KBAT: Linear Programming Modelling

Q: How do I turn 'at least as many A as B' into an inequality?

Read it as a comparison: the number of AAA is greater than or equal to the number of BBB, so A≥BA\ge BA≥B. A useful test is a sample case, if you may have 666 of AAA and 666 of BBB but not 333 of AAA and 666 of BBB, the boundary is A=BA=BA=B with AAA on the larger side, confirming A≥BA\ge BA≥B.

Q: How is Add Math Paper 2 marked on these questions?

Paper 2 is 2 hours 30 minutes and 100 marks, and marking is analytic, method marks are awarded line by line. Defining variables, writing each inequality, shading the region, finding the corners and stating the optimum in context each earn credit, so show every step rather than only the final numbers.

What makes this a KBAT question

A routine linear-programming question already gives you the inequalities and the region to shade. A modelling KBAT question gives you only the story,'at most twelve cards', 'at least as many of one as the other', 'a profit to maximise', and asks you to build the whole model yourself.

That is higher-order thinking, Kemahiran Berfikir Aras Tinggi: the graphing and the corner-point test are familiar, but you must decide what the variables are, translate each sentence into the correct inequality with the right $\ge$ or $\le$ , and write the objective function. Nothing labels which line is which.

In Add Math this rewards students who can turn words into a correct system and then read a real answer off the region, not only students who can shade a graph they were handed.

One worked problem, in the style of Paper 2

This is an original question written in the style of SPM Paper 2. Try it before reading the solution.

Q1[10 marks]

A school craft club makes two kinds of card each week: $x$ greeting cards and $y$ gift cards. The club works within three conditions: it can make at most $12$ cards in a week; it makes at least as many greeting cards as gift cards; and it makes at least $3$ gift cards.

Each greeting card gives a profit of RM2 and each gift card gives a profit of RM5. (a) Write three inequalities, other than $x\ge 0$ and $y\ge 0$ , that model the conditions.

(b) Find how many of each card the club should make to maximise its weekly profit, and state that profit.

Show worked solution

Understand. We must decide two numbers, greeting cards $x$ and gift cards $y$ , subject to three limits, and make the profit $P=2x+5y$ as large as possible.

This is a linear programming problem, so the greatest profit is reached at a corner of the region that satisfies every inequality.

Plan. Translate each sentence into one inequality, list the corner points where two boundary lines meet, and test $P=2x+5y$ at each corner.

The largest value is the answer.

Execute and check. (a) 'At most $12$ cards' gives a limit on the total; 'at least as many greeting cards as gift cards' compares the two; 'at least $3$ gift cards' gives a floor on $y$ :

x+y\le 12, \qquad x\ge y, \qquad y\ge 3

(b) The objective is the profit $P=2x+5y$ . The corners of the region are where the boundary lines cross.

Solving them in pairs:

x=y \text{ and } y=3 \Rightarrow (3,3); \quad y=3 \text{ and } x+y=12 \Rightarrow (9,3)

x=y \text{ and } x+y=12 \Rightarrow 2x=12 \Rightarrow (6,6)

Now test the profit at each corner:

Corner $(x,y)$	Profit $P=2x+5y$ (RM)
$(3,3)$	$2(3)+5(3)=21$
$(9,3)$	$2(9)+5(3)=33$
$(6,6)$	$2(6)+5(6)=42$

The greatest profit is RM42, at $(6,6)$ . So the club should make 6 greeting cards and 6 gift cards each week.

Check. The point $(6,6)$ satisfies every condition: $6+6=12\le 12$ , $6\ge 6$ , and $6\ge 3$ .

A gift card earns more than a greeting card, so the club wants as many gift cards as it can, but the rule $x\ge y$ stops it going past equal numbers, and $x+y\le 12$ caps the total. Equal numbers at the total cap is exactly the best allowed mix, which agrees with the table.

Finding a sensible first step

When a modelling question looks like a wall of words, the first step is to define the two variables in one line, with units,'let $x$ be the number of greeting cards and $y$ the number of gift cards per week'. Everything else hangs off those two letters.

Then read the passage one sentence at a time and convert each into a single inequality, watching the direction: 'at most' and 'no more than' give $\le$ ; 'at least' and 'a minimum of' give $\ge$ ; 'at least as many $A$ as $B$ ' becomes $A\ge B$ . Keep the profit or cost sentence aside, that is the objective function, not a constraint.

Once the system is written, the graph and the corner test are routine, so spend your care on the translation.

What markers reward

Marking is analytic, so method marks are awarded line by line. On a linear-programming modelling question a marker looks for:

The two variables defined clearly, with units,'let $x$ be the number of greeting cards'.
Each condition written as a correct inequality with the right direction, such as $x+y\le 12$ and $x\ge y$ .
The objective function stated separately, for example $P=2x+5y$ .
The region drawn and shaded correctly, with the boundary lines labelled.
The corner points found by solving pairs of lines, not read off approximately.
The objective tested at every corner (or the objective line slid), then the optimum stated in context with its value.

How a teacher helps

The marks in a modelling question are won at the translation, so that is what our teachers rehearse most. In a one-to-one lesson we take a passage apart sentence by sentence, deciding together which is a constraint and which is the objective, and pinning the direction of every $\ge$ and $\le$ .

We practise the phrase students trip on,'at least as many $A$ as $B$ ', until $A\ge B$ is automatic. Teachers at spmaddmath.com.my are experienced, and lessons are online and taught in English.

Message us on WhatsApp to arrange a one-hour paid class from RM50/hr at the teacher's rate.

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Frequently asked questions

How do I turn 'at least as many A as B' into an inequality?

Read it as a comparison: the number of $A$ is greater than or equal to the number of $B$ , so $A\ge B$ . A useful test is a sample case, if you may have $6$ of $A$ and $6$ of $B$ but not $3$ of $A$ and $6$ of $B$ , the boundary is $A=B$ with $A$ on the larger side, confirming $A\ge B$ .

Do I always have to test the corners?

For a linear objective on a polygon, the maximum and minimum always occur at a corner (and possibly all along an edge). Testing every corner is a reliable method.

Alternatively you can slide the objective line across the region until it last touches a corner, both are accepted, so use whichever you are surer of.

Why define the variables with units first?

Because every inequality and the objective function refer back to them. Writing 'let $x$ be the number of greeting cards per week' fixes what $x$ means, prevents you from mixing up the two products, and earns the first method mark.

Undefined variables cost marks even when the algebra is right.

How is Add Math Paper 2 marked on these questions?

Paper 2 is 2 hours 30 minutes and 100 marks, and marking is analytic, method marks are awarded line by line. Defining variables, writing each inequality, shading the region, finding the corners and stating the optimum in context each earn credit, so show every step rather than only the final numbers.

Source:SRC-DSKP-EN SRC-FORMAT

Written by the spmaddmath.com.my editorial team.· Last updated 5 September 2026

KBAT: Modelling a Linear Programming Problem