Source: CSSA Past Papers
Tags: optimisation, maxima, minima, differentiation, area maximisation, exponential growth, simple harmonic motion, SHM, printed area, railway track, trapezium, inscribed square, calculus, HSC, CSSA
These questions require you to form an expression for the quantity to be optimised (usually area), differentiate, set the derivative to zero, and verify whether you have a maximum or minimum. Several involve constraints that link two variables, reducing the problem to a single-variable calculus exercise. The final questions cover exponential growth and motion (SHM).
Constraint equation
A relationship between variables imposed by the geometry or physics of the problem (e.g. a fixed perimeter or fixed total area). Used to eliminate one variable before differentiating.
Objective function
The quantity you want to maximise or minimise, expressed as a function of a single variable after substitution from the constraint.
Stationary point
A point where dy/dx = 0. Could be a local maximum, local minimum, or point of inflection.
Second derivative test
If d²y/dx² < 0 at a stationary point, it is a local maximum. If d²y/dx² > 0, it is a local minimum.
Exponential growth model
P = P₀ eᵏᵗ, where P₀ is the initial population, k is the growth constant, and t is time.
Simple harmonic motion (SHM)
Motion where x = c + A sin(nt + α) or x = c + A cos(nt + α). Velocity is ẋ = dx/dt. The particle oscillates about the centre with amplitude A and angular velocity n.
A miniature railway of total length 1000 m consists of a rectangle with semicircles on each short end. The rectangle has length y and width x (the diameter of each semicircle).
Part (i): Show y = 500 − πx/2
The total perimeter consists of:
Two lengths of y (top and bottom of the rectangle)
Two semicircles of diameter x, which together form one full circle of circumference πx
So: 2y + πx = 1000, giving y = 500 − πx/2 ✓
Part (ii): Show A = (2000x − πx²)/4
The enclosed area is the rectangle plus two semicircles (one full circle):
A = xy + π(x/2)² / 2 × 2 = xy + πx²/4
Wait, let me re-read. The area enclosed by the track:
Rectangle area: xy
Two semicircles = one circle of diameter x, so area = π(x/2)² = πx²/4
A = xy + πx²/4
Substitute y = 500 − πx/2:
A = x(500 − πx/2) + πx²/4
= 500x − πx²/2 + πx²/4
= 500x − πx²/4
= (2000x − πx²) / 4 ✓
Part (iii): Maximum area
dA/dx = (2000 − 2πx) / 4
Set to zero: 2000 − 2πx = 0, so x = 1000/π
d²A/dx² = −2π/4 < 0, confirming a maximum.
A = (2000 × 1000/π − π × (1000/π)²) / 4 = (2000000/π − 1000000/π) / 4 = (1000000/π) / 4 = 250000/π ≈ 79577 m²
Converting: 1 hectare = 10000 m², so A ≈ 79577/10000 ≈ 8 hectares
A rectangular sheet has area 300 cm². Margins: 2 cm top, 1 cm bottom, 2.5 cm each side. Maximise the printed area.
Let the sheet have width w and height h. Then wh = 300, so h = 300/w.
Printed width = w − 2(2.5) = w − 5
Printed height = h − 2 − 1 = h − 3
Printed area P = (w − 5)(h − 3)
Substitute h = 300/w:
P = (w − 5)(300/w − 3)
= 300 − 3w − 1500/w + 15
= 315 − 3w − 1500/w
dP/dw = −3 + 1500/w²
Set to zero: 1500/w² = 3, so w² = 500, giving w = 10√5
Then h = 300/(10√5) = 30/√5 = 6√5
Printed dimensions:
Width: 10√5 − 5 = 5(2√5 − 1) cm
Height: 6√5 − 3 = 3(2√5 − 1) cm
Verify it is a maximum: d²P/dw² = −3000/w³ < 0 for positive w. Confirmed.
P = 5000 eᵏᵗ, where P is population after t days.
Part (a): Find k given P = 8500 when t = 1
8500 = 5000 eᵏ
eᵏ = 1.7
k = ln 1.7 ≈ 0.5 (to 1 decimal place)
Part (b): When does the colony triple?
Triple means P = 15000.
15000 = 5000 eᵏᵗ → eᵏᵗ = 3 → kt = ln 3
t = ln 3 / k = ln 3 / ln 1.7 ≈ 1.0986 / 0.5306 ≈ 2 days (to the nearest day)
Part (c): Growth rate after 2 days
dP/dt = 5000k eᵏᵗ
At t = 2: dP/dt = 5000 × ln 1.7 × e^(2 ln 1.7) = 5000 × ln 1.7 × 1.7²
= 5000 × 0.5306 × 2.89 ≈ 6796 insects/day
x = 2 − 2 sin 2t for 0 ≤ t ≤ 2π
Part (a): Velocity
ẋ = dx/dt = −4 cos 2t
Part (b): Initial position and velocity (t = 0)
x(0) = 2 − 2 sin 0 = 2 − 0 = 2 m
ẋ(0) = −4 cos 0 = −4 m/s
The particle starts 2 m from the origin, moving in the negative direction.
Part (c): Describe the motion
The displacement equation x = 2 − 2 sin 2t is of the form x = c − A sin(nt), which describes SHM.
Centre of motion: x = 2 (the constant term)
Amplitude: A = 2 (the coefficient of sin 2t)
Angular velocity: n = 2
Phase: 0 (no phase shift)
Period: T = 2π/n = π seconds
The particle oscillates between x = 0 and x = 4, centred on x = 2.
ABCD is a unit square (AB = 1 cm). PQRS is a square inscribed inside it with DP = CS = x.
Part (i): Show A = 2x² − 2x + 1
By Pythagoras, the side length of PQRS is:
SP² = DS² + DP² (where S is on CD, P is on DA)
Wait, let me set this up carefully. ABCD is a square with side 1. Points P on DA and S on CD with DP = CS = x.
DS = DC − CS = 1 − x
Side SP of the inner square: SP² = DP² + DS² = x² + (1 − x)² = x² + 1 − 2x + x² = 2x² − 2x + 1
Area of PQRS = SP² = 2x² − 2x + 1 ✓
Part (ii): Minimum area
dA/dx = 4x − 2
Set to zero: x = 1/2
d²A/dx² = 4 > 0, so this is a minimum.
A(1/2) = 2(1/4) − 2(1/2) + 1 = 1/2 − 1 + 1 = 1/2 cm²
ABCD is an isosceles trapezium inscribed in a semicircle of centre O, diameter 20 cm. BC = x, OE is the altitude.
Part (i): Prove △BOE ≅ △COE
OB = OC = 10 (radii)
OE is common
BE = CE (E is the midpoint of BC since the trapezium is isosceles)
By SSS (or by the perpendicular from centre to chord bisects the chord), △BOE ≅ △COE.
Part (ii): Show A = (1/4)(x + 20)√(400 − x²)
Using coordinates: place O at the origin, AD along the diameter (length 20, so A = (−10, 0), D = (10, 0)).
B and C are on the semicircle x² + y² = 100 (radius 10), with BC = x. By symmetry B = (−x/2, h) and C = (x/2, h) where h = √(100 − x²/4).
The trapezium ABCD has:
Parallel sides: AD = 20, BC = x
Height: h = √(100 − x²/4) = √((400 − x²)/4) = √(400 − x²)/2
Area = (1/2)(AD + BC) × h = (1/2)(20 + x) × √(400 − x²)/2
= (x + 20)√(400 − x²) / 4 ✓
Part (iii): Find BC for maximum area
A = (1/4)(x + 20)(400 − x²)^(1/2)
Use the product rule to differentiate:
dA/dx = (1/4)[(400 − x²)^(1/2) + (x + 20) × (−2x)/(2(400 − x²)^(1/2))]
= (1/4)[(400 − x²)^(1/2) − x(x + 20)/(400 − x²)^(1/2)]
= (1/4) × [(400 − x²) − x² − 20x] / (400 − x²)^(1/2)
= (1/4) × [400 − 2x² − 20x] / √(400 − x²)
Set numerator to zero: 400 − 2x² − 20x = 0, so x² + 10x − 200 = 0.
Using the quadratic formula: x = (−10 ± √(100 + 800))/2 = (−10 ± 30)/2
x = 10 or x = −20. Since 0 < x < 20, we take x = 10.
BC = 10 cm for maximum area.
General optimisation procedure:
Draw a diagram and label variables
Write the objective function (quantity to optimise)
Use the constraint to eliminate one variable
Differentiate, set equal to zero, solve
Verify max/min using the second derivative test
Answer in context (units, rounding)
Exponential growth: P = P₀ eᵏᵗ, growth rate = dP/dt = kP₀ eᵏᵗ
SHM displacement: x = c + A sin(nt + α), period T = 2π/n
⚠️ Always state and use the constraint equation before differentiating. A common error is differentiating a two-variable expression without substitution.
⚠️ Do not forget to verify maximum vs minimum (second derivative test, or test values either side). Marks are given for this step.
⚠️ In perimeter/area problems with semicircles, take care whether "two semicircles" means one full circle or two separate half-circles. The circumference of two semicircles of diameter x is πx (one full circumference), not 2πx.
⚠️ For exponential growth, "growth rate" means dP/dt (the derivative), not k. Evaluate at the given time.
⚠️ When describing SHM, examiners expect: centre of motion, amplitude, angular velocity (or period), and direction of initial motion.
⚠️ The printed-area problem is a classic template. If you see a fixed-area sheet with margins, think: express printed area as a function of one side length, substitute the area constraint, differentiate.
Q: A rectangle with two semicircular ends has perimeter 1000 m. Write the total enclosed area as a function of the width x only.
A: The constraint gives y = 500 − πx/2. The area is xy + πx²/4 = 500x − πx²/4, or equivalently (2000x − πx²)/4.
Q: For P = 5000 eᵏᵗ, if P = 8500 at t = 1, what is k?
A: eᵏ = 8500/5000 = 1.7, so k = ln 1.7 ≈ 0.53 (or 0.5 to 1 d.p.).
Q: A particle moves with x = 2 − 2 sin 2t. What are the extreme positions?
A: sin 2t ranges from −1 to 1, so x ranges from 2 − 2(1) = 0 to 2 − 2(−1) = 4. The particle oscillates between x = 0 and x = 4.
Q: In the inscribed-square problem where A = 2x² − 2x + 1, why is the minimum at x = 1/2 and not at a boundary?
A: dA/dx = 4x − 2 = 0 at x = 1/2, and d²A/dx² = 4 > 0 confirms a minimum. At the boundaries (x = 0 or x = 1), A = 1 (the inner square coincides with the outer square rotated), which is larger than 1/2.
Q: What does the second derivative test tell you, and when might it fail?
A: If f″(x₀) < 0 at a stationary point, it is a local maximum; if f″(x₀) > 0, a local minimum. It fails (is inconclusive) when f″(x₀) = 0, in which case you need to check values either side or use higher derivatives.
optimisation, optimization, maxima and minima, stationary points, second derivative test, constraint equation, area maximisation, perimeter constraint, exponential growth, population model, growth rate, simple harmonic motion, SHM, velocity, displacement, angular velocity, amplitude, inscribed shapes, isosceles trapezium, semicircle, CSSA past papers, HSC calculus