L47 — Applying Functions to Real Problems — Solutions

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1. Q1 — Domain and range in context

   Fluency
   • (a) C(h) = 3h + 4.
   • (b) Domain: 0 ≤ h ≤ 8 (max 8 hours). Range: $4 ≤ C ≤ $28.
   • (c) C(5) = 15 + 4 = $19.
   • (d) 3h + 4 = 19 ⇒ 3h = 15 ⇒ h = 5 hours.

2. Q2 — Cost, revenue and profit

   Fluency
   • (a) R(n) = 15n. C(n) = 5n + 120.
   • (b) P(n) = 15n − (5n + 120) = 10n − 120.
   • (c) 10n − 120 = 0 ⇒ n = 12. Break-even at 12 candles.
   • (d) Domain: n ≥ 0, n ∈ ℤ (whole candles only, at least 0).

3. Q3 — Function notation in context

   Fluency
   • (a) d(2.5) = 80 × 2.5 = 200 km — the car travels 200 km in 2.5 hours.
   • (b) d⁻¹(d) = d/80. This gives the time in hours to travel d kilometres.
   • (c) d⁻¹(340) = 340/80 = 4.25 hours (4 h 15 min).
   • (d) Domain of d⁻¹: d ≥ 0 km. Range of d⁻¹: t ≥ 0 hours.

4. Q4 — Composite functions in context

   Fluency
   • (a) t(r(a)) = 1.05 × 0.64a = 0.672a USD.
   • (b) t(r(120)) = 0.672 × 120 = $80.64 USD.
   • (c) Let p = 0.672a. Inverse: a = p/0.672. [t(r)]⁻¹(p) = p/0.672 (AUD from USD price).
   • (d) a = 67.20/0.672 = $100 AUD.

5. Q5 — Identifying model type and domain

   Understanding
   • (a) Quadratic model. 100 = initial height (m). 5 = half the acceleration due to gravity (g/2 ≈ 5 m/s²).
   • (b) Ball hits ground when h = 0: 100−5t²=0 ⇒ t=√20≈4.47 s. Domain: 0 ≤ t ≤ √20 ≈ 4.47 s.
   • (c) Range: 0 ≤ h ≤ 100 m (starts at 100, falls to 0).
   • (d) 100 − 5t² = 55 ⇒ 5t² = 45 ⇒ t² = 9 ⇒ t = 3 seconds.

6. Q6 — Break-even with two products

   Understanding
   • (a) R(n) = 12n + 5(2n) = 12n + 10n = 22n. C(n) = 4n + 1(2n) + 200 = 6n + 200.
   • (b) P(n) = 22n − 6n − 200 = 16n − 200.
   • (c) 16n − 200 = 0 ⇒ n = 12.5 ⇒ minimum 13 burgers (whole number, to exceed break-even).
   • (d) P(40) = 16(40) − 200 = 640 − 200 = $440 profit.

7. Q7 — Inverse in context

   Understanding
   • (a) T(30) = 4(30)+20 = 140°C. T(60) = 4(60)+20 = 260°C.
   • (b) T⁻¹(T) = (T−20)/4. This gives the number of minutes for the oven to reach temperature T.
   • (c) T⁻¹(220) = (220−20)/4 = 200/4 = 50 minutes.
   • (d) The model assumes constant heating rate. In reality, ovens slow their heating as they approach the set temperature (thermostatic control).

8. Q8 — Population with restricted domain

   Understanding
   • (a) P(0) = 50. P(5) = 50 × 1.12&sup5; ≈ 50 × 1.762 = 88. P(10) ≈ 50 × 3.106 = 155.
   • (b) P(t) = 500 when 1.12^t = 10. Trial: 1.12²⁰ ≈ 9.65 (still <10), 1.12²¹ ≈ 10.80 >10. After 21 years.
   • (c) Once at capacity, P stays at 500. Restricted domain: 0 ≤ t ≤ 21 years for the growth model.
   • (d) P(t) → ∞ as t → ∞ in this model — meaning the model breaks down long-term. A logistic model would be more realistic.

9. Q9 — Multi-step function problem

   Problem Solving
   • (a) C(m) = 0.30m + 25 for 0 ≤ m ≤ 100; C(m) = 30 + 0.15(m−100) + 25 = 0.15m + 40 for m > 100.
   • (b) C(80) = 0.30(80)+25 = $49. C(100) = 30+25 = $55. C(150) = 0.15(150)+40 = 22.50+40 = $62.50.
   • (c) $52 is below $55 (the 100-min cost), so usage ≤ 100 min: 0.30m+25=52 ⇒ 0.30m=27 ⇒ m = 90 minutes.
   • (d) Two-segment line: from (0,$25) with slope $0.30 to (100,$55), then slope $0.15 onwards. Kink (change of slope) at m=100 minutes, C=$55.

10. Q10 — Extended: design a pricing model

    Problem Solving
    • (a) C(n) = 30n + 8000.
    • (b) P(n, f) = nf − 30n − 8000 = n(f−30) − 8000.
    • (c) 200(f−30) − 8000 ≥ 2000 ⇒ 200(f−30) ≥ 10000 ⇒ f−30 ≥ 50 ⇒ f ≥ $80/month.
    • (d) n(80−30) − 8000 ≥ 2000 ⇒ 50n ≥ 10000 ⇒ n ≥ 200 members.