12th maths NCERT Linear Programming Exercise 12.1 detailed solution with explanation

Solve the following Linear Programming Problems graphically

Question (1)

Maximise Z = 3x + 4y
subject to the constraints: x +y ≤ 4, x ≥ x, y ≥ 0

Solution

Shaded region is feasible region of given equation
line l₁ : x + y = 4

x	0	4
y	4	0

corner points	z= 3x+4y
A(0, 4)	3(0)+4(4)=0+16 = 16 max
B(4, 0)	4(3) +4(0) = 12+0 = 12
O(0, 0)	0 + 0 = 0

Maximum z = 16 at5 (0,.4)

Question (2)

Minimise Z = -3x + 4y
subject to x + 2y ≤ 8, 3x + 2y ≤ 12, x ≥ 0, y ≥ 0

Solution

Shaded region is feasible region of given equation
line l₁ : x + 2y = 8

x	0	8
y	4	0

and line l₂ : 3x + 2y = 12

x	0	4
y	6	0

corner points	z= - 3x + 4y
A(0, 4)	-3(0)+4(4)=0+16 = 16
B(2, 3)	-3(2) +4(3) = -6 + 12 = 6
C(4, 0)	-3(4) +4(0) = -12 + 0 = -12 minimum
O(0, 0)	0 + 0 = 0

Minimum value = - 12 at (4, 0)

Question (3)

Maximise Z = 5x + 3y
subject to 3x + 5y ≤ 15, 5x + 2y ≤ 10, x ≥ 0 , y ≥ 0

Solution

Shaded region is feasible region of given equation
line l₁ 3x +5y = 15

x	0	5
y	3	0

line₂: 5x + 2y = 10

x	0	2
y	5	0

Cornaer points	z
A(0, 3)	0 + 9 = 9
B($\frac{{20}}{{19}}$,$\frac{{45}}{{19}}$)	$5\left( {\frac{{20}}{{19}}} \right) + 3\left( {\frac{{45}}{{19}}} \right) = \frac{{235}}{{19}}$
C(2, 0)	10 + 0 = 10
O(0, 0)	0 + 0

Maximum $Z = \frac{{235}}{{19}} \quad \text{at} \quad \left( {\frac{{20}}{{19}},\frac{{45}}{{19}}} \right)$

Question (4)

Minimise Z = 3x + 5y
such that x+3y ≥ 3, x + y ≥ 2, x,y≥0

Solution

Shaded region is feasible region of given equation
Line l₁ : x + 3y = 3

x	0	3
y	1	0

Line₂: x + y = 2

x	0	2
y	2	0

corner points	Z = 3x + 5y
A(0, 2)	3(0) + 5(2) = 0 +10 = 10
B(1.5, 0.5)	3(1.5)+5(0.5)= 4.5 +2.5 = 7 minimum
C(3, 0)	3(3) + 5(0) = 9

Minimum value of Z = 7 at (1.5, 0.5)

Question (5)

Maximise Z = 3x +2y
subject to x +2y ≤10, 3x + y ≤ 15 , x,y≥0

Solution

Shaded region is feasible region of given equation
line l₁: x + 2y = 10

x	0	10
y	5	0

line l₂: 3x + y = 15

x	0	5
y	15	0

corner points	Z=3x + 2y
A(5, 0)	3(5) + 2(0) = 15 + 0 = 15
B(4, 3)	3(4) + 2(3)= 12 + 6 = 18 maximum
C(5, 0)	3(5) + 2(0) = 15+0 = 15
O(0, 0)	0 + 0 = 0

Maximum value = 18 at ( 4,3)

Question (6)

Minimise Z = x + 2y
such that 2x+y ≥ 3, x + 2y ≥ 6, x,y≥0

Solution

Shaded region is feasible region of given equation
Line l₁: 2x + y = 3

x	0	1.5
y	3	0

Line l₂: x + 2y = 6

x	0	6
y	3	0

corner points	Z=x+2y
A(0, 3)	0 + 6 = 6
B(6, 0)	6 + 0= 6

Minimum Z = 6, at all points on line joining (3,0) and (0, 6)

Show that the minimum of Z occurs at more than two points

Question (7)

Minimise and maximise Z = 5x + 10 y
subject to x + 2y ≤ 120, x + y ≥ 60, x - 2y ≥ 0, x, y ≥0

Solution

Shaded region is feasible region of given equation
Line l₁ : x + 2y = 120

x	0	120
y	60	0

line l₂: x + y = 60

x	0	60
y	60	0

line l₃: x -2y = 0

x	0	20
y	0	10

corner points	Z= 5x+10y
A(40, 20)	5(40) + 10(20) =400
B(60, 30)	5(60) + 10(30) =600 max
C(120, 0)	5(120) + 10(0) =600 max
D(60, 0)	60(5) + 10(0) =300 min

Maximise Z = 120 at all points of line segment joining points (60, 30) and ( 120 , 0)
Minimium Z = 300 at (60, 0)

Question (8)

Minimise and maximise Z = x + 2y
subject to x + 2y ≥ 100, 2x - y ≤ 0, 2x + y ≤ 200; x,y ≥ 0

Solution

Shaded region is feasible region of given equation
line₁: x + 2y = 100

x	0	100
y	50	0

line₂: 2x - y = 0

x	0	10
y	0	20

line₃: 2x + y = 200

x	0	100
y	200	0

corner points	Z= x + 2y
A(0, 50)	0 +100=100 mini.
B(0, 200)	0 + 400 =400 max
C(50, 100)	50 + 200 = 250
D(20, 40)	20 + 80 = 100 min

MAx Z= 400 at ( 0, 200)
Min Z = 100 at all the points of line segment joining points (0, 50) and (20, 40)

Question (9)

Maximise Z = -x +2y, subject to the constraints

Solution

x ≥ 3, x + y ≥5, x + 2y ≥ 6, y≥ 0 Shaded region is feasible region of given equation
line l₁ : x = 3
line l₂ : x + y = 5

x	0	5
y	5	0

line l₃ : x + 2y = 6

x	0	6
y	3	0

corner points	Z= -x + 2y
A(3, 2)	-3 + 4 = 1 maxi.
B(4, 1)	-4 + 2 = -2
C(6, 0)	-6 + 0 = -6

Maxi Z = 1 at (3, 2)

Question (10)

Maximise Z = x + y, subject to x-y ≤ -1, -x + y ≤ 0. x,y ≥ 0

Solution

line l₁ : x - y= -1

x	0	-1
y	1	0

line l₂ : -x + y = 0

x	0	1
y	0	1

As lines do not intersect , do not have feasible solution
So minimum value