A formula can be algebraically correct and still describe a shape that cannot exist. A calculated maximum can be mathematically correct and still lie outside the dimensions allowed by the question. Geometric algebra becomes reliable only when the equation, the shape and the permitted values are kept together.
This Secondary 3 Mathematics Learning Guide connects perimeter, area, diagonals, borders, fencing and sectors with quadratic expressions and constraints. It distinguishes finding a shape with a specified area from finding the greatest possible area. The maximum-area sections use completing the square and elementary inequalities, not differentiation.
Scope and source: the 2027 SEC G3 Mathematics syllabus connects equations, quadratic graph features and mensuration. The design and optimisation examples here are guided consolidation and enrichment using those tools, not a claim that they form a separate compulsory Secondary 3 chapter. Follow the depth assigned by your school.
Learning route: build the model · maximum area · fencing constraints · paths and borders · sector design · practice · explained answers.
Return to the Secondary Mathematics Hub. Prerequisites: Quadratic Equations and Word Problems and Similarity, Scale Factors and Mensuration.
The Three Questions Behind a Geometric Model
First ask what quantity is fixed. It might be perimeter, area, diagonal length, available fencing or the combined boundary of a sector. Next ask what quantity is required: a missing dimension, a possible area, a maximum or a minimum. Finally ask which dimensions are permitted.
A fixed perimeter creates a sum relationship between lengths. A fixed area creates a product relationship. A fixed rectangle diagonal creates a sum of squares through Pythagoras. These conditions are not interchangeable even when the same shape appears in every question.
Write the domain before solving. Positive side lengths exclude zero and negative values. A border must leave positive inner dimensions. Whole-centimetre dimensions create an integer restriction. A finite wall may exclude a length that an unconstrained formula would recommend. These are mathematical conditions, not extra comments added after the answer.
A Diagnostic That Separates the Constraint From the Target
A rectangle has perimeter 50 cm. If one side is x cm, express the adjacent side and area in terms of x. State the basic domain. Then explain whether the perimeter alone determines one unique rectangle.
The adjacent side is 25 − x, the area is x(25 − x), and the basic domain is 0 < x < 25. The perimeter does not determine a unique rectangle: 5 by 20 and 10 by 15 both have perimeter fifty. An area condition, a side ratio or an instruction to maximise would add a different kind of information.
A learner who writes x(50 − x) has used the whole perimeter as the sum of only two adjacent sides. A learner who immediately sets the area expression equal to fifty has confused the fixed boundary length with the target area. The first repair is to identify what each number measures.
Worked Example 1: Build an Area Expression From a Fixed Perimeter
A rectangle has perimeter 72 cm. Let one side be x cm. Express its area A in terms of x. With adjacent side y, the perimeter equation is 2x + 2y = 72. Divide by two to obtain x + y = 36, so y = 36 − x.
Area is the product of adjacent sides: A = x(36 − x) = 36x − x². Both sides must be positive, so 0 < x < 36. At this stage, the expression describes a family of rectangles. It is not yet an equation selecting a particular member.
The formula returns to the diagram cleanly: x labels one pair of opposite sides, and 36 − x labels the other. If those labels are substituted into the perimeter, 2x + 2(36 − x) = 72 for every permitted x. That identity verifies that the expression respects the fixed condition.
Worked Example 2: A Specified Area Produces an Equation
The same perimeter is 72 cm, and the area is 299 cm². Find the dimensions. Now the area expression must equal the given target: x(36 − x) = 299. Rearranging gives x² − 36x + 299 = 0.
Factorisation gives (x − 13)(x − 23) = 0, so x = 13 or x = 23. Both satisfy the positivity domain. The rectangle has dimensions 13 cm by 23 cm, with perimeter 2(13 + 23) = 72 and area 13 × 23 = 299.
The two algebraic roots do not necessarily describe two different geometric designs. We chose x as either side without distinguishing orientation. Choosing x = 13 makes the adjacent side 23; choosing x = 23 makes it 13. The same unlabeled rectangle has simply been described in the opposite order.
This is different from rejecting an inadmissible root. Both values are mathematically valid, but the final geometric interpretation combines them into one pair of side lengths. Root interpretation depends on how the variable was defined.
Worked Example 3: Find the Maximum Without Calculus
Among all rectangles with perimeter 72 cm, find the greatest area. Start from A = 36x − x². Complete the square: A = −(x² − 36x) = −[(x − 18)² − 324] = 324 − (x − 18)².
A real square is nonnegative, so subtracting it cannot make the result exceed 324. Therefore A ≤ 324. Equality is attained when x − 18 = 0, giving x = 18 and y = 18. The greatest area is 324 cm², achieved by an 18 cm square.
A complete maximum argument needs both parts: a bound that applies to every feasible shape, and a feasible shape that attains it. Merely obtaining the number 324 is not enough. The dimensions must be positive and satisfy the original perimeter; here they do.
| One side x | Adjacent side | Area |
|---|---|---|
| 6 cm | 30 cm | 180 cm² |
| 10 cm | 26 cm | 260 cm² |
| 13 cm | 23 cm | 299 cm² |
| 18 cm | 18 cm | 324 cm² |
| 23 cm | 13 cm | 299 cm² |
The table illustrates the symmetry, but the completed-square form proves the bound for all real permitted dimensions. Checking several rectangles is not a proof that no untested rectangle has a larger area. The algebra covers the whole domain at once.
Worked Example 4: Recognise an Impossible Target Area
Can a rectangle with perimeter 72 cm have area 330 cm²? The completed-square model would require 324 − (x − 18)² = 330, so (x − 18)² = −6. No real square is negative, so the answer is no.
The impossibility follows from the geometric constraint, not from failure to find convenient factors. A calculator error message would not explain it. The bound A ≤ 324 tells us immediately why every proposed 330 cm² design must fail.
This is a useful design habit even in an idealised classroom model: check whether the target is feasible before spending time searching for dimensions. Some problems ask for a number, while others ask whether the requested number can exist under the stated conditions.
Worked Example 5: A Restricted Dimension Changes the Answer
A rectangle still has perimeter 72 cm, but a specified side x must satisfy 10 ≤ x ≤ 14. Find the greatest possible area. The same expression A = 324 − (x − 18)² applies, but the unrestricted maximiser x = 18 is not permitted.
On the allowed interval, x = 14 is closest to eighteen, so it makes (x − 18)² smallest. The area is 14 × 22 = 308 cm². The optimal permitted rectangle is not a square because the extra condition prevents it.
The phrase “a square gives maximum area” needs its constraint attached: it describes rectangles with a fixed full perimeter and no further restriction excluding the square. It is not a universal answer to every rectangle-design question.
Endpoint wording can matter too. If the requirement were x < 14 rather than x ≤ 14, areas could approach 308 cm² but would not attain it. Do not call an excluded endpoint an achieved maximum. This distinction is an extension in reading mathematical conditions precisely.
Worked Example 6: Three Fenced Sides Against a Long Wall
A rectangular enclosure uses a straight wall as one side and exactly 60 m of fencing for the other three sides. The wall is long enough for any candidate considered. Find the greatest area. Let the two perpendicular sides each be x m, and the side parallel to the wall be y m.
Fencing gives 2x + y = 60, not 2x + 2y = 60. Hence y = 60 − 2x and A = x(60 − 2x) = 60x − 2x². The domain is 0 < x < 30.
Complete the square: A = 450 − 2(x − 15)². The maximum is 450 m² at x = 15, giving y = 30. Check the fencing: 15 + 30 + 15 = 60. The enclosed rectangle is 15 m by 30 m.
The optimal side parallel to the wall is twice either perpendicular side. A square is not optimal for this three-side constraint. The best geometry changed because a different boundary was counted, even though the word “rectangle” remained the same.
Worked Example 7: Include an Internal Divider
The same long-wall arrangement now needs one internal dividing fence parallel to the two perpendicular sides. Exactly 60 m of fencing covers all three outer fenced sides and the divider. Maximise the total enclosed area.
There are now three fence segments of length x and one of length y. Thus 3x + y = 60. The area is A = x(60 − 3x) = 60x − 3x² = 300 − 3(x − 10)².
The maximum total area is 300 m², obtained with x = 10 and y = 30. Check the material length: 3(10) + 30 = 60. The divider uses fencing but does not create additional land area, so it belongs in the boundary constraint, not as an extra area term.
This is why a labelled sketch is valuable even when the algebra is straightforward. Count every physical segment once. A missing divider in the equation would produce a plausible-looking but unaffordable design under the stated fence-length limit.
Worked Example 8: A Wall That Is Too Short for the Unrestricted Design
Return to the three-side enclosure with no divider and sixty metres of fencing, but now the available wall segment is only twenty metres long. The side y along the wall must satisfy 0 < y ≤ 20.
Since 2x + y = 60, write x = (60 − y)/2. Then A = y(60 − y)/2 = 450 − (y − 30)²/2. The unrestricted best y is thirty, but the wall excludes it.
Among permitted y-values, twenty is closest to thirty. Thus y = 20 and x = 20, giving 400 m². The square is now the constrained optimum, but for a different reason from the full-perimeter case. It occurs at the end of the permitted wall-length interval.
The model assumes straight boundaries, negligible fence thickness and complete use of the stated fencing. It does not provide a construction specification for real land. The learning task is to see how one new physical limit changes the feasible domain and therefore the selected dimensions.
Worked Example 9: Whole-Number Dimensions
A rectangle has perimeter 54 cm, and both side lengths must be whole numbers of centimetres. Find the greatest area. Adjacent sides satisfy x + y = 27, so A = x(27 − x).
Completing the square gives A = 182.25 − (x − 13.5)². The continuous maximum occurs at x = 13.5, but that is not a whole number. The nearest permitted integers are thirteen and fourteen. Both give area 182 cm², for a 13 cm by 14 cm rectangle.
This is not ordinary rounding of a final measurement. The whole-number rule is a constraint on the design itself. Comparing the squared distances from 13.5 identifies the best integer candidates and proves why more distant integers give smaller areas.
Worked Example 10: Fixed Area and Minimum Perimeter
A rectangle has area 96 cm². Find its least possible perimeter when positive real side lengths are allowed. Let sides be x and y, so xy = 96. Since (x − y)² ≥ 0, we have (x + y)² ≥ 4xy = 384.
Because x + y is positive, x + y ≥ 2√96. Therefore perimeter P = 2(x + y) ≥ 4√96 = 16√6 cm, approximately 39.2 cm. Equality occurs when x = y = √96 = 4√6 cm, which is a feasible square.
If the side lengths must instead be positive integers, the factor pairs of 96 are 1 and 96, 2 and 48, 3 and 32, 4 and 24, 6 and 16, and 8 and 12. Their smallest perimeter is 40 cm, for sides eight and twelve.
The continuous and integer answers differ because they optimise over different sets of shapes. A correct proof must specify the permitted dimensions. The square-root side length is not “wrong” in the continuous question, and it is not an allowed integer design in the second question.
Worked Example 11: Fixed Diagonal and a Perimeter Condition
A rectangle has diagonal 10 cm and perimeter 28 cm. Find its dimensions. The diagonal gives x² + y² = 100. The perimeter gives x + y = 14.
Square the sum: 196 = x² + y² + 2xy = 100 + 2xy. Hence xy = 48. The sides therefore have sum fourteen and product forty-eight, giving 6 cm and 8 cm. Check: 6² + 8² = 100, so the diagonal is ten.
A related maximum question gives only the fixed diagonal ten. Since (x − y)² ≥ 0, x² + y² ≥ 2xy. Therefore 100 ≥ 2A and A ≤ 50. Equality occurs for a square with side 5√2 cm. Its area is 50 cm².
That square does not satisfy the additional perimeter-28 condition. This is the key distinction: an optimum under one constraint need not remain feasible when another is added. A solver must not drop a condition simply because a familiar maximum-area rule becomes available.
Worked Example 12: A Uniform External Path
A 10 m by 6 m rectangular garden is surrounded externally by a path of uniform width w. The combined area is 140 m². Find w. The path appears on both sides of each dimension, so the outer dimensions are 10 + 2w and 6 + 2w.
Set (10 + 2w)(6 + 2w) = 140. Expand to obtain 4w² + 32w + 60 = 140. Divide the rearranged equation by four: w² + 8w − 20 = 0, or (w − 2)(w + 10) = 0.
The candidates are two and negative ten. Since the path width is positive, w = 2 m. Outer dimensions are fourteen by ten, giving the required 140 m². The path itself has area 140 − 60 = 80 m².
Using 10 + w and 6 + w would model a total dimension increase of w, not a path of width w around every side. Labelling the two added strips before expanding is a more effective repair than doing additional factorisation exercises.
Worked Example 13: A Positive Root Can Still Be Geometrically Impossible
A rectangular sheet is 20 cm by 12 cm. A uniform border of width w is removed inside all four edges, leaving a central rectangle of area 128 cm². Find w. The inner dimensions are 20 − 2w and 12 − 2w.
The shorter inner side must remain positive, so 0 < w < 6. The area equation is (20 − 2w)(12 − 2w) = 128. This simplifies to 4w² − 64w + 112 = 0, or w² − 16w + 28 = 0.
Factorising gives (w − 2)(w − 14) = 0. Both candidates are positive, but only w = 2 cm lies in the permitted interval. It leaves an inner rectangle 16 cm by 8 cm, with area 128 cm².
At w = 14, the two algebraic expressions for inner lengths are −8 and −16. Their product is still positive 128, which explains why the area equation alone admits the root. Negative lengths do not form the required rectangle. The domain, not the sign of the area, rejects the false design.
Worked Example 14: Maximum Area of a Sector With Fixed Boundary
A minor sector has a total boundary length of 20 cm, including its two radii and arc. Find its greatest possible area. Let the radius be r and arc length be s. The boundary condition is 2r + s = 20.
With angle θ in radians, s = rθ and sector area A = 1/2r²θ. Substitute s = rθ to get A = rs/2. Then s = 20 − 2r gives A = r(20 − 2r)/2 = 10r − r² = 25 − (r − 5)².
The algebraic upper bound is 25 cm², attained at r = 5 and s = 10. Check the shape condition: θ = s/r = 2 radians, which is positive and less than π, so it is a valid minor sector. Hence the maximum is 25 cm².
The sector restriction is not automatic for every positive radius in the quadratic expression. Since θ = 20/r − 2 and a minor sector has 0 < θ ≤ π, feasible radii satisfy 20/(π + 2) ≤ r < 10. The candidate r = 5 satisfies that interval. Checking it completes the maximum argument.
The boundary contains an arc and two radii, not a chord. Confusing a sector with a segment would change the perimeter model. The related measurement distinctions are developed in Arc Length, Sector Area, Radians and Composite Mensuration.
Not Every Request for a Maximum Has a Finite Answer
A rectangle with fixed width 6 cm and unrestricted positive length has area 6L. Making L larger keeps increasing the area. Without a perimeter, length or other limiting condition, there is no finite greatest area.
This is different from a maximum lying at an excluded endpoint or a requested area being impossible. In one case the feasible areas grow without bound. In another, a finite bound exists but is not attained. In the third, the target exceeds what the constraints permit. The wording and domain determine which conclusion applies.
Independent Practice: State the Domain and the Attaining Shape
These are original practice and enrichment questions. Use exact forms unless a decimal is requested. For a maximum or minimum, show both the bound and a permitted set of dimensions attaining it.
1. A rectangle has perimeter 50 cm. Find its greatest area and the dimensions that attain it.
2. A rectangle has perimeter 50 cm and area 144 cm². Find its dimensions and interpret the two quadratic roots.
3. Can a rectangle with perimeter 50 cm have area 160 cm²? Explain without trial-and-error dimensions.
4. Exactly 80 m of fencing encloses three sides of a rectangle against a sufficiently long straight wall. Find the greatest area.
5. The arrangement in Question 4 now includes one internal divider parallel to the two perpendicular sides. The same total 80 m covers the three outer fenced sides and the divider. Find the greatest total area.
6. A rectangle has perimeter 54 cm and positive integer side lengths. Find its greatest area.
7. A rectangle has area 60 cm² and positive integer side lengths. Find its least perimeter.
8. A rectangle has diagonal 13 cm and perimeter 34 cm. Find its dimensions.
9. A 12 m by 8 m garden has an external uniform path. Garden and path together occupy 192 m². Find the path width.
10. An 18 cm by 10 cm sheet has an internal uniform border removed, leaving area 84 cm². Find the border width and explain why the other root is invalid.
11. A minor sector has total boundary length 24 cm, including both radii and the arc. Find its maximum area and check its angle.
12. A rectangle has perimeter 48 cm. A specified side x must satisfy 5 ≤ x ≤ 9. Find the maximum permitted area.
Explained Answers
1. A = x(25 − x) = 156.25 − (x − 12.5)². The maximum is 156.25 cm², attained at sides 12.5 cm and 12.5 cm. They satisfy perimeter fifty.
2. x(25 − x) = 144 gives x² − 25x + 144 = (x − 9)(x − 16) = 0. The dimensions are 9 cm by 16 cm. The roots exchange which side is called x; they describe the same unlabeled rectangle.
3. No. Question 1 established A ≤ 156.25 cm² for every rectangle with that perimeter. The target 160 exceeds the bound.
4. Let perpendicular sides be x. Then y = 80 − 2x and A = 80x − 2x² = 800 − 2(x − 20)². Maximum area is 800 m² at x = 20 m, y = 40 m.
5. Now 3x + y = 80. Hence A = 80x − 3x² = 1600/3 − 3(x − 40/3)². Maximum area is 1600/3 m² at x = 40/3 m, y = 40 m. Check the fencing: 3(40/3) + 40 = 80.
6. Adjacent sides sum to twenty-seven. The nearest integers to the continuous best value 13.5 are thirteen and fourteen. Maximum integer area is 182 cm².
7. Integer factor pairs of sixty are 1 and 60, 2 and 30, 3 and 20, 4 and 15, 5 and 12, and 6 and 10. Their smallest sum is sixteen, so the least perimeter is 32 cm, for sides 6 cm and 10 cm.
8. x + y = 17 and x² + y² = 169. Therefore 289 = 169 + 2xy, giving xy = 60. The positive sides with sum seventeen and product sixty are 5 cm and 12 cm.
9. (12 + 2w)(8 + 2w) = 192 gives w² + 10w − 24 = (w + 12)(w − 2) = 0. Positive width is 2 m, giving outer dimensions sixteen by twelve.
10. The domain is 0 < w < 5. The equation (18 − 2w)(10 − 2w) = 84 reduces to (w − 2)(w − 12) = 0. Only w = 2 cm is feasible; twelve would make the inner dimensions negative. The remaining rectangle is fourteen by six.
11. s = 24 − 2r gives A = rs/2 = 12r − r² = 36 − (r − 6)². Maximum is 36 cm² at r = 6 cm and s = 12 cm. The angle is s/r = 2 radians, so the minor-sector condition is satisfied.
12. A = x(24 − x) = 144 − (x − 12)². The unrestricted maximum at twelve is excluded. The allowed x closest to twelve is nine, giving adjacent side fifteen and maximum permitted area 135 cm².
Checking a Maximum Requires More Than Substitution
Substitution can verify that a proposed shape meets the original measurements, but it does not by itself prove that no better shape exists. For a quadratic maximum, the completed-square expression supplies the global bound. The equality case supplies the candidate. The domain check establishes whether that candidate is allowed.
Keep those tasks distinct. “The dimensions give area 450” verifies an area. “Every feasible area is at most 450” establishes a bound. “These feasible dimensions attain 450” finishes the maximum argument. Missing any one can leave an incomplete solution even when the final number happens to be correct.
For a specified-area equation, the checking task is different. Find all algebraic roots, reject those outside the geometric domain, and identify whether remaining roots describe different shapes or merely exchanged labels. Do not apply a maximum argument when the question asks for every design meeting a target.
Diagnose the First Incorrect Geometric Decision
If a fencing answer ignores the divider, the first error is boundary counting, not completing the square. If an internal-border answer accepts a width wider than half the sheet, the error is the domain. If a constrained problem recommends an excluded vertex, the error is selecting the optimum over the wrong interval.
Repair that decision with a changed example. Add a second divider, limit the wall length, require whole-number dimensions or exchange an external path for an internal border. The student should predict which equation or domain changes before calculating again. This checks whether the model is understood rather than remembered from one drawing.
Teacher and Parent Prompts
Ask what is fixed, what is being optimised and which values are permitted. Ask the learner to trace every length included in a perimeter. When an answer is described as “the maximum”, ask which line proves that all other feasible values are no larger.
For support, begin with a rectangle and a small table, then show how the square term explains the pattern across all dimensions. For extension, change one boundary condition and require a fresh derivation. More demanding work should involve a new mathematical decision, not merely larger coefficients.
Questions Students Often Ask
Does maximum-area work always require calculus? No. The quadratic models in this guide are handled by completing the square and showing a square is nonnegative. The method is appropriate when that structure is present.
Is a square always the best rectangle? Not under every constraint. Three fenced sides, an internal divider or a restricted dimension can produce a different optimum. State the boundary condition before using a remembered result.
Why reject a positive root? Positivity alone may be insufficient. An internal border must leave positive inner dimensions, and a radius or side length may have additional upper or lower limits.
What does “best” mean in a design question? The mathematical target must define it, such as greatest area or least perimeter. A real design can have other requirements, but those cannot be silently inserted into or removed from the classroom model.
Continue the Secondary 3 Learning Route
Continue with Distance-Time and Speed-Time Graphs for formulas restricted to journey stages; Probability with Unknown Quantities and Changing Sample Spaces for admissible integer roots; and Missing Values, Combined Means and Statistical Data Corrections for totals constrained by several conditions.
A geometric answer is secure when the formula preserves the boundary, the candidate belongs to the domain and the final shape satisfies every original condition. Return to the Secondary Mathematics Hub.
