5ach13(linear Inequalities In Two Unknowns)

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Linear Inequalities in Two Unknowns

Linear Inequalities in Two Unknowns

• • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • •

Activity Activity 13.1 (p. 104) 1.

x+y=1+4=5 x+y=4+2=6 x + y = 2 + (–2) = 0

2.

(a) P, Q (b) R

3.

(a) (1, 5), (2, 3), (2, 4), (3, 2), (3, 3). (or any other reasonable answers)

4. Maximum value of P = 3

Follow-up Exercise p. 103

(b) Yes

1.

(a) Draw the graphical representations of the two inequalities on the same number line, we have

(c) All the points that lie above the straight line x + y = 4 satisfy the inequality x + y > 4. 4.

Yes

Activity 13.2 (p. 120)

Therefore, the solutions of the compound inequality are:

1.

(b) Draw the graphical representations of the two inequalities on the same number line, we have

(c) Draw the graphical representations of two inequalities on the number line, we have 2.

(a) Yes. They have the same slope. (b) Yes

3.

It shifts to the right-hand side.

There are no solutions.

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Certificate Mathematics in Action Full Solutions 5A

2.

x + ( x −1) > 4 2 3x >5 2 …… (2) 10 x> 3

(a) Solving 5x + 2 ≥ 7, we have

5 x +2 ≥ 7

5x ≥5 x ≥1 …… (1) Solving 9 – 2x ≤ 1, we have

9 − 2 x ≤1

∵ ∴

∵ ∴

2 x ≥8 x ≥ 4 …… (2)

10 7 <x< . 3 2

x must satisfy both (1) and (2). The solutions of the compound inequality are x ≥ 4. Graphical representation:

(b) Solving 2x – 6 > 5x – 1, we have

2 x −6 > 5 x −1 3 x < −5 5 …… (1) x <− 3

Solving 5 – 3x < 15 – 5x, we have

5 −3 x <15 −5 x 2 x <10 x <5

…… (2) ∵ x must satisfy both (1) and (2). ∴ The solutions of the compound inequality are

x <−

5 . 3

Graphical representation:

(c) Solving x – 3(x – 3) > 2, we have

x −3( x −3) > 2 2x <7 7 …… (1) x< 2

Solving

x + ( x −1) > 4 , we have 2

x must satisfy both (1) and (2). The solutions of the compound inequality are

Graphical representation:

3.

(a) Rewrite the compound inequality as −1 ≤

2 −4x 3

2 − 4x <2 . 3 2 −4x Solving −1 ≤ , we have 3 2 −4 x −1 ≤ 3 −3 ≤ 2 − 4 x 4x ≤5 …… (1) 5 x≤ 4 2 −4x < 2 , we have Solving 3 2 − 4x <2 3 2 − 4x < 6 …… (2) 4 x > −4 x > −1 and

∵ ∴

x must satisfy both (1) and (2). The solutions of the compound inequality are

−1 < x ≤

5 . 4

Graphical representation:

(b) Rewrite the compound inequality as

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1 −3x 1 − 3x and <7 . 4 4 1 −3x Solving 2 < , we have 4 1 − 3x 2< 4 8 <1 − 3x 3 x < −7 …… (1) 7 x <− 3 1 −3x Solving < 7 , we have 4 2<

1 − 3x <7 4 1 − 3 x < 28 3 x > −27 …… (2) x > −9

∵ ∴

∴ 2.

3.

Evaluate the value of –2x + y + 1 for the point (0, 0). ∵ –2(0) + 0 + 1 = 1 < / 0 ∴ The solution is the lower half-plane.

Evaluate the value of x – y for the point (0, 0). ∵ 0–0=0<2 ∴ The solution is the upper half-plane. Alternative Solution x–y<2 y>x–2 ∴ The solution is the upper half-plane.

4.

7 . 3

Evaluate the value of 2x + y + 2 for the point (0, 0). ∵ 2(0) + 0 + 2 = 2 ≤ / 0 ∴ The solution is the lower half-plane. Alternative Solution 2x + y + 2 ≤ 0 y ≤ –2x – 2 ∴ The solution is the lower half-plane.

Graphical representation:

(c) Rewrite the compound inequality as 1 – 2x ≥ 3x – 4 and 3x – 4 ≥ x + 3. Solving 1 – 2x ≥ 3x – 4, we have

y ≤ –3x – 3 The solution is the lower half-plane.

Alternative Solution –2x + y + 1 < 0 y < 2x – 1 ∴ The solution is the lower half-plane.

x must satisfy both (1) and (2). The solutions of the compound inequality are

−9 < x < −

Linear Inequalities in Two Unknowns

5.

Evaluate the value of 2x + 3y for the point (0, –1). ∵ 2(0) + 3(–1) = –3 < –1 ∴ The inequality is 2x + 3y ≤ –1.

6.

Evaluate the value of 3x – 2y for the point (0, 0). ∵ 3(0) – 2(0) = 0 < 2 ∴ The inequality is 3x – 2y < 2.

1 −2 x ≥ 3x −4 5x ≤5 x ≤1

…… (1) Solving 3x – 4 ≥ x + 3, we have

3x − 4 ≥ x +3 2x ≥7 7 x≥ 2

…… (2)

∵ x must satisfy both (1) and (2). ∴ There are no solutions.

7.

p. 109 1.

Evaluate the value of 3x + y for the point (0, 0). ∵ 3(0) + 0 = 0 ≤ / –3 ∴ The solution is the lower half-plane. Alternative Solution 3x + y ≤ –3 51

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8.

2. 3.

9.

Evaluate the values of 3x – y and x – 2y for the point (–1, – 2). ∵ 3x – y = 3(–1) – (–2) = –1 < 2 ∴ One inequality is 3x – y ≤ 2. ∵ x – 2y = (–1) – 2(–2) = 3 > 1 ∴ One inequality is x – 2y ≥ 1.

∴ The system of inequalities is

4. 10.

p. 114

 3x − y ≤ 2 .   x − 2y ≥ 1

Evaluate the values of x, x + y and x + 3y for the point (1.5, 2). ∵ x = 1.5 > 1 ∴ One inequality is x ≥ 1. ∵ x + y = 1.5 + 2 = 3.5 < 4 ∴ One inequality is x + y < 4. ∵ x + 3y = 1.5 + 3(2) = 7.5 > 4 ∴ One inequality is x + 3y ≥ 4.

x≥ 1  ∴ The system of inequalities is x + y < 4 .   x + 3y ≥ 4 

1. 5.

Evaluate the values of y, 1 – x and 2x – y for the point (2, 0). ∵ y=0<1 ∴ One inequality is y ≤ 1. ∵ 1 – x = 1 – 2 = –1 < 0 ∴ One inequality is y > 1 – x. ∵ 2x – y = 2(2) – (0) = 4 < 5 ∴ One inequality is 2x – y < 5.

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∴

6.

∵ ∴ ∵ ∴

y≤1  The system of inequalities is y > 1 − x .   2x − y < 5 

∵ ∴

Evaluate the values of x – y and x + 2y for the point (0, 1). ∵ x – y = 0 – 1 = –1 < 0 ∴ One inequality is x – y ≤ 0. ∵ x + 2y = 0 + 2(1) = 2 < 4 ∴ One inequality is x + 2y ≤ 4.

∴

∴

 x− y ≤ 0 The system of inequalities is .   x + 2y ≤ 4

Linear Inequalities in Two Unknowns

2x – 6 = 2(1) – 6 = –4 < 1 One inequality is y ≥ 2x – 6. 6–x=6–1=5>1 One inequality is y ≤ 6 – x.

1 1 7 x+2= (1) + 2 = >1 3 3 3 1 One inequality is y ≤ x + 2. 3

The system of inequalities is

x ≥ 0 y ≥ 0  y ≥ 2x − 6  . y ≤ 6 − x  1 y ≤ x + 2 3 

7. (b) Draw the line 2x + 3y = 0. Translate the line 2x + 3y = 0 in the positive direction of the x-axis to obtain increasing values of P. From the graph, P attains its maximum at (3, 3). ∴

Maximum value of P = 2(3) + 3(3) = 15

Translate the line 2x + 3y = 0 in the negative direction of the x-axis to obtain decreasing value of P. From the graph, P attains its minimum at (0, 0). ∴ 2.

8.

Minimum value of P = 2(0) + 3(0) = 0

(a) Evaluate the values of x, 4x – 5y + 5 and x + y – 1 for the point (1, 1). ∵ x=1<5 ∴ One inequality is x ≤ 5. ∵ 4x – 5y + 5 = 4(1) – 5(1) + 5 = 4 > 0 ∴ One inequality is 4x – 5y + 5 ≥ 0. ∵ x+y–1=1+1–1=1>0 ∴ One inequality is x + y – 1 ≥ 0.

p. 126 1.

(a) Evaluate the values of 2x – 6, 6 – x and

1 x + 2 for 3

the point (1, 1).

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Certificate Mathematics in Action Full Solutions 5A

x≤ 5   4x − 5 y + 5 ≥ 0 ∴ The system of inequalities is  .  x + y − 1≥ 0  y ≥ 0

(ii)

(b) Check the values of P = 3x + y at A, B, C and D respectively. At A(0, 1), P = 3(0) + 1 = 1 At B(1, 0), P = 3(1) + 0 = 3 At C(5, 0), P = 3(5) + 0 = 15 At D(5, 5), P = 3(5) + 5 = 20 ∴ Maximum value of P = 20

The black dots in the above figure show all the feasible solutions with integral x-and y-coordinates. Translate the line 2x + y = 0 in the positive direction of the x-axis to obtain increasing values of P. From the graph, P attains its maximum at (3, 0). ∴ Maximum value of P = 2(3) + 0 = 6

Minimum value of P = 1 3.

(a)

Translate the line 2x + y = 0 in the negative direction of the x-axis to obtain decreasing values of P. From the graph, P attains its minimum at (0, 0). ∴

Minimum value of P = 2(0) + 0 = 0

p. 136 1.

The constraints are:

 5x + 9 y ≥ 5 0   5 8x + 4 0y ≥ 3 0 0  x a n dy a ren o n- n e g a tivine te g e rs . 

(b) (i)

After simplification, we have:

 5x + 9 y ≥ 5 0   2 9x + 2 0y ≥ 1 5 0  x a n dy a ren o n- n e g a tivine te g e rs . 

Draw the line 2x + y = 0. Translate the line 2x + y = 0 in the positive direction of the x-axis to obtain increasing values of P. From the graph, P attains its maximum at (4, –0.5). ∴

Maximum value of P = 2(4) + (–0.5) = 7.5

Translate the line 2x + y = 0 in the negative direction of the x-axis to obtain decreasing values of P. From the graph, P attains its minimum at (0, –0.5). ∴

0.5 Minimum value of P = 2(0) + (–0.5) = −

2.

(a)

1 2 0x + 6 0y ≥ 3 0 0 0  x+ y ≥ 40 The constraints are:  x≥ 0  y ≥ 0 54

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After simplification, we have:

 2x + y ≥ 5 0  x+ y≥ 40  x≥ 0  y ≥ 0

Linear Inequalities in Two Unknowns

From the graph, M attains its minimum at (10, 30). Machines P and Q should operate 10 hours and 30 hours every week respectively.

Exercise Exercise 13A (p. 115) Level 1 1.

Draw the graphical representations of the two inequalities on the same number line, we have

2.

Draw the graphical representations of the two inequalities on the same number line, we have

(b)

Therefore, the solutions of the compound inequality are:

3.

(c) Let $M be the total production cost, then M = 120x + 100y.

Draw the graphical representations of the two inequalities on the same number line, we have

There are no solutions. 4.

Solving 3 ≤ 4x – 3, we have

3 ≤ 4x − 3 4x ≥ 6 3 …… (1) x≥ 2

Solving 4x – 3 > 7, we have

4x −3 > 7 4 x >10 5 …… (2) x> 2

∵ x must satisfy both (1) and (2). ∴ The solutions of the compound inequality are

x>

5 . 2

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Certificate Mathematics in Action Full Solutions 5A

2 x −5 ≤ 9

Graphical representation:

2 x ≤14 …… (2) x ≤7

∵ x must satisfy both (1) and (2). ∴ The solutions of the compound inequality are 3 ≤ x ≤ 7. 5.

Solving 4x + 2 > 2x + 4, we have

Graphical representation:

4 x +2 > 2 x +4 2x > 2 x >1

…… (1) Solving 2x – 3 ≥ 4x – 9, we have

2 x −3 ≥ 4 x −9 2x ≤6 x ≤3

…… (2)

8.

2 x +2 x +2 4 ≤ ≤ . and 3 2 2 3 2 x +2 ≤ Solving , we have 3 2 2 x +2 ≤ 3 2 4 ≤ 3x + 6 3 x ≥ −2 2 …… (1) x ≥− 3 x +2 4 ≤ , we have Solving 2 3 x +2 4 ≤ 2 3 3x + 6 ≤ 8 3 x ≤ 2 …… (2) 2 x≤ 3

∵ x must satisfy both (1) and (2). ∴ The solutions of the compound inequality are 1 < x ≤ 3. Graphical representation:

6.

Rewrite the compound inequality as

Solving 3x – 2 ≤ 7, we have

3 x −2 ≤ 7

3x ≤9 x ≤ 3 …… (1) Solving 8 – 5x ≤ –2, we have

8 −5 x ≤ −2

5 x ≥10 …… (2) x ≥2

∵ x must satisfy both (1) and (2). ∴ The solutions of the compound inequality are 2 ≤ x ≤ 3.

∵ x must satisfy both (1) and (2). ∴ The solutions of the compound inequality are

Graphical representation:

−

2 2 ≤x≤ . 3 3

Graphical representation:

7.

Rewrite the compound inequality as 1 ≤ 2x – 5 and 2x – 5 ≤ 9. Solving 1 ≤ 2x – 5, we have

1 ≤ 2x −5

2x ≥ 6 x ≥3

9.

2x – 3y < 1

…… (1) Solving 2x – 5 ≤ 9, we have

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10. 3x + 2y ≥ 5

Linear Inequalities in Two Unknowns

13. Evaluate the values of x + 2y and x for the point (2, 0). ∵ x + 2y = 2 + 2(0) = 2 < 4 ∴ One inequality is x + 2y < 4. ∵ x=2>1 ∴ One inequality is x ≥ 1.

∴ The system of inequalities is

11. 3x – y ≥ 0

.

14. Evaluate the values of x + 2y – 1 and x – 2y + 1 for the point (–2, 0). ∵ x + 2y – 1 = –2 + 2(0) – 1 = –3 < 0 ∴ One inequality is x + 2y – 1 ≤ 0. ∵ x – 2y + 1 = –2 – 2(0) + 1 = –1 < 0 ∴ One inequality is x – 2y + 1 ≤ 0.

∴

12. –2x + y > 3

 x + 2y < 4   x≥ 1

The system of inequalities is

 x + 2y − 1≤ 0 .   x − 2y + 1≤ 0

15. Evaluate the values of 2x + 3y and x + y – 1 for the point (1, 1). ∵ 2x + 3y = 2(1) + 3(1) = 5 < 6 ∴ One inequality is 2x + 3y ≤ 6. ∵ x+y–1=1+1–1=1>0 ∴ One inequality is x + y – 1 ≥ 0. ∴

The system of inequalities is

2 x + 3 y ≤ 6  x + y −1 ≥ 0 .  x ≥0  16. Evaluate the value of 3x – y + 6 for the point (–1, 2). ∵ 3x – y + 6 = 3(–1) – 2 + 6 = 1 > 0 ∴ One inequality is 3x – y + 6 ≥ 0. ∴ The system of inequalities is

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Certificate Mathematics in Action Full Solutions 5A

 3x − y + 6 ≥ 0  . x≤ 0 y≥ 0  20.

17.

21.

18.

22.

19.

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23. Draw the line –x + y = 1.

Linear Inequalities in Two Unknowns

Graphical representation:

x −5 > 2 x −1 , we have 3 x−5 > 2x − 1 3 x − 5 > 6x − 3 5 x < −2 …… (1) 2 x<− 5 x x +1 −1 ≤ Solving , we have 2 3 x x +1 −1 ≤ 2 3 3 x − 6 ≤ 2 x + 2 …… (2) x ≤8

26. Solving

24. Draw the line x + 2y = 8.

∵ x must satisfy both (1) and (2). ∴ The solutions of the compound inequality are

x <−

2 . 5

Graphical representation:

Level 2 25. Solving –2(x + 2) > –12, we have

−2( x +2) >−12 −2 x −4 >−12 2 x <8 …… (1) x <4

Solving 4 – 3x ≥ 9 + 2x, we have

4 −3 x ≥ 9 + 2 x 5 x ≤ −5 x ≤ −1

…… (2)

∵ x must satisfy both (1) and (2). ∴ The solutions of the compound inequality are x ≤ –1.

27. Rewrite the compound inequality as x + 1 < 5 and 5 < 2x + 7. Solving x + 1 < 5, we have

x +1 <5 x < 4 …… (1)

Solving 5 < 2x + 7, we have

5 < 2x + 7

2 x > −2 x > −1

…… (2)

∵ x must satisfy both (1) and (2). ∴ The solutions of the compound inequality are –1 < x < 4. Graphical representation:

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Certificate Mathematics in Action Full Solutions 5A

5 < x ≤8 . 2 Graphical representation:

28. Rewrite the compound inequality as

3x 3x + 5 < 2x + 7 . + 5 and 4 4 3x + 5 , we have Solving 2 < 4 3x 2< +5 4 8 < 3 x + 20 …… (1) 3 x > −12 x > −4 3x + 5 < 2 x + 7 , we have Solving 4 3x +5 < 2x +7 4 5x > −2 4 …… (2) 5 x > −8 8 x >− 5 2<

30. 3x + 1 ≤ y – 2 3x – y ≤ –3 Draw the line 3x – y = –3.

31. 2(2x – 1) < 3(y – x) + 1 4x – 2 < 3y – 3x + 1 7x – 3y < 3 Draw the line 7x – 3y = 3.

∵ x must satisfy both (1) and (2). ∴ The solutions of the compound inequality are

x >−

8 . 5

Graphical representation:

32. 3x – 4y > –2x + 2y 5x – 6y > 0 Draw the line 5x – 6y = 0. 29. Rewrite the compound inequality as 3x – 2 > x + 3 and x + 3 ≥ 2x – 5. Solving 3x – 2 > x + 3, we have

3 x − 2 > x +3 2x >5 5 x> 2

…… (1)

Solving x + 3 ≥ 2x – 5, we have

x +3 ≥ 2 x −5 …… (2) x ≤8

∵ x must satisfy both (1) and (2). ∴ The solutions of the compound inequality are

33. –2x + 3y ≥ 3x – y 5x – 4y ≤ 0 Draw the line 5x – 4y = 0.

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Linear Inequalities in Two Unknowns

37.

34.

38.

35.

39.

Exercise 13B (p. 127) Level 1 1.

36.

(a) The system of inequalities is

0≤ x≤ 5 .  1≤ y ≤ 4 61

Certificate Mathematics in Action Full Solutions 5A

∴ ∵ ∴ ∵ ∴ ∴

One inequality is 3x – 2y ≥ –6. x + 2y = 1 + 2(1) = 3 > 2 One inequality is x + 2y ≥ 2. 2y + 9x = 2(1) + 9 = 11 ≤ 18 One inequality is 2y ≤ –9x + 18. The system of inequalities is

 3x − 2 y ≥ − 6   x + 2y ≥ 2 .  2 y ≤ − 9x + 1 8 

(b) Draw the line x + y = 0. Translate the line x + y = 0 in the positive direction of the x-axis to obtain increasing values of P. From the graph, P attains its maximum at (5, 4). ∴ Maximum value of P = 5 + 4 = 9 Translate the line x + y = 0 in the negative direction of the x-axis to obtain decreasing values of P. From the graph, P attains its minimum at (0, 1). ∴ 2.

Minimum value of P = 0 + 1 = 1

(a) Evaluate the value of 2x + 3y for the point (1, 1). ∵ 2x + 3y = 2(1) + 3(1) = 5 < 6 ∴ One inequality is 2x + 3y ≤ 6. ∴ The system of inequalities is

(b) Draw the line x + y = 0. Translate the line x + y = 0 in the positive direction of the x-axis to obtain increasing values of P. From the graph, P attains its maximum at (1, 4.5). ∴ Maximum value of P = 1 + 4.5 = 5.5 Translate the line x + y = 0 in the negative direction of the x-axis to obtain decreasing values of P. From the graph, P attains its minimum at (–1, 1.5). ∴ Minimum value of P = –1 + 1.5 = 0.5

 2x + 3y ≤ 6  . x≥ 0 y≥ 0  4.

(b) Draw the line x + y = 0. Translate the line x + y = 0 in the positive direction of the x-axis to obtain increasing values of P. From the graph, P attains its maximum at (3, 0). ∴ Maximum value of P = 3 + 0 = 3 Translate the line x + y = 0 in the negative direction of the x-axis to obtain decreasing values of P. From the graph, P attains its minimum at (0, 0). ∴ Minimum value of P = 0 + 0 = 0 3.

(a) Evaluate the values of x – 2y, 2x + 3y and 4x – y at the point (3, 2). ∵ x – 2y = 3 – 2(2) = –1 > –5 ∴ One inequality is x – 2y ≥ –5. ∵ 2x + 3y = 2(3) + 3(2) = 12 > 11 ∴ One inequality is 2x + 3y ≥ 11. ∵ 4x – y = 4(3) – 2 = 10 < 15 ∴ One inequality is 4x – y ≤ 15. ∴ The system of inequalities is

 x − 2y ≥ − 5   2 x + 3 y ≥ 1 1.  4x − y ≤ 1 5 

(a) Evaluate the values of 3x – 2y, x + 2y and 2y + 9x at the point (1, 1). ∵ 3x – 2y = 3(1) – 2(1) = 1 > –6

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Linear Inequalities in Two Unknowns

(b) Draw the line x + y = 0. Translate the line x + y = 0 in the positive direction of the x-axis to obtain increasing values of P. From the graph, P attains its maximum at (5, 5). ∴ Maximum value of P = 5 + 5 = 10 Translate the line x + y = 0 in the negative direction of the x-axis to obtain decreasing values of P. From the graph, P attains its minimum at (1, 3). ∴ Minimum value of P = 1 + 3 = 4

6. Draw the line 4x + y + 2 = 0. Translate the line 4x + y + 2 = 0 in the positive direction of the x-axis to obtain increasing values of P. From the graph, P attains its maximum at (2, 9). ∴ Maximum value of P = 4(2) + 9 + 2 = 19 Translate the line 4x + y + 2 = 0 in the negative direction of the x-axis to obtain decreasing values of P. From the graph, P attains its minimum at (–1, 7). ∴ Minimum value of P = 4(–1) + 7 + 2 = 5

5. Draw the line x + 3y = 0. Translate the line x + 3y = 0 in the positive direction of the x-axis to obtain increasing values of P. From the graph, P attains its maximum at (0, 5). ∴ Maximum value of P = 0 + 3(5) = 15 Translate the line x + 3y = 0 in the negative direction of the x-axis to obtain decreasing values of P. From the graph, P attains its minimum at (0, –4). 12 ∴ Minimum value of P = 0 + 3(–4) = −

7. Draw the line 2x + 3y + 2 = 0. Translate the line 2x + 3y + 2 = 0 in the positive direction of the x-axis to obtain increasing values of P. From the graph, P attains its maximum at (2, 4). ∴ Maximum value of P = 2(2) + 3(4) + 2 = 18 Translate the line 2x + 3y + 2 = 0 in the negative direction of the x-axis to obtain decreasing values of P. From the graph, P attains its minimum at (2, 1). ∴ Minimum value of P = 2(2) + 3(1) + 2 = 9

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Certificate Mathematics in Action Full Solutions 5A

10. Draw the line 2x + y + 1 = 0. Translate the line 2x + y + 1 = 0 in the positive direction of the x-axis to obtain increasing values of P. From the graph, 2x + y + 1 = P coincides with the straight line 2x + y = 6. ∴ Maximum value of P = 6 + 1 = 7 Translate the line 2x + y + 1 = 0 in the negative direction of the x-axis to obtain decreasing values of P. From the graph, P attains its minimum at (0, 0). ∴ Minimum value of P = 2(0) + 0 + 1 = 1

8. Draw the line x + 2y + 1 = 0. Translate the line x + 2y + 1 = 0 in the positive direction of the x-axis to obtain increasing values of P. From the graph, P attains its maximum at (3, 5). ∴ Maximum value of P = 3 + 2(5) + 1 = 14 Translate the line x + 2y + 1 = 0 in the negative direction of the x-axis to obtain decreasing values of P. From the graph, P attains its minimum at (1, 2). ∴ Minimum value of P = 1 + 2(2) + 1 = 6

Level 2

11. Draw the line x + 2y + 3 = 0. Translate the line x + 2y + 3 = 0 in the positive direction of the x-axis to obtain increasing values of P. From the graph, P attains its maximum at (3, 3). ∴ Maximum value of P = 3 + 2(3) + 3 = 12 Translate the line x + 2y + 3 = 0 in the negative direction of the x-axis to obtain decreasing values of P. From the graph, P attains its minimum at (5, –3). ∴ Minimum value of P = 5 + 2(–3) + 3 = 2

9. Draw the line 2x + y – 5 = 0. Translate the line 2x + y – 5 = 0 in the positive direction of the x-axis to obtain increasing values of P. From the graph, P attains its maximum at (4, 2). ∴

Maximum value of P = 2(4) + 2 – 5 = 5

Translate the line 2x + y – 5 = 0 in the negative direction of the x-axis to obtain decreasing values of P. From the graph, P attains its minimum at (0, 0). 5 ∴ Minimum value of P = 2(0) + 0 – 5 = −

12.

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Draw the line x + y = 0. Translate the line x + y = 0 in the positive direction of the x-axis to obtain increasing values of P. From the graph, P attains its maximum at (5, 6). ∴ Maximum value of P = 5 + 6 = 11 Translate the line x + y = 0 in the negative direction of the x-axis to obtain decreasing values of P. From the graph, P attains its minimum at (1, 0). ∴ Minimum value of P = 1 + 0 = 1

Linear Inequalities in Two Unknowns

From the graph, P attains its minimum at

1 1   −5 ,−5  . 4 2  ∴

Minimum value of

1 1 107   P = 3 − 5  + 2 − 5  = − 4 2 4  

15. (a)

13. Draw the line 2x + 4y = 0. Translate the line 2x + 4y = 0 in the positive direction of the x-axis to obtain increasing values of P. From the graph, P attains its maximum at (4, 7). ∴ Maximum value of P = 2(4) + 4(7) = 36 Translate the line 2x + 4y = 0 in the negative direction of the x-axis to obtain decreasing values of P. From the graph, P attains its minimum at (3, 0). ∴ Minimum value of P = 2(3) + 4(0) = 6

(b) (i) Draw the line x + 3y = 0. Translate the line x + 3y = 0 in the positive direction of the x-axis to obtain increasing values of P. From the graph, P attains its maximum at (1.5, 2.5). ∴ Maximum value of P = 1.5 + 3(2.5) = 9 Translate the line x + 3y = 0 in the negative direction of the x-axis to obtain decreasing values of P. From the graph, P attains its minimum at (–2, – 1). 5 ∴ Minimum value of P = –2 + 3(–1) = −

(ii)

14. Draw the line 3x + 2y = 0. Translate the line 3x + 2y = 0 in the positive direction of the x-axis to obtain increasing values of P. From the graph, 3x + 2y = P coincides with the straight line 3x + 2y = 7. ∴ Maximum value of P = 7 Translate the line 3x + 2y = 0 in the negative direction of the x-axis to obtain decreasing values of P.

The black dots in the above figure are feasible solutions with integral x- and y- coordinates. Translate the line x + 3y = 0 in the positive direction of the x-axis to obtain increasing values of P. From the graph, P attains its maximum at (2, 2). ∴ Maximum value of P = 2 + 3(2) = 8 Translate the line x + 3y = 0 in the negative direction of the x-axis to obtain decreasing values

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Certificate Mathematics in Action Full Solutions 5A

of P. From the graph, P attains its minimum at (–2, – 1). 5 ∴ Minimum value of P = (–2) + 3(–1) = −

of P. From the graph, P attains its minimum at (–2, – 2). ∴ Minimum value of P = (–2) + 2(–2) – 1 = − 7

Exercise 13C (p. 137) Level 1 1.

The constraints are:

x≥ 0  y≥ 0   2x + y ≥ 6  x + 3 y ≥ 8

16. (a) (b) (i) Draw the line x + 2y – 1 = 0. Translate the line x + 2y – 1 = 0 in the positive direction of the x-axis to obtain increasing values of P. From the graph, P attains its maximum at (1.5, 5). ∴ Maximum value of P = 1.5 + 2(5) – 1 =

2.

 4x + 6 y ≤ 1 5 0  x+ y ≤ 20  x a n dy a ren o n-n e g a tivine te g e rs . 

10 .5

Translate the line x + 2y – 1 = 0 in the negative direction of the x-axis to obtain decreasing values of P. From the graph, P attains its minimum at (–2, –2). ∴ Minimum value of P = (–2) + 2(–2) – 1 =

After simplification, we have:

 2x + 3y ≤ 7 5  x+ y ≤ 20  x a n dy a ren o n-n e g a tivine te g e rs . 

− 7

3.

(ii) The black dots in the above figure are feasible solutions with integral x- and y- coordinates. Translate the line x + 2y – 1 = 0 in the positive direction of the x-axis to obtain increasing values of P. From the graph, P attains its maximum at (1, 4). ∴ Maximum value of P = 1 + 2(4) – 1 = 8 Translate the line x + 2y – 1 = 0 in the negative direction of the x-axis to obtain decreasing values

The constraints are:

The constraints are:

 2x + 5 y ≤ 1 0 0 14 30   x+ y≥ 7 60 60  x a n dy a ren o n-n e g a tivine te g e rs. After simplification, we have:

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13

Linear Inequalities in Two Unknowns

 2x + 5 y ≤ 1 0 0   7 x + 1 5y ≥ 2 1 0  x a n dy a ren o n-n e g a tivine te g e rs .  4.

(a) The constraints are:

 3 0x + 5 0y ≥ 4 5 0   6 0 0x + 3 0 0y ≥ 6 0 0 0  x a n dy a ren o n-n e g a tivine te g e rs . 

From the graph, C attains its maximum at (10, 3). ∴ The company should hire 10 bus A and 3 bus B. ∴ The minimum cost = $[150(10) + 200(3)] = $2100

After simplification, we have:

 3x + 5 y ≥ 4 5   2x + y ≥ 2 0  x a n dy a ren o n-n e g a tivine te g e rs . 

5.

(a) The constraints are:

 5 x + 2 0y ≤ 6 0 0  x+ y ≤ 60  x a n dy a ren o n-n e g a tivine te g e rs .  After simplification, we have:

 x + 4y ≤ 1 2 0  x+ y≤ 60  x a n dy a ren o n-n e g a tivine te g e rs .  (b)

(c) Let $C be the cost, then C = 150x + 200y

(b)

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Certificate Mathematics in Action Full Solutions 5A

(c) Let $P be the income, then P = 15x + 45y

(b) (c) Let $P be the profit, then P = 50x + 30y

From the graph, P attains its maximum at (40, 20). ∴ The car park can obtain maximum income when there are 40 cars and 20 trucks in the car park. 6.

(a) The constraints are:

 6x + 4 y ≤ 8 0   2x + y ≤ 2 4  x a n dy a ren o n-n e g a tivine te g e rs .  After simplification, we have:

 3x + 2 y ≤ 4 0   2x + y ≤ 2 4  x a n dy a ren o n-n e g a tivine te g e rs . 

From the graph, P attains its maximum at (8, 8). ∴ The maximum profit = $[50(8) + 30(8)] = $640

7.

(a) The constraints are:

 4 5 0x + 7 5 0y ≤ 1 14 0 0  x+ y ≤ 20  x a n dy a ren o n-n e g a tivine te g e rs .  After simplification, we have:

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13

 3x + 5 y ≤ 7 6  x+ y ≤ 20  x a n dy a ren o n-n e g a tivine te g e rs . 

 5x + 2 y ≥ 2 5  y≥ x  x a n dy a ren o n-n e g a tivine te g e rs . 

(b)

(b)

(c) Let $P be the profit, then P = 240x + 300y

(c) Let $C be the cost, then C = 160x + 75y

From the graph, P attains its maximum at (12, 8). ∴ The farmer should buy 12 sheep and 8 pigs. ∴ The maximum profit = $[240(12) + 300(8)] = $5280 8.

Linear Inequalities in Two Unknowns

(a) The constraints are:

 5 0x + 2 0y ≥ 2 5 0  y≥ x  x a n dy a ren o n-n e g a tivine te g e rs .  After simplification, we have:

From the graph, C attains its minimum at (3, 5). ∴ The lorry should make 3 journeys and the van should make 5 journeys. ∴ The minimum cost = $[160(3) + 75(5)] = $855

Level 2 9.

(a) The constraints are:

 2x + y ≥ 9   x + y ≥ 7  x + 2y ≥ 1 0 x ≥ 0   y ≥ 0

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Certificate Mathematics in Action Full Solutions 5A

 5x + y ≥ 1 0   x + y ≥ 6  x + 4y ≥ 1 2 x ≥ 0   y ≥ 0

(b) (c) Let $C be the cost, then C = 50x + 70y

(b) (c) Let $C be the cost, then C = 9x + 6y

From the graph, C attains its minimum at (4, 3). ∴ Weight of food P : weight of food Q = 4 : 3 ∴ There should be

4 3 kg of food P and kg of 7 7

food Q in each kg of the mixture so as to minimize the cost. 10. (a) The constraints are:

10x + 2 y ≥ 2 0   4 x + 4 y ≥ 2 4  2x + 8 y ≥ 2 4 x ≥ 0   y ≥ 0

From the graph, C attains its minimum at (1, 5). ∴ Weight of ingredient A : weight of ingredient B =1:5 ∴ There should be

1 5 kg of ingredient A and kg 6 6

of ingredient B in 1 kg of fertilizer so as to minimize the cost. 11. (a) The constraints are:

After simplification, we have:

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Linear Inequalities in Two Unknowns

From the graph, C attains its minimum at (60, 20). ∴ 60 g of ore P and 20 g of ore Q should be combined in order to minimize the cost of a unit of alloy G.

 0.2 x + 0.1 y ≥ 10   0.1x + 0.1 y ≥ 8  0.1x + 0.3 y ≥ 12 x ≥ 0   y ≥ 0

12. Let x L and y L be the volumes of drink P and drink Q respectively. The constraints are:

1  0 . 4 x + y ≤ 12  3   0.6 x + 2 y ≤ 2 0  3  x ≥ 0 y≥ 0 

After simplification, we have:

 2x + y ≥ 1 0 0   x + y ≥ 8 0  x + 3y ≥ 1 2 0 x ≥ 0   y ≥ 0

After simplification, we have:

 6x + 5 y ≤ 1 8 0   9 x + 1 0y ≤ 3 0 0  x≥ 0  y ≥ 0 Let $M be the profit, then M = 8.2x + 8y.

(b)

(c) Let $p be the cost of 1 g of ore P and $

3 p be the 2

cost of 1 g of ore Q. Let $C be the cost, then C = px +

3 py 2 From the graph, M attains its maximum at (20, 12). ∴ 20 L of drink P and 12 L of drink Q should be made to maximize the profit. 13. (a) ∵ ∴

Paul receives x tonnes of medicine from East Depot. Paul receives (100 – x) tonnes of medicine from West Depot. ∵ William receives y tonnes of medicine from East Depot. ∴ William receives (50 – y) tonnes of medicine from West Depot. The constraints are:

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Certificate Mathematics in Action Full Solutions 5A

x+ y ≤ 240   (1 0 0− x) + (5 0− y ) ≤ 4 0  0 ≤ x ≤ 1 0 0  0 ≤ y ≤ 5 0

Depot, while William receives 10 tonnes of medicine from East Depot and 40 tonnes of medicine from West Depot.

Revision Exercise 13 (p. 142) Level 1 1.

4 x + 6 ≥ 1 , we have

4x + 6 ≥ 1

After simplification, we have:

x+  x+  0 ≤  0 ≤

Solving

4 x ≥ −5 5 ……(1) x≥− 4

y≤ 240 y≥ 110 x≤100 y≤ 50

Solving 3x – 2 > 5, we have

3x − 2 > 5 3x > 7 7 ……(2) x> 3

∵ x must satisfy both (1) and (2). ∴ The solutions of the compound inequality are x>

7 . 3

Graphical representation:

(b) (c) Cost of transportation from East Depot = $(60x + 80y) Cost of transportation from West Depot = $[30(100 – x) + 20(50 – y)] = $(4000 – 30x – 20y) Let $C be the total transportation cost. C = (60x + 80y) + (4000 – 30x – 20y) = 30x + 60y + 4000 ∴ The total transportation cost is $(30x + 60y + 4000).

2.

Solving 5 x −3 > x +1 , we have

5 x −3 > x +1 4x >4 x >1

……(1)

Solving 4x + 1> 2x – 5, we have

4 x +1 > 2 x −5 2 x > −6 x > −3

……(2)

∵ x must satisfy both (1) and (2). ∴ The solutions of the compound inequality are x > 1. Graphical representation:

(d)

3.

Solving 2x – 3 ≤ 5, we have

From the graph, C attains its minimum at (100, 10). ∴ Paul receives 100 tonnes of medicine from East

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13

2 x −3 ≤ 5 2 x ≤8 x ≤4

2 x +3 ≤11 2 x ≤8 x ≤4

……(1)

……(2)

∵ x must satisfy both (1) and (2). ∴ The solutions of the compound inequality are −4 < x ≤ 4. Graphical representation:

Solving 3x – 2 > 4, we have

3 x −2 > 4 3x > 6 x >2

Linear Inequalities in Two Unknowns

……(2)

∵ x must satisfy both (1) and (2). ∴ The solutions of the compound inequality are 2 < x ≤ 4. Graphical representation: 6.

Rewrite the compound inequality as

2 − 3x 2 − 3x and <4 . 2 2 2 − 3x Solving − 3 ≤ , we have 2 2 −3 x −3 ≤ 2 − 6 ≤ 2 −3 x 3x ≤ 8 ……(1) 8 x≤ 3 2 − 3x Solving < 4 , we have 2 2 −3 x <4 2 2 −3 x < 8 ……(2) 3 x > −6 x > −2 −3 ≤

Solving 4.

3≤

3≤

1 x − 3 , we have 2

1 x−3 2

1 x≥6 ……(1) 2 x ≥ 12 3 Solving x – 6 > 0, we have 4 3 x −6 > 0 4 3 x > 6 ……(2) 4 x >8

∵ x must satisfy both (1) and (2). ∴ The solutions of the compound inequality are

∵ x must satisfy both (1) and (2). ∴ The solutions of the compound inequality are x ≥ 12. Graphical representation:

−2 < x ≤

8 . 3

Graphical representation:

5.

Rewrite the compound inequality as −5 < 2x + 3 and 2x + 3 ≤ 11. Solving −5 < 2x + 3, we have

−5 < 2 x +3

2 x > −8 x > −4

……(1)

Solving 2x + 3 ≤ 11, we have 7.

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Certificate Mathematics in Action Full Solutions 5A

8. 9.

14. Evaluate the value of 5x – 3y + 1 for the point (0, 0). ∵ 5(0) – 3(0) + 1 = 1 > 0 ∴ The inequality is 5x – 3y + 1 ≥ 0.

10. Evaluate the value of −2x – 3y + 4 for the point (0, 0). ∵ –2(0) – 3(0) + 4 = 4 > 0 ∴ The inequality is –2x – 3y + 4 > 0. 11. Evaluate the values of y and x for the point (2, 1). ∵ y=1<2=x ∴ One inequality is y ≤ x. ∵ x=2>1 ∴ One inequality is x ≥ 1.

15.

y≤ x  ∴ The system of inequalities is x ≥ 1 .  y≥ 0  12. Evaluate the values of x + y + 1 and y for the point (–1, 2). ∵ x + y + 1 = –1 + 2 + 1 = 2 > 0 ∴ One inequality is x + y + 1 ≥ 0. ∵ y=2>1 ∴ One inequality is y ≥ 1.

 x + y + 1≥ 0  ∴ The system of inequalities is y ≥ 1 .  x≤ 0  13.

16.

17. Draw the line x + 2y = 0. Translate the line x + 2y = 0 in the negative direction of the x-axis to obtain decreasing values of P. From the graph, P attains its minimum at (–2, 0). 2 ∴ Minimum value of P = –2 + 2(0) = −

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13

Linear Inequalities in Two Unknowns

x-axis to obtain decreasing values of P. From the graph, P does not attain any minimum value. 21. (a) The constraints are:

 2x + y ≤ 3 0   3x + 4 y ≤ 6 0  x a n dy a ren o n-n e g a tivine te g e rs . 

18. Draw the line x + 2y = 0. Translate the line x + 2y = 0 in the negative direction of the x-axis to obtain decreasing values of P. From the graph, P attains its minimum at (−1, 1). ∴ Minimum value of P = −1 + 2(1) = 1 .

(b) (c) Let $P be the profit, then P = 30x + 20y

19. Draw the line 3x + y = 0. Translate the line 3x + y = 0 in the positive direction of the x-axis to obtain increasing values of P. From the graph, P attains its maximum at (7, 6). ∴ Maximum value of P = 3(7) + 6 = 27 Translate the line 3x + y = 0 in the negative direction of the x-axis to obtain decreasing values of P. From the graph, P attains its minimum at (2, 6). ∴ Minimum value of P = 3(2) + 6 = 12

From the graph, P attains its maximum at (12, 6). ∴ The carpenter should make 12 cupboards and 6 bookshelves. ∴ The maximum profit = $[30(12) + 20(6)] = $480 22. (a) The constraints are:

20. Draw the line x – 2y = 0. Translate the line x – 2y = 0 in the positive direction of the x-axis to obtain increasing values of P. From the graph, P attains its maximum at (2, –1). ∴ Maximum value of P = 2 – 2(–1) = 4 Translate the line x – 2y = 0 in the negative direction of the

 2 0 0 0x + 1 0 0 0y ≤ 1 00 0 0  1  x ≥ y  2  x an dy are n o n-n eg a tivine teg e rs. After simplification, we have:

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Certificate Mathematics in Action Full Solutions 5A

 2x + y ≤ 1 0   2x ≥ y  x a n dy a ren o n-n e g a tivine te g e rs .  (b) (c) Let $P be the profit, then P = 24x + 20y (b)

From the graph, P attains its maximum at (8, 8). ∴ 8 pieces of toy A and 8 pieces of toy B should be produced weekly. ∴ The maximum profit = $[24(8) + 20(8)] = $352

(c) The number of monitors bought is x + y. From the graph, it attains its maximum at (3, 4). ∴ The company can bought a maximum of 7 monitors. 23. (a) The constraints are:

 3x + 2 y ≤ 4 0   x + 2y ≤ 2 4  x a n dy a ren o n-n e g a tivine te g e rs . 

Level 2

1 ≥ 5x + 1, we have 2 1 3 x + ≥ 5 x +1 2 1 2x ≤ − 2 ……(1) 1 x ≤− 4 3 x +1 2 x +1 ≤ Solving , we have 4 3

24. Solving 3x +

3 x +1 2 x +1 ≤ 4 3 9 x +3 ≤8 x +4 ……(2) x ≤1 ∵ ∴

x must satisfy both (1) and (2). The solutions of the compound inequality are

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13

x ≤−

x − 3 < x +1 2 x −6 < 2x + 2 x > −8

1 . 4

Graphical representation:

∵ ∴

Linear Inequalities in Two Unknowns

……(2)

x must satisfy both (1) and (2). The solutions of the compound inequality are

9 −8 < x < − . 8 Graphical representation:

25. Solving 4(x + 2) ≤ 3(2x − 1), we have

4( x +2) ≤3( 2 x −1) 4 x +8 ≤ 6 x −3 2 x ≥11 x≥

11 2

Solving 2( x −1) +

……(1)

1 ( x −3) ≥1 , we have 2

1 ( x −3) ≥1 2 4( x −1) + ( x −3) ≥ 2 5 x −7 ≥ 2 5 x ≥ 9 ……(2) 9 x≥ 5

2( x −1) +

∵ ∴

27. Solving 4x + 6 ≤ 2(x – 1) + 4, we have

4 x +6 ≤ 2( x −1) +4 4 x +6 ≤ 2 x +2 2 x ≤ −4 x ≤ −2

11 . 2

Graphical representation:

x x 5x + ≥3+ , we have 2 3 4 x x 5x + ≥3 + 2 3 4 6 x + 4 x ≥ 36 +15 x 5 x ≤ −36 36 ……(2) x ≤− 5

Solving

x must satisfy both (1) and (2). The solutions of the compound inequality are

x≥

……(1)

∵ ∴

x must satisfy both (1) and (2). The solutions of the compound inequality are

x ≤−

36 . 5

Graphical representation:

26. Solving

3x + 2 <

x −1 3 9 x + 6 < x −3 8 x < −9

x −1 , we have 3

3x + 2 <

28. Rewrite the compound inequality as ……(1)

9 x <− 8 x Solving − 3 < x +1 , we have 2

x −1 3 3 < ( x + 2) and ( x + 2) ≤ 2 x −3 . 2 8 8 x −1 3 < ( x + 2) , we have Solving 2 8

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Certificate Mathematics in Action Full Solutions 5A

x −1 3 < ( x + 2) 2 8 4( x −1) < 3( x + 2) ……(1)

4 x −4 < 3 x +6 x <10 Solving

3 ( x + 2) ≤ 2 x −3 , we have 8

3 ( x + 2) ≤ 2 x −3 8 3( x + 2) ≤ 8( 2 x −3) 3 x + 6 ≤16 x − 24 13 x ≥ 30 x≥ ∵ ∴

31.

……(2)

30 13

x must satisfy both (1) and (2). The solutions of the compound inequality are

30 ≤ x <10 . 13

32.

Graphical representation:

29. Rewrite the compound inequality as 3 – 2x > 2 – x and 2 – x ≥ 4 – 3x. Solving 3 – 2x > 2 – x, we have

3 −2 x > 2 − x x <1

33.

……(1)

Solving 2 – x ≥ 4 – 3x, we have

2 − x ≥ 4 −3 x

∵ ∴

2x ≥2 x ≥1

……(2)

x must satisfy both (1) and (2). There are no solutions.

34.

30.

Draw the line 4x + 6y = 0. Translate the line 4x + 6y = 0 in the positive direction of the x-axis to obtain increasing values of P. From the graph, P attains its maximum at (1, 4). ∴ Maximum value of P = 4(1) + 6(4) = 28 Translate the line 4x + 6y = 0 in the negative direction of

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13

the x-axis to obtain decreasing values of P. From the graph, P does not attain any minimum.

Linear Inequalities in Two Unknowns

37. (a) The constraints are:

10x + 5 y ≥ 45   0.1x + 0.9 y ≥ 1.8 10x + 30y ≥ 120 x ≥ 0   y ≥ 0 After simplification, we have:

 2x + y ≥ 9   x + 9 y ≥ 1 8  x + 3y ≥ 1 2 x ≥ 0   y ≥ 0

35. Draw the line 4x + y = 0. Translate the line 4x + y = 0 in the positive direction of the x-axis to obtain increasing values of P. From the graph, P does not attain any maximum value. Translate the line 4x + y = 0 in the negative direction of the x-axis to obtain decreasing values of P. From the graph, P attains its minimum at (1, 4). ∴ Minimum value of P = 4(1) + 4 = 8

(b) 36. (a)

(c) Let $C be the cost, then C = 50x + 30y.

(b) The black dots in the above figure show all the feasible solutions with integral x- and y-coordinates. Draw the line x + y = 0. Translate the line x + y = 0 in the positive direction of the x-axis to obtain increasing value of P. From the graph, P coincides with the straight line x + y = 4. ∴ Maximum value of P = 4

From the graph, C attains its minimum at (3, 3). ∴ The minimum cost = $[50(3) + 30(3)] = $240 38. (a) The constraints are:

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Certificate Mathematics in Action Full Solutions 5A

 0.6 x + 0.5 y ≤ 16.5   0.2 x + 0.3 y ≤ 7.5  0.2 x + 0.2 y ≤ 6 x ≥ 0   y ≥ 0 After simplification, we have:

 6x + 5 y ≤ 1 6 5   2 x + 3 y ≤ 7 5 x + y ≤ 30 x ≥ 0   y ≥ 0

= $210 39. Let x and y be the numbers of refrigerators A and B produced respectively. The constraints are:

 2x + 3y ≤ 1 2   4x + 3y ≤ 1 8  x a n dy a ren o n-n e g a tivine te g e rs . 

Let $P be the profit, then P = 300x + 300y From the graph, P attains its maximum at (3, 2). ∴ The number of refrigerators A and B should be produced are 3 and 2 respectively. ∴ The maximum profit = $[300(3) + 300(2)] = $1500 40. (a) The constraints are: (b) (c) Let $P be the profit, then P = 6x + 8y

1 0 8x + 7 2y ≥ 2 1 6 0  x+ y ≥ 25  0 ≤ x ≤ 2 4  0 ≤ y ≤ 2 4 After simplification, we have:

 3x + 2 y ≥ 6 0  x+ y≥ 25  0 ≤ x ≤ 2 4  0 ≤ y ≤ 2 4 From the graph, P attains its maximum at (15, 15). ∴ The maximum profit = $[6(15) + 8(15)]

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Linear Inequalities in Two Unknowns

 4 x + 5 y ≤ 34  2 x + 3 y ≤ 18   x ≥ 3  x and y are non-negativeintegers. (b)

(c) Let $C be the cost, then C = 100x + 80y

(b)

From the graph, C attains its minimum at (10, 15) ∴ The minimum daily production cost = $[100(10) + 80(15)] = $2200

(c) Let N be total number of seats provided, then N = 4x + 8y

41. (a) The constraints are:

 200x + 250y ≤ 1700 1.2 x + 1.8 y ≤ 10.8   x ≥ 3  x and y are non-negativeintegers.

After simplification, we have:

From the graph, N attains its maximum at (3, 4). ∴ The man should buy 3 table A and 4 table B. ∴ The maximum number of seats = 4(3) + 8(4) = 44 42. (a) ∵ ∴ ∵

x tonnes of coals are sent from P to A. (20 – x) tonnes of coals are sent from Q to A. y tonnes of coals are sent from P to B.

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Certificate Mathematics in Action Full Solutions 5A

∴ (15 – y) tones of coals are sent from Q to B. The constraints are:

x+ y ≤ 30   (2 0− x) + (1 5− y ) ≤ 1 5  0 ≤ x ≤ 2 0  0 ≤ y ≤ 1 5

= $[4(5) + 2(15) + 90] = $140

Multiple Choice Questions (p. 148) 1.

2 x −3 ≤ 7

2 x ≤10 x ≤5 ∴ ∴

After simplification, we have:

x+  x+  0 ≤  0 ≤

y≤ 30 y≥ 20

Answer : B For option B: Solving 2x + 3 > x + 1, we have x > –2 ……(1) Solving 5 – x > 2x – 4, we have 5 – x > 2x – 4 3x < 9 x < 3 ……(2) ∵ x must satisfy both (1) and (2). ∴ The solutions of the compound inequality are –2 < x < 3. ∴ B is the answer.

3.

Answer: D ∵ –1 + 0 = –1 < 0 ∴ The region containing (–1, 0) is not the solution. ∴ A is not the answer. ∵ 0 – 1 = –1 < 0 ∴ The region containing (0, 1) is not the solution. ∴ B is not the answer. ∵ 0 + (–1) = –1 < 0 ∴ The region containing (0, –1) is not the solution. ∴ C is not the answer. ∵ 1 + 0 = 1 > 0 and 1 – 0 = 1 > 0 ∴ The region containing (1, 0) is the solution. ∴ D is the answer. Answer: D Draw the line 2x – y = 0.

(b)

From the graph, C attains its minimum at (5, 15). ∴ The minimum transportation cost

x > 2 and x ≤ 5 2<x≤ 5

2.

x≤ 20 y≤ 15

(c) Cost of transporting coal to customer A = $[7x + 3(20 – x)] =$(4x + 60) Cost of transporting coal to customer B = $[4y +2(15 −y)] =$(2y + 30) Let $C be the total transportation cost. C = (4x + 60) + (2y + 30) = 4x + 2y + 90

Answer : C Solving 2x – 3 ≤ 7, we have

4.

Translate the line 2x – y = 0 in the positive direction of the x-axis to obtain increasing values of P.

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From the graph, P attains its maximum at (3, 1). ∴ Maximum value of P = 2(3) – 1 = 5 5.

∵ ∴ ∵ ∴ ∵ ∴ ∴

Answer: B Draw the line x – y = 0.

9.

Translate the line x – y = 0 in the negative direction of the x-axis to obtain decreasing values of P. From the graph, P attains its minimum at (0, 3). 6.

Answer : B

Linear Inequalities in Two Unknowns

x = 1.5 ≥ /2 I is false. x + y = 1.5 + 0 = 1.5 ≤ /1 II is false. x + y = 1.5 + 0 = 1.5 > 1 (1.5, 0) satisfies x + y ≥ 1 III is true.

Answer : D Consider the point (–1.5, 2) in the shaded region. ∵ y=2<3 ∴ One inequality is y ≤ 3. ∵ x + y = –1.5 + 2 = 0.5 > 0 ∴ One inequality is x + y ≥ 0, i.e. y ≥ –x ∵ x – y + 3 = –1.5 – 2 + 3 = –0.5 < 0 ∴ One inequality is x – y + 3 ≤ 0, i.e. x – y ≤ –3 ∴ The system of the inequalities is

y≤ 3  y≥ −x . x− y ≤ −3  10. Answer: A Check the values of C = px + qy – 2 at the vertices. At (–p, –q), C = p(–p) + q(–q) – 2 = –(p2 + q2 + 2) At (p, q), C = p(p) + q(q) – 2 = p2 + q2 – 2 At (–p, q), C = p(–p) + q(q) – 2 = –p2 + q2 – 2 At (–q, p), C = p(–q) + q(p) – 2 = –2 ∵ For fixed values of p and q, –(p2 + q2 + 2) is the minimum. ∴ C attains its minimum at (–p, –q).

The region that satisfies the constraints is shaded in the figure above. Draw the line x + 4y = 0. Translate the line x + 4y = 0 in the positive direction of the x-axis to obtain increasing values of P. From the graph, P attains its maximum at (0, 2). ∴ Maximum value of P = 0 + 4(2) = 8 7.

Answer : D From the inequality x ≥ 0, any region containing point (a, b) with a < 0 is not the solution of x ≥ 0. ∴ Any region containing point (a, b) with a < 0 is not the solution of the system of inequalities. ∵ Only region IV does not contain any point (a, b) with a < 0. ∴ Region IV represents the solutions of the system of inequalities.

8.

Answer: C Consider the point (1.5, 0) in the shaded region. 83