L48 — Optimisation and Decision Making
Key Terms
- Optimisation
- Finding the largest or smallest value of a quantity (area, profit, cost) subject to given constraints.
- Maximum / minimum
- The highest or lowest value a function achieves; for a quadratic, it occurs at the vertex.
- Vertex
- The turning point of a parabola, at x = −b/(2a); gives the maximum when a < 0, minimum when a > 0.
- Constraint
- A condition restricting the allowable values (e.g. total fencing ≤ 100 m, quantity ≥ 0).
- Feasible domain
- The set of input values that satisfy all constraints — the vertex must lie here to be the valid optimum.
- Endpoint check
- When the vertex lies outside the feasible domain, evaluate the function at both endpoints to find the constrained optimum.
Finding the optimal value
| Problem type | Strategy | Key formula |
|---|---|---|
| Maximise area with fixed perimeter | Write area as quadratic in one variable, find vertex | x = −b/(2a) |
| Minimise cost with fixed output | Write cost in terms of one variable, find vertex or check endpoints | Complete the square or vertex formula |
| Maximise profit (revenue − cost) | P(n) = R(n) − C(n); find vertex of P | nopt = −b/(2a) |
| Optimal dimensions | Set up constraint, substitute into area/volume, optimise | Differentiation (senior) or vertex (Year 10) |
Worked Example — Maximum Area
Situation: A rectangular paddock is to be fenced using 80 m of fencing on three sides (one side is a river, so no fencing needed there). What dimensions give the maximum area?
Step 1 — Define variables and constraint.
Let width = x, length = y. Fencing on three sides: 2x + y = 80 ⇒ y = 80 − 2x.
Step 2 — Write area as a function of x.
A(x) = xy = x(80 − 2x) = 80x − 2x².
Step 3 — Find the vertex.
x = −80/(2 × −2) = 20 m. y = 80 − 2(20) = 40 m.
Step 4 — State maximum area and check domain.
A(20) = 20 × 40 = 800 m². Domain: 0 < x < 40 (both x and y must be positive). Vertex is inside domain. ✓
What Is Optimisation?
Optimisation problems ask: "what is the best possible outcome, given the constraints?" Businesses want maximum profit. Engineers want minimum material used. Athletes want maximum performance. In Year 10, we solve these using quadratic functions and the vertex.
Setting Up an Optimisation Problem
The key steps are always:
- Identify what to optimise (area, cost, profit, volume…).
- Write a constraint equation that links the variables.
- Substitute to reduce to one variable.
- Find the vertex of the resulting quadratic.
- Check the domain — is the vertex inside the feasible region?
Worked Example 2 — Maximum Revenue
A cinema seats 400 people. When the ticket price is $10, all 400 seats are sold. For each $1 increase in price, 20 fewer people attend.
Solution
Let the price increase be $p. Ticket price = (10 + p). Attendance = 400 − 20p.
Revenue: R(p) = (10 + p)(400 − 20p) = 4000 − 200p + 400p − 20p² = −20p² + 200p + 4000.
Vertex: p = −200/(2 × −20) = 5. Price = $15. R(5) = −20(25) + 200(5) + 4000 = −500 + 1000 + 4000 = $4 500.
Domain: 0 ≤ p ≤ 20 (at p = 20, attendance = 0). Vertex at p = 5 is inside domain. ✓
Worked Example 3 — Minimum Cost
A box with a square base and no lid must have a volume of 32 m³. Base side = x, height = h. The base costs $3/m² and each side costs $2/m². Find x that minimises cost.
Solution
Volume constraint: x²h = 32 ⇒ h = 32/x².
Base area = x². Side area = 4 × xh = 4x(32/x²) = 128/x.
Cost: C(x) = 3x² + 2(128/x) = 3x² + 256/x.
This is not a quadratic — minimising it uses calculus at senior level, but numerically: C(4) = 48 + 64 = $112. Test nearby values to confirm x ≈ 4 m gives the minimum.
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Maximum area with fixed perimeter. Fluency
A rectangle has a perimeter of 60 cm. Let the width be x cm.
- (a) Show that the length is (30 − x) cm.
- (b) Write the area A(x) as a quadratic in x.
- (c) Find the value of x that maximises A.
- (d) What are the dimensions of the maximum-area rectangle? What is the maximum area?
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Vertex of a profit function. Fluency
A company's profit (dollars) for producing n items is P(n) = −2n² + 80n − 300.
- (a) Find the number of items that maximises profit.
- (b) Find the maximum profit.
- (c) Find the values of n where the company makes a loss (P < 0).
- (d) State a sensible domain for n in context.
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Maximum revenue. Fluency
A market stall sells bags. At $8/bag, they sell 200 bags/day. For each $1 price rise, 10 fewer bags are sold.
- (a) Let p be the price rise. Write expressions for price and quantity sold.
- (b) Write revenue R(p) as a quadratic.
- (c) Find the price that maximises revenue.
- (d) What is the maximum daily revenue?
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Fence problem — one wall given. Fluency
A farmer has 120 m of fencing to make a rectangular paddock alongside a straight river (no fence needed on the river side).
- (a) Let the width (perpendicular to river) be x. Show that the length is (120 − 2x).
- (b) Write A(x) and find the maximum area.
- (c) What are the optimal dimensions?
- (d) State the domain of x in context.
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Ticket pricing optimisation. Understanding
A concert venue holds 1 000 people. At $25 per ticket all seats are sold. For every $5 price increase, 50 fewer tickets are sold.
- (a) Let n be the number of $5 increases. Write expressions for ticket price and number sold.
- (b) Write revenue R(n) as a quadratic.
- (c) Find the ticket price that maximises revenue.
- (d) Check that the optimal n is within the feasible domain.
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Minimum material for a box. Understanding
An open-top rectangular box has a square base with side length x cm and height h cm. The volume must be 250 cm³.
- (a) Express h in terms of x using the volume constraint.
- (b) Write the total surface area S(x) = base + 4 sides in terms of x only.
- (c) Evaluate S at x = 5, x = 6, x = 7. Which is smallest?
- (d) What does minimising surface area achieve in a real-world packaging context?
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Comparing options. Understanding
A landscaper can design two shapes for a garden bed with perimeter 24 m: a square or a rectangle with length twice the width.
- (a) Find the dimensions and area of the square.
- (b) Find the dimensions and area of the rectangle (length = 2 × width).
- (c) Which design has the larger area?
- (d) In general, for a fixed perimeter, which shape gives the larger area — a square or a rectangle? Explain.
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Endpoint check. Understanding
A factory's profit for producing x thousand units is P(x) = −x² + 8x − 7, with the constraint 1 ≤ x ≤ 5.
- (a) Find the unconstrained maximum (vertex) of P.
- (b) Is the vertex within the domain 1 ≤ x ≤ 5?
- (c) Evaluate P at x = 1, x = 4, and x = 5. What is the maximum profit on the feasible domain?
- (d) Why must we always check the endpoints when the domain is restricted?
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Optimisation in context. Problem Solving
A company manufactures and sells x thousand units per week. The selling price is p(x) = 50 − 2x dollars per unit. The cost to produce x thousand units is C(x) = 10x + 30 (in $thousands).
- (a) Write the revenue function R(x) = x × p(x) (in $thousands). Expand fully.
- (b) Write the profit function P(x) = R(x) − C(x).
- (c) Find the production level that maximises profit.
- (d) Find the maximum profit and the selling price at that production level.
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Extended optimisation problem. Problem Solving
A school is designing a running track with a rectangular inner field and two semicircular ends. The total perimeter of the track must be 400 m. Let the straight length of the rectangle be L and the radius of each semicircle be r.
- (a) Show that the perimeter constraint gives L = (400 − 2πr)/2 = 200 − πr.
- (b) Write the total area enclosed (rectangle + two semicircles) as A(r) = 2rL + πr². Substitute for L.
- (c) Simplify A(r) and find the value of r that maximises A.
- (d) Find the maximum area (to the nearest m²). What special shape does the track enclose when A is maximised?