Did you know that every circular sector with a fixed perimeter has the largest possible area at the same central angle—exactly 2 radians, or approximately 114.59°? The radius changes with the perimeter, but the optimal angle does not. To prove this, express the perimeter as the sum of two radii and the arc, then write the area in one variable and differentiate.
For example, suppose a sector has fixed perimeter 24 units, counting both radii and its arc. If its radius is r and arc length is s, then 2r + s = 24 and its area is A = rs/2 = 12r − r². This downward-opening quadratic is largest at r = 6, giving arc length 12, central angle 2 radians and maximum area 36 square units. This guide explains the geometry, calculus, valid angle range, equality case and common wrong approaches.
First define what the sector perimeter contains
A circular sector is bounded by two radii and the arc joining their outer endpoints. When the sector angle is strictly between zero and a full turn, those three pieces form its boundary.
If the radius is r and the arc length is s, its perimeter is P = r + r + s = 2r + s. Every term has units of length.
This perimeter is not the circumference of the complete circle unless the problem specifically concerns the whole circle. Using P = 2πr for a sector discards both radial edges and changes the problem.
A small labelled diagram should show the two equal straight radii, curved arc and central angle. The first mathematical decision is to identify precisely which boundary pieces the given perimeter includes.
Use radians for the arc and area formulas
For a central angle θ measured in radians, the arc length is s = rθ. The sector area is A = r²θ/2.
These formulas are simplest in radian measure. When a question gives degrees, convert with θ_rad = θ_degπ/180 before substituting.
The perimeter can therefore also be written P = 2r + rθ = r(2 + θ). The number 2 represents the two radii, while θ represents the arc-length contribution per unit radius.
Do not put a degree number such as 120 directly into rθ. One hundred and twenty degrees is 2π/3 radians, not the number 120 radians. Unit mistakes can overwhelm an otherwise correct optimisation.
Why the perimeter creates a trade-off
Keeping P fixed means that increasing the radius uses more length along the two straight sides and leaves less for the curved arc.
Conversely, shortening the radii leaves more of the available perimeter for the arc, but it may reduce the scale of the area.
There is no reason to expect that the sector with the longest arc or largest possible radius automatically has the largest area. Area depends on two quantities together.
The most useful strategy is to eliminate one variable using P = 2r + s, then optimise the resulting single-variable area expression. This is the same general method used in many constrained geometry problems.
Express arc length in terms of radius
From P = 2r + s, rearrange to s = P − 2r. For fixed P, this is a linear relationship between arc length and radius.
A sector’s area can be written A = rs/2 because s = rθ and A = r²θ/2. Substitute s = P − 2r.
Then A(r) = r(P − 2r)/2 = Pr/2 − r². This is a quadratic in r with a negative coefficient on r².
The entire geometric optimisation has been reduced to maximising a downward-opening parabola over the admissible radius interval. That is much easier than trying arbitrary angle and radius combinations.
Work the numerical example with perimeter twenty-four
Take P = 24. Then s = 24 − 2r and A = r(24 − 2r)/2.
Simplify to A(r) = 12r − r². Complete the square: A(r) = 36 − (r − 6)².
A real square is non-negative, so A(r) ≤ 36. Equality occurs exactly when r − 6 = 0, giving r = 6.
Thus the greatest sector area is 36 square units. The completed-square form already proves the maximum without differentiation, while differentiation gives a second method and helps connect the result to calculus.
Recover the arc length and central angle
At the optimum r = 6 for P = 24, the arc length is s = 24 − 2(6) = 12.
Because s = rθ in radians, θ = s/r = 12/6 = 2 radians.
The angle is approximately 2 × 180/π = 114.59°, which lies between 0° and 360°, so it describes a valid ordinary sector.
Check the perimeter: two radii contribute 12 units and the arc contributes another 12, totaling 24. Check the area directly: A = r²θ/2 = 36·2/2 = 36 square units.
Prove the general result by completing the square
For a positive fixed perimeter P, the area is A(r) = Pr/2 − r².
Complete the square: A(r) = P²/16 − (r − P/4)². Expanding the right side recovers Pr/2 − r² exactly.
Therefore A(r) ≤ P²/16, with equality at r = P/4.
At that radius, the arc length is s = P − 2(P/4) = P/2. The angle is θ = s/r = (P/2)/(P/4) = 2 radians.
Hence the maximum sector area is P²/16. The optimal angle is independent of P, while the optimal radius and arc length scale with P.
Prove it again using differentiation
Differentiate the general area function A(r) = Pr/2 − r², treating P as a fixed positive constant.
The derivative is A′(r) = P/2 − 2r. Set it equal to zero: P/2 − 2r = 0, so r = P/4.
The second derivative is A″(r) = −2, which is strictly negative. Thus the stationary point is a maximum.
The downward quadratic is also increasing for r < P/4 and decreasing for r > P/4. These signs explain how area changes as the radius passes the optimal value rather than simply supplying an answer from a memorised formula.
Check the allowable angle range
A nondegenerate ordinary sector has positive radius, positive arc length and a central angle strictly between zero and a complete turn: 0 < θ < 2π.
Because θ = (P − 2r)/r = P/r − 2, positivity requires 0 < r < P/2. The upper angle restriction also gives r > P/(2 + 2π).
The optimal radius P/4 is admissible because P/(2 + 2π) < P/4 < P/2. Its central angle is 2 radians, less than 2π.
This domain check matters. An unconstrained calculus stationary point should not be accepted in a geometry problem unless it actually represents a permissible shape.
The optimal arc consumes half the perimeter
At the maximum, r = P/4 and s = P/2. Therefore the two straight sides together have length 2r = P/2.
The other half of the perimeter is the curved arc. Thus the perimeter is equally divided between the combined radial edges and the arc.
This provides an intuitive way to remember the geometry without memorising the final numerical angle. Since s = 2r at the optimum and s = rθ, the central angle must equal 2 radians.
The equality refers to arc length and the sum of the two radii, not to arc length equalling a single radius. Confusing those two statements would give an incorrect angle of one radian.
Why the angle is always two radians
Using the alternative variable θ, the perimeter relation P = r(2 + θ) gives r = P/(2 + θ).
Substitute into A = r²θ/2. The area becomes A(θ) = P²θ/[2(2 + θ)²].
Differentiate: A′(θ) = P²(2 − θ)/[2(2 + θ)³]. Its sign is positive for θ < 2 and negative for θ > 2, within the valid sector range.
Therefore the area increases until θ = 2 radians and decreases afterwards. The fixed perimeter P affects the size but not the angle at the maximum. This directly confirms the radius-based result.
The angle method is a useful independent check
For P = 24 and θ = 2, the angle-based radius is r = 24/(2 + 2) = 6.
The angle-based area is 24²·2/[2(4)²] = 1152/32 = 36 square units.
These agree with the area function A(r) = 12r − r² evaluated at r = 6.
The two routes are mathematically equivalent: one eliminates arc length to work with radius, while the other eliminates radius to work with angle. Knowing both allows students to select the more convenient variable when an examination question is worded differently.
A narrow sector is not necessarily efficient
When θ becomes very small while P stays fixed, the radius approaches P/2, because the two radial sides consume almost all the perimeter.
But the arc length approaches zero. The area A = rs/2 therefore approaches zero, even though the radius is comparatively large.
A very narrow sector is long and thin. It cannot achieve the maximum area because it uses nearly all its fixed boundary length on straight sides with little angular opening.
This gives a geometric explanation of why maximum area is not obtained by choosing the largest possible radius.
An extremely wide sector is not automatically optimal
Increasing θ far beyond 2 radians makes the radius P/(2 + θ) smaller. Although the angular opening becomes wider, the overall scale of the sector shrinks.
For example, at a half-circle angle θ = π, with P = 24, the radius is 24/(2 + π). The sector area is 288π/(2 + π)², approximately 34.23 square units.
This is less than the optimum 36. A semicircle may look wide, but it does not maximise area when both its arc and two radii count toward a fixed sector perimeter.
The comparison highlights why visual intuition alone is insufficient: curvature, scale and the boundary constraint all matter.
A quarter-circle comparison
A quarter-circle sector has central angle θ = π/2. With P = 24, its radius is 24/(2 + π/2).
Its area is P²(π/2)/[2(2 + π/2)²], approximately 35.48 square units. It is close to but still smaller than the maximum 36.
This makes geometric sense because π/2 ≈ 1.571 is not far below the optimal 2 radians. Area rises as the angle increases through that part of the permitted range.
The comparison is a check, not a proof of the maximum. The derivative or completed-square argument establishes the global result for every admissible angle, including values not tested numerically.
Scaling the perimeter scales area quadratically
The maximum-area formula is A_max = P²/16. If P doubles, A_max is multiplied by four.
For P = 24, the optimum has radius 6 and area 36. For P = 48, the optimum has radius 12 and area 144.
The angle remains 2 radians because enlarging the whole shape scales every length but preserves its angles.
This is dimensional reasoning: perimeter is a length, while area depends on length squared. An answer linear in P would be suspicious for geometrically similar maximum-area sectors.
Recover perimeter from a stated maximum area
Suppose a problem says a circular sector can achieve maximum area 100 square units under a fixed-perimeter condition.
Since A_max = P²/16, we have P²/16 = 100. Therefore P² = 1600 and P = 40, choosing the positive root because perimeter is a positive length.
The optimal radius is P/4 = 10, the optimal arc length is P/2 = 20 and the central angle is 2 radians.
Check the area: rs/2 = 10·20/2 = 100. This reverse question tests whether the student can use the derived relationship rather than only solve the original optimisation direction.
Why a fixed radius gives a different task
If radius r is fixed and the central angle can change up to a full turn, sector area A = r²θ/2 increases with θ.
There is no interior optimum at θ = 2 when radius, rather than perimeter, is fixed. The largest permitted angle gives the largest area under the corresponding angle constraint.
The special 2-radian optimum depends on the perimeter P = 2r + rθ remaining constant while both r and θ are allowed to vary.
This distinction is crucial. Optimisation results belong to specific constraints, not merely to a shape category. A sector can obey different constraints in different questions.
Why fixing arc length alone is also different
Suppose only the arc length s is fixed and radius is free. The sector area is A = rs/2, which increases with r.
To maintain the same arc length as the radius increases, the angle θ = s/r becomes smaller. For arbitrarily large r, the area becomes arbitrarily large, so no finite maximum follows from fixing arc length alone.
A separate radius bound or perimeter condition would be required to create a finite optimum.
This demonstrates why every geometrical optimisation problem must state what remains fixed. Without that constraint, looking for a stationary point can be unnecessary or even conceptually wrong.
A sector is not the same as a circular segment
A sector is the region between two radii and an arc. Its area is r²θ/2.
A circular segment is the region between a chord and the corresponding arc. For a minor angle, its area is r²(θ − sinθ)/2.
Their boundaries and perimeter formulas are different. A segment’s boundary contains a chord and an arc, not two radial sides and an arc.
Therefore the fixed-perimeter result θ = 2 radians does not automatically apply to a fixed-perimeter circular segment. A student must identify the requested region before selecting the appropriate formulas and constraint.
The derivative check also identifies the global maximum
For A(r) = Pr/2 − r², the graph is a downward-opening parabola in r. Its vertex is at r = P/4.
The area is zero at r = 0 and r = P/2 in the algebraic limiting expression, corresponding to degenerate sector limits. Inside the geometric radius range, A is positive.
The maximum is therefore at the vertex, with no competing interior stationary points. The angle-domain restriction still contains that vertex.
This is stronger than simply noting A″(P/4) < 0. It establishes the maximum over the full physically permitted set and aligns with the simple complete-square upper bound.
A frequent error: confuse sector perimeter with arc length
Some students read “perimeter 24” and set rθ = 24. That treats the entire available boundary as though it were only the curved arc.
The actual condition is 2r + rθ = 24. Forgetting 2r changes the relationship between radius and angle, so all later differentiation describes a different problem.
A helpful repair is to draw and label the three visible boundary pieces, even with a rough sketch. Write P = radius + radius + arc before simplifying it.
Geometry determines the algebra. If the algebraic objective does not match the boundaries in the drawing, a perfect derivative calculation will still answer the wrong question.
A frequent error: calculate area from the wrong formula
The area of the full circle is πr². A sector occupies only a fraction θ/(2π) of that area when θ is in radians.
Therefore A_sector = [θ/(2π)]πr² = r²θ/2. Using πr² directly would calculate the complete disk area instead.
Another error is using A = rs rather than rs/2. The factor 1/2 comes from the sector formula: substitute θ = s/r into r²θ/2 to obtain rs/2.
A quick unit check cannot detect a missing factor 1/2 because both formulas have area units. Deriving from the sector fraction or radian definition is the stronger conceptual repair.
A concise exam solution in six decisions
First define r, θ and s, making clear that θ is in radians. State the fixed perimeter P = 2r + s and the area A = rs/2.
Second eliminate s: s = P − 2r, giving A(r) = Pr/2 − r². Third state the admissible radius and angle conditions.
Fourth differentiate or complete the square to find r = P/4. Fifth justify the maximum using the negative second derivative, derivative signs or quadratic vertex.
Finally calculate s = P/2, θ = 2 and A_max = P²/16, with appropriate units. This is a complete optimisation solution rather than an unexplained memorised result.
Practice questions with checked answers
Question one: a sector has perimeter 24. Find the maximum area. Answer: 36 square units.
Question two: in Question one, find the optimal radius, arc length and angle. Answer: r = 6, s = 12 and θ = 2 radians.
Question three: sector perimeter P = 40. Find the maximum area and optimal radius. Answer: area 100 square units, radius 10.
Question four: derive A(r) for any fixed sector perimeter P. Answer: A(r) = Pr/2 − r².
Question five: derive the optimal sector area in terms of P. Answer: P²/16.
Question six: if the fixed perimeter doubles, how does the maximum area change? Answer: it quadruples.
Question seven: why does a sector angle of 2 radians not maximise area when only the radius is fixed? Because sector area grows with angle at fixed radius; the fixed-perimeter trade-off is absent.
Parent questions and targeted teaching
A child who writes the right area formula but the wrong perimeter constraint needs help reading and labelling geometric boundaries. Another who models the constraint correctly but cannot differentiate a quadratic needs an algebra or calculus repair.
A third may reach r = P/4 but forget to prove it is the maximum or check that θ = 2 radians is admissible. These are separate stages of mathematical reasoning.
In a maximum three-student lesson, a tutor can identify the first stage that fails and use a changed perimeter value to see whether the learner can repeat the structure independently.
The aim is understanding why half the perimeter becomes the arc and why the optimal angle stays fixed as the shape scales, not merely copying r = P/4 from one worked example.
Syllabus alignment and connected reading
This optimisation combines circular sectors, radians, area formulas, algebra and differentiation. Some fixed-perimeter sector extensions may be enrichment depending on the Singapore school’s taught Additional Mathematics sequence.
For official reference, see the relevant SEAB syllabus pages for 2026 O-Level Additional Mathematics, 2027 SEC G3 or 2027 SEC G2, as appropriate.
Read related guides How to Find the Central Angle and Arc Length from a Chord, How to Find the Maximum Area of a Rectangle Under a Parabola and How to Check the Meaning of a Calculus Answer. Further resources are in the Additional Mathematics archive and the eduKate tutorial reference.
Families can enquire about suitable eduKate Singapore small-group Mathematics lessons near Sixth Avenue MRT. Properly taught kids shine a bright light into the future.