Lesson 8: Optimization Problems | Level 3 Calculus

Lesson at a Glance | He Tirohanga Whakamua

Do NowOptimising a farm paddock boundary with 100m of fencing against a river10 min

Optimisation ProtocolSetting Objective Function

f (x)

& Constraint Equation15 min

Solving & VerificationSolving

f^{'} (x) = 0

and proving maximum with

f^{″} (x) < 0

15 min

Portfolio EntryWrite Section 8: Optimisation Modelling & Engineering Max/Min10 min

Exit DrillFind dimensions of maximum volume open box cut from a 12cm x 12cm sheet5 min

Ngā Whāinga Ako | Learning Intentions

Students will know

The 4-Step Optimisation Protocol:
1. Define variables & sketch diagram.
2. Formulate Objective & Constraint equations.
3. Express Objective in 1 variable $f (x)$ .
4. Solve $f^{'} (x) = 0$ and verify with $f^{″} (x)$ .
Common optimisation applications: Maximising enclosed area, volume of open/closed boxes, cylindrical cans, and minimising construction costs.

Students will demonstrate

By solving 3 complete engineering optimisation problems from initial diagram to second-derivative verification.
By completing Section 8 of their Level 3 Calculus Differentiation Portfolio.

Do Now | Tīmatanga Whakaaro (10 min)

Fencing Paddock Prompt:

"A farmer has $100 m$ of fencing wire to build a rectangular paddock along a straight riverbank (no fence needed along the river). What dimensions $(x, y)$ enclose the maximum possible area?"

Unpack: Constraint: $x + 2 y = 100 \to x = 100 - 2 y$ . Objective: $A = x y = (100 - 2 y) y = 100 y - 2 y^{2}$ . Derivative: $\frac{d A}{d y} = 100 - 4 y = 0 \to y = 25 m, x = 50 m$ ! Maximum Area $= 1250 m^{2}$ .

Optimisation Protocol & Box Volume Example (15 min)

1. Open Box Cutout Problem

Cut corners of size $x$ from $12 cm \times 12 cm$ card.
$V (x) = x (12 - 2 x)^{2} = 4 x^{3} - 48 x^{2} + 144 x$
$V^{'} (x) = 12 x^{2} - 96 x + 144 = 0 \to x = 2 cm$ .

2. Second Derivative Verification

$V^{″} (x) = 24 x - 96$ . At $x = 2$ : $V^{″} (2) = - 48 < 0$ , confirming a true local maximum volume ( $V_{m a x} = 128 {cm}^{3}$ )!

📁 Calculus Differentiation Portfolio — Section 8: Optimisation Problems

Students open their Level 3 Calculus Portfolio and complete Section 8:

Section 8 Requirements:

1. Optimisation Protocol Flowchart: Draw the 4-step workflow from geometric diagram to $f^{″} (x)$ verification.

2. Cylindrical Can Optimisation Solver: A closed tin can must hold $330 {cm}^{3}$ volume ( $V = π r^{2} h = 330$ ). Find radius $r$ and height $h$ that minimise surface area $A = 2 π r^{2} + 2 π r h$ .

3. Extension: Optimal Cylinder Aspect Ratio: Prove mathematically that for ANY closed cylindrical can of fixed volume, minimum surface area occurs when height equals diameter ( $h = 2 r$ ).

Exit Verification | Ka Mutu Hoki (5 min)

Exit Check:

"My Section 8 proves minimum tin can surface area occurs when h = 2r (r = 3.74 cm for 330 mL) using f''(r) > 0 verification."

Teacher Planning & NCEA Alignment

NCEA Level 3 Calculus Alignment (6 Credits External):

Optimisation Applications: Formulate mathematical models for physical situations and apply differentiation to solve optimisation problems.

Vocabulary: Optimisation, objective function, constraint equation, stationary point ( $f^{'} = 0$ ), second derivative test ( $f^{″} < 0$ ), local maximum, local minimum.

Lesson 8: Optimisation Problems (Max/Min Applications)