Mathematics Advanced • Year 11 • Module 3 • Lesson 9
Optimisation
Apply the optimisation routine to multi-step problems in geometry, business, packaging and design.
Problem 1 — Rectangle inscribed in a semicircle (geometric)
A rectangle is inscribed in a semicircle of radius 5 m, with its base lying along the diameter. Let the half-width of the rectangle be x, so the full width is 2x and the height is y = √(25 − x²) (from the semicircle equation x² + y² = 25). The rectangle's area is
A(x) = 2x · √(25 − x²), 0 < x < 5
Working with A² is easier than working with A.
Set up: What are we solving for?
(i) Show that A² = 4x²(25 − x²) = 100x² − 4x⁴, and let F(x) = A². 2 marks
(ii) Find F′(x) and solve F′(x) = 0 in the domain 0 < x < 5. 3 marks
(iii) State the dimensions and area of the largest such rectangle. Briefly explain why maximising F (= A²) automatically maximises A. 2 marks
Stuck on (iii)? A is non-negative, so A is increasing in A². Maximising A² maximises A.Problem 2 — Pricing a streaming subscription (business)
A streaming service charges $p per month per subscriber. At $p, the number of subscribers is approximately
N(p) = 8000 − 200p, 0 ≤ p ≤ 40
(when p = 0 the service is free with 8000 subscribers; when p = 40 nobody subscribes).
Set up: What are we solving for?
(i) Write monthly revenue R(p) as a function of p and expand. 2 marks
(ii) Use calculus to find the price p* that maximises revenue, and the maximum monthly revenue. 3 marks
(iii) The fixed running cost of the service is $50 000/month. Find the monthly profit at p = p*. Is the service viable at this price? 2 marks
Problem 3 — Cylindrical drinks can (packaging)
An aluminium can is designed as a closed cylinder of radius r cm and height h cm with fixed volume 330 cm³ (a standard soft-drink can size).
Set up: What are we solving for?
(i) Express h in terms of r using the volume constraint πr²h = 330, then write surface area S(r) = 2πr² + 2πrh as a function of r only. 3 marks
(ii) Find S′(r) and the value of r that minimises S. Show that S″(r) > 0 at this r. 3 marks
(iii) Find the corresponding height h. State the optimal r : h ratio. (You should find h = 2r — a satisfying classical result.) 2 marks
Stuck on (i)? h = 330/(πr²), so 2πrh = 2πr · 330/(πr²) = 660/r. So S(r) = 2πr² + 660/r.Problem 4 — Closest point on a line (distance)
Find the point on the line y = 2x + 1 that is closest to the origin. Let the point be (x, 2x + 1); the squared distance to the origin is
D²(x) = x² + (2x + 1)²
Set up: What are we solving for?
(i) Expand D²(x) into a single polynomial in x. 2 marks
(ii) Differentiate D²(x) and solve (D²)′ = 0. State the x-coordinate of the closest point and find its y-coordinate. 3 marks
(iii) Compute D itself (the shortest distance). Explain in one sentence why minimising D² is equivalent to minimising D. 2 marks
Problem 5 — Sheep paddock with internal divider (modelling)
A farmer has 120 m of fencing for a rectangular sheep paddock divided into two equal halves by an internal fence parallel to one pair of sides. So fencing covers the four outer sides plus one internal divider (three "lengths" parallel to one direction, two "widths" parallel to the other).
Set up: What are we solving for?
(i) Let the side parallel to the divider be y and the perpendicular side be x. Write the fencing constraint as an equation in x and y. 2 marks
(ii) Write the total enclosed area A as a function of x only, and find A′(x). 3 marks
(iii) Find x* and y*, verify A″(x*) < 0, and state the maximum area. 2 marks
Stuck on (i)? Total fence = 2y + 3x (the divider runs the perpendicular direction). Or, in another labelling: 2x + 3y = 120. Be consistent.How did this worksheet feel?
What I'll revisit before next class:
Problem 1 — Inscribed rectangle in semicircle of radius 5
Set up. We are maximising area by working with the algebraically friendlier A² to avoid the square root.
(i) A² = (2x)² · (25 − x²) = 4x²(25 − x²) = 100x² − 4x⁴.
(ii) F′(x) = 200x − 16x³ = 8x(25 − 2x²). Setting = 0 in 0 < x < 5: 25 − 2x² = 0 ⇒ x² = 25/2 ⇒ x = 5/√2 = 5√2/2 ≈ 3.54 m.
(iii) y = √(25 − 25/2) = √(25/2) = 5/√2 also; full width 2x = 5√2 m. A = 2x · y = (5√2)(5/√2) = 25 m². Because A ≥ 0 on the domain, A is an increasing function of A², so any maximiser of A² also maximises A.
Problem 2 — Streaming pricing
Set up. We are pricing a service so revenue (not subscriber count) is maximised, then checking viability.
(i) R(p) = p · N(p) = p(8000 − 200p) = 8000p − 200p².
(ii) R′(p) = 8000 − 400p = 0 ⇒ p* = $20. R″(p) = −400 < 0 (max). R(20) = 160 000 − 80 000 = $80 000/month.
(iii) Profit = R − cost = 80 000 − 50 000 = $30 000/month. Profit > 0, so the service is viable at p = $20.
Problem 3 — Cylindrical can, V = 330 cm³
Set up. We are finding the radius that minimises aluminium use for a fixed-volume can.
(i) h = 330/(πr²); S(r) = 2πr² + 2πr · 330/(πr²) = 2πr² + 660/r.
(ii) S′(r) = 4πr − 660/r² = 0 ⇒ 4πr³ = 660 ⇒ r³ = 165/π ⇒ r = (165/π)^(1/3) ≈ 3.745 cm. S″(r) = 4π + 1320/r³ > 0 always (so min).
(iii) h = 330/(πr²) = 330/(π · (165/π)^(2/3)). Using V = πr²h ⇒ h = V/(πr²). Compare 2r vs h: 2r/h = 2r · πr²/V = 2πr³/V = 2 · 165/π · π / 330 = 330/330 = 1. So h = 2r exactly. Numerically h ≈ 7.49 cm. The optimal ratio is diameter = height.
Problem 4 — Closest point on y = 2x + 1 to origin
Set up. We minimise squared distance because the algebra (no √) is cleaner; D and D² have minima at the same point because D ≥ 0.
(i) D²(x) = x² + (2x + 1)² = x² + 4x² + 4x + 1 = 5x² + 4x + 1.
(ii) (D²)′ = 10x + 4 = 0 ⇒ x = −2/5. y = 2(−2/5) + 1 = −4/5 + 1 = 1/5. Closest point: (−2/5, 1/5).
(iii) D²(−2/5) = 5(4/25) + 4(−2/5) + 1 = 4/5 − 8/5 + 5/5 = 1/5. D = √(1/5) = 1/√5 = √5/5 ≈ 0.447. Because D ≥ 0 on the entire line, the function D is an increasing function of D², so minimising D² minimises D.
Problem 5 — Divided paddock
Set up. We are maximising enclosed area when an internal fence eats into the budget.
(i) Fencing = 2y (the two sides parallel to the divider, each length y) + 3x (the two outer ends plus the divider, each length x) = 3x + 2y = 120. (An equivalent labelling 2x + 3y = 120 is also acceptable, but be consistent.)
(ii) From the constraint: y = (120 − 3x)/2 = 60 − 1.5x. A(x) = x · y = x(60 − 1.5x) = 60x − 1.5x². A′(x) = 60 − 3x.
(iii) A′(x) = 0 ⇒ x* = 20 m, y* = 60 − 30 = 30 m. A″(x) = −3 < 0 (max confirmed). Maximum area A(20) = 20 · 30 = 600 m².