KBAT: Maximum & Minimum with Quadratics

Q: Why does the constraint change the best price?

The unrestricted profit peaks at p=5p=5p=5, but that price sells only 75 packets. Requiring at least 100 packets forces p≤4p\le 4p≤4. Because profit rises towards p=5p=5p=5, the best allowed price on p≤4p\le 4p≤4 is the boundary p=4p=4p=4, giving RM200, RM25 less than the unrestricted maximum.

Q: Can I earn marks if my final profit is wrong?

Yes. Add Math Paper 2 is 2 hours 30 minutes and 100 marks, and marking is analytic, method marks are awarded line by line. A correct profit model, a valid completing-the-square step and a sound constraint can score well even if an arithmetic slip changes the final figure.

What makes this a KBAT question

A routine quadratic question gives you $y=ax^2+bx+c$ and asks for the minimum. A KBAT question gives you a situation, a price, a demand, a cost, and expects you to build the quadratic first, then optimise it, then decide whether the answer is allowed.

That is Kemahiran Berfikir Aras Tinggi, higher-order thinking: completing the square is ordinary Form 4 work, but forming 'profit = (price − cost) × quantity', reading the vertex as the best price, and testing it against a real-world limit are steps no one lists for you. In Add Math this rewards students who understand that the vertex of a quadratic is the maximum or minimum, and who can move between a story, an equation and a graph.

The modelling and the interpretation, not the algebra, carry the difficulty.

One worked problem, in the style of Paper 2

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

Q1[8 marks]

A stall sells packets of nasi lemak. A market survey shows that at a selling price of RM $p$ per packet, the number sold each day is $n=200-25p$ .

Each packet costs the stall RM2 to make. (a) Show that the daily profit, in ringgit, is $P=-25p^{2}+250p-400$ , and by completing the square find the price that maximises the profit and the maximum daily profit.

(b) The owner also wants to sell at least 100 packets a day. Find the highest daily profit possible under this condition, and comment on the trade-off.

Show worked solution

Understand. Profit per packet is (price − cost) = $p-2$ .

The number sold is $n=200-25p$ . Daily profit is profit per packet times number sold.

We must build $P$ , complete the square, then re-optimise under a sales constraint.

Plan. Write $P=(p-2)(200-25p)$ , expand, complete the square to read the vertex.

For part (b), turn 'at least 100 packets' into an inequality in $p$ and find the best allowed price.

Execute and check. (a) Form the profit:

P=(p-2)(200-25p)=200p-25p^{2}-400+50p=-25p^{2}+250p-400

Complete the square. Factor $-25$ from the first two terms:

P=-25\left(p^{2}-10p\right)-400=-25\left[(p-5)^{2}-25\right]-400

P=-25(p-5)^{2}+625-400=-25(p-5)^{2}+225

Since $-25(p-5)^{2}\le 0$ , the maximum occurs when $(p-5)^{2}=0$ , i.e. $p=5$ . The maximum daily profit is RM225, at a price of RM5 per packet.

(At $p=5$ , $n=200-25(5)=75$ packets, a sensible positive number.)

(b) 'At least 100 packets' means $n\ge 100$ :

200-25p\ge 100 \;\Rightarrow\; 25p\le 100 \;\Rightarrow\; p\le 4

From the vertex form, $P$ increases as $p$ rises towards $5$ . On the allowed range $p\le 4$ , the profit is therefore largest at $p=4$ :

P(4)=-25(4-5)^{2}+225=-25(1)+225=200

So under the condition the best price is RM4, giving a daily profit of RM200 with exactly $200-25(4)=100$ packets sold. The trade-off: insisting on at least 100 packets costs the stall RM25 a day (RM225 − RM200) compared with the unrestricted optimum, in exchange for higher volume and reach.

Check. Test the vertex a different way: $P(5)=(5-2)(200-125)=3\times 75=225$ ✓, and $P(4)=(4-2)(200-100)=2\times 100=200$ ✓.

Both agree with the completed-square form, so the readings are consistent.

Finding a sensible first step

When an optimisation question is dressed as a story, the reliable first step is to name the quantity you want to make as large or small, here, profit, and write it as a product of simpler pieces before expanding. Profit is 'money per packet times packets sold', so $P=(p-2)(200-25p)$ .

Getting that single line right is worth more than any later algebra, because everything else depends on it. Only after the model is formed do you expand and complete the square; the vertex then hands you both the best price and the best value at once.

Resist the urge to substitute numbers early, keep $p$ as a variable so the completed-square form can reveal the turning point. Naming the target quantity and writing it as a product turns a wordy problem into a standard quadratic.

What markers reward

Marking is analytic, so method marks are awarded line by line. On a maximum/minimum application a marker looks for:

The profit modelled correctly as $(p-2)(200-25p)$ , price minus cost, times quantity.
A clean expansion to $-25p^{2}+250p-400$ .
Completing the square with the $-25$ factored out first, reaching $-25(p-5)^{2}+225$ .
The maximum read from the vertex, value RM225 at $p=5$ , not from trial and error.
The constraint translated into an inequality, $200-25p\ge 100\Rightarrow p\le 4$ .
A justified choice of $p=4$ on the allowed range and a clear statement of the RM25 trade-off.

How a teacher helps

Optimisation improves fastest when a student can explain why the vertex is the answer, not just compute it. In a one-to-one lesson our teachers ask you to build the profit line in words first, to complete the square carefully, the factored $-25$ is where marks are often lost, and to say what the maximum means for the stall.

We pay special attention to constraints, because a KBAT question usually turns on whether the ideal answer is actually allowed. 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

Why complete the square instead of using calculus for a Form 4 quadratic?

Because the maximum or minimum of a quadratic is exactly its vertex, and completing the square gives the vertex directly: $-25(p-5)^{2}+225$ shows the maximum value 225 at $p=5$ in one line. It is the intended Form 4 method for this chapter and needs no differentiation.

How do I know whether the turning point is a maximum or a minimum?

Look at the sign of the squared term after completing the square. Here the coefficient is $-25$ , which is negative, so $-25(p-5)^{2}$ is at most zero, the curve opens downward and the vertex is a maximum.

A positive coefficient would give a minimum.

Why does the constraint change the best price?

The unrestricted profit peaks at $p=5$ , but that price sells only 75 packets. Requiring at least 100 packets forces $p\le 4$ .

Because profit rises towards $p=5$ , the best allowed price on $p\le 4$ is the boundary $p=4$ , giving RM200, RM25 less than the unrestricted maximum.

Can I earn marks if my final profit is wrong?

Yes. Add Math Paper 2 is 2 hours 30 minutes and 100 marks, and marking is analytic, method marks are awarded line by line.

A correct profit model, a valid completing-the-square step and a sound constraint can score well even if an arithmetic slip changes the final figure.

Source:SRC-DSKP-EN SRC-FORMAT

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

KBAT: Maximum and Minimum with Quadratics