Exercise 12.1 Class 12 Maths Linear Programming

NCERT Solutions Class 12 Maths Chapter 12 Ex 12.1 | LearnCBSEHub

💡 Corner Point Method

Graph the constraints (inequalities) to find the Feasible Region.
Determine the coordinates of the Corner Points (vertices) of this region.
Evaluate the Objective Function $Z = ax + by$ at each corner point.
The largest/smallest value is the Maximum/Minimum (for bounded regions).

Question 01

Maximise $Z = 3x + 4y$ subject to $x + y \le 4, x \ge 0, y \ge 0$.

1. Graph the Constraints

Line $x + y = 4$ passes through $(0, 4)$ and $(4, 0)$.
Region is the triangle bounded by axes and the line.

2. Corner Points & Value of Z

Corner Point (x, y)	Value of Z = 3x + 4y
(0, 0)	0
(4, 0)	12
(0, 4)	16 (Maximum)

Maximum Z = 16 at (0, 4)

Question 02

Minimise $Z = -3x + 4y$ subject to $x + 2y \le 8, 3x + 2y \le 12, x \ge 0, y \ge 0$.

1. Graph the Constraints

Line 1 ($x+2y=8$): passes $(0,4), (8,0)$.
Line 2 ($3x+2y=12$): passes $(0,6), (4,0)$.
Intersection: Solving equations gives $x=2, y=3$. Point $B(2,3)$.

2. Evaluate Z

Corner Point	Z = -3x + 4y
(0, 0)	0
(4, 0)	-12 (Minimum)
(2, 3)	-6 + 12 = 6
(0, 4)	16

Minimum Z = -12 at (4, 0)

Question 03

Maximise $Z = 5x + 3y$ subject to $3x + 5y \le 15, 5x + 2y \le 10, x \ge 0, y \ge 0$.

1. Intersections & Graph

Line 1: $(0,3), (5,0)$. Line 2: $(0,5), (2,0)$.
Intersection of $3x+5y=15$ and $5x+2y=10$:
Multiply (ii) by 3, (i) by 5 $\to 19x = 20 \implies x = \frac{20}{19}, y = \frac{45}{19}$.

2. Evaluate Z

Corner Point	Z = 5x + 3y
(0, 0)	0
(2, 0)	10
(20/19, 45/19)	235/19 $\approx$ 12.37 (Maximum)
(0, 3)	9

Maximum Z = 235/19

Question 04

Minimise $Z = 3x + 5y$ subject to $x + 3y \ge 3, x + y \ge 2, x,y \ge 0$.

1. Unbounded Region

Intersection of $x+3y=3$ and $x+y=2$ is $(1.5, 0.5)$.
Since constraints are $\ge$, the region is unbounded away from origin.

2. Evaluate Z

Corner Point	Z = 3x + 5y
(3, 0)	9
(1.5, 0.5)	4.5 + 2.5 = 7 (Minimum)
(0, 2)	10

Check: Does $3x+5y < 7$ have common points with region?
Graphing $3x+5y < 7$ shows no overlap with feasible region.

Minimum Z = 7 at (1.5, 0.5)

Question 05

Maximise $Z = 3x + 2y$ subject to $x + 2y \le 10, 3x + y \le 15, x,y \ge 0$.

1. Intersections

Intersection of $x+2y=10$ and $3x+y=15$:
From (i) $x=10-2y$. Sub into (ii): $3(10-2y)+y=15 \implies 30-5y=15 \implies y=3, x=4$. Point (4,3).

2. Evaluate Z

Corner Point	Z = 3x + 2y
(0, 0)	0
(5, 0)	15
(4, 3)	12 + 6 = 18 (Maximum)
(0, 5)	10

Maximum Z = 18 at (4, 3)

Question 06

Minimise $Z = x + 2y$ subject to $2x + y \ge 3, x + 2y \ge 6, x,y \ge 0$.

1. Unbounded Region

Line 1 ($2x+y=3$) meets axes at $(1.5, 0), (0, 3)$.
Line 2 ($x+2y=6$) meets axes at $(6, 0), (0, 3)$.
They intersect at $(0, 3)$. The region is unbounded.

2. Evaluate Z

Corner Point	Z = x + 2y
(6, 0)	6
(0, 3)	6

Since Z has the same minimum value at two corner points, the minimum occurs at all points on the line segment joining (6,0) and (0,3).

Minimum Z = 6 at multiple points

Question 07

Minimise & Maximise $Z = 5x + 10y$ subject to $x+2y \le 120, x+y \ge 60, x-2y \ge 0$.

1. Identify Region

$x-2y \ge 0 \implies x \ge 2y$ (Below/Right of line $y=x/2$).
Intersection of $x=2y$ and $x+y=60 \implies (40, 20)$.
Intersection of $x=2y$ and $x+2y=120 \implies (60, 30)$.
Axis points: $(60, 0)$ and $(120, 0)$.

2. Evaluate Z

Corner Point	Z = 5x + 10y
(60, 0)	300 (Minimum)
(120, 0)	600 (Maximum)
(60, 30)	300 + 300 = 600 (Maximum)
(40, 20)	200 + 200 = 400

Min Z = 300 at (60,0). Max Z = 600 at (120,0) and (60,30).

Question 08

Minimise & Maximise $Z = x + 2y$ subject to $x+2y \ge 100, 2x-y \le 0, 2x+y \le 200$.

1. Vertices Calculation

$2x-y \le 0 \implies y \ge 2x$.
Intersection $y=2x$ & $x+2y=100 \implies x+4x=100 \implies (20, 40)$.
Intersection $y=2x$ & $2x+y=200 \implies 2x+2x=200 \implies (50, 100)$.
Y-axis intercepts: $(0, 50)$ from first line, $(0, 200)$ from third line.

2. Evaluate Z

Corner Point	Z = x + 2y
(0, 50)	100 (Minimum)
(20, 40)	20 + 80 = 100 (Minimum)
(50, 100)	50 + 200 = 250
(0, 200)	400 (Maximum)

Min = 100 at segment (0,50)-(20,40). Max = 400 at (0,200).

Question 09

Maximise $Z = -x + 2y$ subject to $x \ge 3, x+y \ge 5, x+2y \ge 6$.

1. Analyze Region

The region is unbounded (open upwards).
Corner points: $(6, 0), (4, 1), (3, 2)$.
$Z(6,0) = -6$. $Z(4,1) = -2$. $Z(3,2) = 1$.

2. Unbounded Check

Since the region is unbounded in the positive y-direction, and the coefficient of $y$ in $Z$ is positive (+2), $Z$ can increase indefinitely.

No Maximum Value

Question 10

Maximise $Z = x + y$ subject to $x – y \le -1, -x + y \le 0$.

1. Analyze Constraints

Constraint 1: $x – y \le -1 \implies y \ge x + 1$. (Region above line $y=x+1$)
Constraint 2: $-x + y \le 0 \implies y \le x$. (Region below line $y=x$)

2. Conclusion

It is impossible for a point to be both above $y=x+1$ and below $y=x$ simultaneously, as these are parallel lines with gap.
There is no common feasible region.

No Feasible Region (No Solution)