Maximum Volume Surface: Optimisation Exam Method Explained

Maximum Volume Surface: Optimisation Explained

Optimisation: Maximum Volume with Fixed Surface Area

🧭 Why this question is really about balance, not brute calculus

Optimisation questions that ask for the maximum volume given a fixed surface area often look familiar. Cylinders appear, formulas come to mind, and many students instinctively reach for differentiation as quickly as possible.

That instinct is understandable — and it’s exactly where marks start to leak.

This question is not really about calculus. It is about recognising what is fixed, what is allowed to vary, and how those two ideas interact. Students who rush tend to differentiate expressions that are not yet meaningful, or optimise the wrong quantity entirely.

This kind of thinking is where A Level Maths understanding for mocks really shows up. The calculus only works once the modelling does.

At this point in an exam, this is usually where I stop students. Not to check differentiation, but to ask a quieter question: what exactly is fixed here, and what is still allowed to change?
If that answer isn’t clear, the rest of the solution doesn’t really matter — even perfect calculus won’t rescue it.

A secure understanding of this question depends on the optimisation framework developed in Optimisation — Method & Exam Insight, especially handling constraints carefully.

🔙 Previous topic:

Finding the maximum volume with a fixed surface area builds directly on earlier optimisation work, where calculus was used to minimise the surface area of a cylinder by forming and differentiating the relevant expressions.

📘 What the examiner is actually testing

When surface area is fixed and volume is to be maximised, the examiner is deliberately reversing the usual roles students are comfortable with. Normally, volume feels like the given quantity. Here, it becomes the target.

Surface area is the constraint.

That reversal is not accidental. It allows examiners to see whether students can:

treat a physical condition as a mathematical restriction,
express it correctly,
and use it to reduce the problem to a single variable.

Students who differentiate before doing this almost always lose method marks, even if the calculus itself is correct.

🧠 The modelling decision that controls everything

Before writing any formulas, there is one decision that determines whether the solution will work:

Which variable will the final expression depend on?

Volume depends on more than one dimension. Differentiation only makes sense once that dependence has been reduced.

This is not a technical step — it is a reasoning step. Optimisation questions reward thinking before writing.

✏️ Setting the geometry up carefully

Consider a closed cylinder with radius $r$ and height $h$ .

The volume of the cylinder is:

$V = \pi r^2 h$

This is the quantity we want to maximise.

The surface area of the closed cylinder is:

$S = 2\pi r^2 + 2\pi r h$

Here, $S$ is fixed. This equation is not optional — it is the restriction that limits how large the volume can become.

At this stage, nothing should be differentiated. We still have two variables, and differentiation cannot tell us anything useful yet.

🔁 Using the surface area constraint properly

The constraint equation:

$S = 2\pi r^2 + 2\pi r h$

can be rearranged to express $h$ in terms of $r$ :

$h = \frac{S – 2\pi r^2}{2\pi r}$

This is the crucial substitution — and it’s also where students often feel slightly uncomfortable. The expression gets messier before it gets simpler. That discomfort is normal. In fact, it’s a good sign that the modelling is doing its job.

Once this step is done correctly, the rest of the question becomes routine. If it is done badly, no amount of correct differentiation will rescue the solution. This is exactly the sort of careful setup that A Level Maths revision guidance is designed to reinforce.

🧮 Building the single-variable volume function

Substitute this expression for $h$ into the volume formula:

$V = \pi r^2\left(\frac{S – 2\pi r^2}{2\pi r}\right)$

Simplifying carefully gives:

$V = \frac{Sr}{2} – \pi r^3$

This line is the turning point of the entire solution.

We now have volume written entirely in terms of one variable, $r$ . Only now does differentiation make sense.

It’s worth pausing here and noticing something slightly odd. We’ve done very little calculus, and yet the question is essentially finished.
That’s deliberate. Optimisation questions look like calculus questions, but they are really thinking questions that happen to use differentiation at the end.

📉 Differentiation — briefly and deliberately

Differentiate $V$ with respect to $r$ :

$\frac{dV}{dr} = \frac{S}{2} – 3\pi r^2$

To find the maximum volume, set this equal to zero:

$\frac{S}{2} – 3\pi r^2 = 0$

Rearranging gives:

$r^2 = \frac{S}{6\pi}$

This is the radius at which the volume is maximised.

🔍 Interpreting the result properly

Using the earlier expression for $h$ and substituting this value of $r$ , it follows that:

$h = 2r$

This relationship matters far more than the algebra itself.

It tells us that the cylinder with maximum volume for a fixed surface area has height equal to its diameter. Examiners reward this conclusion because it shows understanding rather than just manipulation.

🧠 Why students usually lose marks

Most lost marks here do not come from differentiation errors. They come from earlier decisions.

Common issues include treating volume as the constraint instead of surface area, differentiating before eliminating a variable, or reaching the correct algebraic result but failing to interpret it in words.

Optimisation questions are decided by modelling, not speed.

🧮 Worked Exam Example

🧪 Worked Exam Example
A closed cylinder has surface area $600\pi$ . Find the maximum possible volume.

The surface area constraint is:

$2\pi r^2 + 2\pi r h = 600\pi$

So:

$h = \frac{600\pi – 2\pi r^2}{2\pi r} = \frac{300}{r} – r$

Volume:

$V = \pi r^2 h$

Substitute:

$V = \pi r^2\left(\frac{300}{r} – r\right)$

Simplifying gives:

$V = 300\pi r – \pi r^3$

Differentiate:

$\frac{dV}{dr} = 300\pi – 3\pi r^2$

Set equal to zero:

$300\pi = 3\pi r^2$

So:

$r^2 = 100 \quad \Rightarrow \quad r = 10$

Then:

$h = \frac{300}{10} – 10 = 20$

Confirming again that $h = 2r$ .

Author Bio – S. Mahandru

When students struggle with optimisation, it’s rarely the calculus. It’s the modelling. In lessons, I deliberately delay differentiation and make students explain what is fixed and what is free to vary. Once that is clear, the maths usually behaves itself.

🧭 Next topic:

After solving optimisation problems involving maximum volume, the focus shifts to Integration by Parts, where products such as $\int x e^x,dx$ are evaluated using structured calculus techniques.

🎯 Final exam takeaway

If there is one habit to take from this question, it is this: identify the constraint, reduce to one variable, then differentiate. Practising this consistently — supported by a A Level Maths Revision Course trusted by teachers — turns optimisation from a stressful topic into a predictable one.

❓ Quick FAQs

🧭 Why does a fixed surface area limit the volume so strongly?

Because surface area controls how much material is available to enclose space. If too much surface area is used on the ends, there isn’t enough left to build height. If too much is used on height, the ends become inefficient. The maximum volume occurs when these contributions are balanced. This balance is exactly what the calculus reveals, and it is what examiners expect students to interpret.

🧠 Do I need to prove the stationary point is a maximum?

Often the context makes this clear, but you should always read the question carefully. In longer problems, a brief justification using the second derivative or by reasoning from the geometry may be expected. Examiners do not want unnecessary work, but they do expect correct interpretation. A short explanation is usually enough.

⚖️ Why does the result keep turning into

h = 2r

Because that ratio represents the most efficient geometric balance. If the cylinder is too tall, surface area is wasted on the curved surface. If it is too wide, surface area is wasted on the ends. The optimum occurs when those effects are balanced. This is a genuine geometric principle, not a coincidence, which is why it appears so often in exams.