12th NCERT Linear Programming Exercise 12.1 Questions 10
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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
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⇒ Exercise 12.2