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Advanced Optimisation
🔥 Advanced Optimisation: Maximum Volume, Minimum Cost Exam Problems
Right, let’s get into this one — and honestly, these multi-step optimisation questions are the ones where students either grin or groan. They look innocent, but the moment the question starts mixing cost, constraints and volume… hang on— that’s when the algebra starts throwing elbows. 🎯 Anyway, I’ll talk it through the way I would in class, nice and steady, with a few pauses where people usually get stuck. 📘
And if you’re doing A Level Maths Revision, this type of question is absolutely core — it pops up in every exam series.
🔙 Previous topic:
And if you want a cleaner warm-up before tackling mixed-cost and maximum-volume setups, step back to Optimisation Problems: Minimising Surface Area (Water Tank Example) — same core method, just a gentler entry point.
📘 Exam Context
These “maximum volume, minimum cost” questions appear across all the major A Level Maths exam boards, usually in the latter half of Section B. They test whether you can build a model from a messy description and not panic when the algebra gets tangled. The mark schemes reward clear constraints and correct use of calculus — even if your final number is slightly off.
📐 Problem Setup
Suppose you’re designing a closed cylindrical container. You want maximum volume but you’re working with a fixed total cost for the metal sheets. The cost per unit area of the curved surface is different from the cost of the circular ends. The volume is V=\pi r^{2}h.
You’ll be asked to form a cost equation, substitute the constraint, and optimise.
🖼 Required Diagram
🔹 Subsection 1: Converting the words into algebra
Alright, this is the true hurdle. Students assume the modelling part is “just reading the question,” but actually it’s where the marks leak away. The cost is usually something like: curved area multiplied by one cost rate, ends multiplied by another. And curved area uses the lateral surface formula 2\pi rh.
Let me pause here — because this is where the question loves hiding a twist: sometimes the top is not required (an open container) or the cost rates differ slightly. Exams absolutely adore that tiny difference.
🟦 Subsection 2: Building the cost constraint
So once you’ve got the curved cost and the end cost, add them together and set them equal to the fixed budget. The constraint might look like C = a(2\pi rh)+b(2\pi r^{2}).
But — and this is where I tell my class to slow down — your aim is to express one variable in terms of the other. Lots of students jump straight to volume first, but eliminating h or r earlier makes the differentiation job a whole lot kinder.
🟢 Subsection 3: Substituting into the volume (or cost) function
Once you’ve used the constraint to tidy the variables, you plug the relationship into your main target function. If we’re maximising volume, then substitute h from the cost equation into V=\pi r^{2}h.
The algebra always balloons here. It’s normal — the examiners expect it. Just keep the steps consistent and neat enough to follow.
🟥 Subsection 4: Differentiating — but without losing the plot
When everything is in a single-variable form, that is when you differentiate.
Sometimes you’ll end up with something like V'(r)=0 to find the stationary point.
What really matters is actually saying what you’re doing:
“I’m differentiating with respect to r to find maximum volume.”
It’s a tiny sentence. It’s also often a mark.
🟨 Subsection 5: Checking the nature of the stationary point
We’re not done until you show it’s a maximum or minimum.
The second derivative, or checking sign changes, works fine.
Something like V''(r)<0 tells the examiner exactly what they want to see — that the value you found is genuinely a maximum.
🔸 Subsection 6: Interpreting the result properly
After finding the optimised r (or h), interpret it.
Does that radius make sense?
Does it conflict with the cost condition?
This part is strangely overlooked, but it’s where examiners tuck the last two method marks in these long questions.
🟣 Subsection 7: Connecting cost and volume simultaneously
Sometimes the question flips: instead of maximising volume with fixed cost, it asks you to minimise cost with a fixed volume.
Same logic, same modelling, just a different constraint.
In that case, you’d use something like h=V/(\pi r^{2}) to remove the height.
🧷 Subsection 8: Choosing which variable to eliminate
This is sneakily important. Always eliminate the variable that leads to simpler algebra.
Students often eliminate r “because it feels natural,” but eliminating h often produces cleaner expressions.
No penalty for the “harder” route — but it wastes exam time.
🔻 Subsection 9: Spotting when the model needs adjusting
Questions sometimes include different material costs, or even a thickness condition. This makes your cost formula asymmetric.
Just treat each surface separately.
Messy algebra doesn’t lose marks — missing a term does.
⭐ Subsection 10: Avoiding assumptions about symmetry
Classic myth: “maximum volume shapes are always symmetric.” No.
Different material costs break symmetry.
If h turns out surprisingly small, that’s often exactly what the cost model demands.
❗ Common Errors & Exam Traps
Thinking curved surface area is \pi r^{2} (happens every year)
Mixing up fixed cost and fixed volume
🎯 Leaving both variables in the expression and trying to differentiate
Forgetting the max/min check
Rounding too early and losing accuracy
📘 Ignoring context (negative height, unrealistic proportions)
🌍 Real-World Link
Manufacturers genuinely solve these problems — though using software, not tired students. Ever wondered why drinks cans or food tubs look oddly proportioned? Cost optimisation. A tiny reduction in material per unit becomes millions saved per year. This is basically the A Level Maths version of industrial design.
🚀 Next Steps
If you want more structured practice on modelling and constraint-heavy questions, check out the full A Level Maths Revision Course — especially useful if you’re deep into your A Level Maths Revision and want calm, guided examples instead of last-minute panic.
📏 Recap Table
Build the cost/volume constraint
Express one variable in terms of the other
Substitute into V=\pi r^{2}h or cost
Differentiate single-variable expression
Set derivative to zero
Verify max/min
Interpret in context
👤Author Bio – S Mahandru
I’m a maths teacher who’s spent too many evenings marking optimisation attempts and quietly muttering about missing radius–height checks. I write these breakdowns to make the tricky bits feel more human — and to help students secure those last few marks in A Level Maths exams.
🧭 Next step:
If you want to strengthen the engine behind these optimisation problems, move next to Differentiation Techniques Every A Level Student Must Master, because product/chain/quotient fluency is what keeps the algebra behaving when the modelling gets heavy.
❓ FAQ Section
Q1: Should I always use the second derivative test?
Not necessarily. Checking the sign on either side works just as well.
Q2: What if my algebra becomes a mess?
Totally normal. As long as the structure is correct, examiners will award method marks.
Q3: Can I leave my answers in terms of π?
Usually yes, unless the context or instructions say otherwise.