A fixed-volume cylinder reveals how to minimise surface area by differentiating a single-variable objective and checking positive domains and design conditions.
Read the cylinder optimisation guide
- The quick answer: optimal height equals the diameter
- Define the cylinder’s dimensions
- Fix the volume before changing dimensions
- Why a radius increase forces the height down
- Substitute the volume constraint into the surface area
- Interpret the two surface-area contributions
- Compare several exact designs
- Differentiate the surface-area function
- Find the stationary radius
- Find the corresponding height
- Evaluate the minimum surface area
- Use the second derivative test
- Prove it is the global minimum
- Why the extreme shapes are inefficient in this model
- What balance occurs at the optimum?
- Derive the general result for any fixed volume
- Why this is a shape rule rather than a size rule
- An independent algebraic check through a new variable
- A counterexample: an open-top cylinder has a different optimum
- Differentiate the open-top model
- Find the open-top height
- Why “least surface area” is not the same as “lowest manufacturing cost”
- What if the radius must be no more than four centimetres?
- Why an endpoint can beat every admissible stationary candidate
- What if the container height must be at least twelve centimetres?
- Why volume and surface area have different units
- A six-minute independent check
- Common mistake: counting only one circular end
- Common mistake: treating radius and height as independent
- Common mistake: minimising the wrong quantity
- Common mistake: stopping after S′ = 0
- Common mistake: ignoring the admissible domain
- Common mistake: assuming the cylinder ratio works for every container
- What parents can ask without doing the calculus
- How focused A-Math tuition can diagnose the real difficulty
- Where this fits in Singapore G3 Additional Mathematics
- Frequently asked questions
- Other Additional Mathematics applications
- Official reference and evidence boundaries
What is Additional Mathematics when a container must hold a fixed volume but can be made tall and narrow, short and wide, or somewhere between? A closed cylinder gives a clear optimisation problem.
Imagine a hypothetical container shaped like a circular cylinder with a flat circular top and bottom. The amount it must hold is fixed. We can vary its radius and height, but changing either dimension affects the other.
A very narrow container needs to be tall to hold the required volume. A very wide container can be short, but its two circular ends become large. Somewhere between those extremes is a shape that minimises the idealised total surface area.
Additional Mathematics can prove where that minimum occurs using functions, differentiation, domain restrictions and a second-derivative test.
This article uses an original mathematical example. It does not provide manufacturing, food-storage or engineering specifications. Real materials require thickness, seams, openings, stability, strength and cost information not included in the ideal model.
The quick answer: optimal height equals the diameter
For an ideal closed circular cylinder with fixed positive volume and unrestricted positive radius and height, the minimum surface area occurs when:
Height = 2 × radius.
In other words:
h = 2r.
The height equals the diameter of the circular ends.
That result may seem surprising. But it follows from the formula for volume, the formula for surface area and a valid optimisation argument.
We will prove it with a particular target volume first, then derive the general relationship.
The crucial assumption is that the cylinder has both a top and a bottom. An open-top container has a different surface-area formula and a different optimum.
Define the cylinder’s dimensions
Let:
r be the cylinder’s radius in centimetres.
h be its perpendicular height in centimetres.
The radius must be positive:
r > 0.
The height must also be positive:
h > 0.
For an ideal circular cylinder, the volume is:
V = πr²h.
The total surface area of a closed cylinder is:
S = 2πr² + 2πrh.
The first term gives the areas of the two circular ends.
The second term gives the curved side’s area.
We cannot minimise surface area correctly unless we include all the surfaces required by the question.
Fix the volume before changing dimensions
Suppose our hypothetical container must have volume:
V = 250π cm³.
This is approximately 785 cm³, but keeping the exact form is more convenient for algebra.
The volume condition is:
πr²h = 250π.
Divide by π, which is positive and nonzero:
r²h = 250.
Solve for the height:
h = 250/r².
Now height is determined by radius.
We no longer have two independently adjustable dimensions. Any selected positive radius fixes the height needed to preserve the volume.
This is the central modelling step.
Why a radius increase forces the height down
The constraint is:
h = 250/r².
If r doubles, r² becomes four times as large.
To maintain the same product r²h, the height must become one-quarter as large.
For example, at radius 5 cm:
h = 250/25 = 10 cm.
At radius 10 cm:
h = 250/100 = 2.5 cm.
The radius doubled and height became one-quarter.
The volume remains:
π(5²)(10) = 250π.
And:
π(10²)(2.5) = 250π.
Both containers hold the same idealised volume but have different shapes and surface areas.
Substitute the volume constraint into the surface area
Start with the closed-cylinder surface area:
S = 2πr² + 2πrh.
Substitute:
h = 250/r².
Then:
S(r) = 2πr² + 2πr(250/r²).
Simplify the second term:
2πr(250/r²) = 500π/r.
Therefore:
S(r) = 2πr² + 500π/r, for r > 0.
This function contains one variable.
That is important because we can now differentiate with respect to radius and locate the minimum.
The first term increases as the radius grows. The second decreases as the radius grows.
Optimisation balances those two competing effects.
Interpret the two surface-area contributions
The term:
2πr²
represents the combined area of the two circular ends.
When the radius increases, these ends become larger.
The term:
500π/r
represents the curved side surface under the fixed-volume condition.
A larger radius allows a shorter container, so the side contribution decreases as r increases.
Neither trend alone determines the optimum.
If we focus only on the ends, we might make r almost zero.
If we focus only on the curved side, we might make r arbitrarily large.
The minimum of the total requires accounting for both terms simultaneously.
Compare several exact designs
| Radius r | Height h = 250/r² | Closed surface area |
|---|---|---|
| 2.5 cm | 40 cm | 212.5π cm² |
| 4 cm | 15.625 cm | 157π cm² |
| 5 cm | 10 cm | 150π cm² |
| 6 cm | 125/18 cm | 466π/3 cm² |
| 10 cm | 2.5 cm | 250π cm² |
The table suggests that radius five centimetres gives a smaller area than the nearby listed choices.
But checking a handful of radii is not proof of a global minimum.
There are infinitely many positive possible radius values.
We need a mathematical argument that no other permitted radius produces a smaller surface area.
Differentiation supplies that argument.
Differentiate the surface-area function
Write:
S(r) = 2πr² + 500πr⁻¹.
Using the power rule:
d(2πr²)/dr = 4πr.
And:
d(500πr⁻¹)/dr = −500πr⁻².
Therefore:
S′(r) = 4πr − 500π/r².
This derivative measures how the surface area changes for a small change in radius while the volume remains fixed.
Its unit is square centimetres per centimetre, which simplifies dimensionally to centimetres.
The derivative is not itself the container’s surface area.
It is the sensitivity of that area to radius.
Find the stationary radius
A stationary point occurs where:
S′(r) = 0.
Set:
4πr − 500π/r² = 0.
Move the second term:
4πr = 500π/r².
Since r > 0, multiply by r²:
4πr³ = 500π.
Divide by 4π:
r³ = 125.
Take the positive real cube root:
r = 5 cm.
This is the stationary radius.
Because r must be positive, we select the positive cube root.
We still need to establish that this point is a minimum rather than some other stationary behaviour.
Find the corresponding height
Return to the volume constraint:
h = 250/r².
At r = 5:
h = 250/25.
So:
h = 10 cm.
The relationship is:
h = 2r.
The height is twice the radius, which is the diameter.
Check the volume:
π(5²)(10) = 250π cm³.
The container holds the required idealised volume.
Both dimensions are strictly positive, so the proposed solution belongs to the permitted domain.
Evaluate the minimum surface area
Substitute r = 5 into:
S(r) = 2πr² + 500π/r.
The two-end contribution is:
2π(5²) = 50π.
The curved-side contribution is:
500π/5 = 100π.
Add:
S(5) = 150π cm².
Numerically:
150π ≈ 471.24 cm².
The exact answer is 150π cm².
The decimal approximation is appropriate only when numerical precision is requested.
Our mathematical task is now to prove that this is indeed the smallest possible area under the stated model.
Use the second derivative test
Differentiate the first derivative:
S′(r) = 4πr − 500πr⁻².
The second derivative is:
S″(r) = 4π + 1000πr⁻³.
Therefore:
S″(r) = 4π + 1000π/r³.
For every r > 0, both terms are positive.
So:
S″(r) > 0.
In particular:
S″(5) > 0.
Therefore the stationary point at r = 5 is a local minimum.
The fact that the second derivative is positive throughout the positive domain also tells us the surface-area curve is strictly convex there.
That supports a unique minimum.
Prove it is the global minimum
A local minimum is not automatically the smallest value over an entire domain.
We should examine the derivative sign and limiting behaviour.
Factor:
S′(r) = 4πr − 500π/r².
Write it over a common denominator:
S′(r) = 4π(r³ − 125)/r².
Since r > 0, the denominator is positive.
Therefore the derivative sign depends on r³ − 125.
For 0 < r < 5:
r³ − 125 < 0, so S′(r) < 0.
At r = 5:
S′(r) = 0.
For r > 5:
r³ − 125 > 0, so S′(r) > 0.
The function decreases up to r = 5 and increases afterwards.
Thus r = 5 gives the unique global minimum on r > 0.
Why the extreme shapes are inefficient in this model
As r approaches zero from the positive side, the term:
500π/r
becomes arbitrarily large.
The container becomes extremely tall and narrow to preserve volume.
As r becomes arbitrarily large, the term:
2πr²
becomes arbitrarily large.
The container becomes very wide and shallow, with enormous circular ends.
The surface-area function therefore grows without bound at both extremes.
The minimum occurs between those extremes, where the competing effects balance.
This provides a graphical reasonableness check on the calculus result.
What balance occurs at the optimum?
At r = 5 and h = 10:
The two circular ends together have area:
2πr² = 50π.
The curved side has area:
2πrh = 100π.
So the curved-side area is twice the combined circular-end area.
This ratio is a consequence of the optimum condition.
It is not a statement that each individual end equals the side, nor is it a universal rule for arbitrary cylinder dimensions.
The cylinder’s height-to-radius relationship h = 2r is the key result.
The area contribution ratio gives another way to understand the balance.
Derive the general result for any fixed volume
The previous calculation used the special volume 250π cm³.
Now let the fixed positive volume be any constant V.
The volume condition is:
V = πr²h.
Solve for height:
h = V/(πr²).
The closed surface area is:
S = 2πr² + 2πrh.
Substitute h:
S(r) = 2πr² + 2V/r.
Differentiate:
S′(r) = 4πr − 2V/r².
Set to zero:
4πr = 2V/r².
Multiply by r²:
4πr³ = 2V.
Therefore:
V = 2πr³.
But the cylinder’s volume is also:
V = πr²h.
Equate:
πr²h = 2πr³.
Divide by πr², positive because r > 0:
h = 2r.
The optimal ratio does not depend on the specific target volume.
Why this is a shape rule rather than a size rule
The general optimisation establishes:
h/r = 2.
It does not say all optimal cylinders have radius five centimetres.
Radius five arose because our selected volume was 250π cm³.
If the fixed volume changes, the optimal dimensions scale.
But the height-to-radius ratio remains two for this ideal closed-cylinder surface-area problem.
The result belongs to the stated shape and surface-area formula.
Change the container’s top, bottom or required dimensions and the optimal relation may change.
This is why modelling assumptions must accompany a mathematical optimisation claim.
An independent algebraic check through a new variable
Set:
u = r/5.
Then r = 5u and u > 0.
Our original surface-area function becomes:
S = 2π(25u²) + 500π/(5u).
So:
S = 50πu² + 100π/u.
Factor out 50π:
S = 50π(u² + 2/u).
The optimum should occur at u = 1.
Differentiate the bracket:
d(u² + 2u⁻¹)/du = 2u − 2u⁻².
Set to zero:
2u = 2/u².
Thus:
u³ = 1,
so u = 1 on the positive domain.
Therefore r = 5u = 5.
This scaled-variable check gives the same optimum and shows the structure independent of the original numeric units.
A counterexample: an open-top cylinder has a different optimum
Now change the problem.
Suppose the container has no lid. Only the base and curved side require material.
The volume is still fixed at 250π cm³.
But the open-top surface area is:
S_open = πr² + 2πrh.
Substitute h = 250/r²:
S_open(r) = πr² + 500π/r.
Notice the first term is now πr², not 2πr².
The area model has changed because one circular end was removed.
We must not reuse the closed-cylinder optimum automatically.
Differentiate the open-top model
Start with:
S_open(r) = πr² + 500πr⁻¹.
Differentiate:
S_open′(r) = 2πr − 500π/r².
Set the derivative to zero:
2πr = 500π/r².
Multiply by r²:
2πr³ = 500π.
Divide by 2π:
r³ = 250.
Therefore:
r = ∛250 cm.
This is approximately 6.30 cm.
The positive radius is different from the closed-cylinder optimum of five centimetres.
Find the open-top height
The volume constraint remains:
h = 250/r².
At the stationary open-top radius:
r³ = 250.
Therefore:
250/r² = r³/r² = r.
So:
h = r.
The open-top optimum has height equal to radius, whereas the closed-cylinder optimum has height equal to twice the radius.
The change of one geometric condition changes the mathematical optimum.
The open-top second derivative is:
S_open″(r) = 2π + 1000π/r³ > 0
for r > 0, so the stationary point is a minimum.
This comparison teaches a vital lesson: never transfer an optimisation answer without confirming that the objective function still represents the new question.
Why “least surface area” is not the same as “lowest manufacturing cost”
Our model assumes that each square centimetre of surface requires equal material and that making all surfaces has no additional fixed cost.
Real cylinders may require lids, rims, tabs, seams, supports, openings or thicker material in one direction.
The material may have different costs on the curved side and circular ends.
There may also be minimum or maximum dimensions imposed by transport, handling or stability.
Those changes would alter the objective function or the permissible domain.
The calculus result is exact for the defined ideal model, not a universal industrial design recommendation.
What if the radius must be no more than four centimetres?
Return to the closed cylinder with fixed volume 250π.
Suppose an additional mathematical constraint requires:
0 < r ≤ 4.
The unconstrained optimum r = 5 is not allowed.
Our derivative sign analysis showed S′(r) < 0 for all 0 < r < 5.
So the area decreases throughout the permitted interval.
The smallest allowed surface area occurs at the largest permitted radius:
r = 4.
Then:
h = 250/16 = 15.625 cm.
And:
S(4) = 157π cm².
That is greater than the unrestricted minimum 150π.
A new restriction can move the best feasible design to the boundary of the domain.
Why an endpoint can beat every admissible stationary candidate
On the constrained interval 0 < r ≤ 4, there is no stationary point of S inside the interval.
The stationary radius r = 5 lies outside it.
But the surface area still has a minimum over the permitted set because the function is decreasing and r = 4 is included.
This is a general lesson in constrained optimisation.
A student who only solves S′(r) = 0 and reports “no solution” misses the fact that the best permitted input may occur at an endpoint.
The original domain and any added constraints must be checked after differentiation.
What if the container height must be at least twelve centimetres?
Again keep the fixed volume:
h = 250/r².
Require:
h ≥ 12.
Then:
250/r² ≥ 12.
Since r > 0, multiply by r²:
250 ≥ 12r².
Thus:
r² ≤ 125/6.
So:
0 < r ≤ √(125/6).
This upper bound is about 4.56 cm, smaller than the unrestricted optimal radius five.
The closed-cylinder surface area decreases for every r below five.
Therefore the best feasible radius occurs at the upper bound:
r = √(125/6),
with h = 12 cm.
The exact optimal constrained area can be evaluated by substituting this radius into S(r). The key reasoning is recognising that the boundary, not the excluded stationary point, determines the result.
Why volume and surface area have different units
Volume is measured in cubic centimetres:
cm³.
Surface area is measured in square centimetres:
cm².
Radius and height are measured in centimetres.
Our selected volume is 250π cm³.
Our minimum closed surface area is 150π cm².
The units are not interchangeable, even when numerical expressions both contain π.
The derivative dS/dr has units cm²/cm = cm.
Reading the requested quantity and units can detect an otherwise plausible but mislabelled answer.
A six-minute independent check
For a closed cylinder with fixed volume 250π cm³:
1. State the volume constraint. 2. Express height as a function of radius. 3. Write the closed surface-area formula. 4. Substitute the volume constraint to obtain S(r). 5. Differentiate S. 6. Find the positive stationary radius. 7. Find the corresponding height. 8. Verify the minimum using the derivative sign or second derivative. 9. State the minimum surface area. 10. Explain why an open-top container needs a different calculation.
Answers: πr²h = 250π; h = 250/r²; S = 2πr² + 2πrh; S(r) = 2πr² + 500π/r; S′ = 4πr − 500π/r²; r = 5 cm; h = 10 cm; S′ changes negative to positive and S″ > 0; 150π cm²; and because the open-top cylinder has one fewer circular end.
The explanation of the first four steps is often more important than memorising the final radius.
Common mistake: counting only one circular end
The closed cylinder has two circular ends.
Therefore its circular-area contribution is 2πr².
Using πr² would instead model an open-top container.
The resulting stationary radius would solve a different question.
Check the diagram and wording before writing the objective function.
Common mistake: treating radius and height as independent
The fixed volume requires:
πr²h = 250π.
Increasing r changes h.
The student must use the constraint to express the objective as a function of one variable before differentiating.
Differentiating S = 2πr² + 2πrh while pretending h remains fixed would not solve the fixed-volume optimisation problem.
Common mistake: minimising the wrong quantity
A problem may ask for surface area, volume, height, or total cost.
The formula being minimised must match that request.
Here volume is a fixed condition and surface area is the objective.
We do not minimise volume when it has already been prescribed as 250π cm³.
Common mistake: stopping after S′ = 0
A stationary point is a candidate for a maximum or minimum.
It must be classified.
Here S″ > 0 and the derivative changes from negative to positive at r = 5, confirming a minimum.
Checking behaviour at the positive-domain extremes establishes the global result.
Common mistake: ignoring the admissible domain
Radius and height must both be positive.
A new constraint, such as r ≤ 4, can exclude the unconstrained optimum.
Then the best feasible solution may be a boundary value.
The calculation must answer the constrained problem, not the easier one it resembles.
Common mistake: assuming the cylinder ratio works for every container
For an ideal closed cylinder, the minimum occurs at h = 2r.
For an ideal open-top cylinder of the same fixed volume, the minimum occurs at h = r.
Real manufacturing constraints may change the objective again.
The ratio belongs to the defined model.
What parents can ask without doing the calculus
Ask your child why a very narrow cylinder needs to be tall when volume is fixed.
Ask why making the radius huge increases the area of the two ends.
Ask what calculation balances the end area with the curved-side area.
Then ask why removing the lid changes the best proportions.
These questions test the mathematical modelling behind the derivative.
A student who can explain the changed objective is building the ability to solve unfamiliar optimisation tasks, not merely copying one remembered formula.
How focused A-Math tuition can diagnose the real difficulty
One learner may know how to differentiate but accidentally omit a circular end.
Another may write both geometry formulas correctly but fail to eliminate height using the volume constraint.
Another may find the stationary radius and forget to prove it is a minimum.
A fourth may give the correct unrestricted optimum even though the problem imposes an upper limit on radius.
These are different weaknesses.
A useful small-group tutorial should identify the first wrong assumption, repair that specific step and ask the learner to solve a changed volume or open-top problem independently.
The aim is reliable method selection and interpretation.
Where this fits in Singapore G3 Additional Mathematics
The official 2027 Singapore–Cambridge Secondary Education Certificate G3 Additional Mathematics K341 syllabus includes differentiation, stationary points, maxima and minima, and applications of differentiation to optimisation under Calculus C1.
The cylinder example combines those skills with the earlier mathematics of circle area, cylinder volume, negative indices and algebraic substitution.
It is an original teaching model, not an official SEAB examination question.
The 2027 G3 Additional Mathematics code is K341. SEAB lists the earlier GCE O-Level 4049 code as its reference for 2026 and before, while G2 Additional Mathematics K232 is separately specified.
Frequently asked questions
Why does the optimum closed cylinder have h = 2r?
The fixed-volume surface-area derivative becomes zero when V = 2πr³. Combining that with V = πr²h gives h = 2r.
Is the smallest surface area found by minimising the volume?
No. The volume is fixed. We minimise the surface area while preserving that constraint.
Why does the surface-area formula contain two circular ends?
The cylinder is closed at both top and bottom, so both circular disks require material.
Does the second derivative prove the result is a minimum?
A positive second derivative at a stationary point establishes a local minimum. Here derivative signs and limiting behaviour also establish the global minimum on r > 0.
Would an open-top container have the same optimum?
No. Removing one circular end changes the objective function. In the ideal open-top model, the optimum has h = r.
Can the best radius lie at a constraint boundary?
Yes. If the allowed radius does not include the unrestricted stationary point, the best feasible design may occur at an endpoint.
Is 150π a surface area or a volume?
It is the minimum ideal closed surface area in cm² for the specified volume 250π cm³.
Does the model guarantee minimum real manufacturing cost?
No. Real manufacturing costs depend on material thickness, seams, lids, waste, structural needs and other constraints that are not included.
Is optimisation included in G3 Additional Mathematics?
Yes. The official 2027 K341 syllabus explicitly includes maxima, minima and differentiation applications under Calculus C1.
Other Additional Mathematics applications
The three-sided garden fence shows how a fixed length creates a quadratic maximum.
The packaging box made by corner cuts uses a different geometric volume function and explains why some cubic roots fall outside physical constraints.
The growing rectangle and product rule connects two changing dimensions to a changing area.
For the wider subject, see What Is Additional Mathematics? and How Additional Mathematics Works.
The immutable Clementi small-group Mathematics tutor reference explains how strong algebra foundations, explicit checks and targeted feedback prepare students for upper-secondary optimisation.
Official reference and evidence boundaries
The primary source is SEAB’s 2027 G3 Additional Mathematics K341 syllabus (PDF), especially Calculus C1, together with the official G3 subject listing.
The volume, dimensions and container comparisons are fictional educational inputs. The analysis assumes an ideal mathematical cylinder with no material thickness, waste, seams or engineering restrictions.
The central learning outcome is to identify the correct fixed-volume constraint, form a single-variable objective, prove a minimum and recheck the result when the design assumptions change.