Odisha State Board CHSE Odisha Class 12 Math Solutions Chapter 3 Linear Programming Ex 3(b) Textbook Exercise Questions and Answers.

## CHSE Odisha Class 12 Math Solutions Chapter 3 Linear Programming Exercise 3(b)

Question 1.

Maximize Z = 5x_{1}+ 6x_{2}

Subject to: 2x_{1} + 3x_{2} ≤ 6

x_{1}, x_{2} ≥ 0

Solution:

Step – 1 Treating the constraint as equation, we get 2x_{1} + 3x_{2} = 6

Step – 2 Let us draw the graph

x_{1} |
3 | 0 |

x_{2} |
0 | 0 |

Step – 3 Clearly (0,0) statisfies 2x_{1} + 3x_{2} ≤ 6

The shaded region is the feasible region with vertices 0(0,0), A(3,0), B(0,2).

Step – 4

Corner point | Z = 5x_{1}+ 6x_{2} |

0(0.0) | 0 |

A(3,0) | 15 → maximum |

B(0,2) | 12 |

Z is maximum at A (3,0)

∴ The solution of LPP is x_{1} = 3, x_{2} = 0

Z_{max} = 15

Question 2.

Minimize: Z = 6x_{1} + 7x_{2}

Subject to: x_{1} + 2x_{2} ≥ 4

x_{1}, x_{2} ≥ 0

Solution:

Step – 1 Treating the constraint as equation we get x_{1} + 2x_{2} = 0

Step – 2 Let us draw the graph of x_{1} + 2x_{2} = 4

x_{1} |
0 | 4 |

x_{2} |
2 | 0 |

Step – 3 Clearly 0(0,0) does not satisfy

x_{1} + 2x_{2} > 4, x_{1 }> 0, x_{2 }> 0 is the first quadrant.

The feasible region is the shaded region with vertices A(4, 0), B(0, 2).

Step – 4 Z (4, 0) = 24

Z (0, 2) = 14 → minimum

Step – 5 As the feasible region is unbounded we cannot immediately decide Z is minimum at B (0, 2).

Let us draw the half-plane 6x_{1} + 7x_{2} < 14

x_{1} |
0 | 3.5 |

x_{2} |
2 | -1 |

As this half-plane has no point common with the feasible region, we have Z is minimum for x_{1}= 0, x_{2} = 2 and the minimum value of Z = 14.

Question 3.

Maximize Z = 20x_{1}+ 40x_{2}

Subject to: x_{1} + x_{2} ≤ 1

6x_{1} + 2x_{2} ≤ 3

x_{1}, x_{2} ≥ 0.

Solution:

Step – 1 Treating the constraints as equations

x_{1} + x_{2} = 1 …. (1)

6x_{1} + 2x_{2} = 3 …. (2)

x_{1}, x_{2} ≥ 0

Step – 2 Let us draw the graph:

Table – 1

x_{1} |
0 | 1 |

x_{2} |
1 | 0 |

Table – 2

x_{1} |
0 | 0.5 |

x_{2} |
1.5 | 0 |

Step – 3 As (0, 0) satisfies both the inequations the shaded region is the feasible region.

Step – 4 Solving

x_{1} + x_{2} = 1

6x_{1} + 2x_{2} = 3

we have x_{1} = ¼ x_{2} = ¾

The vertices are O(0, 0), A(0.5, 0), B(0,1) and C(¼, ¾)

Now Z(O) = 0

Z(A) = 10

Z(B) = 40

Z(C) = 20 × ¼ + 40 × ¾ = 35

∴ Z attains maximum at B for x_{1}= 0, x_{2} = 1

Z_{max} = 40

Question 4.

Minimize: Z = 30x_{1} + 45x_{2}

Subject to: 2x_{1} + 6x_{2} ≥ 4

5x_{1} + 2x_{2} ≥ 5

x_{1}, x_{2} ≥ 0

Solution:

Step – 1 Consider the constraints as equations

2x_{1} + 6x_{2} = 4

5x_{1} + 2x_{2} = 5

Step – 2

Table – 1

x_{1} |
2 | -1 |

x_{2} |
0 | 1 |

Table – 2

x_{1} |
1 | 0 |

x_{2} |
0 | 2.5 |

Step – 3 Clearly 0(0,0) does not satisfy 2x_{1} + 6x_{2} ≥ 4 and 5x_{1} + 2x_{2} ≥ 5.

Thus the shaded region is the feasible region.

Solving the equations we get

x_{1} = \(\frac{11}{13}\), x_{2} = \(\frac{5}{13}\).

∴ The vertices are A(2, 0)

B(\(\frac{11}{13}\), \(\frac{5}{13}\)) and C(0, \(\frac{5}{2}\)).

Step – 4 Z(A) = 60

Z(B) = \(\frac{555}{13}\) → minimum

Z(C) = \(\frac{225}{2}\)

Step – 5 As the feasible region is unbounded we cannot immediately decide Z is minimum at B(\(\frac{11}{13}\), \(\frac{5}{13}\))

Let us draw the half plane

30x_{1} + 45x_{2} < \(\frac{555}{13}\)

x_{1} |
\(\frac{11}{13}\) | 0 |

x_{2} |
\(\frac{5}{13}\) | \(\frac{27}{39}\) |

As this half plane and the feasible region has no point in common we have Z is minimum for x_{1} = \(\frac{11}{13}\), x_{2} = \(\frac{5}{13}\), and Z_{min} = \(\frac{555}{13}\)

Question 5.

Maximize: Z = 3x_{1}+ 2x_{2}

Subject to: -2x_{1} + x_{2} ≤ 1

x_{1} ≤ 2

x_{1}+ x_{2} ≤ 3

x_{1}, x_{2} ≥ 0

Solution:

Step – 1 Treating the constraints as equations

-2x_{1} + x_{2} = 1 …..(1)

x_{1} = 2 …..(2)

x_{1}+ x_{2} = 3 …..(3)

Step – 2 Let us draw the lines.

Table – 1

x_{1} |
0 | -1 |

x_{2} |
1 | -1 |

Table – 2

x_{1} |
2 | 2 |

x_{2} |
0 | 1 |

Table – 3

x_{1} |
0 | 3 |

x_{2} |
3 | 0 |

Step – 3 (0, 0) satisfies all the constraints and x_{1}, x_{2} > 0 is the 1st quadrant the shaded region is the feasible region.

Step – 4 Solving -2x_{1} + x_{2} = 1

x_{1}+ x_{2} = 3

we have 3x_{1} = 2

⇒ x_{1} = \(\frac{2}{3}\), x_{2} = 3 – \(\frac{2}{3}\) = \(\frac{7}{3}\)

From x_{1}+ x_{2} = 3 and x_{1} = 2 we have x_{1} = 2, x_{2} = 1

∴ The vertices are 0(0, 0), A(2, 0), B(2, 1), C(\(\frac{2}{3}\), \(\frac{7}{3}\)), D(0, 1)

Z(0) = 0, Z(A) = 6, Z(B) = 8, Z(C) = 3.\(\frac{2}{3}\) + 2.\(\frac{7}{3}\) = \(\frac{20}{3}\), Z(D) = 2

Z is maximum at B.

∴ The solution of given LPP is x_{1} = 2, x_{2} = 1, Z_{(max)} = 8.

Question 6.

Maximize: Z = 50x_{1}+ 60x_{2}

Subject to: x_{1} + x_{2} ≤ 5

x_{1}+ 2x_{2} ≤ 4

x_{1}, x_{2} ≥ 0

Solution:

Step – 1 Treating the constraints as equations we get

x_{1} + x_{2} = 5 ….(1)

x_{1}+ 2x_{2} = 4 ….(2)

Step – 2 Let us draw the graph

Table – 1

x_{1} |
5 | 5 |

x_{2} |
0 | 0 |

Table – 2

x_{1} |
4 | 0 |

x_{2} |
0 | 2 |

Step – 3 0(0,0) satisfies x_{1} + x_{2} ≤ 5 and does not satisfy x_{1}+ 2x_{2} ≤ 4

Thus the shaded region is the feasible region.

Step – 4 The corner points are A(4,0), B(5,0), C(0,5) , D(0,2)

∴

Corner point | z = 50x_{1}+ 60x_{2} |

A(4,0) | 200 |

B (5,0) | 250 → maximum |

C(0,5) | 300 |

D(0,2) | 120 |

Z is maximum for x_{1} = 0, x_{2} = 5, Z_{(max)} = 300.

Question 7.

Maximize: Z = 5x_{1}+ 7x_{2}

Subject to: x_{1} + x_{2} ≤ 4

5x_{1}+ 8x_{2} ≤ 30

10x_{1}+ 7x_{2} ≤ 35

x_{1}, x_{2} ≥ 0

Solution:

Step – 1 Treating the constraints as equations we get,

x_{1} + x_{2} = 4 …. (1)

5x_{1}+ 8x_{2} = 30 …. (2)

10x_{1}+ 7x_{2} = 35 …. (3)

Step – 2 Let us draw the graph

Table – 1

x_{1} |
4 | 0 |

x_{2} |
0 | 4 |

Table – 2

x_{1} |
6 | 2 |

x_{2} |
0 | 2.5 |

Table – 3

x_{1} |
0 | 3.5 |

x_{2} |
5 | 0 |

Step – 3 0(0,0) satisfies all the constraints.

Thus the shaded region is the feasible region.

From (1) and (2) we get (\(\frac{2}{3}\), \(\frac{10}{3}\))

From (1) and (3) we get

x_{1} = \(\frac{7}{3}\), x_{1} = \(\frac{5}{3}\)

∴ The corner points are 0(0,0), A(\(\frac{7}{2}\), 0), B(\(\frac{7}{3}\), \(\frac{5}{3}\)), C(\(\frac{2}{3}\), \(\frac{10}{3}\)), D(0, \(\frac{15}{4}\))

Step – 4

Corner point | z = 5x_{1}+ 7x_{2} |

0(0,0) | 0 |

A(\(\frac{7}{2}\), 0) | \(\frac{35}{2}\) |

B(\(\frac{7}{3}\), \(\frac{5}{3}\)) | \(\frac{70}{3}\) |

C(\(\frac{2}{3}\), \(\frac{10}{3}\)) | \(\frac{80}{3}\) |

D(0, \(\frac{15}{4}\)) | \(\frac{105}{4}\) |

Z attains its maximum value \(\frac{80}{3}\) for x_{1} = \(\frac{2}{3}\) and x_{2} = \(\frac{10}{3}\).

Question 8.

Maximize: Z = 14x_{1} – 4x_{2}

Subject to: x_{1} + 12x_{2} ≤ 65

7x_{1 }– 2x_{2} ≤ 25

2x_{1}+ 3x_{2} ≤ 10

x_{1}, x_{2} ≥ 0

Also find two other points which maximize Z.

Solution:

Step – 1 Treating the constraints as equations we get

x_{1} + 12x_{2} = 65 …. (1)

7x_{1 }– 2x_{2} = 25 …. (2)

2x_{1 }+ 3x_{2} = 10 …. (3)

Step – 2 Let us draw the graph

Table – 1

x_{1} |
65 | 5 |

x_{2} |
0 | 5 |

Table – 2

x_{1} |
5 | 10 |

x_{2} |
5 | 22.5 |

Table – 3

x_{1} |
5 | 2 |

x_{2} |
0 | 2 |

Step – 3 Clearly 0(0,0) satisfies x_{1} + 12x_{2} ≤ 65 and 7x_{1 }– 2x_{2} ≤ 25 but does not satisfy 2x_{1}+ 3x_{2} ≤ 10. Thus shaded region is the feasible region.

Equation (1) and (2) meet at (5, 5).

From (2) and (3)

∴ The corner points of the feasible region are A(0, \(\frac{10}{3}\)), B(\(\frac{19}{5}\), \(\frac{4}{5}\)), C(5, 5), D(0, \(\frac{65}{12}\)).

Step – 4

Corner point | z = 14x_{1} – 4x_{2} |

A(0, \(\frac{10}{3}\)) | \(\frac{-40}{3}\) |

B(\(\frac{19}{5}\), \(\frac{4}{5}\)) | 50 → maximum |

C(5, 5) | 50 → maximum |

D(0, \(\frac{65}{12}\)) | \(\frac{65}{3}\) |

Z is maximum for x_{1} = \(\frac{19}{5}\), x_{2} = \(\frac{4}{5}\) or x_{1} = 5, x_{2} = 5 and Z_{max} = 50

There is no other point that maximizes Z.

Question 9.

Maximize: Z = 10x_{1} + 12x_{2} + 8x_{3}

Subject to: x_{1} + 2x_{2} ≤ 30

5x_{1 }– 7x_{3} ≤ 12

x_{1} + x_{2} + x_{3} = 20

x_{1}, x_{2} ≥ 0

[Hints: Eliminate x_{3} from all expressions using the given equation in the set of constraints, so that it becomes an LPP in two variables]

Solution:

Eliminating x_{3} this LPP can be written as Maximize Z = 2x_{1} + 4x_{2} + 160

Subject to: x_{1} + 2x_{2} ≤ 30

5x_{1 }– 7x_{3} ≤ 12

x_{1}, x_{2} ≥ 0

Step – 1 Treating the consraints as equations we get

x_{1} + 2x_{2} = 30 …..(1)

5x_{1 }– 7x_{3} = 12 …..(2)

Step – 2 Let us draw the graph

Table – 1

x_{1} |
30 | 0 |

x_{2} |
0 | 15 |

Table – 2

x_{1} |
8 | 1 |

x_{2} |
8 | 20 |

Step – 3 Clearly 0(0,0) satisfies x_{1} + 2x_{2} ≤ 30 and does not satisfy 12x_{1} + 7x_{2} ≤ 152

∴ The shaded region is the feasible region.

Step – 4

Z is maximum for x_{1} = 30, x_{2} = 0 and Z_{max} = 220

Question 10.

Maximize: Z = 20x_{1} + 10x_{2}

Subject to: x_{1} + 2x_{2} ≤ 40

3x_{1 }+ x_{2} ≥ 30

4x_{1}+ 3x_{2} ≥ 60

x_{1}, x_{2} ≥ 0

Solution:

Step – 1 Treating the constraints as equalities we have:

x_{1} + 2x_{2} = 40 ….(1)

3x_{1 }+ x_{2} = 30 ….(2)

4x_{1}+ 3x_{2} = 60 ….(3)

Step – 2 Let us draw the graph

Step – 3 (0, 0) satisfies x_{1} + 2x_{2} ≤ 40 and does not satisfy 3x_{1 }+ x_{2} ≥ 30 and 4x_{1}+ 3x_{2} ≥ 60, x_{1}, x_{2} ≥ 0 is the first quadrant.

∴ The shaded region is the feasible region.

Step – 4 x_{1} + 2x_{2} = 40 and 3x_{1 }+ x_{2} = 30

∴ The vetices are A(15, 0), B(10, 0), C(4, 18) and D(6, 12)

Z(A) = 300, Z(B) = 800

Z (C) = 20 x 4 + 10 x 18 = 260

Z (D) = 120 + 120 = 240

Z attains minimum at D(6 ,12).

∴ The required solution x_{1 }= 6, x_{2} =12 and Z_{min} = 240

Question 11.

Maximize: Z = 4x_{1} + 3x_{2}

Subject to: x_{1} + x_{2} ≤ 50

x_{1 }+ 2x_{2} ≥ 80

2x_{1}+ x_{2} ≥ 20

x_{1}, x_{2} ≥ 0

Solution:

Step – 1 Treating the constraints as equations

x_{1} + x_{2} ≤ 50 ….(1)

x_{1 }+ 2x_{2} ≥ 80 ….(2)

2x_{1}+ x_{2} ≥ 20 ….(3)

Step – 2 Let us draw the graph

Step – 3 (0, 0) satisfies x_{1} + x_{2} < 50, x_{1} + 2x_{2} < 80 but does not satisfy

2x_{1} + x_{2} > 20, x_{1} > 0, x_{2} > 0 is the 1st quadrant.

Hence the shaded region is the feasible region.

Step – 4 x_{1} + x_{2} = 50

x_{1} + 2x_{2} = 80

=> x_{2} = 30, x_{1} = 20

The vertices of feasible region are

A(10, 0), B(50, 0), C(20, 30), D (0, 40) and E (0, 20)

Point | Z = 4x_{1} + 3x_{2} |

A(10,0) | 40 |

5(50,0) | 200 |

C(20,30) | 170 |

D(0,40) | 120 |

E(0,120) | 60 |

Question 12.

Optimize: Z = 5x_{1} + 25x_{2}

Subject to: -0.5x_{1} + x_{2} ≤ 2

x_{1 }+ x_{2} ≥ 2

-x_{1}+ 5x_{2} ≥ 5

x_{1}, x_{2} ≥ 0

Solution:

Step – 1 Treating the constraints as equations

-0.5x_{1} + x_{2} = 2 ….(1)

x_{1 }+ x_{2} = 2 ….(2)

-x_{1}+ 5x_{2} = 5 ….(3)

Step – 2 Let us draw the graph.

Step – 3 (0, 0) satisfies -0.5x_{1} + x_{2} ≤ 2, but does not satisfy x_{1 }+ x_{2} ≥ 2 and -x_{1}+ 5x_{2} ≥ 5, x_{1} > 0, x_{2} > 0 is the 1st quadrant.

The shaded region is the feasible region with vertices A(\(\frac{5}{6}\), \(\frac{7}{6}\)) and B(0, 2).

Step – 4 Z can be made arbitrarily large.

∴ Problem has no maximum.

But Z(A) = \(\frac{100}{3}\), Z(B) = 50

Z is minimum at A(\(\frac{5}{6}\), \(\frac{7}{6}\)).

But the feasible region is unbounded.

Hence we cannot immediately decide, Z is minimum at A.

Let us draw the half plane

5x_{1} + 25x_{2} < \(\frac{100}{3}\)

⇒ 3x_{1 }+ 15x_{2} < 20

As there is no point common to this half plane and the feasible region.

we have Z is minimum for x_{1} = \(\frac{5}{6}\), x_{2} = \(\frac{7}{6}\) and the minimum value = \(\frac{100}{3}\)

Question 13.

Optimize: Z = 5x_{1} + 2x_{2}

Subject to: -0.5x_{1} + x_{2} ≤ 2

x_{1 }+ x_{2} ≥ 2

-x_{1}+ 5x_{2} ≥ 5

x_{1}, x_{2} ≥ 0

Solution:

Step – 1 Treating the constraints as equations

-0.5x_{1} + x_{2} = 2 ….(1)

x_{1 }+ x_{2} = 2 ….(2)

-x_{1}+ 5x_{2} = 5 ….(3)

Step – 2 Let us draw the graph.

Step – 3 The shaded regian is feasible region which is unbounded, thus Z does not have any maximum.

As Z can be made arbitrarily large, the given LPP has no maximum.

Z is minimum at B (0, 2). But we cannot immediately decide, Z is minimum at B.

Let us draw the half plane 5x_{1} + 2x_{2} < 4

x_{1} |
0 | 4/5 |

x_{2} |
2 | 0 |

As there is no point common to this half plane and the feasible region,

we have Z is minimum for x_{1} = 0, x_{2} = 2 and the minimum value of Z = 4.

Question 14.

Optimize: Z = -10x_{1} + 2x_{2}

Subject to: -x_{1} + x_{2} ≥ -1

x_{1 }+ x_{2} ≤ 6

x_{2} ≤ 5

x_{1}, x_{2} ≥ 0

Solution:

Step – 1 Treating the constraints as equations

-x_{1} + x_{2} = -1 ….(1)

x_{1 }+ x_{2} = 6 ….(2)

x_{2} = 5 ….(3)

Step – 2 Let us draw the graph

Table – 1

x_{1} |
1 | 0 |

x_{2} |
0 | -1 |

Table – 2

x_{1} |
6 | 0 |

x_{2} |
0 | 1 |

Step – 3 Clearly 0(0,0) satisfies all the constraints.

Thus the shaded region is the feasible region.

The vertices are 0(0,0) , A(1,0), B(\(\frac{7}{2}\), \(\frac{5}{2}\)) ,C(1, 5) and D (0, 5)

Step – 4 Z(O) = 0

Z(A) = -10

Z(B) = – 30

Z(C) = 0

Z(D) = 10

∴ Z is maximum for x_{1}= 0, x, = 5 and Z_{(max)} = 10

Z is minimum for x_{1} = \(\frac{7}{2}\) x_{2} = \(\frac{5}{2}\) and Z_{(min)} = -30

Question 15.

Solve the L.P.P.s obtained in Exercise 3(a) Q.1 to Q. 9 by graphical method.

(1) Maximise: Z = 1500x + 2000y

Subject to: x + y < 20

x + 2y < 24

x, y ≥ 0

Solution:

Step – 1 Treating the constraints as equations we get

x + y = 20

x + 2y = 24

Step – 2 Let us draw of graph.

Step – 3 Clearly 0(0,0) satisfies all the constraints.

Thus the shaded region is the feasible region.

From (1) and (2) we get

y = 14

x = 16

With vertices 0(0, 0), A(20, 0), B(16, 4), C(0, 12).

Step – 4 Z(0) = 0

Z(A) = 30,000

Z(B) = 32,000 → Maximum

Z(C) = 24000

Z is maximum for x = 16, y = 4 with Z = 32000

To get maximum profit he must keep 16 sets of model X and 4 sets of model Y.

Maximum profit = 1500 × 16 + 2000 × 4 = ₹32,000

(2) Maximize: 15x + 10y

Subject: x + 3y ≤ 600

2x + y ≤ 480

x, y ≥ 0

Solution:

Step – 1 Treating the constraints as equations we get

2x +3y = 600

2a + y = 480

Step – 2 Let us draw the graph

Step – 3 Clearly 0(0,0) satisfies all the constraints.

The corner point are 0(0, 0), A (240, 0) B(210, 60),C(0, 200)

Step – 4 Z(0) = 6

Z(A) = 3600

Z(B) = 3150 + 600

= 3750 → maximum

Z(C) = 2000

Thus Z is maximum for x = 210 and y = 60

and Z_{(max)} = 3750

(3) Maximize: Z = 20x + 30y

Subject to: x + 2y ≤ 10

x + y ≤ 6

x ≤ 4

x, y ≥ 0.

Solution:

Step – 1 Treating the constraints as equations we get

x + 2y = 10 …(1)

x + y = 6 …(2)

x = 4

Step – 2 Let us draw the graph

Step – 3 As 0(0,0) satisfies all the constraints the shaded region is the feasible region.

Solving (1) and (2) we get x = 2, y = 4.

The vertices and 0(0, 0) , A(4, 0), B(4, 2), C(2, 4), D (0, 5).

Step – 4 Z(0) =0

Z(A) = 80

Z (B) =140

Z(C) = 1 60 → maximum

Z (D) = 150

∴ Z is Maximum when x = 2, y = 4 and Z_{(max)} = 160

(4) Maximize: Z = 15x + 17y

Subject to: 4x + 7y ≤ 150

x + y ≤ 30

15x + 17y > 300

x, y ≥ 0

Solution:

Step – 1 Treating the constraints as equations we get

4x + 7y = 150 ….(1)

x + y = 30 ….(2)

15x + 17y = 300 ….(3)

Step – 2 Let us draw the graph

Step – 3 Clearly 0(0,0) satisfies all the constraints.

4x + 7y ≤ 150, x + y ≤ 30, but does not satisfy 15x + 17y ≥ 300.

∴ The shaded region is the feasible region.

From (1) and (2) we get

∴ Z is maximum for x = 20. y = 10 and Z_{(max)} = 470

(5) Maximize: Z = 2x + 4y

Subject to: 3x + 2y ≤ 10

2x + 5y ≤ 15

5x + 6y ≤ 21

x, y ≥ 0

Solution:

Step – 1 Treating the constraints as equations we get

3x + 2y = 10 …(1)

2x + 5y = 15 …(2)

5x + 6y = 21 …(3)

Step – 2 Let us draw the graph

Step – 3 As 0(0,0) satisfies all the constraints the shaded region is the feasible region.

From (1) and (3) we get

From (2) and (3) we get

Step-4 Z(O) = 0

(6) Maximize: Z = 1000x + 800y

Subject to: x + y ≤ 5

2x + y ≤ 9

x, y ≥ 0

Solution:

Step – 1 Treating the constraints as equations we get

x + y = 5 ….(1)

2x + y = 9 ….(2)

Step – 2 Let us draw the graph

Step – 3 Clearly 0(0,0) satisfies all the constraints.

∴ Thus the shaded region is the feasible region.

From (1) and (2) we get x = 4, y = 1.

∴ The vertices are A(0, 0), A(4.5, 0), B(4, 1) and C(0, 5).

Step – 4 Z(0) =0

Z (A) = 4500

Z (B) = 4800 → Maximum

Z (C) = 4000

Z is maximum for x = 4 and y = 1, Z_{(max)} = 4800

(7) Minimize: Z = 4960 – 70x – 130y

Subject to: x + y ≤ 12

x + y ≥ 6

x ≤ 8

y ≤ 8

x, y ≥ 0

Solution:

Step – 1 Treating the constraints as equations we get

x + y = 12 ….(1)

x + y = 6 ….(2)

x = 8 ….(3)

y = 4 ….(4)

Step – 2 Let us draw the graph

Step – 3 Clearly 0(0,0) satisfies all the constraints except x + y > 6.

The shaded region is the feasible region.

The vertices are A(6, 0), B(8, 0), C(8, 4), D(4, 8), E(0, 8) and F(0, 6).

Step – 4 Z (A) = 4540

Z (B) = 4400

Z (C) = 3880

Z (D) = 3640 → Minimum

Z (E) = 3920

Z (F) = 4180

∴ Z is maximum for x = 4 and y = 8 and Z_{(min)} = 3640.

(8) Minimize: Z = 16x + 20y

Subject to x + 2y ≥ 10

x + y ≥ 6

3x + y ≥ 8

x, y ≥ 0

Solution:

Step – 1 Treating the constraints as equations we get

x + 2y = 10 ….(1)

x + y = 6 …(2)

3x + y = 8 …(3)

Step – 2 Let us draw the graph

Step – 3 Clearly 0(0,0) satisfies all the constraints. Thus the shaded region is the feasible region.

From (1) and (2) we get y = 4, x = 2.

From (2) and (3) we get x = 1, y = 5.

The vertices are A(10, 0), B(2, 4), C(1, 5), D(0, 8).

Step – 4 Z (A) = 160

Z (B) = 112 → Minimum

Z (C) =116

Z (D) = 160

As the region is unbounded, let us draw the half plane Z < Z_{(min)}

⇒ 16x + 20y < 112

⇒ 4x + 5y < 28

x_{1} |
7 | 0 |

x_{2} |
0 | 5.6 |

There is no point common to the shaded region and the half plane 4x + 5y ≤ 28 other than B(2, 4).

∴ Z is minimum for x = 2, y = 4 and Z_{(min)} = 112.

(9) Minimize: Z = (512.5)x + 800y

Subject to: 5x + 4y = 40

x ≤ 7

x ≤ 3

x, y ≥ 0

Solution:

Step – 1 Let us draw the graph of

5x + 4y = 40

x = 7, y = 3

x_{1} |
8 | 0 |

x_{2} |
0 | 10 |

Step – 1 Let us draw the graph of

5x + 4y = 40

x = 7, y = 3

Step – 2 The line segment AB is the feasible region.

Step – 3 Z (A) = 3587.5 + 1000 = 4587.5

Z (B) = 2870 + 2400 = 5270

Clearly Z is minimum for

x = 7, y = \(\frac{5}{4}\) and Z_{(min)} = 4587.5