Solutions — Break-even Analysis
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Q1 — TC and R equations given directly
Fluency(a) Break-even quantity:
Set R = TC: 14x = 5x + 900
9x = 900
x = 100
Break-even at 100 units.
Check: R = 14(100) = $1400; TC = 5(100) + 900 = $1400 ✓(b) Profit/loss at x = 80:
R = 14(80) = $1120; TC = 5(80) + 900 = 400 + 900 = $1300
Profit = R − TC = 1120 − 1300 = −$180 (a loss of $180)
(80 < 100, so we’re below break-even — as expected, the business is making a loss.)(c) Profit at x = 200:
R = 14(200) = $2800; TC = 5(200) + 900 = 1000 + 900 = $1900
Profit = 2800 − 1900 = $900 profit -
Q2 — Writing TC and R from a description
Fluency(a) Equations:
TC = 6n + 2 400 (FC = $2400, VC = $6/unit)
R = 18n (selling price = $18/unit)(b) Break-even quantity:
Set R = TC: 18n = 6n + 2400
12n = 2400
n = 200
Break-even at 200 units.
Check: R = 18(200) = $3600; TC = 6(200) + 2400 = $3600 ✓(c) Contribution margin:
Contribution margin = selling price − variable cost = $18 − $6 = $12 per unit.
This means each unit sold contributes $12 toward covering the $2400 fixed costs. Once fixed costs are covered (at 200 units), each additional unit sold generates $12 of profit. -
Q3 — Deducing the TC equation from graph information
Fluency(a) Revenue at break-even (150 units):
R = selling price × n = $20 × 150 = $3 000(b) Finding variable cost per unit:
At break-even: TC = R = $3 000.
TC = VC×n + FC → 3000 = VC × 150 + 1800
VC × 150 = 3000 − 1800 = 1200
VC = 1200 ÷ 150 = $8 per unit.
TC = 8n + 1 800(c) Verification at n = 150:
TC = 8(150) + 1800 = 1200 + 1800 = $3000
R = 20(150) = $3000
TC = R = $3000 ✓ Break-even confirmed. -
Q4 — School canteen sandwiches
Understanding(a) Equations:
TC = 2.80n + 120 (FC = $120/day, VC = $2.80/sandwich)
R = 6.50n (selling price = $6.50/sandwich)(b) Break-even sandwiches:
6.50n = 2.80n + 120
3.70n = 120
n = 120 ÷ 3.70 ≈ 32.43
Break-even at 33 sandwiches (round up).
Check at 33: R = 6.50(33) = $214.50; TC = 2.80(33) + 120 = 92.40 + 120 = $212.40. R > TC ✓(c) Profit at 60 sandwiches:
R = 6.50(60) = $390; TC = 2.80(60) + 120 = 168 + 120 = $288
Profit = 390 − 288 = $102 profit(d) Sandwiches for $250 profit:
Profit = R − TC = 6.50n − (2.80n + 120) = 3.70n − 120
Set profit = 250: 3.70n − 120 = 250
3.70n = 370
n = 100
100 sandwiches must be sold to earn a $250 daily profit.
Check: Profit = 3.70(100) − 120 = 370 − 120 = $250 ✓ -
Q5 — Band venue hire
Understanding(a) Equations:
TC = 800 (no variable costs; fixed venue hire only)
R = 25n (ticket price $25 each)(b) Break-even tickets:
25n = 800
n = 800 ÷ 25 = 32
Break-even at 32 tickets.(c) Profit at 60 tickets:
Profit = R − TC = 25(60) − 800 = 1500 − 800 = $700 profit(d) Maximum profit (full house, 120 tickets):
Profit = 25(120) − 800 = 3000 − 800 = $2 200 profit(e) New break-even at $18 ticket price:
R = 18n; 18n = 800; n = 800 ÷ 18 ≈ 44.4
New break-even at 45 tickets.
At full house (120 tickets): Profit = 18(120) − 800 = 2160 − 800 = $1360. Yes, a full house is still profitable (120 > 45 break-even). -
Q6 — Manufacturing business analysis
Understanding(a) Break-even quantity:
Contribution margin = $35 − $12.50 = $22.50/unit
Break-even = FC ÷ CM = $5500 ÷ $22.50 ≈ 244.4
Break-even at 245 units.(b) Profit at 400 units:
R = 35(400) = $14 000; TC = 12.50(400) + 5500 = 5000 + 5500 = $10 500
Profit = 14 000 − 10 500 = $3 500(c) Selling price to halve break-even:
Original break-even ≈ 245 units. Halved break-even = 122.5 units (target).
Using: break-even = FC ÷ (price − VC)
122.5 = 5500 ÷ (price − 12.50)
price − 12.50 = 5500 ÷ 122.5 ≈ 44.90
price ≈ 44.90 + 12.50 = $57.40
A selling price of approximately $57.40/unit would halve the break-even quantity.
Check: CM = 57.40 − 12.50 = $44.90; Break-even = 5500 ÷ 44.90 ≈ 122.5 ✓ -
Q7 — Two business options compared
Understanding(a) Equations (both sell at $7/item):
Option A: TC⊂A = 2n + 8 000; R = 7n
Option B: TC⊂B = 4.50n (no fixed cost); R = 7n(b) Break-even quantities:
Option A: 7n = 2n + 8000 → 5n = 8000 → n = 1 600 units.
Option B: 7n = 4.50n → 2.50n = 0 → n = 0 units (break-even immediately; every unit is profitable from the first).(c) Profit at 2 000 units:
Option A: Profit = 7(2000) − [2(2000) + 8000] = 14000 − 12000 = $2 000
Option B: Profit = 7(2000) − 4.50(2000) = 14000 − 9000 = $5 000
At 2 000 units, Option B is more profitable by $3 000.(d) Quantity where Option A becomes more profitable than Option B:
Set Profit⊂A = Profit⊂B:
(7−2)n − 8000 = (7−4.50)n
5n − 8000 = 2.50n
2.50n = 8000
n = 3 200 units.
Option A becomes more profitable than Option B at 3 200 units (and beyond).
Check at 3200: A profit = 5(3200) − 8000 = $8000; B profit = 2.50(3200) = $8000 ✓ -
Q8 — Reading a break-even graph
Understanding(a) Gradients:
TC line: passes through (0, 1200) and (400, 2800).
Gradient (VC) = (2800 − 1200) ÷ (400 − 0) = 1600 ÷ 400 = $4 per unit
R line: passes through (0, 0) and (400, 3200).
Gradient (price) = 3200 ÷ 400 = $8 per unit(b) Equations:
TC = 4n + 1 200
R = 8n(c) Break-even quantity:
8n = 4n + 1200
4n = 1200
n = 300
Break-even at 300 units.
Check: R = 8(300) = $2400; TC = 4(300) + 1200 = $2400 ✓(d) Contribution margin:
CM = price − VC = $8 − $4 = $4 per unit(e) Profit at 600 units:
Profit = R − TC = 8(600) − [4(600) + 1200] = 4800 − [2400 + 1200] = 4800 − 3600 = $1 200 -
Q9 — Food truck analysis
Problem Solving(a) Daily equations:
TC = 5.80n + 180 (FC = $180/day, VC = $5.80/meal)
R = 16n (selling price = $16/meal)(b) Daily break-even meals:
16n = 5.80n + 180
10.20n = 180
n = 180 ÷ 10.20 ≈ 17.6
Break-even at 18 meals per day.(c) Total profit over 4 weeks (50 meals/day, 6 days/week):
Daily profit at 50 meals: Profit = R − TC = 16(50) − [5.80(50) + 180] = 800 − [290 + 180] = 800 − 470 = $330/day
Operating days in 4 weeks: 6 × 4 = 24 days
Total profit = 24 × $330 = $7 920(d) Additional meals needed with increased fixed costs:
Original daily profit at 50 meals = $330 (from above).
With new FC = $220: TC = 5.80n + 220; R = 16n.
Profit = 16n − (5.80n + 220) = 10.20n − 220
Set equal to $330: 10.20n − 220 = 330
10.20n = 550
n = 550 ÷ 10.20 ≈ 53.9 → 54 meals.
Additional meals = 54 − 50 = 4 additional meals per day needed to maintain the same profit.
Check: Profit = 10.20(54) − 220 = 550.80 − 220 = $330.80 ≈ $330 ✓ -
Q10 — Marketing campaign decision
Problem Solving(a) Current monthly profit (300 units, $45 each):
R = 45 × 300 = $13 500
TC = 18 × 300 + 4200 = 5400 + 4200 = $9 600
Current profit = 13500 − 9600 = $3 900/month(b) Profit with marketing campaign (420 units, extra $1500 fixed cost):
New fixed costs = 4200 + 1500 = $5 700/month
R = 45 × 420 = $18 900
TC = 18 × 420 + 5700 = 7560 + 5700 = $13 260
Campaign profit = 18900 − 13260 = $5 640/month(c) Is the campaign worthwhile?
Without campaign: $3 900/month profit. With campaign: $5 640/month profit.
Increase in profit = 5640 − 3900 = $1 740/month.
Yes, the campaign is worthwhile. Even accounting for the extra $1 500 cost, the business gains $1 740 more profit per month — a net improvement of $1 740.(d) Minimum price increase (no campaign, 300 units) to achieve $5 640 profit:
Let the new selling price = p.
Profit = p×300 − (18×300 + 4200) = 300p − 9600
Set equal to $5640: 300p − 9600 = 5640
300p = 15 240
p = 15240 ÷ 300 = $50.80
Price increase needed = 50.80 − 45 = $5.80 per unit increase (new price: $50.80)
Check: Profit = 50.80(300) − 9600 = 15240 − 9600 = $5640 ✓