L46 — Mathematical Modelling in Context — Solutions

← Back to Problem Solving and Modelling

1. Q1 — Formulating a linear model

   Fluency
   • (a) Let k = distance in km, C = total cost ($). C(k) = 2.20k + 3.50.
   • (b) C(15) = 2.20(15) + 3.50 = 33 + 3.50 = $36.50.
   • (c) 2.20k + 3.50 = 30.10 ⇒ 2.20k = 26.60 ⇒ k = 12.09 km ≈ 12.1 km.
   • (d) e.g. "The cost per km is constant regardless of speed or traffic." (Any reasonable assumption accepted.)

2. Q2 — Reading a model from a table

   Fluency
   • (a) Each extra GB costs $10 — constant rate. Linear model.
   • (b) C(g) = 10g + 25.
   • (c) C(7) = 70 + 25 = $95.
   • (d) The y-intercept ($25) is the base monthly fee — charged even if no data is used.

3. Q3 — Linear model — two rates

   Fluency
   • (a) A(h) = 30h + 50. B(h) = 40h + 20.
   • (b) 30h + 50 = 40h + 20 ⇒ 30 = 10h ⇒ h = 3 hours. Both cost $140.
   • (c) A(4) = 120+50 = $170. B(4) = 160+20 = $180. Company A is cheaper for a 4-hour job.
   • (d) A(1) = 80. B(1) = 60. Company B is cheaper for a 1-hour job.

4. Q4 — Quadratic model — projectile

   Fluency
   • (a) h(1) = −4+16 = 12 m. h(2) = −16+32 = 16 m. h(3) = −36+48 = 12 m.
   • (b) t = −16/(2×−4) = 2 s. Maximum height = 16 m.
   • (c) −4t²+16t=0 ⇒ −4t(t−4)=0 ⇒ t=0 or t=4 s.
   • (d) Assumption: no air resistance. Real balls experience drag, so they travel lower and land sooner.

5. Q5 — Exponential model — savings

   Understanding
   • (a) B(3) = 2000 × 1.05³ = 2000 × 1.157625 = $2 315.25. B(10) = 2000 × 1.05¹⁰ ≈ 2000 × 1.6289 = $3 257.79.
   • (b) Trial: 1.05⁸ = 1.477 ⇒ $2954. 1.05⁹ = 1.551 ⇒ $3101 > $3000. After 9 years.
   • (c) 1.05 is the growth factor — each year the balance grows to 105% of the previous year (a 5% increase).
   • (d) The model assumes the interest rate stays at 5% every year. In reality rates can change.

6. Q6 — Model validation

   Understanding
   • (a) Linear model. The gradient (12) represents 12 km per litre — the fuel efficiency.
   • (b) d(60) = 12 × 60 = 720 km.
   • (c) Predicted: 720 km. Actual: 680 km. Error = (720−680)/680 × 100 ≈ 5.9%. Model overestimates.
   • (d) e.g. Driving conditions (highway vs city), speed, air conditioning usage, tyre pressure all affect real fuel efficiency.

7. Q7 — Building a model from a description

   Understanding
   • (a) V(t) = 300t + 2000 (litres), for t ≥ 0.
   • (b) 300t + 2000 = 20000 ⇒ 300t = 18000 ⇒ t = 60 minutes.
   • (c) Graph: straight line from (0, 2000) to (60, 20000). Gradient = 300. y-intercept = 2000. Label axes: t (min) and V (L).
   • (d) Piecewise: V(t) = 300t + 2000 for 0 ≤ t ≤ 30; V(t) = 11000 for t > 30.

8. Q8 — Piecewise model

   Understanding
   • (a) C(u) = 0.28u for 0 ≤ u ≤ 500; C(u) = 0.28(500) + 0.35(u−500) = 140 + 0.35(u−500) for u > 500.
   • (b) C(400) = 0.28 × 400 = $112.00. C(700) = 140 + 0.35(200) = 140 + 70 = $210.00.
   • (c) C(500) = 0.28 × 500 = $140.00.
   • (d) For u > 500: 140 + 0.35(u−500) = 238 ⇒ 0.35(u−500) = 98 ⇒ u−500 = 280 ⇒ u = 780 kWh.

9. Q9 — Choosing and justifying a model

   Problem Solving
   • (a) The increases get smaller each day (5, 4, 3, 2, 1, 1), suggesting a non-linear (concave) model is more appropriate than linear.
   • (b) Differences: 5, 4, 3, 2, 1, 1. The differences are decreasing — growth is slowing, approaching a limit.
   • (c) Day 14: 2.67(14)+39.3 ≈ 76. Day 30: 2.67(30)+39.3 ≈ 119. Day 14 is plausible. Day 30 (119 customers) is likely unrealistic since the actual data shows growth slowing.
   • (d) A linear model doesn't capture the slowing growth. The coffee shop will eventually reach a capacity limit — a logistic or square-root model would be more appropriate.

10. Q10 — Full modelling problem

    Problem Solving
    • (a) 4 internal dividers + 2 outer lengths: 4x + 2y = 120. ⇒ 2y = 120 − 4x ⇒ y = 60 − 2x.
    • (b) Total area = 3 pens each of area xy: A = 3x(60−2x) = 180x − 6x². A(x) = −6x² + 180x.
    • (c) Vertex at x = −180/(2×−6) = 15 m. A(15) = −6(225) + 180(15) = −1350 + 2700 = 1350 m².
    • (d) x must be positive and y must be positive: y = 60−2x > 0 ⇒ x < 30. So 0 < x < 30.