IB Maths AA HL
Topic 5 — Calculus
Paper 1 & 2
~11 min read
Modelling with Differentiation
Optimisation problems — where a question asks for the MAXIMUM or MINIMUM of something — are differentiation in disguise. Set up the quantity to optimise as a function of ONE variable, differentiate, solve f′(x) = 0, and use the second derivative (or context) to confirm max vs min. The skill is in the SETUP: rewriting the quantity in one variable using the question’s constraint.
📘 What you need to know
- Optimisation = stationary points: max/min of a function occur where f′(x) = 0.
- One variable only: the quantity to be optimised must be written as a function of ONE variable. Use the constraint to eliminate any others.
- Different letters allowed: optimisation problems often use V (volume), S (surface area), C (cost), A (area), P (perimeter), r (radius), and so on — instead of f, x, y.
- Differentiate by the variable you’ve chosen — e.g., dV/dr if V is in terms of r.
- Second derivative test: f′′(x) > 0 → minimum; f′′(x) < 0 → maximum.
- Domain constraints matter: dimensions must be positive; sometimes there’s a natural upper bound. Reject solutions outside the valid domain.
- “Show that” Q parts are common — the given formula lets you proceed with later parts even if you can’t derive it yourself.
- Interpret the answer: state the max/min value AND the value of the variable that achieves it, with correct units.
The 5-step optimisation workflow
🧭 Recipe — solve an optimisation problem
- Express the quantity to optimise in ONE variable — use the constraint to eliminate any others. Sometimes this is given as “show that…”
- Differentiate with respect to that variable; set the derivative equal to zero; solve for the stationary point(s).
- Reject solutions outside the valid domain (negative lengths, dimensions wider than the available material, etc.).
- Use the second derivative test: f′′ > 0 = minimum, f′′ < 0 = maximum. Confirm you’ve got the right one for the question.
- Compute the max/min value and interpret in context — include units, and state both the optimum variable value AND the optimum quantity.
The pivotal step is #1. Once the quantity is a function of one variable, the rest is mechanical. Most marks (and most mistakes) are in setting up the constraint to eliminate variables.
Common optimisation contexts
| Context | Quantity | Typical constraint |
|---|
| Box from sheet | volume V | folded from given-size sheet |
| Cylinder / cube / box | surface area S | fixed volume V |
| Rectangle inscribed in curve | area A | vertices on a known function |
| Fencing a plot | area A | fixed perimeter / wall available |
| Cost minimisation | cost C | fixed area or fixed total cost |
| Projectile / motion | height, speed, time | given equation of motion |
Worked examples
WE 1Open box from a square sheet — maximise volume
A square sheet of cardboard, side 12 cm, has equal squares of side x cm cut from each corner. The flaps are folded up to form an open-top box. Find the value of x that maximises the volume of the box, and state this maximum volume.
Step 1 — express V in terms of x
base of box: (12 − 2x) by (12 − 2x); height: x
V(x) = x(12 − 2x)² for 0 < x < 6
= x(144 − 48x + 4x²)
= 4x³ − 48x² + 144x
Step 2 — differentiate and solve V'(x) = 0
V'(x) = 12x² − 96x + 144 = 12(x − 2)(x − 6)
→ x = 2 or x = 6
Step 3 — reject x = 6 (gives zero base, no box)
valid stationary point: x = 2
Step 4 — second derivative test
V”(x) = 24x − 96 → V”(2) = 48 − 96 = −48 < 0
→ x = 2 gives a MAXIMUM ✓
Step 5 — compute V(2)
V(2) = 2 · (12 − 4)² = 2 · 64 = 128
Maximum volume = 128 cm³, achieved when x = 2 cm
domain check: 0 < x < 6 because the cut squares must not eat up the whole sheet
WE 2Closed cylinder — minimise surface area for given volume
A closed cylinder has volume 250π cm³. Find the radius r that minimises its total surface area, and state this minimum surface area.
Step 1 — express S in terms of r alone using V = 250π
V = πr²h = 250π → h = 250/r²
S = 2πr² + 2πrh (top + bottom + side)
= 2πr² + 2πr · (250/r²)
= 2πr² + 500π/r
Step 2 — differentiate
dS/dr = 4πr − 500π/r²
set = 0: 4πr = 500π/r² → r³ = 125 → r = 5
Step 3 — second derivative test
d²S/dr² = 4π + 1000π/r³
at r = 5: 4π + 1000π/125 = 4π + 8π = 12π > 0 → minimum ✓
Step 4 — compute S(5) and h
h = 250/25 = 10 cm
S = 2π(25) + 500π/5 = 50π + 100π = 150π
Minimum S = 150π cm², achieved when r = 5 cm (and h = 10 cm)
interesting result: h = 2r at the optimum — the most efficient closed cylinder has height equal to its diameter
WE 3Open-top square-base box — minimise surface area
An open-top box has a square base of side x cm and height h cm. Its volume must be 500 cm³. Find the value of x that minimises the surface area, and state the minimum surface area.
Step 1 — express S in terms of x using V = 500
V = x² h = 500 → h = 500/x²
S = x² + 4xh (one base + four sides, no top)
= x² + 4x · (500/x²)
= x² + 2000/x
Step 2 — differentiate
dS/dx = 2x − 2000/x²
set = 0: 2x = 2000/x² → x³ = 1000 → x = 10
Step 3 — second derivative test
d²S/dx² = 2 + 4000/x³
at x = 10: 2 + 4 = 6 > 0 → minimum ✓
Step 4 — compute S(10) and h
h = 500/100 = 5 cm
S = 100 + 2000/10 = 100 + 200 = 300
Minimum S = 300 cm², achieved when x = 10 cm (and h = 5 cm)
for an open-top box, h = x/2 at the optimum — height is half the base side
WE 4Rectangle inscribed in parabola — maximise area
A rectangle has its base on the x-axis and its upper two corners on the curve y = 9 − x². Find the maximum area of the rectangle, giving an exact answer.
Step 1 — express A in terms of x
by symmetry, rectangle has corners at (−x, 0), (x, 0), (x, 9 − x²), (−x, 9 − x²)
width = 2x, height = 9 − x²
A(x) = 2x(9 − x²) = 18x − 2x³ for 0 < x < 3
Step 2 — differentiate
dA/dx = 18 − 6x²
set = 0: 6x² = 18 → x² = 3 → x = √3 (positive root)
Step 3 — second derivative test
d²A/dx² = −12x
at x = √3: −12√3 < 0 → maximum ✓
Step 4 — compute A(√3)
A(√3) = 2√3 · (9 − 3) = 2√3 · 6 = 12√3
Maximum area = 12√3 square units (≈ 20.8)
when the rectangle is symmetric about the y-axis, you only optimise x ≥ 0 — the rectangle’s full width is 2x = 2√3
WE 5Fencing along an existing wall — maximise area
A rectangular pen is to be fenced along the back of an existing wall. The wall forms one side of the pen, so only three sides need fencing. A total of 60 m of fencing is available. Find the dimensions that maximise the enclosed area, and state this maximum area.
Step 1 — set up the variables
let x = side parallel to the wall (one of these)
let y = side perpendicular to the wall (two of these)
fencing: x + 2y = 60 → x = 60 − 2y
Step 2 — express A in terms of y
A = xy = (60 − 2y) · y = 60y − 2y²
Step 3 — differentiate
dA/dy = 60 − 4y
set = 0: 4y = 60 → y = 15
Step 4 — second derivative test
d²A/dy² = −4 < 0 → maximum ✓
Step 5 — compute dimensions and area
x = 60 − 2(15) = 30 m
A = 30 · 15 = 450
Maximum area = 450 m², achieved when x = 30 m (along the wall) and y = 15 m
the side along the wall is TWICE the perpendicular side at the optimum — a common pattern in “fence three sides” problems
WE 6Cost-constrained area maximisation — different prices per side
A rectangular enclosure is to be built. The two longer sides are made from a fence costing $4 per metre; the two shorter sides are made from a sturdier fence costing $6 per metre. The total budget is $1200. Find the dimensions that maximise the enclosed area, and state this maximum area.
Step 1 — write the budget constraint
let x = longer side ($4/m), y = shorter side ($6/m)
cost: 2(4)x + 2(6)y = 1200
8x + 12y = 1200
2x + 3y = 300 → x = (300 − 3y)/2
Step 2 — express A in terms of y
A = xy = y · (300 − 3y)/2 = (300y − 3y²)/2 = 150y − (3/2)y²
Step 3 — differentiate
dA/dy = 150 − 3y
set = 0: 3y = 150 → y = 50
Step 4 — second derivative test
d²A/dy² = −3 < 0 → maximum ✓
Step 5 — compute dimensions and check budget
x = (300 − 150)/2 = 75 m
A = 75 · 50 = 3750
budget check: 2(4)(75) + 2(6)(50) = 600 + 600 = 1200 ✓
Maximum area = 3750 m², achieved when x = 75 m (longer) and y = 50 m (shorter)
at the optimum, the total spent on each pair of sides is EQUAL ($600 each) — this is a general principle in cost-constrained optimisation
💡 Top tips
- Sketch the situation — a quick diagram with labelled dimensions makes the constraint and the quantity-to-optimise obvious.
- Use natural letters (V, S, A, P, C, r, h) and be consistent — don’t switch between x and y partway through.
- Domain matters — most optimisation problems have a valid range (e.g., 0 < x < 6 in WE 1). Reject solutions outside it.
- Second derivative test is fast — usually quicker than checking sign of df/dx on both sides of the stationary point.
- Always interpret the final answer — state max/min value, the variable that achieves it, AND units. “x = 2” earns fewer marks than “x = 2 cm gives maximum volume of 128 cm³.”
⚠ Common mistakes
- Not using the constraint to eliminate variables — leaving the quantity as a function of TWO variables means you can’t differentiate it usefully.
- Forgetting to reject invalid stationary points (e.g., negative lengths, points outside the domain) — the answer might be the OTHER stationary point you discarded.
- Stopping at f′(x) = 0 without justifying max vs min — examiners want to see the second derivative test or a sign check.
- Differentiating with respect to the wrong letter — if V is a function of r, you compute dV/dr, NOT dV/dx.
- Forgetting units — area in cm² or m², volume in cm³ or m³, cost in $ etc. Drop units and lose marks.
- Confusing “show that” with “find” — “show that” provides the formula and you must derive it. Even if you can’t, USE the given formula in later parts.
Up next: Kinematics — the special case of optimisation where the variable is TIME. Displacement, velocity, and acceleration are linked by differentiation: v = ds/dt, a = dv/dt. You’ll use these to find max speed, time-to-rest, total distance travelled, and where a particle changes direction.
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