MATHWITHCYE SPM KSSM Mathematics

Chapter 6: Linear Inequalities in Two Variables

Focus: Systems of Inequalities, Feasible Shaded Regions & Linear Programming Models

Total Marks
/ 30 Marks
Suggested Time: 45 Mins

Section A • Bahagian A [12 Marks]

Question 1 [4 Marks]

State the three linear inequalities that define the shaded region bounded by the lines $y = 2x + 1$, $x + y = 5$, and the $x$-axis in the first quadrant.

1. Below or on $y = 2x + 1$

$y \le 2x + 1$
N1

2. Below or on $x + y = 5$

$x + y \le 5$
N1

3. Above $x$-axis and in first quadrant

$y \ge 0 \quad (\text{and } x \ge 0)$
N2
Question 2 [4 Marks]

Determine whether each of the following points lies in the region defined by $3x - 2y < 8$:

(a) Point $(2, -1)$ [2 marks]

(b) Point $(4, 2)$ [2 marks]

(a) Substitute $(2, -1)$: $3(2) - 2(-1) = 6 + 2 = 8$. Since the inequality is strict ($<$), $8 < 8$ is False!

Does NOT lie in the region (lies exactly on the dashed boundary line).
K1 N1

(b) Substitute $(4, 2)$: $3(4) - 2(2) = 12 - 4 = 8$. Again $8 < 8$ is False!

Does NOT lie in the region.
K1 N1
Question 3 [4 Marks]

A manufacturer produces $x$ units of chair $A$ and $y$ units of chair $B$. Write an inequality for each of the following statements:

(i) The number of chairs $B$ produced is at most three times the number of chairs $A$. [2 marks]

(ii) The minimum total production of chairs is 120 units. [2 marks]

(i) "At most three times":

$$y \le 3x$$
N2

(ii) "Minimum 120 units":

$$x + y \ge 120$$
N2

Section B • Bahagian B (HOTS / KBAT) [18 Marks]

Question 4 (Bakery Production Optimization) [9 Marks]

A bakery bakes $x$ chocolate cakes and $y$ butter cakes daily. The daily baking is based on the following constraints:

  • I: The maximum total number of cakes baked is 100.
  • II: The number of chocolate cakes is at least $\frac{1}{3}$ the number of butter cakes.
  • III: The total flour used is at most $24\text{ kg}$, where a chocolate cake uses $300\text{ g}$ and a butter cake uses $200\text{ g}$.

(a) Write three linear inequalities, other than $x \ge 0$ and $y \ge 0$, representing the constraints. [4 marks]
(b) If the bakery bakes 60 butter cakes on a given day, determine the minimum and maximum number of chocolate cakes that can be baked. [3 marks]
(c) The profit from a chocolate cake is RM15 and from a butter cake is RM10. Calculate the maximum profit if 40 chocolate cakes are baked. [2 marks]

(a) Constraints:

I: $x + y \le 100$
II: $x \ge \frac{1}{3}y$ or $y \le 3x$
III: $300x + 200y \le 24000 \implies 3x + 2y \le 240$
N4

(b) When $y = 60$:

From II: $x \ge \frac{1}{3}(60) \implies x \ge 20$
From III: $3x + 2(60) \le 240 \implies 3x \le 120 \implies x \le 40$
Minimum $x = 20$, Maximum $x = 40$
K1 N2

(c) When $x = 40$:

$3(40) + 2y \le 240 \implies 2y \le 120 \implies y \le 60$
$\text{Profit} = 15(40) + 10(60) = 600 + 600 = \mathbf{RM\ 1,200}$
K1 N1
Question 5 (Tour Bus Rental Modeling) [9 Marks]

A travel agency rents $x$ 40-seater buses and $y$ 25-seater vans to transport at least 300 passengers for an eco-tour:

  • The total number of vehicles available cannot exceed 12.
  • The agency must rent at least 2 buses.
  • The rental cost for a bus is RM800 and for a van is RM500. The budget is at most RM8,000.

(a) Write a system of four linear inequalities representing the constraints. [4 marks]
(b) If exactly 6 buses are rented, find the range of values for $y$ that satisfy all conditions. [3 marks]
(c) State the minimum cost to transport all 300 passengers. [2 marks]

(a) Inequalities:

1. $40x + 25y \ge 300 \implies 8x + 5y \ge 60$
2. $x + y \le 12$
3. $x \ge 2$
4. $800x + 500y \le 8000 \implies 8x + 5y \le 80$
N4

(b) For $x = 6$:

$8(6) + 5y \ge 60 \implies 48 + 5y \ge 60 \implies 5y \ge 12 \implies y \ge 2.4 \implies y \ge 3$
$6 + y \le 12 \implies y \le 6$
$8(6) + 5y \le 80 \implies 5y \le 32 \implies y \le 6.4 \implies y \le 6$
Range: $3 \le y \le 6$
K1 N2

(c) Minimum cost: $x = 8, y = 0 \implies 8(800) = \text{RM } 6,400$ (320 passengers $\ge 300$)

$$\text{Minimum Cost} = \mathbf{RM\ 6,000} \quad (x=5, y=4 \implies 200+100=300 \text{ pax}, 5(800)+4(500)=\text{RM } 6,000)$$
K1 N1
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