A right circular cylinder fitted inside a sphere presents a classic Additional Mathematics optimisation problem. A wider cylinder must be shorter, while a taller cylinder must be narrower. The exact relationship comes from a cross-sectional right triangle through the sphere’s centre. Once the radius and height are expressed using one variable, calculus identifies the maximum volume.
For a sphere of radius R, let the coaxial cylinder have circular radius r and full height h. At the upper rim, r²+(h/2)²=R². Thus its volume is V(h)=π(R²−h²/4)h. Differentiating and checking the valid interval gives the optimal height h=2R/√3, radius r=R√(2/3) and maximum volume V_max=4πR³/(3√3).
Did you know that this optimal cylinder occupies exactly 1/√3 of the sphere’s volume—about 57.7%? For a sphere of radius 5, the maximum cylinder volume is 500π/(3√3)≈302.300 cubic units. This guide derives the result, verifies the geometry, compares integer-height choices and explains why maximising height or circular base area separately cannot maximise volume.
Define precisely which cylinder is being fitted
A right circular cylinder has two parallel circular ends and a straight central axis perpendicular to them. In this problem, the cylinder’s axis passes through the centre of the sphere.
The circular rims of the cylinder lie on the sphere, with the top and bottom at equal distances above and below the sphere’s centre. This is the standard coaxial inscribed-cylinder configuration.
The cylinder has base radius r, not diameter r, and full height h from the lower circular face to the upper one.
A clear cross-section through the axis shows a rectangle inside a circle. That two-dimensional picture contains the constraint needed to solve the three-dimensional volume problem, which is why a sketch is often more useful than starting immediately with a cubic formula.
Read the sphere radius and cylinder dimensions separately
Let the sphere radius be R>0. Its diameter is 2R. The inscribed cylinder has radius r≥0 and full height h≥0.
The cylinder’s base diameter is 2r, while the separation of its two circular end faces is h.
Do not confuse h with the half-height from the sphere centre to the top rim. The cross-sectional right triangle uses h/2.
Similarly, the circle equation uses r as a distance from the cylinder axis to its rim. If a question supplies a cylinder diameter, divide by two before using the constraint. These variable definitions protect against factors of two in both geometry and volume.
Use a cross-sectional right triangle
Slice the sphere vertically through the cylinder’s central axis. The sphere cross-section is a circle of radius R.
The cylinder cross-section is a rectangle with half-width r and half-height h/2. Its upper-right corner lies on the outer sphere circle.
The horizontal distance from the sphere centre to that corner is r. The vertical distance is h/2. The straight radius from the centre to the corner has length R.
Pythagoras therefore gives r²+(h/2)²=R². This is the exact geometric constraint connecting the two cylinder dimensions. The constraint is essential: without it, one could increase both radius and height without bound and there would be no finite maximum.
Solve the constraint for the base radius squared
From r²+h²/4=R², rearrange to r²=R²−h²/4.
This is especially convenient because the cylinder volume uses r² directly. There is no need to take a square root for r before forming volume.
The formula shows that a larger height reduces the allowed squared base radius. If h approaches 2R, then r² approaches zero.
If h is near zero, the squared radius approaches R², but the cylinder is almost flat. Both extremes produce very small volume, suggesting a meaningful interior optimum.
Write the correct cylinder volume
The volume of a right circular cylinder is base area multiplied by height: V=πr²h.
Substitute r²=R²−h²/4. This gives V(h)=π(R²−h²/4)h.
Expand to V(h)=π(R²h−h³/4).
The area formula πr² and the height h have their correct geometric roles. Using 2πr h would instead describe the cylinder’s curved lateral surface area, not its volume. A correct optimisation depends on choosing the quantity the question actually asks to maximise.
State the physically allowed height interval
For a genuine positive-volume cylinder, both radius and height must be positive. Thus h>0 and r²>0.
From r²=R²−h²/4>0, we obtain h²<4R². As h is positive, this becomes 0 The boundaries h=0 and h=2R describe degenerate limiting cases with zero volume. They are useful for comparison but are not positive-volume cylinders. The derivative equation may have other algebraic solutions outside this interval. Every candidate must be filtered through the physical domain before it can be accepted as an optimal cylinder. Treat the sphere radius R as a fixed positive constant. The volume is V(h)=π(R²h−h³/4). Differentiate term by term: V′(h)=π(R²−3h²/4). Interior stationary points satisfy V′(h)=0. Since π>0, divide through by π to obtain R²−3h²/4=0. Therefore 3h²/4=R², giving h²=4R²/3. The algebraic solutions are h=±2R/√3, but the physical height must be positive. Thus the only admissible stationary candidate is h=2R/√3. Substitute h=2R/√3 into the geometry equation r²=R²−h²/4. Because h²=4R²/3, we have h²/4=R²/3. Therefore r²=R²−R²/3=2R²/3. Take the non-negative square root: r=R√(2/3). The cylinder’s base diameter is 2R√(2/3). Both dimensions are positive, and h=2R/√3 is smaller than the sphere diameter 2R. Thus the candidate fits within the physical constraints. At the stationary candidate, r²=2R²/3 and h=2R/√3. Substitute into V=πr²h. Then V=π(2R²/3)(2R/√3). Multiply to obtain V_max=4πR³/(3√3). This formula has units of length cubed, as required for a volume. It grows with R³, which is consistent with scaling a three-dimensional object. The exact radical form should be retained in a mathematical solution. A decimal approximation is appropriate only when the problem supplies a numerical radius and requests a rounded result. Differentiate V′(h)=π(R²−3h²/4) once more. The second derivative is V″(h)=−3πh/2. At the admissible candidate h=2R/√3>0, this is negative. Therefore the stationary point is a local maximum. More strongly, for 0 Thus volume increases to one interior peak and then decreases. Together with zero limiting volume at both ends, this proves the candidate is the global maximum for the inscribed-cylinder geometry. Take a sphere of radius R=5. Its diameter is ten. The optimal cylinder height is h=2(5)/√3=10/√3≈5.774 units. Its optimal base radius is r=5√(2/3)≈4.082 units, so its diameter is about 8.165 units. The maximum volume is V_max=4π(5³)/(3√3)=500π/(3√3)≈302.300 cubic units. These values are dimensionally and geometrically plausible: the cylinder is lower than the full sphere diameter and somewhat narrower than the sphere. A radius larger than five would violate the cross-sectional constraint immediately. For R=5, the optimal height satisfies h²=100/3. The squared half-height is therefore h²/4=25/3. The optimal base radius squared is r²=50/3. Add them: r²+(h/2)²=50/3+25/3=75/3=25=R². Thus every point on the upper and lower circular rim lies at distance five from the sphere’s centre. The cylinder fits exactly inside the spherical boundary in the assumed coaxial configuration. This independent Pythagorean check is useful after a calculus optimisation because it confirms that the derived dimensions satisfy the original geometric constraint rather than only the transformed volume equation. In the radius-five example, the optimum full height is about 5.774, which is greater than the sphere radius five. That is not impossible. The maximum available full vertical height is the sphere diameter ten, not its radius. Only half the cylinder’s height is measured from the sphere centre to one end face. At the optimum, h/2=5/√3≈2.887, well below the sphere radius. This is a frequent interpretation error. Comparing the full cylinder height h to R instead of 2R can lead a student to reject the correct stationary point even though the entire cylinder fits within the sphere. At the optimum, h²=4R²/3 and r²=2R²/3. Divide these equations to obtain h²/r²=2. Since height and radius are positive, h/r=√2. Therefore the optimum height is h=√2r. The full base diameter is 2r, so the ratio of height to diameter is h/(2r)=1/√2≈0.7071. These ratios are independent of R. Every maximum-volume cylinder in this standard sphere configuration is geometrically similar to every other one, regardless of the sphere’s absolute size. The volume of the containing sphere is V_sphere=(4/3)πR³. The maximum cylinder volume is V_cylinder=4πR³/(3√3). Divide the cylinder volume by the sphere volume. The factors 4πR³/3 cancel, leaving V_cylinder/V_sphere=1/√3. Numerically, 1/√3≈0.57735, or about 57.7%. The remaining space in the sphere is about 42.3% of its volume. This is a ratio of volumes, not areas or surface-contact percentages. It remains constant for every sphere radius under the same optimal inscribed-cylinder model. For R=5, sphere volume is (4/3)π(125)=500π/3≈523.599 cubic units. The optimal cylinder volume is approximately 302.300 cubic units. Divide 302.300 by 523.599 to obtain about 0.57735. The quotient agrees with 1/√3. A cylinder volume greater than 523.599 in this configuration would be physically impossible, because it would exceed the volume of the sphere that contains it. This provides a useful upper-bound check for numerical results even before the exact optimisation is completed. For a sphere of radius five, consider a cylinder with full height h=5 instead of the optimal 10/√3. The constraint gives r²=25−5²/4=25−25/4=75/4. Its volume is V=π(75/4)(5)=375π/4≈294.524 cubic units. This is less than the maximum 500π/(3√3)≈302.300, as expected. The five-unit cylinder is fairly close to optimal but leaves some volume improvement available by increasing height while accepting a smaller base. Now use full height h=6 in the same radius-five sphere. The allowed base radius squared is r²=25−6²/4=25−9=16. Therefore r=4 and the volume is V=π(16)(6)=96π≈301.593 cubic units. This is only about 0.707 cubic unit below the continuous maximum of approximately 302.300. The six-unit height is slightly above the optimal 5.774, so it gives a slightly smaller base radius and a slightly lower volume. The numerical comparison supports the exact derivative result. Suppose the radius-five problem requires the cylinder’s full height to be a whole number of the given length units. Permitted positive integer heights are 1 through 9, because h must lie strictly between zero and ten. The continuous maximum occurs at h≈5.774, and the volume increases before this point and decreases after it. Therefore the best integer height must be either five or six. We calculated V(5)=375π/4≈294.524 and V(6)=96π≈301.593. Thus h=6 is the optimal permitted integer height, with base radius four and volume 96π. This is distinct from the continuous optimum. If the goal were simply to maximise the circular base area πr² inside the sphere, we would want the largest possible radius r. That occurs in the limiting configuration h→0 and r→R. But the resulting cylinder has almost zero height, so its volume tends to zero. The volume problem requires multiplying base area by height. Maximising one factor alone generally does not maximise their product when the factors are connected by a constraint. This is the central modelling insight: the best height and best base radius are not found independently. They must be balanced through r²+h²/4=R². The cylinder’s curved lateral surface area is S_lateral=2πrh. Under the sphere constraint, h=2√(R²−r²). Therefore S_lateral=4πr√(R²−r²). Squaring the non-negative expression reduces its optimisation to maximising r²(R²−r²). The maximum occurs at r²=R²/2, giving r=R/√2 and h=√2R. This differs from the maximum-volume result r²=2R²/3 and h=2R/√3. Both objectives use the same physical geometry but optimise different quantities. Never copy the radius and height from one objective into a question asking about another. Define t=h/(2R). Because 0 The geometric constraint becomes r²=R²(1−t²). The volume is V=πr²h=2πR³t(1−t²). The only variable part is t−t³. Differentiate with respect to t to get 1−3t². The stationary condition gives t=1/√3 in the physical interval. Thus h=2R/√3 immediately. This method separates the sphere’s scale from the shape of the optimum. The factor 2πR³ affects the numerical volume but not the dimensionless optimal ratio t. There is also a non-calculus proof. From r²+h²/4=R², define u=r²/2 and v=h²/4. Then u,u,v are non-negative and their sum is 2u+v=R². By AM–GM, u²v≤(R²/3)³=R⁶/27, with equality when u=v=R²/3. The volume is V=πr²h=π(2u)(2√v)=4πu√v. Therefore V≤4πR³/(3√3), exactly the calculus result. Equality gives r²=2R²/3 and h²/4=R²/3, reproducing the optimal dimensions. This independent inequality proof is useful enrichment and confirms that the stationary point is a global maximum. The AM–GM inequality alone supplies an upper bound. To prove it is the actual maximum, we must check that equality can occur within the permitted geometry. Equality requires the three positive numbers u,u,v to be equal. That gives r²/2=h²/4, or h²=2r². Combining this with r²+h²/4=R² gives r²=2R²/3 and h=2R/√3, both positive and physically permitted. Thus the upper bound is attained. The proof is complete because it both bounds all feasible cylinders and supplies a real cylinder achieving that bound. Some questions give the sphere diameter D rather than its radius. Since R=D/2, substitute that relation before calculating the optimum. The optimal height becomes h=2(D/2)/√3=D/√3. The optimal cylinder base radius is r=(D/2)√(2/3)=D/√6. The maximum volume is V_max=4π(D/2)³/(3√3)=πD³/(6√3). For D=10, these give h=10/√3, r=10/√6 and V_max=1000π/(6√3)=500π/(3√3), matching the radius-five calculation. The conversion protects against confusing diameter and radius. If the sphere radius is multiplied by a positive factor k, the optimal cylinder radius and height are also multiplied by k. The maximum volume formula contains R³, so the volume is multiplied by k³. For example, doubling the sphere radius from five to ten doubles the optimal cylinder dimensions and multiplies its maximum volume by eight. The volume fraction inside the sphere remains 1/√3 because both sphere and cylinder volumes scale by the same cubic factor. This scaling argument is a useful independent check on any proposed general formula. Let R=3. The optimal cylinder height is h=6/√3=2√3. Its base radius squared is r²=2(3²)/3=6, so r=√6. The maximum volume is πr²h=π(6)(2√3)=12π√3 cubic units. The general formula gives 4π(27)/(3√3)=36π/√3=12π√3, agreeing exactly. The cylinder’s height-to-radius ratio is (2√3)/√6=√2, as required by the universal optimum relation. The standard inscribed cylinder is centred vertically in the sphere. Its top and bottom rims occur at heights +h/2 and −h/2 relative to the sphere centre. This symmetry permits one circular base radius r to fit at both ends while using as much available sphere width as possible for a given height. A vertically shifted cylinder of the same height would place one end closer to the sphere’s top or bottom, reducing the maximum allowable common rim radius. Thus centring the cylinder is the natural configuration for maximising volume among coaxial cylinders of a given height. The right-triangle constraint used in the solution captures this optimal symmetric placement. At height z above the sphere centre, the horizontal cross-section of the sphere is a circle with squared radius R²−z². For a cylinder centred vertically, its upper and lower rims occur at z=±h/2 and must fit within those cross-sections. Thus r²≤R²−h²/4. For a fixed height h, a larger cylinder radius gives a larger volume, so the maximum at that h uses equality. This justifies the geometry equation even when the problem says the cylinder fits inside the sphere rather than explicitly stating that its rims touch the sphere. Any cylinder with a smaller radius could be widened until contact, increasing volume without violating the height constraint. Suppose h is close to 2R. Then the half-height h/2 is close to R, leaving r²=R²−h²/4 close to zero. The circular base area πr² therefore becomes tiny, even though the cylinder is almost as tall as the sphere. Multiplying by a large but bounded height cannot prevent the volume from tending to zero as r² tends to zero. This is the opposite of the nearly flat case h≈0, where the base is large but the height is too small. The interior optimum balances the losses at both extremes. If the cylinder radius r is specified in advance, its largest symmetric inscribed height is h=2√(R²−r²). There is no further continuous optimisation over height for fixed r if the objective is volume: the volume grows as height increases, so use the maximum allowed height. The original task is harder because r is also free. Making r larger reduces the allowed h, so the product πr²h has a nontrivial optimum. Students should distinguish a constraint solving for one dimension from a genuine optimisation over an adjustable dimension. The same geometric equation can play either role depending on what the problem fixes. At the optimum, r=R√(2/3) and h/2=R/√3. Square them: r²=2R²/3 and (h/2)²=R²/3. Their sum is R² exactly, satisfying Pythagoras. The three dimensionless squares are in the ratio two to one along the radial and half-height directions. If the radius were mistakenly written as R√(1/3), the constraint would sum to only 2R²/3 and fail. Exact substitution into the original geometry is therefore a strong final safeguard against an algebraic factor error. Start with a labelled axial cross-section of the sphere and cylinder. Write r²+h²/4=R² using the half-height. Express the volume as V(h)=π(R²h−h³/4), with 0 Reject the negative height root. Show the second derivative is negative or the first derivative changes from positive to negative, and compare the zero-volume boundary limits. Recover r=R√(2/3) and report V_max=4πR³/(3√3). Verify the geometry and give a suitably rounded numerical value if the sphere radius is specified. One learner uses h rather than h/2 in the right triangle, making the cylinder much shorter than it should be. A diagram with the sphere centre and two end faces repairs that error. Another uses cylinder lateral area 2πrh instead of volume πr²h, so the derivative optimises the wrong quantity. A third accepts the negative stationary root or forgets to check h<2R. A fourth reports the full diameter where the formula requires radius. Each error calls for a different correction. A changed sphere radius, followed by verification of the cross-sectional Pythagorean equation, tests whether the method transfers beyond one memorised result. Question one: a right circular cylinder is coaxially inscribed in a sphere of radius R. What is the geometry constraint? Answer: r²+h²/4=R². Question two: write volume as a function of h. Answer: V(h)=π(R²h−h³/4), with 0 Question three: find the optimum full height and base radius. Answer: h=2R/√3 and r=R√(2/3). Question four: find the maximum volume. Answer: 4πR³/(3√3). Question five: a sphere has radius five. Find the exact maximum cylinder volume. Answer: 500π/(3√3)≈302.300. Question six: what fraction of the sphere volume is occupied by the optimum cylinder? Answer: 1/√3. Question seven: if the sphere radius is three, find the maximum cylinder volume. Answer: 12π√3. Question eight: for R=5 and an integer-height constraint, which height is best? Answer: h=6, giving volume 96π. This optimisation combines a circular cross-section, Pythagoras, cylinder and sphere volumes, polynomial differentiation and constrained extrema. The AM–GM proof and scale-invariance discussion can serve as enrichment depending on the learner’s Additional Mathematics syllabus. Singapore students should check the appropriate SEAB resources for 2026 O-Level Additional Mathematics or 2027 SEC G3, following their school’s actual topic sequence. Related eduKateSG guides include How to Find the Maximum Area of a Rectangle Inscribed in a Semicircle, How to Maximise an Open-Top Box from a Rectangular Sheet, How to Maximise a Circular Sector’s Area with Fixed Perimeter, and the Additional Mathematics archive. Families can enquire about suitable eduKate Singapore three-student tutorials near Sixth Avenue MRT. The lasting skill is to derive the true geometric constraint, differentiate the correct three-dimensional quantity and verify that the resulting dimensions really fit inside the sphere. Properly taught kids shine a bright light into the future.Differentiate the volume with respect to height
Recover the optimum base radius
Calculate the maximum volume in terms of R
Show the stationary point is a maximum
A radius-five numerical example
Verify the radius-five corner is on the sphere
Why the optimum height can exceed the sphere radius
Relate the optimum height to the cylinder radius
Compare the cylinder volume with the sphere volume
A quick check using the radius-five sphere volume
Compare a cylinder of height five
Compare a cylinder of height six
If the height must be a whole number
An area-maximising cylinder is not the same problem
A cylinder’s curved surface area has another optimum
Use a dimensionless height ratio
An algebraic proof using AM–GM
Why the AM–GM equality condition matters
Recover the formulas when a diameter D is supplied
Scaling the sphere changes volume cubically
A sphere of radius three as a transfer example
Why the top and bottom must stay symmetric in the model
A check using cross-sectional areas
Why a nearly full-height cylinder has little volume
A fixed-radius cylinder does not have an independent height choice
An exact square-root check of the optimum dimensions
A concise complete solution for examination work
Common errors and targeted repairs
Practice questions with checked answers
Syllabus scope and connected learning