Mathematics Advanced • Year 11 • Module 3 • Lesson 9
Optimisation
Practise HSC-style writing on optimisation — including a structured packaging-and-cost extended response with constraint, derivation and verification.
1. Short-answer questions
1.1 A rectangular garden is to be fenced on all four sides with 60 m of fencing. Find the dimensions that give the maximum area, state the maximum area, and justify it is a maximum using the second derivative. 4 marks Band 3-4
1.2 The cost of producing x items is C(x) = x² + 10x + 50, and the selling price per item is $30. Find the number of items that maximises profit. 3 marks Band 3-4
1.3 Find two positive numbers whose sum is 20 and whose product is as large as possible. 3 marks Band 3
Stuck on 1.3? Let one number be x and the other (20 − x); maximise P(x) = x(20 − x).2. Extended response
2.1 A closed cylindrical can is to be designed to hold a fixed volume of 500π cm³.
(a) Let the can have radius r cm and height h cm. Using the volume constraint, show that the surface area is S(r) = 2πr² + 1000π/r.
(b) Find dS/dr and the value of r that gives a stationary point of S.
(c) Confirm that this stationary point is a minimum by computing d²S/dr² and noting its sign.
(d) Find the corresponding height h, and comment on the ratio of height to diameter at the optimal design. 8 marks Band 5-6
Explicit marking criteria
Part (a) — 2 marks
• 1 mark — uses V = πr²h = 500π to solve h = 500/r².
• 1 mark — substitutes into S = 2πr² + 2πrh and simplifies to S(r) = 2πr² + 1000π/r.
Part (b) — 2 marks
• 1 mark — correctly differentiates: dS/dr = 4πr − 1000π/r².
• 1 mark — sets dS/dr = 0 and solves r³ = 250 to give r = ∛250 = 5∛2 cm.
Part (c) — 2 marks
• 1 mark — computes d²S/dr² = 4π + 2000π/r³.
• 1 mark — observes d²S/dr² > 0 for all r > 0 (in particular at r* = 5∛2), so the stationary point is a minimum.
Part (d) — 2 marks
• 1 mark — h = 500/r² = 500/(5∛2)² = 500/(25 · ∛4) = 20/∛4 = 20 · ∛2/2 = 10∛2 cm; equivalently h = 2r.
• 1 mark — states the optimal design has height equal to diameter (h = 2r), the classical "Coke can" result.
Your response:
Stuck on (d)? After substituting r* into the constraint, you can simplify by noting r³ = 250 ⇒ r² = 250/r.How did this worksheet feel?
What I'll revisit before next class:
1.1 — Garden, perimeter 60 m (4 marks)
Sample response. Let width w m and length l m, with w, l > 0. Perimeter: 2w + 2l = 60 ⇒ l = 30 − w.
Area A(w) = w(30 − w) = 30w − w².
A′(w) = 30 − 2w = 0 ⇒ w = 15 m, so l = 15 m.
A″(w) = −2 < 0, confirming a maximum.
Maximum area = 15 × 15 = 225 m².
Marking notes. 1 — sets up variables and constraint. 1 — writes A in one variable and differentiates. 1 — solves for w and finds l. 1 — verifies using A″ and states the area with units. Missing A″ verification loses 1 mark.
1.2 — Profit, C(x) = x² + 10x + 50 (3 marks)
Sample response. Revenue R(x) = 30x. Profit P(x) = R − C = 30x − (x² + 10x + 50) = −x² + 20x − 50.
P′(x) = −2x + 20 = 0 ⇒ x = 10 items.
P″(x) = −2 < 0, confirming a maximum. Maximum profit = P(10) = −100 + 200 − 50 = $50.
Marking notes. 1 — sets up profit correctly (revenue minus cost). 1 — finds stationary point. 1 — confirms maximum (and ideally states the maximum profit). A response that gives only x = 10 without computing P(x) and without using P″ scores 2/3.
1.3 — Two positive numbers summing to 20, max product (3 marks)
Sample response. Let one number be x; the other is 20 − x. We need x > 0 and 20 − x > 0, so 0 < x < 20.
Product P(x) = x(20 − x) = 20x − x². P′(x) = 20 − 2x = 0 ⇒ x = 10; the other number is also 10.
P″(x) = −2 < 0 ⇒ maximum. The two numbers are 10 and 10; maximum product = 100.
Marking notes. 1 — variables defined with domain. 1 — solves P′ = 0. 1 — verifies maximum (P″ or sign argument) and states both numbers. Common error: students give only one number, forgetting the second.
2.1 — Closed cylindrical can, V = 500π (8 marks): sample Band-6 response with annotations
Sample Band-6 response.
Part (a). Volume constraint: V = πr²h = 500π. Dividing through by π: r²h = 500, so
h = 500/r² (r > 0). [1 mark — h in terms of r.]
Surface area: S = 2πr² + 2πrh = 2πr² + 2πr · (500/r²) = 2πr² + 1000π/r. ✓ [1 mark — substitution and simplification.]
Part (b). dS/dr = 4πr − 1000π · r−2 = 4πr − 1000π/r². [1 mark.]
Set dS/dr = 0: 4πr = 1000π/r² ⇒ 4πr³ = 1000π ⇒ r³ = 250 ⇒ r = ∛250 = 5∛2 cm ≈ 6.30 cm. [1 mark — solved for r.]
Part (c). d²S/dr² = 4π + 2000π · r−3 = 4π + 2000π/r³. [1 mark.]
For all r > 0 we have d²S/dr² > 0 (sum of two positive terms). In particular at r* = ∛250, d²S/dr² = 4π + 2000π/250 = 4π + 8π = 12π > 0. Hence the stationary point is a minimum. [1 mark — minimum confirmed.]
Part (d). h = 500/r² = 500/(5∛2)² = 500/(25 · ∛4) = 20/∛4 = 20 · ∛(2)/2 = 10∛2 cm ≈ 12.60 cm. [1 mark — h computed.]
Diameter d = 2r = 10∛2 cm = h. So at the optimal design h = 2r (height equals diameter). This is the classical "minimum-aluminium" result and explains why drink cans look approximately as tall as they are wide; small deviations in real cans come from labelling, drinking ergonomics, and material thickness considerations not captured by this model. [1 mark — ratio commented.]
Total: 8/8.
Band descriptors for marker.
Band 3: Writes the volume constraint but does not solve for h; surface area expression incomplete; cannot proceed to differentiation. ≈ 2-3 marks.
Band 4: Reaches S(r) = 2πr² + 1000π/r and differentiates correctly, but does not solve cleanly for r or verify minimum. ≈ 4-5 marks.
Band 5: Full algebra; r* found exactly; minimum verified; h computed but ratio comment missing. ≈ 6-7 marks.
Band 6: All parts complete; ratio h : d = 1 : 1 explicitly stated and explained; calculations show both exact and approximate values; minimum verified by sign of S″ across the whole domain, not just at one point. 8/8.