How should a closed cylindrical container be proportioned if its total surface area is fixed but we want the largest possible volume? The two circular end caps and the curved side wall compete for the available material. A broader base provides more area per unit height, but uses more material in the caps and leaves less for the wall.
For a closed right circular cylinder of radius r and height h, total surface area is S=2πr²+2πrh and volume is V=πr²h. Eliminating h gives V(r)=Sr/2−πr³. Differentiation proves the maximum occurs when h=2r: **the optimal cylinder’s height equals its base diameter**.
For example, if S=150π square centimetres, the optimum is r=5 cm, h=10 cm and V=250π cubic centimetres. This guide derives and verifies the result, explains the required radius domain, compares other shapes, handles an integer-radius constraint and shows why an open-top cylinder follows a different rule.
Define the cylinder and the quantity held fixed
A right circular cylinder has two parallel circular end faces and a curved lateral surface joining them. Its radius r is the distance from the centre of a circular face to its rim. Its height h is the perpendicular distance between the two end faces.
For the optimisation, assume r>0 and h>0. The cylinder is closed, so material covers both ends as well as the curved wall.
The total surface area S is fixed. The volume V is the quantity to maximise.
These details determine the mathematics. A question about an open-top container, a fixed radius or only a fixed curved surface area would have a different constraint and potentially a different optimum.
Count the two circular end caps
One circular end has area πr². A closed cylinder has two such ends.
Therefore the combined end-cap area is 2πr². The factor two represents the top and bottom disks, not the circumference or diameter.
The total end-cap area grows quadratically with radius. Doubling r makes it four times as large, before any side-wall material is counted.
A student who writes only πr² for a closed cylinder has omitted one lid and is solving a different problem. Always list every physical surface that counts towards the fixed material before differentiating.
Find the curved lateral surface area
The cylinder’s curved wall can be imagined as a rectangle rolled around the base circle.
Its horizontal unrolled length is the circumference 2πr. Its vertical dimension is the cylinder height h.
Thus the lateral area is (2πr)h=2πrh.
Adding this to both end caps gives the closed-cylinder surface formula S=2πr²+2πrh. The two terms have distinct meanings: circular caps and side wall. This separation will become important when interpreting the final maximum-volume proportions.
Write volume separately from surface area
The volume of a cylinder is the circular base area times its height: V=πr²h.
The formula has units of cubic length, whereas both surface terms have units of square length. That dimensional distinction is a useful guard against accidentally maximising the wrong quantity.
Because S is fixed but r and h may vary, use the surface formula to express one dimension through the other.
The goal is a single-variable volume expression. Differentiation of the correct geometric objective can then identify an interior maximum.
Eliminate the height using the fixed surface
Start with S=2πr²+2πrh. Subtract 2πr² from both sides: 2πrh=S−2πr².
Divide by 2πr. This is valid because r>0. The height is h=S/(2πr)−r.
For a given positive radius, this tells us how tall the cylinder can be while using exactly the fixed surface area.
The height decreases as the radius becomes too large because more material is consumed by the caps. Very small radii can permit large heights, but the narrow base limits the volume. The next equation makes that trade-off exact.
Construct the single-variable volume function
Substitute h=S/(2πr)−r into V=πr²h.
Then V=πr²[S/(2πr)−r]. Multiply the first term: πr²·S/(2πr)=Sr/2.
The second term is πr²·r=πr³. Therefore V(r)=Sr/2−πr³.
This cubic function is the correct objective for a closed cylinder with fixed total surface area S. Its positive linear term rewards increasing radius initially, while its negative cubic term captures the growing surface cost of a broad base.
Find the physically permitted radius interval
Both radius and height must be positive. We already require r>0.
From h=S/(2πr)−r>0, multiply through by positive 2πr to obtain S−2πr²>0.
Thus r²
The upper boundary corresponds to spending the entire surface area on two end disks, leaving zero side-wall height. The lower boundary corresponds to an extremely narrow, extremely tall limiting cylinder. Neither limiting case has positive maximum volume.
Differentiate the volume function
Treat S as a fixed positive constant. Differentiate V(r)=Sr/2−πr³ with respect to r.
The derivative is V′(r)=S/2−3πr².
Interior stationary candidates satisfy S/2−3πr²=0. Rearranging gives 3πr²=S/2.
Therefore r²=S/(6π), and the positive radius candidate is r=√[S/(6π)].
It is smaller than √[S/(2π)], so it lies inside the physical radius interval. Only the positive root is relevant because radius is a length.
Recover the height at the stationary radius
At the candidate r²=S/(6π), rearrange the surface-area equation to find h.
The cap area is 2πr²=S/3. Therefore the remaining lateral area is S−S/3=2S/3.
Since lateral area is 2πrh, we have 2πrh=2S/3. Use S=6πr² to obtain 2πrh=4πr².
Divide by 2πr>0. The result is h=2r.
Thus the height equals the diameter at the maximum-volume proportions. This relationship is independent of the numerical total surface area.
Prove the stationary point is a maximum
The first derivative is V′(r)=S/2−3πr². For radii smaller than √[S/(6π)], this derivative is positive, so volume increases.
For larger permitted radii it is negative, so volume decreases. Therefore the stationary candidate is the only interior maximum.
The second derivative gives a shorter confirmation: V″(r)=−6πr, which is strictly negative for every positive r.
As r approaches either physical boundary, volume approaches zero. Hence the single interior maximum is global for all positive closed cylinders with the stated fixed total surface area.
The general maximum-volume formula
At the optimum, h=2r, so V_max=πr²(2r)=2πr³.
Substitute r=√[S/(6π)]. Then V_max=2π[S/(6π)]^(3/2).
Simplifying the constants gives V_max=S^(3/2)/(3√(6π)). Both expressions are exact and equivalent.
The exponent 3/2 on S is dimensionally correct: surface area has units of length squared, so its three-halves power has units of length cubed.
The optimum radius, height and volume are all determined once S is supplied. The shape ratio h/r=2 is universal for this model.
Apply the result to a surface area of 150π
Let S=150π square centimetres. The stationary radius satisfies r²=S/(6π)=(150π)/(6π)=25.
Thus r=5 cm. The optimal height is h=2r=10 cm.
The volume is V=π(5²)(10)=250π cubic centimetres, approximately 785.398 cubic centimetres.
The result uses the entire specified surface area and gives a physically valid cylinder. The exact form 250π is preferable to rounding prematurely when the problem supplies the surface area exactly.
Verify every part of the example’s surface area
For r=5, each circular end has area 25π square centimetres. Both ends together use 50π.
The curved side area is 2πrh=2π(5)(10)=100π square centimetres.
Adding gives 50π+100π=150π, exactly the fixed total.
The cap allocation is one-third of the total, and the side allocation is two-thirds. These fractions are useful geometric checks and provide a physical interpretation of the optimum rather than only an algebraic radius value.
The optimal surface allocation is one-third versus two-thirds
At the maximum, h=2r. The combined cap area is 2πr².
The lateral area is 2πrh=2πr(2r)=4πr², exactly twice the combined end-cap area.
Thus S=6πr²: one-third of the available surface is used by the two caps and two-thirds by the curved wall.
This relationship is independent of S. It is a useful way to understand the optimum’s balance of height and base area.
The individual caps each consume one-sixth of S, while the side wall consumes four-sixths. These allocations refer to surface area, not to volume fractions.
Compare a narrower radius of four centimetres
Keep the surface area S=150π but choose r=4 cm instead of the optimum five.
The fixed-area height is h=S/(2πr)−r=150π/(8π)−4=18.75−4=14.75 cm.
The volume is V=π(4²)(14.75)=236π cubic centimetres.
That is less than the maximum 250π even though the narrower cylinder is taller. The small base area outweighs the height gain.
This numerical comparison illustrates why the tallest allowed shape is not the volume winner. The volume objective depends on both radius squared and height.
Compare a broader radius of six centimetres
Keep S=150π and choose r=6 cm.
The height is h=150π/(12π)−6=12.5−6=6.5 cm.
Its volume is V=π(6²)(6.5)=234π cubic centimetres, again below 250π.
The broader base gives more area per unit height, but the large circular caps consume so much surface material that the available height falls sharply.
The two comparisons bracket the optimum radius five. They are useful checks, but the derivative proof is what establishes the maximum over every admissible real radius rather than only a few examples.
A quick calculation directly from the cubic
For S=150π, the volume function is V(r)=75πr−πr³=π(75r−r³).
At r=4, this gives π(300−64)=236π. At r=5, it gives π(375−125)=250π. At r=6, it gives π(450−216)=234π.
The derivative is V′(r)=π(75−3r²), which vanishes at r=5 for positive r.
This alternate route agrees with the dimension calculations. It is faster when several candidate radii must be compared, because the height need not be computed separately each time.
What changes if only whole-number radii are allowed?
Suppose the fixed-area condition remains S=150π and radius must be a whole number of centimetres.
The permitted positive integer radii lie below √75, approximately 8.66. The continuous maximum is already at the integer r=5.
Therefore r=5 remains the best choice, with h=10 and volume 250π.
This example happens to have an exact integer optimum, but that is not guaranteed for other surface areas. If the continuous optimum falls between integer choices, compare the nearby permitted integers rather than rounding blindly.
An integer-constrained example with a non-integer optimum
Let S=120π square centimetres. The continuous optimum radius is √[120π/(6π)]=√20, approximately 4.472 cm.
If only whole-centimetre radii are allowed, compare r=4 and r=5 because the volume rises before √20 and falls afterwards.
The cubic is V(r)=60πr−πr³. At r=4, V=240π−64π=176π. At r=5, V=300π−125π=175π.
Thus r=4 wins by π cubic centimetres. Its height is 120π/(8π)−4=11 cm. The best integer radius is not necessarily the nearest integer under an ordinary rounding rule; the objective must decide.
An open-top cylinder follows a different formula
If the lid is absent, the surface consists of one circular base and a curved side. The fixed-area equation is S=πr²+2πrh.
Eliminate h: h=S/(2πr)−r/2. Then volume becomes V=πr²h=Sr/2−(π/2)r³.
Differentiate: V′=S/2−(3π/2)r². The stationary condition gives r²=S/(3π).
At that radius, S=3πr², so the remaining side area is 2πr² and h=r. Thus the open-top optimum has height equal to radius, unlike the closed-cylinder optimum h=2r.
Why a missing lid changes the optimum ratio
A closed cylinder spends surface area on two end disks, whereas an open-top version spends material on only one.
The material budget therefore rewards base width differently. For the open-top shape, the optimal height is h=r; for the closed shape, h=2r.
These are not conflicting results. They solve different constrained optimisation problems.
A student should count material surfaces before choosing a memorised formula. In practical packaging, additional seams, lid thickness or material costs may introduce further constraints; the mathematical rules here apply to the idealised surfaces described explicitly.
If only lateral area were fixed, there is no finite maximum
Suppose the sole constraint were curved side area S_side=2πrh, with no material counted for either cap.
Then h=S_side/(2πr). The volume becomes V=πr²h=S_side r/2.
As r increases, this expression increases without bound. The height decreases, but not enough to prevent volume from growing.
Therefore no finite maximum exists under only a fixed lateral area and unrestricted positive dimensions. This counterexample explains why specifying total closed surface area is essential. A problem without sufficient constraints may not have an optimisation solution at all.
The inverse problem: minimum surface for fixed volume
Instead of fixing S and maximising V, suppose the volume V₀ is fixed and we want to minimise the surface area of a closed cylinder.
From V₀=πr²h, express h=V₀/(πr²). Substitute into S=2πr²+2πrh to obtain S(r)=2πr²+2V₀/r.
Differentiate: S′(r)=4πr−2V₀/r². At a stationary point, 4πr³=2V₀, so V₀=2πr³.
Then h=V₀/(πr²)=2r. The optimal proportions agree with the maximum-volume problem, as expected from the underlying geometry. The condition h=2r is therefore useful in both equivalent idealised optimisation directions.
Check the reverse problem’s minimum
For S(r)=2πr²+2V₀/r, the second derivative is S″(r)=4π+4V₀/r³.
Both terms are positive for r>0, so the stationary point is a strict local minimum.
As r approaches zero, the term 2V₀/r grows without bound. As r grows without bound, 2πr² also grows without bound.
Thus the single stationary point is the global minimum surface for fixed positive volume. The proof is structurally parallel to the original global maximum argument, but the objective and variable constraints have been reversed.
A dimensionless ratio can simplify the model
Let t=h/r be the cylinder’s height-to-radius ratio. Then S=2πr²(1+t), and V=πr³t.
For a given surface S, solve r²=S/[2π(1+t)]. This gives r³=[S/(2π(1+t))]^(3/2).
Therefore V(t)=πt[S/(2π(1+t))]^(3/2), with t>0.
The dependence on S is a positive scale factor; the shape optimisation depends only on t/(1+t)^(3/2). Differentiating its logarithm gives 1/t−3/[2(1+t)]=0, so 2(1+t)=3t and t=2.
This recovers h/r=2 without choosing r as the original decision variable.
An AM–GM proof gives the same volume bound
For a closed cylinder, set X=πr², so the combined cap area is 2X. Let Y=2πrh be the curved side area.
The fixed surface relation is S=X+X+Y. By AM–GM, the product X·X·Y is at most (S/3)³, with equality when X=Y=S/3.
However, the equality condition X=Y would imply 2πrh=πr², or h=r/2, which is not the actual optimum. The product X²Y is not directly proportional to cylinder volume V because V involves the square root of X. Therefore this naive AM–GM choice is not a valid proof of the desired volume bound.
This is an instructive warning: even a correct inequality may optimise the wrong algebraic product. The derivative proof or a carefully constructed weighted inequality is preferable unless the correspondence to V is established exactly.
A correct weighted inequality proof
Let U=2πr² be the combined cap area and W=2πrh be the wall area. The surface constraint is U+W=S.
Since r=√[U/(2π)] and h=W/(2πr), volume is V=πr²h=Wr/2 = W√U/(2√(2π)).
Thus V² is proportional to UW². We want to maximise UW² under U+W=S and U,W>0.
By AM–GM applied to the three nonnegative numbers U,W/2,W/2, we have U(W/2)²≤[(U+W)/3]³=(S/3)³.
Hence UW²≤4S³/27, with equality when U=W/2. Therefore W=2U, so the wall uses two-thirds of S and caps one-third. This is exactly h=2r, confirming the calculus optimum.
Why scaling surface area changes volume to the three-halves power
The maximum-volume formula is V_max=S^(3/2)/(3√(6π)).
If total surface area is multiplied by four, the optimal radius and height are multiplied by two because lengths scale as the square root of area.
The maximum volume is then multiplied by eight, because two cubed equals eight.
More generally, multiplying S by a positive factor k multiplies V_max by k^(3/2). This is the expected dimensional relationship between a surface budget and the volume of geometrically similar optimised containers.
Recover surface area from a desired maximum volume
Suppose the maximum volume achieved by a closed cylinder under its available surface budget is V_max, and we want to infer that budget.
From V_max=S^(3/2)/(3√(6π)), multiply through to get S^(3/2)=3V_max√(6π).
Raise both sides to the power 2/3: S=[3V_max√(6π)]^(2/3).
The expression is positive and has area units. If V_max=250π, it reproduces S=150π.
This reverse calculation is useful when a problem specifies an optimal packaging capacity and asks how much idealised surface material it requires.
A simple surface-area allocation table
At the optimum with S=150π, each individual cap uses πr²=25π, so both together use 50π.
The curved wall uses 100π. Their fractions of the total are one-sixth, one-sixth and two-thirds respectively.
If the cylinder were built with different materials for its top, bottom and wall, the objective would need to reflect material costs rather than merely total geometric area.
Under the present equal-area budget, these fractions summarise the optimum clearly. They can also be checked without recalculating differentiation once h=2r has been established.
Why a broad flat cylinder wastes the material budget for volume
Take r close to √[S/(2π)]. The two caps then consume almost all the available surface.
The height h=S/(2πr)−r approaches zero. Despite the large base area, volume πr²h tends to zero.
At the opposite extreme, r approaches zero and h becomes very large, but r² decreases so rapidly that the volume again tends to zero.
These limiting behaviours show why the optimum must be between the flat and narrow extremes. They are physical explanations of the positive-to-negative derivative sign change around r=√[S/(6π)].
The calculus optimum is more than a convenient ratio
The condition h=2r is derived from the fixed closed-surface constraint and the volume objective. It is not merely a convention used to draw cans.
It follows because changing radius slightly near the optimum causes gains and losses in base area and available height that exactly balance in the first derivative.
If the construction introduces thicker ends, a minimum height, a preferred radius range or a different material cost for the wall, the optimal ratio can change.
A student should therefore understand which assumptions produced the result. Correct application to a new problem begins by checking those assumptions rather than automatically imposing h=2r.
A quick examination solution
Write the correct closed-cylinder equations S=2πr²+2πrh and V=πr²h, with r,h>0. Solve the first for h.
Substitute to obtain V(r)=Sr/2−πr³ and domain 0 Differentiate and solve V′(r)=S/2−3πr²=0, giving r²=S/(6π). Show V″(r)=−6πr<0 and note the zero-volume limiting boundaries. Calculate h=2r and V_max=2πr³, and substitute any numerical S. Verify the total surface-area budget before reporting the final dimensions and volume. Question one: a closed cylinder has S=150π cm². Find its maximum volume. Answer: 250π cm³. Question two: for Question one, find the optimal radius and height. Answer: r=5 cm, h=10 cm. Question three: state the optimum height-to-radius ratio for any fixed closed-cylinder surface budget. Answer: h/r=2. Question four: find the general maximum volume from surface area S. Answer: S^(3/2)/(3√(6π)). Question five: if the optimal cylinder’s wall area is 100π and both caps total 50π, what is its surface area? Answer: 150π. Question six: for S=120π and integer radius, compare r=4 and r=5. Answer: r=4 gives 176π, greater than 175π at r=5. Question seven: what is the corresponding optimal height-to-radius ratio for an ideal open-top cylinder with one base? Answer: h/r=1. For each, begin by counting the actual surfaces present. Fixed-surface cylinder optimisation combines mensuration, cubic functions, differentiation, physical constraints and proportional reasoning. The inverse minimum-surface problem and weighted-inequality proof may be useful enrichment beyond a routine school application. Singapore learners can consult SEAB’s applicable 2026 O-Level Additional Mathematics and 2027 SEC G3 syllabuses according to their school course. Related eduKateSG guides include How to Find the Maximum Volume of a Cylinder Inscribed in a Sphere, How to Maximise an Open-Top Box from a Rectangular Sheet and How to Maximise a Circular Sector’s Area with Fixed Perimeter. Browse the Additional Mathematics archive. Families can enquire about suitable eduKate Singapore small-group Mathematics tuition near Sixth Avenue MRT. The lasting skill is to account for every part of a fixed material budget, choose the correct quantity to optimise and justify why the resulting dimensions give a true global maximum. Properly taught kids shine a bright light into the future.Practice questions with checked answers
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