SPM Paper 2 Section C Priority (10 Marks)

Constraint Modeling, Feasible Region & Optimal Solutions

Linear Programming is an applied mathematical method used in industrial operations research. Master converting verbal constraints into mathematical inequalities, graphing boundary lines accurately to shade feasible region $R$, and using the objective function line $ax + by = k$ to determine maximum profit or minimum cost.

1. Verbal Phrases into Mathematical Inequalities

English Phrasing Bahasa Melayu Phrasing Mathematical Inequality
$y$ is at least $k$ / not less than $k$ $y$ sekurang-kurangnya $k$ / tidak kurang daripada $k$ $y \ge k$
$y$ is at most $k$ / not more than $k$ $y$ selebih-lebihnya $k$ / tidak lebih daripada $k$ $y \le k$
Total of $x$ and $y$ does not exceed $M$ Jumlah $x$ dan $y$ tidak melebihi $M$ $x + y \le M$
The number of $y$ exceeds $x$ by at least $d$ Bilangan $y$ melebihi $x$ sekurang-kurangnya $d$ $y - x \ge d$
Ratio of $y$ to $x$ is at most $p : q$ Nisbah $y$ kepada $x$ selebih-lebihnya $p : q$ $\frac{y}{x} \le \frac{p}{q} \implies qy \le px$
$y$ is at least twice of $x$ $y$ adalah sekurang-kurangnya dua kali $x$ $y \ge 2x$

2. SPM Examiner Pitfalls in Linear Programming

Trap 1: Strict vs Non-Strict Inequalities ($>$ vs $\ge$):

In SPM Linear Programming, nearly all boundary inequalities are non-strict ($\le$ or $\ge$) and drawn with solid lines. If a strict inequality ($>$ or $<$) is explicitly stated (e.g. "more than"), draw a dashed line. Using solid lines for dashed boundaries loses 1 mark ($K1$).

Trap 2: Finding Maximum / Minimum with Objective Line ($k$):

SPM marking schemes strictly reward showing the objective line $ax + by = k$ on the graph paper! Choose an easy value of $k = \text{LCM}(a, b)$, plot the reference line, and draw a parallel dashed arrow shifting to the furthest vertex in $R$. Do not simply guess points!

Trap 3: Integer Constraint for Real-Life Objects:

If the optimal vertex has non-integer coordinates (e.g. $(14.6, 22.3)$) and the problem involves whole units (cakes, trucks, tables), you MUST test the neighboring integer coordinates inside region $R$. Stating a decimal answer for people or chairs results in an automatic deduction!

3. Full 10-Mark SPM Paper 2 Section C Model Worked Example

SPM Paper 2 Section C Full Blueprint Model [10 Marks]
A furniture workshop produces two models of study desks: Model $A$ ($x$ units) and Model $B$ ($y$ units) per week. The production is subject to the following operational constraints:
  • I: The total number of desks produced is at least $30$ units.
  • II: The number of Model $B$ desks produced is at most three times the number of Model $A$ desks.
  • III: The total assembly time available per week is $240$ hours. Model $A$ requires $4$ hours to assemble while Model $B$ requires $2$ hours to assemble.
  • (a) Write three linear inequalities, other than $x \ge 0$ and $y \ge 0$, which satisfy all the above constraints. [3 marks]

    (b) Using a scale of $2\text{ cm to } 10\text{ units}$ on both axes, construct and shade the feasible region $R$ that satisfies all the constraints. [3 marks]

    (c) Using the graph constructed in (b), find:

    (i) The maximum number of Model $B$ desks that can be produced if $15$ units of Model $A$ desks are produced. [1 mark]

    (ii) The maximum weekly profit earned by the workshop if the profit from each Model $A$ desk is RM 80 and from each Model $B$ desk is RM 60. [3 marks]

    Step-by-Step Marking Scheme:
    (a) Constraint I: $x + y \ge 30$ [1m: N1]
    Constraint II: $y \le 3x$ [1m: N1]
    Constraint III: $4x + 2y \le 240 \implies 2x + y \le 120$ [1m: N1]
    (b) At least one line drawn correctly with correct scale ($2\text{ cm to } 10$). [1m: K1]
    All 3 boundary lines ($x+y=30, y=3x, 2x+y=120$) drawn accurately with intercepts. [1m: K1]
    Region $R$ shaded correctly satisfying all 3 inequalities and first quadrant. [1m: N1]
    (c)(i) At $x = 15$, look at vertical line $x = 15$ inside region $R$: upper boundary is $2x + y = 120 \implies 2(15) + y = 120 \implies y = 90$. Check $y \le 3x \implies y \le 45$. Max $y$ constrained by $y = 3(15) = 45$. [1m: N1: Max Model B = 45]
    (c)(ii) Objective Profit function: $P = 80x + 60y$. Plot reference line $80x + 60y = 2400 \implies 4x + 3y = 120$. [1m: K1: Draw objective line $k$]
    Shifting parallel upwards, the furthest vertex is the intersection of $y = 3x$ and $2x + y = 120$: $2x + 3x = 120 \implies 5x = 120 \implies x = 24, y = 72$. [1m: K1: Identify optimal vertex $(24, 72)$]
    Max Profit $= 80(24) + 60(72) = 1920 + 4320 = \text{RM } 6,240.00$. [1m: N1: Correct maximum profit]
    Open Form 5 Chapter 7 Section C Worksheet