Additional Mathematics optimisation questions sometimes ask for the largest rectangle that fits beneath a semicircular arch. When the rectangle’s base lies on the diameter and its two upper corners touch the semicircle, the width and height cannot be chosen independently. The circle equation creates their relationship, and differentiation identifies the exact maximum area.
For the upper semicircle x² + y² = 25, take upper corners (−x,y) and (x,y), where y = √(25 − x²). The rectangle has width 2x and height y, so its area is A(x) = 2x√(25 − x²). The maximum occurs at x = 5/√2, giving height 5/√2, full width 5√2 and area exactly 25 square units.
Did you know that the optimal rectangle has twice as much width as height? It is not a square. This guide explains the geometry, the permitted domain, differentiation, an alternative trigonometric proof, and how to recognise changes in constraints that require a different optimisation model.
Read the geometry before assigning any variable
A semicircle is half a circular disk. In this problem its straight diameter lies on the x-axis, and its curved boundary is the upper half of the circle x²+y²=25.
The rectangle has its lower side along a portion of the diameter and both upper corners on the arc. Its sides are horizontal and vertical, so the upper two corners must have equal y-coordinate.
Because the circle is symmetric about the y-axis, those two upper-corner x-coordinates must be opposites for a non-degenerate rectangle. Use x≥0 for the right upper coordinate and −x for the left.
This first sketch fixes the width as 2x. A solution using x as the complete width would lose half the actual rectangle and produce an incorrect area function.
Find the semicircle radius and available height
The equation x²+y²=25 describes a circle centred at O(0,0) with radius R=5. The upper semicircle consists of its points with y≥0.
Solve for the upper branch: y=√(25−x²). The positive square root is chosen because the rectangle sits above the x-axis.
At the right upper contact (x,y), the height of the rectangle is exactly y. Therefore its height is h(x)=√(25−x²).
As the rectangle widens and x moves toward five, this available height decreases. That trade-off creates the maximum-area problem: a narrow rectangle is tall, and an extremely wide rectangle becomes nearly flat.
Derive the area formula correctly
The upper contact coordinates are (−x,√(25−x²)) and (x,√(25−x²)). Their horizontal separation is 2x.
The rectangle’s lower corners are (−x,0) and (x,0), so its vertical height is √(25−x²).
Area is width multiplied by height. Hence A(x)=2x√(25−x²).
The factor two comes from the two halves of the symmetric rectangle, not from the fact that the boundary is a semicircle. The square root comes from Pythagoras or the circle equation.
Writing and labelling these geometric quantities before differentiating is essential. A perfectly correct derivative of a wrong area formula answers a different problem.
Set the permitted interval for x
The square root √(25−x²) is real only when 25−x²≥0. This gives −5≤x≤5.
But x is a non-negative half-width, so 0≤x≤5. The actual positive-area rectangle requires 0 At x=0 the rectangle has zero width. At x=5 its upper corners lie on the diameter, giving zero height. Both limiting cases have area zero. Therefore a positive global maximum must occur at an interior input. This domain check will also tell us which algebraic solutions of the derivative equation are admissible. Write A(x)=2x(25−x²)^(1/2). Apply the product rule and the chain rule. The derivative of 2x is 2. The derivative of (25−x²)^(1/2) is (1/2)(25−x²)^(−1/2)(−2x)=−x/√(25−x²). Therefore A′(x)=2√(25−x²)−2x²/√(25−x²). Put the terms over a common denominator: A′(x)=2(25−2x²)/√(25−x²). This formula is valid for 0 Set A′(x)=0. Since the denominator is positive inside the permitted interval, the derivative vanishes exactly when its numerator does. Thus 25−2x²=0, so 2x²=25 and x²=25/2. The algebraic roots are x=±5/√2. Because x represents a positive half-width, retain x=5/√2 only. This value lies between zero and five, so it is geometrically admissible. A student who reports both signs as two different rectangles has forgotten the meaning of x. The negative root describes the opposite horizontal coordinate already included as −x in the symmetric construction. At x=5/√2, the full width is w=2x=10/√2=5√2. The height is h=√(25−x²)=√(25−25/2)=√(25/2)=5/√2. Therefore w/h=(5√2)/(5/√2)=2. The rectangle is twice as wide as it is tall. Its area is A=wh=(5√2)(5/√2)=25 square units. All three quantities are exact; no decimal approximation is needed. The upper corners are (−5/√2,5/√2) and (5/√2,5/√2). Substituting either into x²+y²=25 confirms that both lie on the semicircle. The derivative is A′(x)=2(25−2x²)/√(25−x²). Its denominator is positive for every interior permitted x. For 0 Thus area rises toward the stationary candidate and falls after it. The candidate is a strict interior maximum. At the endpoints x=0 and x=5, the area is zero. Since the stationary value is 25, it is the global maximum on the closed permitted interval. This sign-and-boundary argument establishes the optimisation fully. For 0≤x≤5, area A(x) is non-negative. Therefore maximising A is equivalent to maximising A². Square the expression: A²=4x²(25−x²)=100x²−4x⁴. Let u=x². The range of u is 0≤u≤25. Then A²=100u−4u²=625−4(u−25/2)². This is largest when u=25/2. Hence x=5/√2, and A²_max=625, giving A_max=25. This method avoids differentiating the square root altogether. The change of variable remains valid because area is non-negative and u=x² has a known non-negative domain. Squaring a general signed expression does not preserve which value is larger: for instance, −10 is smaller than 3 but has a larger square. In this geometric problem, A represents an area and is non-negative everywhere on the permitted domain. For non-negative numbers, increasing A also increases A². Therefore the squared-area shortcut is mathematically justified. It would not be safe to apply the same reasoning to a signed displacement or arbitrary negative-valued objective without checking its sign. This is an important method-selection lesson. A clever algebraic simplification should always be accompanied by a statement of why it preserves the original optimisation question. For an upper semicircle x²+y²=R² with R>0, an inscribed rectangle of half-width x has height √(R²−x²), for 0≤x≤R. Its area is A(x)=2x√(R²−x²). Differentiating gives A′(x)=2(R²−2x²)/√(R²−x²) on the interior. The stationary condition is x²=R²/2, so x=R/√2. The height is R/√2 and full width is √2R. Thus the maximum area is A_max=(√2R)(R/√2)=R². The result scales with the square of the radius, as area should. The ratio of width to height remains exactly two for every positive R. Let θ be the angle between the positive x-axis and the radius from the centre to the upper-right corner. Then x=Rcosθ and y=Rsinθ, with 0≤θ≤π/2. The rectangle’s area is A=2xy=2R²sinθcosθ. By the double-angle identity, this becomes A=R²sin2θ. Since sin2θ≤1, the largest possible area is R². Equality occurs when 2θ=π/2 within the permitted interval, giving θ=π/4. Thus x=y=R/√2. The trigonometric proof agrees with the derivative and squared-area methods and explains why the corner lies along a radius making a 45° angle to the diameter. At the optimum, the radial coordinates of one upper corner satisfy x=y=R/√2. This makes the right triangle from the centre to that corner an isosceles right triangle. However, x is only half the rectangle’s width. The full width is 2x, while its height is y. Therefore equal values of x and y mean full width equals twice height, not equal width and height. The optimum rectangle is not a square. This subtle distinction is worth checking on a diagram. The 45° radius and the 2:1 rectangle dimension ratio are perfectly consistent once the half-width is labelled correctly. Choose x=3 for the radius-five semicircle. The height is √(25−9)=4, and full width is 6. The area is 6×4=24 square units, slightly below the maximum 25. This is a convenient 3–4–5 right-triangle comparison. Choose x=4 instead. The height is 3 and width is 8, also giving area 24. The two sample rectangles have equal area because exchanging x and y leaves their product 2xy unchanged. Their positions lie on opposite sides of the optimal 45° contact direction. This gives a useful reasonableness check without replacing the proof. At any upper corner (x,y), the circle equation gives x²+y²=R² with x,y≥0. The rectangular area is 2xy. Use the elementary inequality (x−y)²≥0 to obtain x²+y²≥2xy. Since x²+y²=R², it follows that A=2xy≤R². Equality holds exactly when x=y. This is a third proof of the maximum, requiring neither calculus nor trigonometric identities. It is useful for learners who are strengthening algebraic inequalities before formal optimisation lessons. The geometric interpretation remains unchanged: x is half-width, y is height, and equality gives width-to-height ratio two. For R=5, the maximum area is 25. For R=10, the optimal half-width is 10/√2, height 10/√2, full width 10√2 and maximum area 100. Doubling the radius doubles every length and multiplies the area by four. That is the expected scaling behaviour of similar plane figures. If a student obtains a maximum area proportional only to R rather than R², a factor or dimension is probably missing. The general formula A_max=R² is therefore both an exact result and a dimensional consistency check. It also makes it easy to answer reverse problems. Suppose a question states that the greatest possible rectangle area under a particular semicircle is 144 square units. Since A_max=R², the radius must satisfy R²=144. Taking the positive square root gives R=12 units. The optimum height is 12/√2=6√2, while the full width is w=√2R=12√2. Thus the ratio is w/h=(12√2)/(6√2)=2. The direct area check is (12√2)(6√2)=144. A reverse calculation should always use the radius as a positive length and verify the recovered dimensions against the area. For a rectangle inscribed in a full circle of radius R with all four corners on the circumference, the largest area is 2R² and the optimum is a square. That result does not apply here. The present rectangle occupies the upper semicircle with its lower side on the diameter, and only its two upper corners lie on the circular arc. The maximum area here is R², half the full-circle maximum, and the optimal rectangle has width twice height. Students who memorise “the biggest inscribed rectangle in a circle is a square” may incorrectly transfer that answer to a semicircular arch. The differing boundary conditions change the optimisation. The equation A(x)=2x√(R²−x²) relies on the rectangle’s base lying along the horizontal diameter y=0. If the bottom edge is instead fixed at y=c>0, the available height becomes √(R²−x²)−c and the admissible range of x changes. The maximum area need not equal R². If the rectangle may rotate rather than remaining axis-aligned, the geometry must be reconsidered. Similarly, allowing only one upper corner to touch the arc does not provide the same symmetric contact constraint. Optimisation results should always be tied to the exact permissible configuration. A familiar formula is not evidence that a modified problem has the same maximum. If the semicircle diameter is D, then R=D/2. Substitute into the general maximum-area formula. The maximum rectangle area is A_max=(D/2)²=D²/4. The optimal full width is √2(D/2)=D/√2 and the height is (D/2)/√2=D/(2√2). For a diameter of 10 units, this reproduces width 5√2, height 5/√2 and area 25. A common error is to insert the diameter directly where the circle formula requires radius. Writing R=D/2 at the beginning protects against a factor-of-four mistake in the final area. Suppose the semicircle has radius R and an inscribed rectangle must have a specified height h, with 0≤h≤R. The upper corner satisfies x²+h²=R², so its positive half-width is x=√(R²−h²). The full width is therefore w=2√(R²−h²). The area is A(h)=2h√(R²−h²). This expression is symmetric with the earlier formula in the variables x and h. The maximum occurs at h=R/√2, just as before. This version is useful when a word problem gives a building clearance or display height first rather than a chosen horizontal half-width. If the full width is w, then the upper-right x-coordinate is w/2. The semicircle equation gives height h=√(R²−w²/4). The allowable width lies between zero and 2R. Therefore area as a function of full width is A(w)=w√(R²−w²/4). The maximum occurs at w=√2R, producing height R/√2 and area R². This is algebraically the same problem written using a different decision variable. Students should be able to change variables without changing the physical answer, an important check in optimisation and mathematical modelling. Near x=0 the width is very small, so the area approaches zero even though the height is nearly R. Near x=R, the height approaches zero and again the area approaches zero, even though the width is almost the whole diameter. The derivative is positive before x=R/√2 and negative after it, showing a single interior peak. This makes sense because width benefits dominate first, while height loss dominates later. The endpoint behaviour helps students anticipate the shape of the area function before differentiating. A calculation that identifies the maximum at x=0 or x=R would contradict the geometry immediately. First draw and label the semicircle and rectangle, making x the non-negative half-width. Second use x²+y²=R² to write the positive height y=√(R²−x²). Third construct A(x)=2x√(R²−x²), with 0≤x≤R. Fourth differentiate or maximise A² and solve for the admissible stationary input. Fifth confirm the area increases before that input and decreases afterwards, or use an exact inequality. Sixth report full width, height and maximum area with appropriate units. Finally substitute the optimum corner into the circle equation. This check confirms that the proposed rectangle really fits the stated semicircle rather than an imagined alternative region. Question one: an upper semicircle has radius 5. Find the maximum rectangle area with base on its diameter. Answer: 25 square units. Question two: give the optimum width and height for Question one. Answer: 5√2 and 5/√2. Question three: for a semicircle of radius R, prove the maximum area is R². Answer: A=2xy≤x²+y²=R², with equality when x=y. Question four: if the radius is 8, find the optimal width, height and area. Answer: 8√2, 4√2 and 64. Question five: for a diameter D, give the maximum area. Answer: D²/4. Question six: if the maximum area is 144, find the radius. Answer: 12. Question seven: explain why the optimal rectangle is not a square. Answer: the contact coordinates have x=y, but the full rectangle width is 2x and its height is y. The same geometry can be extended to the upper half of an ellipse x²/a²+y²/b²=1, with a>0 and b>0. Its upper arc has height y=b√(1−x²/a²). An axis-aligned rectangle with base on the horizontal diameter and symmetric upper corners has width 2x and area A(x)=2bx√(1−x²/a²). Set u=x/a. Then A=2ab·u√(1−u²), with 0≤u≤1. The same optimisation used for a circle gives u=1/√2. Therefore the optimal full width is √2a, height b/√2 and maximum area ab. The width-to-height ratio is 2a/b, so the 2:1 ratio is special to a circle where a=b. Suppose instead the rectangle lies in the first quadrant with its lower-left corner at the origin, sides along both coordinate axes and only the opposite upper-right corner on x²+y²=R². Its width is x, height is y, and area is xy—not 2xy—because there is no mirrored half to the left. From x²+y²=R², the inequality 2xy≤R² gives xy≤R²/2. Equality occurs at x=y=R/√2. Thus the largest quarter-circle rectangle is a square of area R²/2. This contrasts with the semicircle-diameter problem, where the maximum is R² and the full rectangle is twice as wide as it is tall. Keep the original semicircle and axis-aligned rectangle, but change the objective from area to the rectangle’s perimeter. The width is 2x and the height is y=√(R²−x²), so its perimeter is P_rect=2(2x+y)=4x+2√(R²−x²). Differentiate: P′=4−2x/√(R²−x²). The stationary condition is x=2√(R²−x²), or x=2y. With x²+y²=R², this gives y=R/√5 and x=2R/√5. The optimal rectangle has full width 4R/√5 and height R/√5, a 4:1 ratio. Its perimeter is 2√5R, while its area is only 4R²/5. This shows why an area optimum cannot be reused when a problem changes the quantity being maximised. For radius R=5, the area-optimal rectangle has width 5√2≈7.071, height 5/√2≈3.536 and area 25. At half-width x=3, the width is 6 and height 4, giving area 24. At half-width x=4, the width is 8 and height 3, again giving area 24. For the perimeter-optimal rectangle, the half-width is 2√5≈4.472 and height is √5≈2.236. Its width is 4√5≈8.944, so its area is 20, noticeably below 25. These examples make the trade-off concrete: different valid rectangles can look wide or tall, and the one with the largest perimeter is not the one with the largest area. At a fixed half-width x, the semicircle supplies maximum permitted height y=√(R²−x²). If the rectangle’s upper edge lies strictly below that height, the rectangle can be made taller without changing its width. Therefore a maximum-area rectangle in this family must have its upper corners on the arc. Otherwise it cannot be optimal because increasing the height immediately increases area. This justifies the contact assumption geometrically rather than simply asserting it. The same argument applies to a fixed horizontal width under many downward arches: the upper corners must reach the highest allowed level compatible with the curve and with a horizontal top. Some questions label the height h rather than the half-width x. From the circle equation, x=√(R²−h²), giving area A(h)=2h√(R²−h²). Differentiate on 0 The associated half-width is x=√(R²−R²/2)=R/√2, yielding the same maximum area R². The agreement matters because the physical shape is unchanged by a different parameterisation. A valid optimisation should produce the same rectangle whether its height or half-width is selected as the variable. For R=5, the exact optimal dimensions contain √2. They cannot both be represented precisely by a short decimal. Rounding the half-width to 3.54 and then computing the height from an independently rounded value may create a rectangle whose quoted corners do not satisfy x²+y²=25 exactly. Keep x=5/√2 and y=5/√2 throughout the derivation, so the circle condition is exactly 25/2+25/2=25. Convert to decimals only at the final requested precision. This is a practical lesson for any constrained geometry model: exact expressions preserve constraints, while independent rounding of coordinates may break them slightly. This Additional Mathematics optimisation example joins Pythagoras, circle equations, differentiation, quadratic inequalities and trigonometry. The simplest approach begins with the right triangle, while derivative and algebraic methods provide independent proofs. Singapore students can consult the relevant SEAB syllabus listings for 2026 O-Level Additional Mathematics or 2027 SEC G3, selecting extension exercises according to their taught school programme. Related eduKateSG guides include How to Find the Maximum Area of a Rectangle Under a Parabola, How to Maximise a Circular Sector’s Area with Fixed Perimeter, How to Find a Circle Equation from Diameter Endpoints and the Additional Mathematics archive. Families can enquire about suitable eduKate Singapore maximum three-student Mathematics tutorials near Sixth Avenue MRT. The purpose is not only to obtain the maximum but to explain how geometry creates the trade-off and how several independent methods confirm the result. Properly taught kids shine a bright light into the future.Differentiate the product with a square root
Solve the stationary-area condition
Calculate the optimum dimensions
Prove it is a maximum rather than a minimum
A shorter approach by maximising the square of area
Why squaring the objective works here
The general answer for radius R
An alternative proof using trigonometry
Why a 45-degree contact angle does not make the rectangle square
Verify the maximum using one smaller rectangle
The original circle equation also produces a product constraint
How the radius determines the dimensions
Recover a semicircle radius from the maximum area
A semicircle is not the same as a full circle
Why the base-on-diameter condition matters
What if the diameter is given instead of the radius?
A fixed rectangular height determines its width
A fixed rectangular width determines its height
What the derivative says near the endpoints
A compact six-stage examination solution
Practice questions with checked answers
A semi-ellipse changes the rectangle’s aspect ratio
A quarter-circle rectangle is a different problem
Maximising perimeter gives different dimensions
A numerical comparison of three rectangle shapes
Why both upper corners must actually touch the arc
An alternative derivative in terms of height
Distinguish a geometric maximum from a measurement approximation
Syllabus context and connected learning