1

Additional Mathematics Chapter 21

© Penerbitan Pelangi Sdn. Bhd.

1. (a)

0

y

x < 3

x = 3

x

(b)

02

–4

y

y = 2x – 4

y � 2x – 4

x

(c)

05

5

y

y = –x + 5

y > –x + 5

x

2. (a) y < –2x + 3

(b) y , x + 2

3. (a)

0

y

x = 3

y = 1

y = x

x

(b)

05

1

y

y = –x + 5

y = x + 1

y = 2x

x

4. (a) y . 3, x , 2 and y < 2x

(b) y > 1, y , x and x + y < 6

(c) Equation of OA is y = 2x

Equation of OB is y = 1—2 x

Equation of CD is x + y = 6

Therefore, y < 2x, y > 1—2 x and x + y , 6.

(d) Equation of AB is y – 3 = 3 – 1–––––2 – 4

(x – 2)

y – 3 = –x + 2 y = –x + 5

Equation of BC is y = 1 2 – 1–––––0 – 4 2x + 2

y = – 1—4 x + 2

Therefore, y – x – 1 < 0, y < –x + 5 and

y > – 1—4 x + 2.

5. (a) y . x

(b) y . 2x

(c) y – x . 5

(d) y , x – 2

(e) y < 2x

(f) y > 3x

(g) y < 2x

(h) x + 2y < 10

CHAPTER

21 Linear Programming

2

Additional Mathematics SPM Chapter 21

© Penerbitan Pelangi Sdn. Bhd.

(i) 2x – y > 8

(j) x—y < 3—4 4x < 3y

(k) y . 3x

(l) y < 4x – 1

(m) x—y < 2—5 5x < 2y

(n) (2x + 3y) – (x + y) , 4 x + 2y , 4

6. (a) (i) Maximum x = 4

(ii) Minimum y = 2

(iii) 2 < y < 4

(iv) The maximum value of x + 2y = 4 + 2(4) = 12 The minimum value of x + 2y = 2 + 2(2) = 6

(b) (i) Maximum x = 3

(ii) Minimum y = 0

(iii) 1 < y < 5

(iv) The maximum value of x + 2y = 2 + 2(5) = 12 The minimum value of x + 2y = 0 + 2(0) = 0

(c) (i) Maximum x = 4 (ii) Minimum y = 2 (iii) None (iv) The maximum value of x + 2y = 3 + 2(4) = 11 The minimum value of x + 2y = 4 + 2(2) = 8

7. (a) x > 8, y < 2x, 0.5x + 0.4y < 20 5x + 4y < 200

0

20

30

40

50

10

xx + y = 10

x = 8 y = 2x

5x + 4y = 200

y

10

(16, 30)

3020 40

(b) Total number of fruits = x + y

(c) (i) Maximum number of fruits bought = 16 + 30 = 46

(ii) The range of number of oranges = {y : 0 < y < 26, y [ w}

(iii) The range of number of apples = {x : 8 < x < 40, x [ w}

(iv) Cost = 0.5x + 0.4y The total cost spent = 0.5(20) + 0.4(25) = 10 + 10 = RM20.00

1. (a) I: 30x + 30y < 540 3x + 3y < 54

II: 40x + 30y > 480 4x + 3y > 48

III: x—y > 1—2 2x > y

(b)

0

4

6

8

10

12

14

16

18

2

x2x + y = 2

y = 2x

4x + 3y = 48

3x + 3y = 54

y

R

2 64 8 10 12 14 16 18

(c) (i) Maximum number of cake A produced is 8

(ii) Profit=2x + y Minimumprofit=2(5)+10 = RM20.00

3

Additional Mathematics Chapter 21

© Penerbitan Pelangi Sdn. Bhd.

2. (a) x + y > 100, y < 2x, y + 3x < 240

(b)

0

40

60

80

100

20

x120x + 140y = 1680

y = 2x

x + y = 100

y + 3x = 240

y

R

20 6040 80 100

(c) (i) When y = 45, the maximum value of x = 65 and the minimum value of x = 55

(ii) Charges = 120x + 140y The maximum charges paid = 120(48) + 140(96) = RM19 200

3. (a) x + y < 80, x < 3y, x – y < 10

(b)

0

20

30

40

50

60

70

80

10

x

40x + 20y = 400

x = 3y

x + y = 80 x – y = 10

y

R

10 3020 40 50 60 70 80

(c) (i) If the number of students for subject P is 20, then the range for subject C is 10 < y < 60.

(ii) Fees = 40x + 20y The maximum total fees per month = 40(45) + 20(35) = 1800 + 700 = RM2500

4. (a) Let x and y represent the number of science and mathematics reference books respectively.

I: x + y 80 II: y 2x III: y – x 10(b)

y

010 20 30 40 50

x

R

x + y = 80

4x + 2y = 40

y = 2x

y – x = 10

10

20

40

60

70

80

90

30

50

60 70 80

(c) (i) The maximum number of mathematics books that can be bought is 53.

(ii) Cost = 4x + 2y Let 4x + 2y = 40 At point (10, 20), the total cost is minimum. Total minimum cost = 4(10) + 2(20) = 40 + 40 = RM80

4

Additional Mathematics SPM Chapter 21

© Penerbitan Pelangi Sdn. Bhd.

1. (a) x < 80, y < 60, x + y . 30(b)

0

20

40

30

50

60

10

x

20x + 10y = 200

y = 60

x = 80

5x + 2y = 100

x + y = 30

y

R

302010 5040 60 70 80

(c) (i) Cost for a piece of alloy = 20x + 10y The minimum cost of production = 20(1) + 10(30) = RM320 (ii) Profit=5x + 2y Maximumprofit=5(80)+2(60) = RM520

2. (a) x > 45, y . 40, x + y > 90

(b)

0

20

40

30

50

60

70

80

90

10

x

y = 40

x + y = 90

x = 45

y

302010 5040 60 70 80 90

(c) (i) 50 ways (ii) 10 ways

3. (a) x + y < 200,

x—y < 1—8 , that is, 8x < y

100x + 500y > 10 000, that is, x + 5y > 100

(b)

0

100

200

150

50

x

y = 8x

x + y = 200

100x + 50y = 2500

x + 5y = 100

y

R

15010050 200

(c) (i) 160 < y < 180

(ii) Profit=100x + 50y Themaximumprofit=100(22)+50(178) = RM11 100

1. (a)

0

B(0, 1)

A(0, 2)

D(2, 8)

C(4, 1)

y

R

x

Equation of AD is

y = 1 8 – 2–––––2 – 0 2x + 2

y = 3x + 2

5

Additional Mathematics Chapter 21

© Penerbitan Pelangi Sdn. Bhd.

Equation of CD is

y – 1 = 8 – 1–––––2 – 4

(x – 4)

y – 1 = 7––––2

(x – 4)

–2y + 2 = 7x – 28 –2y = 7x – 30 2y = –7x + 30

Therefore, the three inequalities are y . 1, y < 3x + 2, 2y + 7x < 30.

(b) {(0, 2), (1, 5), (2, 8)}

2. Equation of line DEFG is

y – 1 = 2 – 1–––––4 – 3

(x – 3)

y – 1 = x – 3 y = x – 2

Equation of line ABCD is

y = 1 4 – 3–––––0 – 1 2x + 4

y = –x + 4

Therefore, the three inequalities are y < 4, y > x – 2, y > –x + 4.

Maximum value = 4(x + y) = 4(6 + 4) = 40

3. (a) x + y < 60, x > 10, x < 2y(b)

0

30

40

50

60

20

10

x

R

y

x = 10

10

20x

+ 1

0y =

200

x + y =

60

x = 2

y

3020 40 50 60

(c) Transportation collection = 20x + 10y Maximum transportation collection = 20(40) + 10(20) = 800 + 200 = RM1000

4. (a) x + y < 40, 50x + 100y > 3000, x—y < 1—2

x + 2y > 60 2x < y (b)

0

30

40

20

10

x

R

y

10

2x = y

20x + 80y = 640

x + y =

40

x + 2y = 60

3020 40 50 60

(c) (i) From the graph, the minimum number of adults = 24 people

(ii) Profit=20x + 80y Minimumprofit=20(12)+80(24) = 240 + 1920 = RM2160

5. (a) x < 40, x + y > 50, y < 2x, 50x + 100y < 5000

x + 2y < 100(b)

0

30

40

50

60

70

80

20

10

x

R

y

x = 40

y = 2x

x + y = 50

10

x + 2y = 100

3020 40 50 60 70 80 90 100

(c) (i) From the graph, when x = 30, 20 < y < 35

(ii) From the graph, if x = y, 25 < x < 33 and 25 < y < 33

(iii) From the graph, the maximum number of commodity B is 40.