KBAT: Optimisation In Context

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. A correct model, sound differentiation and a proper second-derivative justification can score well even if a final arithmetic step slips.

What makes this a KBAT question

A routine differentiation question hands you a function and asks for its stationary point. An optimisation KBAT question does the opposite: it describes a real object, an open box, a fenced field, a production cost, and expects you to construct the function yourself, use a constraint to reduce it to a single variable, then differentiate, solve and justify.

That is Kemahiran Berfikir Aras Tinggi, higher-order thinking: differentiating, solving $\frac{dS}{dx}=0$ and testing with the second derivative are all familiar, but you apply them in an unfamiliar setting and choose the whole plan. In Add Math this rewards students who understand that calculus is a tool for finding best-possible values in the real world, not only a set of rules for handling given expressions.

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 manufacturer designs an open-topped box (no lid) with a square base of side $x$ cm and height $h$ cm. The box must hold a volume of $500$ cm $^3$ .

The material used is measured by the total external surface area $S$ cm $^2$ . (a) Show that $S=x^2+\dfrac{2000}{x}$ .

(b) Find the value of $x$ for which $S$ is a minimum, and justify that it gives a minimum. (c) Hence find the minimum surface area and the corresponding height $h$ .

Show worked solution

Understand. The box has a square base and four sides but no top.

Two quantities can vary, the base side $x$ and the height $h$ , but they are tied together by the fixed volume $500$ cm $^3$ . We want the surface area as a function of $x$ alone, then its minimum.

Plan. Use the volume constraint to write $h$ in terms of $x$ , substitute into the surface-area expression to eliminate $h$ , then differentiate, set $\frac{dS}{dx}=0$ , and confirm a minimum with the second derivative.

Execute and check. (a) The volume gives $x^2h=500$ , so $h=\frac{500}{x^2}$ .

The open box has one square base and four rectangular sides, so its surface area is $S=x^2+4xh$ . Substituting for $h$ :

S=x^2+4x\left(\frac{500}{x^2}\right)=x^2+\frac{2000}{x}

which is the required expression. (b) Differentiate with respect to $x$ :

\frac{dS}{dx}=2x-\frac{2000}{x^2}

Set $\frac{dS}{dx}=0$ : $2x=\dfrac{2000}{x^2}$ , so $2x^3=2000$ , giving $x^3=1000$ and $x=10$ . To justify a minimum, use the second derivative:

\frac{d^2S}{dx^2}=2+\frac{4000}{x^3}

At $x=10$ , $\frac{d^2S}{dx^2}=2+\frac{4000}{1000}=6>0$ , so $S$ is a minimum at $x=10$ cm.

(c) The minimum surface area is $S=10^2+\frac{2000}{10}=100+200=300$ cm $^2$ , and the height is $h=\frac{500}{10^2}=5$ cm.

Check. The volume is $10\times10\times5=500$ cm $^3$ , as required.

Nearby values confirm the minimum: at $x=8$ , $S=314$ cm $^2$ ; at $x=12$ , $S\approx310.7$ cm $^2$ , both larger than $300$ .

Finding a sensible first step

When an optimisation question looks unfamiliar, do not reach for a derivative straight away, you cannot differentiate a function you have not written yet. The reliable first step is to name every quantity and separate the one that is fixed from the ones that can change.

Here the volume is fixed at $500$ cm $^3$ , while $x$ and $h$ can vary. That fixed quantity is your constraint: use it to express one variable in terms of the other, so the thing you are optimising depends on a single variable.

Only then differentiate. A quick sketch with $x$ and $h$ labelled almost always makes the constraint visible.

Once you have $S$ as a function of $x$ alone, the question becomes ordinary Form 5 calculus.

What markers reward

Marking is analytic, so method marks are awarded line by line. On an optimisation-in-context question a marker looks for:

The constraint written explicitly, here $x^2h=500$ , and used to eliminate a variable.
The quantity to be optimised built correctly, such as $S=x^2+4xh$ before substitution.
A clean function of one variable, $S=x^2+\frac{2000}{x}$ , ready to differentiate.
Correct differentiation and the equation $\frac{dS}{dx}=0$ solved, with working shown.
A justification that the stationary point is a minimum, the second derivative positive, or a sign test.
The answer interpreted back into context, with units and the matching height: $300$ cm $^2$ at $h=5$ cm.

How a teacher helps

Optimisation improves fastest with feedback on the modelling, the step where students freeze because there are two variables and no obvious function. In a one-to-one lesson our teachers slow that step down: we label the diagram, write the constraint, and use it to reduce to one variable before any calculus begins, so the habit transfers to new contexts.

We also insist on the justification and the interpretation, because markers reward the second-derivative test and the answer in context, not a bare value of $x$ . 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.

Get 1-to-1 help.

Book a Trial Class

Frequently asked questions

How do I start an optimisation question with two variables?

Find the constraint, the quantity that is fixed. Here the volume $500$ cm $^3$ gives $x^2h=500$ , so $h=\frac{500}{x^2}$ .

Substitute this into the quantity you are optimising so it depends on one variable only; then, and only then, differentiate.

Do I have to use the second derivative to justify a minimum?

You must justify it somehow. The second derivative is cleanest: $\frac{d^2S}{dx^2}=2+\frac{4000}{x^3}=6>0$ at $x=10$ confirms a minimum.

A sign test of $\frac{dS}{dx}$ on either side of $x=10$ is also accepted, markers reward a clear, valid justification.

Why substitute the constraint before differentiating?

Differentiation needs a function of a single variable. With both $x$ and $h$ present you cannot apply $\frac{dS}{dx}$ cleanly.

Using $h=\frac{500}{x^2}$ to eliminate $h$ leaves $S=x^2+\frac{2000}{x}$ , which is ready to differentiate.

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. A correct model, sound differentiation and a proper second-derivative justification can score well even if a final arithmetic step slips.

Source:SRC-DSKP-EN SRC-FORMAT

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

KBAT: Optimisation Problems In Context