
Once a problem is written as "optimise Z=ax+by subject to linear inequalities in x and y," this is the standard CBSE technique for finding the optimum.
Steps
Step 1: Draw every constraint as a line.
Replace each inequality by an equation and plot the line from its axis intercepts. Include x=0 and y=0.
Step 2: Shade the correct half-plane.
Test the origin (0,0) in each inequality: if it is satisfied, keep the origin's side; if not, take the other side. Non-negativity confines you to the first quadrant.
Step 3: Identify the feasible region.
It is the single region where all the shaded half-planes overlap.
Step 4: Find the corner points.
Each vertex is the intersection of two boundary lines, found by solving those two equations together. Keep a candidate only if it satisfies every other constraint -- an intersection that breaks a third inequality is not a corner of the region.
Step 5: Evaluate Z at each corner. …
Common Mistakes
Mistake 1: Omitting the "large vans cannot exceed small vans" constraint.
Why it's wrong: dropping x≤y enlarges the feasible region and can give a spuriously lower cost at an inadmissible point. Correct approach: include x≤y alongside the capacity and budget constraints.
Mistake 2: Minimising the budget line instead of the cost. …
- KEAM 2026Set eng-2026-04194 marksMCQQ.Consider the Linear Programming Problem (LPP): Maximize z=30x+60y subject to the constraints x+2y≤12;2x+y≤12;4x+5y≥20;x≥0;y≥0. Then the number of corner points of the feasible region is (A) 8 (B) 6 (C) 3 (D) 4 (E) 5
›Reveal solutionSolution
Bounding the region by x+2y≤12, 2x+y≤12, 4x+5y≥20, x,y≥0 gives a pentagon with 5 corner points.
The constraint lines are L1:x+2y=12, L2:2x+y=12, L3:4x+5y=20, plus the axes.
Find the vertices of the feasible region (satisfying all constraints):
- L3∩x-axis: (5,0) — check: 5≤12, 10≤12 ✓
- L2∩x-axis: (6,0) — check: 24≥20, 6≤12 ✓
- L1∩L2: solving gives (4,4) — check: 36≥20 ✓
- L1∩y-axis: (0,6) — check: 6≤12, 30≥20 ✓ …
- KEAM 2026Set eng-2026-04204 marksMCQQ.Consider the Linear Programming Problem (LPP): Maximize z=20x+40y subject to the constraints 3x+y≤14; x+3y≤10; x≥0; y≥0. The number of corner points of the feasible region is (A) 5 (B) 4 (C) 3 (D) 2 (E) 6
›Reveal solutionSolution
Find the vertices of the region bounded by the two lines and the axes.
Constraints: 3x+y≤14, x+3y≤10, x,y≥0.
- Origin: (0,0).
- x-axis intercept of 3x+y=14: (14/3,0).
- y-axis intercept of x+3y=10: (0,10/3). …
- KEAM 2026Set eng-2026-04214 marksMCQQ.Which one of the following point is not in a feasible region bounded by the inequalities x≤4, y≤6, x+y≤6, x≥0, y≥0 (A) (0,0) (B) (4,0) (C) (4,2) (D) (0,6) (E) (6,0)
›Reveal solutionSolution
(6,0) violates x≤4, so it is the point not in the feasible region.
Constraints: x≤4, y≤6, x+y≤6, x≥0, y≥0.
- (0,0): all satisfied.
- (4,0): x=4≤4, x+y=4≤6 — OK.
- (4,2): x=4, y=2, x+y=6≤6 — OK. …
- KEAM 2025Set eng-2025-04274 marksMCQQ.Consider the linear programming problem. Minimize z=x+y Subject to the constraint 2x+3y≥6, x≥0, y≥0. Then the solution of L.P.P. is (A) 0 (B) 2 (C) 3 (D) 5 (E) 6
›Reveal solutionSolution
Evaluating z=x+y at the corner points (3,0) and (0,2) of the feasible region gives the minimum value 2.
The constraints are 2x+3y≥6, x≥0, y≥0. The boundary line 2x+3y=6 meets the axes at (3,0) and (0,2); the feasible region is the unbounded region above this line in the first quadrant. …
- KEAM 2022Set eng-2022-P2-B14 marksMCQQ.Consider the linear programming problem: Maximize z=10x+5y subject to the constraints 2x+3y≤120, 2x+y≤60, x,y≥0. Then the coordinates of the corner points of the feasible region are (A) (0,0),(30,0),(0,40) and (15,30) (B) (0,0),(60,0),(0,40) and (15,30) (C) (0,0),(30,0),(0,60) and (15,30) (D) (0,0),(30,0),(0,40) and (30,40) (E) (0,0),(60,0),(0,40) and (30,40)
›Reveal solutionSolution
The corner points are (0,0),(30,0),(0,40),(15,30).
Concept and Intuition
The feasible region is a polygon bounded by the constraint lines and axes; its vertices are the corner points.
Step-by-Step Solution
- Origin (0,0) is a corner.
- On y=0: 2x≤60 (binding) and 2x≤120, so x=30⇒(30,0).
- On x=0: 3y≤120 (binding) and y≤60, so y=40⇒(0,40).
- Intersect 2x+3y=120 and 2x+y=60: subtract to get 2y=60⇒y=30, then 2x=30⇒x=15, giving (15,30). …
🎓Unlock everything free for 14 days
- ✓Full step-by-step solutions
- ✓Concept-first explanations
- ✓Methods, shortcuts & mistakes
- ✓PYQ mapping + timed mock tests
Full access for 14 days. No credit card required.