Solutions — Piecewise Linear Graphs
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Q1 — Electricity tariff costs
Fluency(a) 300 kWh:
300 ≤ 500, so use rate $0.18/kWh.
C = 0.18 × 300 = $54.00(b) 500 kWh:
500 ≤ 500, so still use rate $0.18/kWh (the boundary belongs to the first tier).
C = 0.18 × 500 = $90.00(c) 750 kWh:
750 > 500, so first 500 kWh at $0.18, remaining 250 kWh at $0.25.
C = 0.18 × 500 + 0.25 × (750 − 500)
C = 90.00 + 0.25 × 250
C = 90.00 + 62.50 = $152.50 -
Q2 — Piecewise rule for electricity cost
FluencyIdentify breakpoint: u = 500 kWh.
At u = 500: cost = 0.18 × 500 = $90 (value at breakpoint).Rule for each piece:
• For 0 ≤ u ≤ 500: C = 0.18u
• For u > 500: C = 90 + 0.25(u − 500)C = { 0.18u if 0 ≤ u ≤ 500
{ 90 + 0.25(u − 500) if u > 500Check continuity at u = 500: Rule 2: 90 + 0.25(500 − 500) = 90 + 0 = $90. Both rules give $90 at u = 500 ✓ (continuous function)
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Q3 — Gym visit costs
Fluency(a) Piecewise rule:
Breakpoint at v = 10. At v = 10: C = 8 × 10 = $80.
• For 0 ≤ v ≤ 10: C = 8v (gradient = $8/visit)
• For v > 10: C = 80 + 4(v − 10) (gradient = $4/visit after 10)C = { 8v if 0 ≤ v ≤ 10
{ 80 + 4(v − 10) if v > 10(b) Cost for 7 visits:
7 ≤ 10, so use rule 1: C = 8(7) = $56(c) Cost for 15 visits:
15 > 10, so use rule 2: C = 80 + 4(15 − 10) = 80 + 4(5) = 80 + 20 = $100 -
Q4 — Taxi fare piecewise model
Understanding(a) Piecewise rule:
Flagfall: $3.50. First 10 km at $2.20/km, then $1.80/km beyond 10 km.
At d = 10: C = 3.50 + 2.20(10) = 3.50 + 22 = $25.50 (value at breakpoint).C = { 2.20d + 3.50 if 0 ≤ d ≤ 10
{ 25.50 + 1.80(d − 10) if d > 10(b)(i) 6 km: 6 ≤ 10, use rule 1: C = 2.20(6) + 3.50 = 13.20 + 3.50 = $16.70
(b)(ii) 10 km: Use rule 1: C = 2.20(10) + 3.50 = 22 + 3.50 = $25.50
(b)(iii) 18 km: 18 > 10, use rule 2: C = 25.50 + 1.80(18 − 10) = 25.50 + 1.80(8) = 25.50 + 14.40 = $39.90
(c) Fare of $29.30 — find distance:
Since $29.30 > $25.50 (the breakpoint fare), the passenger travelled beyond 10 km. Use rule 2:
29.30 = 25.50 + 1.80(d − 10)
3.80 = 1.80(d − 10)
d − 10 = 3.80 ÷ 1.80 ≈ 2.11
d ≈ 12.11 km
Check: 25.50 + 1.80(2.11) = 25.50 + 3.80 = $29.30 ✓
Distance: approximately 12.1 km -
Q5 — Evaluating and interpreting a piecewise function
Understanding(a) Evaluating C at each time:
t = 0: 0 ≤ 2, use rule 1: C = 20. So C = $20
t = 2: 0 ≤ 2 ≤ 2, use rule 1: C = 20. So C = $20
t = 5: 2 < 5 ≤ 6, use rule 2: C = 20 + 8(5 − 2) = 20 + 8(3) = 20 + 24 = C = $44
t = 8: 8 > 6, use rule 3: C = 52 + 12(8 − 6) = 52 + 12(2) = 52 + 24 = C = $76
Note: value at breakpoint t = 6: rule 2 gives C = 20 + 8(4) = 52, and rule 3 starts at 52. Continuous ✓
(b) Possible real-world context:
This model could represent a car hire or equipment rental cost: a flat rate of $20 for the first 2 hours (minimum charge), then $8 per hour for the next 4 hours (hours 2–6), then a higher rate of $12 per hour after 6 hours (premium rate for extended hire).(c) Gradient in each section:
• Section 1 (0 ≤ t ≤ 2): gradient = 0 (cost is flat). In context: no additional charge during the minimum hire period.
• Section 2 (2 < t ≤ 6): gradient = $8/hr. In context: $8 charged for each additional hour beyond the first 2.
• Section 3 (t > 6): gradient = $12/hr. In context: the rate increases to $12/hr for extended hire beyond 6 hours. -
Q6 — Internet data plan
Understanding(a) Piecewise rule:
Breakpoint at d = 50 GB. Value at breakpoint: $60.
Beyond 50 GB: $1 per GB extra (i.e. $10 per 10 GB = $1/GB).B = { 60 if 0 ≤ d ≤ 50
{ 60 + 1(d − 50) if d > 50(b) Bill for 42 GB:
42 ≤ 50, use rule 1: B = $60(c) Bill for 75 GB:
75 > 50, use rule 2: B = 60 + 1(75 − 50) = 60 + 25 = $85(d) Bill for 120 GB:
120 > 50, use rule 2: B = 60 + 1(120 − 50) = 60 + 70 = $130 -
Q7 — Daily temperature model
Understanding(a) Piecewise model:
Temperature rises at 3°C/hr from 8 am (h = 0). Maximum at 2 pm = h = 6.
Maximum temperature: T = 12 + 3(6) = 12 + 18 = 30°C (value at breakpoint h = 6).
After 2 pm: drops at 2°C/hr.T = { 3h + 12 if 0 ≤ h ≤ 6
{ 30 − 2(h − 6) if h > 6(b) Maximum temperature:
At h = 6 (2 pm): T = 3(6) + 12 = 30. Maximum = 30°C at 2 pm.(c) Temperature at 5 pm:
5 pm = h = 9. Since 9 > 6, use rule 2: T = 30 − 2(9 − 6) = 30 − 6 = 24°C(d) When does temperature return to 12°C?
Using rule 2: 12 = 30 − 2(h − 6)
2(h − 6) = 18
h − 6 = 9
h = 15 hours after 8 am = 11 pm
(The temperature returns to 12°C at 11 pm that evening.) -
Q8 — Comparing delivery pricing
Understanding(a) Company A (single linear rule):
A = 0.30d + 50 (flat $50 + $0.30/km for all distances)(b) Company B (piecewise rule):
At d = 100: B = 0.40(100) = $40 (value at breakpoint).B = { 0.40d if 0 ≤ d ≤ 100
{ 40 + 0.20(d − 100) if d > 100(c) Cost for 80 km:
Company A: A = 0.30(80) + 50 = 24 + 50 = $74
Company B: 80 ≤ 100, use rule 1: B = 0.40(80) = $32
Company B is cheaper by $42 at 80 km.(d) Cost for 200 km:
Company A: A = 0.30(200) + 50 = 60 + 50 = $110
Company B: 200 > 100, use rule 2: B = 40 + 0.20(200 − 100) = 40 + 20 = $60
Company B is cheaper by $50 at 200 km.(e) Equal cost for d > 100:
Set A = B (using rule 2 for Company B):
0.30d + 50 = 40 + 0.20(d − 100)
0.30d + 50 = 40 + 0.20d − 20
0.30d + 50 = 0.20d + 20
0.10d = −30
d = −300
This gives a negative distance, which is impossible. This means the two companies are never equal for d > 100 km. Company B is always cheaper beyond 100 km (its increasing cost per km is lower than Company A’s).
Interpretation: Company A charges $50 flagfall (high fixed cost); Company B has no fixed cost, so for longer distances Company B is always cheaper. -
Q9 — Overtime pay structure
Problem Solving(a) Overtime rates per hour:
Normal rate: $22/hr.
1.5× rate: 22 × 1.5 = $33/hr
2× rate: 22 × 2 = $44/hr(b) Piecewise rule for weekly pay P:
Breakpoint 1: h = 38 (end of standard hours). Pay at 38 hrs: 22 × 38 = $836.
Breakpoint 2: h = 42 (end of 1.5× overtime). Pay at 42 hrs: 836 + 33(4) = 836 + 132 = $968.P = { 22h if 0 ≤ h ≤ 38
{ 836 + 33(h − 38) if 38 < h ≤ 42
{ 968 + 44(h − 42) if h > 42(c) Weekly pay for each scenario:
(i) 38 hours: P = 22(38) = $836
(ii) 42 hours: 38 < 42 ≤ 42, use rule 2: P = 836 + 33(42 − 38) = 836 + 33(4) = 836 + 132 = $968
(iii) 50 hours: 50 > 42, use rule 3: P = 968 + 44(50 − 42) = 968 + 44(8) = 968 + 352 = $1 320
(d) Minimum hours for $1 200/week:
At h = 42: P = $968. At h = 50: P = $1 320. So the target of $1 200 requires working beyond 42 hours.
Use rule 3: 1200 = 968 + 44(h − 42)
232 = 44(h − 42)
h − 42 = 232 ÷ 44 ≈ 5.27
h ≈ 47.27
Since hours must be whole hours, minimum 48 hours per week is needed to earn at least $1 200.
Check: P = 968 + 44(48 − 42) = 968 + 264 = $1 232 ≥ $1 200 ✓ -
Q10 — Sales commission structure
Problem Solving(a) Piecewise rule for commission earnings E:
Tier 1 (S ≤ $1 000): no commission. E = 0.
Tier 2 ($1 000 < S ≤ $5 000): 5% on amount above $1 000. E = 0.05(S − 1000).
Tier 3 (S > $5 000): 8% on the full amount. E = 0.08S.E = { 0 if 0 ≤ S ≤ 1000
{ 0.05(S − 1000) if 1000 < S ≤ 5000
{ 0.08S if S > 5000(b) Earnings for each sales amount:
(i) $800: S ≤ 1000, use rule 1: E = $0
(ii) $3 500: 1000 < 3500 ≤ 5000, use rule 2:
E = 0.05(3500 − 1000) = 0.05 × 2500 = $125(iii) $7 200: S > 5000, use rule 3:
E = 0.08 × 7200 = $576(c) When does 8% full become more beneficial?
At S = $5 000:
Rule 2 (5% above $1000): E = 0.05(5000 − 1000) = 0.05 × 4000 = $200
Rule 3 (8% full): E = 0.08 × 5000 = $400
At exactly $5 000, the 8% rule gives more ($400 > $200).
The 8% rule first becomes applicable (and more beneficial) at sales just above $5 000. In fact, there is a discontinuity (jump) at S = $5 000 — a salesperson earning $4 999 gets 5% on $3 999 = $199.95, but at $5 001 they get 8% on $5 001 = $400.08. The jump in earnings of ~$200 provides strong incentive to reach $5 000+ in sales.(d) Strategic implications:
At sales just below $5 000 (e.g. $4 999): E ≈ $200. At sales just above $5 000 (e.g. $5 001): E ≈ $400. This creates a discontinuity: there is a sudden jump of about $200 in earnings at the $5 000 threshold. A salesperson would be strongly motivated to either push sales past $5 000 (to benefit from the higher rate on the full amount) or, if unlikely to reach $5 000, not bother pushing beyond it since the extra effort below $5 000 only earns 5% on the marginal amount. The range $4 000–$5 000 is worth pushing through quickly to reach the higher tier.