Mathematics Advanced • Year 11 • Module 3 • Lesson 9
Optimisation
Build procedural fluency in the optimisation routine: variables → objective function → constraint → derivative → verify.
1. Quick recall
Answer each in the space provided. 1 mark each
Q1.1 Number the five steps of the optimisation routine in the correct order:
____ Differentiate and set the derivative to zero.
____ Verify the stationary point is the required maximum (or minimum).
____ Define variables (and draw a diagram if relevant).
____ Write the objective function in one variable using the constraint.
____ Write the constraint equation linking the variables.
Q1.2 For each term, write its role.
Objective function: ______________________________________________
Constraint: ______________________________________________
Q1.3 A length and an area are both physical quantities that must be ____________ (positive / negative). Always state the domain of your variable.
2. Worked example — rectangle of perimeter 40 m, maximum area
Problem. A rectangle has perimeter 40 m. Find the dimensions that give the maximum area.
Step 1 — Define variables.
Let width = w m, length = l m, with w > 0, l > 0.
Step 2 — Write the constraint.
2w + 2l = 40 ⇒ l = 20 − w. Domain: 0 < w < 20.
Step 3 — Write the objective in one variable.
A(w) = w · l = w(20 − w) = 20w − w²
Step 4 — Differentiate and set to zero.
A′(w) = 20 − 2w = 0 ⇒ w = 10
Step 5 — Verify maximum and state answer.
A″(w) = −2 < 0 ⇒ maximum. l = 20 − 10 = 10. A(10) = 100 m².
Conclusion: a 10 m × 10 m square gives the maximum area of 100 m².
3. Faded example — find two positive numbers with sum 20 and maximum product
Fill in each blank. 4 marks
Step 1 — Define: let the numbers be x and ______, with x > 0 and the other > 0.
Step 2 — Constraint: x + ______ = 20, so the other number is ______________.
Step 3 — Objective (the thing to maximise): P(x) = x · ( ______________ ) = ____________________
Step 4 — Differentiate and solve: P′(x) = ____________ = 0 ⇒ x = ______
Step 5 — Verify and state: P″(x) = ____________; this is ______ 0, so the stationary point is a ____________. The two numbers are ______ and ______, with maximum product ______.
4. Graduated practice — optimisation problems
Show your variables, constraint, objective, derivative, and verification for every Standard and Extension question.
Foundation — direct optimisation (4 questions)
| Q | Function or scenario | x* (where extremum occurs) | Extreme value |
|---|---|---|---|
| 4.1 1 | Maximise f(x) = x(10 − x), 0 ≤ x ≤ 10 | ||
| 4.2 1 | Minimise g(x) = x² − 8x + 20 | ||
| 4.3 1 | Maximise h(x) = −3x² + 18x | ||
| 4.4 1 | Profit P(x) = −2x² + 40x − 100 (find x*) |
Standard — set up + solve (6 questions)
4.5 A rectangular garden has perimeter 60 m. Find the dimensions giving maximum area, and the maximum area. 3 marks
4.6 A farmer has 200 m of fencing to enclose a rectangular paddock against an existing river (so only three sides need fencing). Find the maximum area. 3 marks
4.7 A box with a square base and open top has volume 32 cm³. Find the dimensions that minimise its surface area. 4 marks
4.8 Cost C(x) = x² + 10x + 50 to produce x items; selling price $30 per item. Find x that maximises profit. 3 marks
4.9 The sum of two positive numbers is 12. Find the numbers that minimise the sum of their squares. 3 marks
4.10 Find the maximum of f(x) = x²(6 − x) on the interval 0 ≤ x ≤ 6. 3 marks
Extension — cylinder & page-margins (2 questions)
4.11 A closed cylindrical can must have volume 500π cm³. Show that the surface area is S(r) = 2πr² + 1000π/r, then find the value of r that minimises S. 4 marks
4.12 A rectangular page must contain 24 cm² of print, with 2 cm margins all round. Find the page dimensions that minimise total page area. 4 marks
5. Self-check the easy 3
Tick the first three once you have checked.
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What I'll revisit before next class:
Q1.1 — Order of the optimisation routine
1: Define variables (and diagram). 2: Write the constraint. 3: Write the objective function in one variable. 4: Differentiate and set to zero. 5: Verify the stationary point is the required extremum.
Q1.2 — Roles
Objective function: the quantity to be maximised or minimised. Constraint: the condition that links the variables and reduces the problem to one unknown.
Q1.3 — Sign of physical quantities
Lengths and areas must be positive.
Q3 — Faded example: two positive numbers summing to 20, max product
Step 1: numbers x and (20 − x). Step 2: x + (20 − x) = 20, so the other is 20 − x. Step 3: P(x) = x · (20 − x) = 20x − x². Step 4: P′(x) = 20 − 2x = 0 ⇒ x = 10. Step 5: P″(x) = −2, which is < 0, so maximum. The two numbers are 10 and 10; maximum product = 100.
Q4.1 — f(x) = x(10 − x)
f(x) = 10x − x²; f′(x) = 10 − 2x = 0 ⇒ x* = 5; f(5) = 25. f″(x) = −2 < 0 (max).
Q4.2 — g(x) = x² − 8x + 20
g′(x) = 2x − 8 = 0 ⇒ x* = 4; g(4) = 16 − 32 + 20 = 4. g″(x) = 2 > 0 (min).
Q4.3 — h(x) = −3x² + 18x
h′(x) = −6x + 18 = 0 ⇒ x* = 3; h(3) = −27 + 54 = 27. h″(x) = −6 < 0 (max).
Q4.4 — P(x) = −2x² + 40x − 100
P′(x) = −4x + 40 = 0 ⇒ x* = 10. P″(x) = −4 < 0 (max). (Maximum profit P(10) = −200 + 400 − 100 = 100.)
Q4.5 — Garden, perimeter 60 m
2w + 2l = 60 ⇒ l = 30 − w; A(w) = w(30 − w) = 30w − w². A′(w) = 30 − 2w = 0 ⇒ w = 15 ⇒ l = 15. A″ = −2 < 0 (max). Dimensions 15 m × 15 m; max area 225 m².
Q4.6 — Three-sided paddock, 200 m fencing
Let the two perpendicular sides be x and the parallel side be y. Fencing: 2x + y = 200 ⇒ y = 200 − 2x. Area A(x) = x(200 − 2x) = 200x − 2x². A′(x) = 200 − 4x = 0 ⇒ x = 50 ⇒ y = 100. A″ = −4 < 0 (max). Maximum area 5000 m² with sides 50 m, 100 m, 50 m.
Q4.7 — Open-top box, volume 32 cm³
Let base side = x, height = h. Volume: x²h = 32 ⇒ h = 32/x². Surface area (no lid): S = x² + 4xh = x² + 4x · 32/x² = x² + 128/x. S′(x) = 2x − 128/x² = 0 ⇒ 2x³ = 128 ⇒ x³ = 64 ⇒ x = 4. h = 32/16 = 2. S″(x) = 2 + 256/x³ > 0 (min). Dimensions base 4 cm × 4 cm, height 2 cm; minimum surface area 48 cm².
Q4.8 — Profit from items
Revenue R = 30x; profit P(x) = 30x − (x² + 10x + 50) = −x² + 20x − 50. P′(x) = −2x + 20 = 0 ⇒ x = 10. P″(x) = −2 < 0 (max). Max profit P(10) = −100 + 200 − 50 = $50.
Q4.9 — Sum 12, minimise sum of squares
Numbers x and (12 − x). S(x) = x² + (12 − x)² = x² + 144 − 24x + x² = 2x² − 24x + 144. S′(x) = 4x − 24 = 0 ⇒ x = 6, other = 6. S″(x) = 4 > 0 (min). Both numbers are 6; minimum sum of squares = 72.
Q4.10 — Maximise f(x) = x²(6 − x) on [0, 6]
f(x) = 6x² − x³; f′(x) = 12x − 3x² = 3x(4 − x) = 0 ⇒ x = 0 or x = 4. Both endpoints x = 0 and x = 6 give f = 0. At x = 4: f(4) = 16 · 2 = 32. f″(x) = 12 − 6x; f″(4) = −12 < 0 (max). Maximum value 32 at x = 4.
Q4.11 — Closed cylinder, volume 500π
πr²h = 500π ⇒ h = 500/r². S = 2πr² + 2πrh = 2πr² + 2πr · 500/r² = 2πr² + 1000π/r. S′(r) = 4πr − 1000π/r² = 0 ⇒ 4πr³ = 1000π ⇒ r³ = 250 ⇒ r = ∛250 = 5 ∛2 cm ≈ 6.30 cm. S″(r) = 4π + 2000π/r³ > 0 (min).
Q4.12 — Page with 24 cm² of print and 2 cm margins
Let printed area be x cm wide × y cm tall, so xy = 24 ⇒ y = 24/x. Page is (x + 4) wide × (y + 4) tall. Page area A(x) = (x + 4)(24/x + 4) = (x + 4)(24 + 4x)/x = (24x + 4x² + 96 + 16x)/x = 4x + 96/x + 40. A′(x) = 4 − 96/x² = 0 ⇒ x² = 24 ⇒ x = 2√6 cm ≈ 4.90 cm. y = 24/x = 24/(2√6) = 12/√6 = 2√6 cm. Page: (x + 4) = (2√6 + 4) cm ≈ 8.90 cm wide; (y + 4) = (2√6 + 4) cm ≈ 8.90 cm tall. A″(x) = 192/x³ > 0 (min). Minimum page is a square of side (2√6 + 4) cm.