Q.Refer to Exercise 12. What will be the minimum cost?
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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. …
- Higher Secondary (+2 Stage) Examination 2025Set ANNUAL5 marksQ.Solve the following linear programming problem graphically: Find the maximum and minimum value of Z, where Z=x+2y, subject to the constraints: x+2y≥100, 2x−y≤0, 2x+y≤200, x≥0, y≥0.
›Reveal solutionSolution
Plot the feasible region from the four constraints, find its corner points, then evaluate Z=x+2y at each corner — the maximum/minimum of a linear objective over a bounded polygon always occurs at a corner.
Constraints: x+2y≥100, 2x−y≤0 (i.e. y≥2x), 2x+y≤200, x≥0, y≥0.
Finding the corner points by intersecting the boundary lines in pairs:
- x=0 with x+2y=100: gives (0,50).
- x+2y=100 with y=2x: substituting, x+4x=100⇒x=20,y=40: gives (20,40).
- y=2x with 2x+y=200: substituting, 2x+2x=200⇒x=50,y=100: gives (50,100).
- 2x+y=200 with x=0: gives (0,200).
(One can check each of these four points satisfies all the constraints simultaneously, and that they are the actual vertices of the bounded feasible region — a quadrilateral.)
Evaluating Z=x+2y at each corner:
Corner Z=x+2y (0,50) 0+100=100 (20,40) 20+80=100 (50,100) 50+200=250 (0,200) 0+400=400 … - Higher Secondary (+2 Stage) Examination 2023Set ANNUAL5 marksQ.A businessman plans to sell two types of special computers: a desktop model and a portable model, whose current prices are Rs 25,000 and Rs 40,000 respectively. He estimates that the total monthly demand for computers will not exceed 250 units. If the profit is Rs 4,500 on each desktop model and Rs 5,000 on each portable model, and he does not want to invest more than Rs 70,00,000 (seventy lakh rupees) in total, determine, by the graphical method, how many units of each type of computer the businessman should stock to obtain the maximum profit. What is the maximum profit?
›Reveal solutionSolution
This is a linear programming problem: set up the demand and investment constraints, plot the feasible region, and evaluate the profit at each corner point — the maximum occurs at (200,50) giving a profit of Rs 11,50,000.
Let x = number of desktop models, y = number of portable models to stock.
Objective (maximize profit): Z=4500x+5000y
Constraints:
- Demand: x+y≤250
- Investment: 25000x+40000y≤70,00,000, which simplifies (dividing by 5000) to 5x+8y≤1400
- Non-negativity: x≥0, y≥0
Corner points of the feasible region:
- (0,0)
- x-intercept of demand line: (250,0) — check investment: 5(250)=1250≤1400 ✓, so this is a genuine corner.
- y-intercept of investment line: 8y=1400⇒y=175, giving (0,175) — check demand: 175≤250 ✓ (this is the binding constraint here, since the demand line would allow y=250 but investment caps it at 175).
- Intersection of x+y=250 and 5x+8y=1400: from the first, x=250−y; substitute: 5(250−y)+8y=1400⇒1250+3y=1400⇒y=50, x=200. Point: (200,50). …
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