T4T3 Review Solutions — Problem Solving & Modelling

1. Formulating a model. Fluency

   A plumber charges a $75 call-out fee plus $55 per hour.

   • (a) C(h) = 55h + 75.
   • (b) C(2) = 110+75 = $185.
   • (c) 55h+75=350 ⇒ 55h=275 ⇒ h=5 hours.
   • (d) e.g. The hourly rate is constant regardless of the type of work or time of day.

2. Profit function. Fluency

   A bakery sells loaves for $7 each. Fixed costs are $140/day. Variable cost is $3/loaf.

   • (a) R(n) = 7n, C(n) = 3n + 140.
   • (b) P(n) = 4n − 140.
   • (c) 4n − 140 = 0 ⇒ n = 35 loaves.
   • (d) P(60) = 240−140 = $100.

3. Quadratic model. Fluency

   A stone is thrown upward from a 5 m ledge. Its height h(t) = −5t² + 15t + 5.

   • (a) h(1)=−5+15+5=15 m. h(2)=−20+30+5=15 m.
   • (b) Vertex: t=−15/(2×−5)=1.5 s. h(1.5)=−5(2.25)+22.5+5=16.25 m.
   • (c) −5t²+15t+5=0 ⇒ t²−3t−1=0 ⇒ t=(3+√13)/2 ≈ 3.30 s.
   • (d) Domain: 0 ≤ t ≤ 3.30 s. Range: 0 ≤ h ≤ 16.25 m.

4. Optimisation — maximum area. Fluency

   A rectangle with perimeter 100 m. Let the width be x m.

   • (a) Length = 50−x. A(x) = x(50−x) = −x²+50x.
   • (b) x = 25 m.
   • (c) A(25) = 25 × 25 = 625 m².
   • (d) A square (25 m × 25 m).

5. Composite and inverse. Understanding

   A taxi converts km to minutes using m(k) = 2k + 5 (average journey). The cost is c(m) = 0.5m + 4 dollars.

   • (a) c(m(k)) = 0.5(2k+5)+4 = k+2.5+4 = k + 6.5 dollars.
   • (b) c(m(12)) = 12+6.5 = $18.50.
   • (c) k+6.5=24.5 ⇒ k = 18 km.
   • (d) Inverse: k = C−6.5. [c∘m]⁻¹(C) = C−6.5 — gives the journey distance (km) for a total cost of $C.

6. Model validation. Understanding

   A population model predicts P(t) = 5000 + 200t for a town (t = years from 2020).

   • (a) P(5)=5000+1000=6 000. P(10)=5000+2000=7 000.
   • (b) Error = |6000−6400|/6400 × 100 = 400/6400 × 100 ≈ 6.25%.
   • (c) e.g. The growth rate may be increasing (not constant) — perhaps due to new infrastructure attracting more residents.
   • (d) Exponential model. 1.04 represents a 4% annual growth rate.

7. Piecewise pricing. Understanding

   A theme park charges: $30 for adults, $15 for children. Group discount: groups of 10+ get 20% off.

   • (a) 2(30)+3(15)=60+45=$105 (no discount, group of 5).
   • (b) 12 people, so 20% discount. Full price = 4(30)+8(15)=120+120=$240. With 20% off: 0.8×240=$192.
   • (c) C(a,c) = 0.8(30a + 15c).
   • (d) 10 people, discount applies. C = 0.8(180+60) = 0.8(240) = $192.

8. Ticket pricing for maximum revenue. Understanding

   A theatre sells 500 tickets at $20 each. For each $2 price rise, 25 fewer tickets are sold.

   • (a) R(n) = (20+2n)(500−25n) = 10000+1000n−500n−50n² = −50n² + 500n + 10000.
   • (b) n = −500/(2×−50) = 5 price increases. Price = $20+$10 = $30.
   • (c) R(5) = −50(25)+500(5)+10000 = −1250+2500+10000 = $11 250.
   • (d) 500−25(5) = 375 tickets.

9. Multi-step modelling. Problem Solving

   A company produces x units with cost C(x) = x² − 10x + 50 dollars and sells each for $(30 − x).

   • (a) R(x) = x(30−x) = −x² + 30x.
   • (b) P(x) = −x²+30x − x²+10x−50 = −2x² + 40x − 50.
   • (c) Vertex: x=−40/(2×−2)=10. P(10)=−200+400−50=$150 at x=10 units.
   • (d) −2x²+40x−50=0 ⇒ x²−20x+25=0 ⇒ x=(20±√300)/2=10±5√3. x≈1.34 or 18.66 units.

10. Extended: choosing the best model. Problem Solving

    A scientist records temperature T (°C) of a cooling liquid every 5 minutes:

    • (a) Drops: 16, 13, 10, 8. The drops are decreasing — not constant. Linear model is inappropriate.
    • (b) 64/80=0.80, 51/64≈0.797, 41/51≈0.804, 33/41≈0.805. Ratios are approximately constant at ≈0.80 — suggesting an exponential model.
    • (c) T(n) = 80 × 0.80ⁿ.
    • (d) T(6) = 80 × 0.80⁶ ≈ 80 × 0.262 ≈ 21°C. As n→∞, T→0°C. In reality the liquid cools to room temperature (e.g. 20°C), not absolute zero — so the model has a long-term limitation.