Solutions — Linear Models and Applications

Q1 — Taxi fare model

(a) Flag-fall = $3.50 (y-intercept); rate = $2.20/km (gradient).

C = 2.20d + 3.50 (where d = distance in km, C = cost in $)

(b) Substitute d = 15: C = 2.20(15) + 3.50 = 33.00 + 3.50 = $36.50

(c) Set C = 25.50: 2.20d + 3.50 = 25.50

2.20d = 22.00 → d = 22.00 / 2.20 = 10 km

Q2 — Phone plan model

(a) Monthly base = $30 (y-intercept); rate = $0.15/text (gradient).

C = 0.15n + 30 (where n = number of texts, C = monthly cost in $)

(b) C = 0.15(80) + 30 = 12 + 30 = $42

(c) 0.15n + 30 = 51 → 0.15n = 21 → n = 21 / 0.15 = 140 texts

Q3 — Linear model from table

(a) At t = 0, D = 12 (y-intercept c = 12). Gradient: m = (22−12)/(1−0) = 10 km/hr.

D = 10t + 12 (where t = time in hours, D = distance in km)

(b) Gradient m = 10: the car travels at 10 km per hour. y-intercept c = 12: the car started 12 km from the reference point (already 12 km along the route at t = 0).

(c) D = 10(5) + 12 = 50 + 12 = 62 km

Q4 — Fuel consumption model

(a) Two points: (0, 65) and (120, 50).

m = (50 − 65)/(120 − 0) = −15/120 = −0.125 L/km

c = 65

F = −0.125d + 65 (where d = distance in km, F = fuel in litres)

(b) m = −0.125 L/km: the car uses 0.125 litres of fuel per kilometre (or 1 L per 8 km).

(c) Set F = 0: 0 = −0.125d + 65 → 0.125d = 65 → d = 65/0.125 = 520 km

Assumption: the car consumes fuel at a constant rate (no idling, constant speed, flat terrain).

Q5 — Leaking water tank

(a) Initial volume = 4800 L; leak rate = −12 L/min.

V = −12t + 4800 (where t = time in minutes, V = volume in litres)

(b) Set V = 1200: −12t + 4800 = 1200 → −12t = −3600 → t = 300 minutes = 5 hours

(c) Set V = 0: 0 = −12t + 4800 → 12t = 4800 → t = 400 min = 6 hours 40 minutes

(d) Domain: 0 ≤ t ≤ 400 minutes (cannot have negative volume; cannot have negative time).

Q6 — Comparing internet plans

(a) Plan A: C_A = 0.05g + 29 Plan B: C_B = 0.02g + 45

(b) Set equal: 0.05g + 29 = 0.02g + 45

0.03g = 16 → g = 16/0.03 ≈ 533.3 GB/month

Both plans cost the same at approximately 533.3 GB/month.

(c) At g = 300 GB:

Plan A: C_A = 0.05(300) + 29 = 15 + 29 = $44

Plan B: C_B = 0.02(300) + 45 = 6 + 45 = $51

Since 300 GB < 533.3 GB (break-even), Plan A is cheaper at $44. Plan A is better for 300 GB/month, saving $7/month.

Note: Plan B becomes cheaper only for usage above 533.3 GB/month.

Q7 — Linear cooling model

(a) From 6 pm to midnight = 6 hours. Temperature drops from 22°C to 10°C = drop of 12°C.

Rate = −12/6 = −2°C per hour

(b) At h = 0 (6 pm), T = 22. Gradient m = −2.

T = −2h + 22 (h = hours after 6 pm)

Check: at h = 6 (midnight): T = −12 + 22 = 10 ✓

(c) 2 am = 8 hours after 6 pm (h = 8):

T = −2(8) + 22 = −16 + 22 = 6°C

(d) Set T = 5: −2h + 22 = 5 → −2h = −17 → h = 8.5 hours after 6 pm.

8.5 hours after 6 pm = 2:30 am

Q8 — Hooke’s Law spring model

(a) At F = 0: L = 8 cm (natural length, so c = 8).

At F = 5: L = 13. Gradient m = (13 − 8)/(5 − 0) = 5/5 = 1 cm/N.

L = F + 8 (or L = 1×F + 8)

(b) F = 8: L = 8 + 8 = 16 cm

(c) Set L = 20: 20 = F + 8 → F = 12 N

(d) It is not reasonable to use this model at F = 100 N. In reality, springs have an elastic limit (Hooke’s Law only applies within the elastic range). At very large forces, the spring would permanently deform or break. This would be a case of unsafe extrapolation.

Q9 — Comparing gym memberships

(a) Gym A: C_A = 15n + 200 Gym B: C_B = 30n + 50

(b) Set equal: 15n + 200 = 30n + 50

200 − 50 = 30n − 15n → 150 = 15n → n = 10 months

At n = 10: C = 15(10) + 200 = $350 (check: 30(10) + 50 = $350 ✓)

(c) Gym A starts higher ($200 y-intercept) but has a gentler slope ($15/month). Gym B starts lower ($50 y-intercept) but climbs more steeply ($30/month). The two lines cross at 10 months. Before 10 months, Gym B is cheaper; after 10 months, Gym A is cheaper.

(d) 2 years = 24 months:

Gym A: C = 15(24) + 200 = 360 + 200 = $560

Gym B: C = 30(24) + 50 = 720 + 50 = $770

Gym A is cheaper for 2 years by $770 − $560 = $210.

Q10 — Car rental comparison

(a) Standard car: C = 0.22k + 35 (k = km, C = daily cost in $)

(b) Set C ≤ 200: 0.22k + 35 ≤ 200 → 0.22k ≤ 165 → k ≤ 750

Maximum 750 km per day within budget.

(c) Daily cost at 180 km: C = 0.22(180) + 35 = 39.60 + 35 = $74.60

Total 5-day cost: 5 × $74.60 = $373.00

(d) Upgraded car: C = 0.12k + 55

Daily cost at 180 km: C = 0.12(180) + 55 = 21.60 + 55 = $76.60

Total 5-day cost: 5 × $76.60 = $383.00

The standard car is cheaper for this trip ($373.00 vs $383.00, saving $10 over 5 days).

Note: The upgraded car would become cheaper at higher daily distances: set 0.22k + 35 = 0.12k + 55 → 0.10k = 20 → k = 200 km/day. Above 200 km/day, the upgraded car is cheaper.