Break-even analysis and linear cost or revenue models for HSC Maths Standard 2

Syllabus dot point

“Model practical problems with linear cost and revenue functions and find the break-even point”

HSCMaths Standard 2Year 12: Algebra13 min readUpdated 20 Jun 2026

Quick answer

Set up $C = F + vn$ and $R = pn$ ; break-even is where they cross ( $C = R$ ), so $n_{\text{be}} = \dfrac{F}{p - v}$ , the fixed cost over the contribution margin. Use $P = (p - v)n - F$ for profit at any level, and read profit on a graph as the vertical gap between the lines (loss to the left of break-even, profit to the right). Round break-even up to the next whole unit if it is not exact.

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What this dot point is asking
The answer
Exam-style questions
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What this dot point is asking

NESA wants you to set up linear cost and revenue equations for a small business or stall. You then find the break-even quantity, the number of units where cost equals revenue, and work out the profit at any level of production. This is one of the most predictable Section II questions in the whole paper. It is really a simultaneous-equations problem in disguise: the break-even point is just where the cost line and the revenue line cross.

Break-even matters in the real world because a business does not start earning the moment it makes its first sale. First it has to claw back its fixed costs, the money it owes whether it sells anything or not (rent, a gas bottle, a market-stall fee). Break-even is the moment those fixed costs are finally covered. After that, every extra sale turns into profit. Once you see that "fixed costs first, profit after" story, every part of the question falls into place.

The answer

Cost, revenue and profit

For a linear model with $n$ units produced or sold:

Cost $C = F + v n$ , where $F$ is the fixed cost (paid regardless of sales) and $v$ is the variable cost per unit (the extra cost of making one more).
Revenue $R = p n$ , where $p$ is the selling price per unit. Revenue starts at the origin: no sales, no money in.
Profit $P = R - C = (p - v) n - F$ . Profit can be negative, in which case it is a loss.

The quantity $(p - v)$ is the contribution margin per unit: the part of each sale left over after paying that unit's own variable cost. It is what each sold unit "contributes" first towards covering the fixed costs, and then, once those are covered, straight to profit. Almost every break-even shortcut comes from this one idea.

Break-even

Break-even is the quantity at which profit is exactly zero, that is, where cost equals revenue:

F + v n = p n \implies F = (p - v) n \implies n_{\text{be}} = \frac{F}{p - v}.

Read the middle step in words: the fixed cost $F$ has to be paid off at the contribution margin $(p - v)$ per unit, so the number of units needed is fixed cost divided by contribution margin. Below the break-even quantity the business runs at a loss; above it, the business is in profit.

Key fact

Break-even is where cost equals revenue ( $C = R$ ), so profit is zero. With $C = F + vn$ and $R = pn$ , the break-even quantity is $n_{\text{be}} = \dfrac{F}{p - v}$ : the fixed cost divided by the contribution margin $(p - v)$ . Each unit sold first chips away at the fixed cost; once break-even is passed, each further unit adds its full $(p - v)$ to profit. If the break-even quantity is not a whole number, round up, because at the rounded-down value the fixed costs are not yet fully covered.

Why you round break-even up

The break-even formula often gives a fraction, such as $n_{\text{be}} = 11.4$ items. You cannot sell $0.4$ of an item, so the answer must be a whole number. It also has to be the whole number that actually covers costs. At $11$ items the business is still a few dollars short, so it is below break-even and making a loss. At $12$ items it is over the line and making a profit. So break-even quantities round up, never down. This is the opposite of normal rounding, and it is a favourite place for the HSC to dock half a mark. Always justify it: "round up to $12$ , since $11$ items would still be a loss."

Build the break-even chart, stage by stage

A graph turns the algebra into a picture you can read at a glance. The four panels below build one up for a charity sausage-sizzle stall outside a Bunnings in Western Sydney. The stall has fixed costs of $120 (gas bottle, site fee, esky hire), a variable cost of $1.50 per sausage sandwich (bread, sausage, onion, sauce), and a selling price of $3.50 each.

Stage 1, plot the cost line. Cost is $C = 120 + 1.5n$ . Unlike revenue, the cost line does not start at the origin: at $n = 0$ , before a single sandwich is sold, the stall already owes its $120 fixed cost. So the cost line starts high, at the fixed-cost intercept of $120 on the dollar axis, and rises gently by $1.50 for each sandwich.

Stage 2, plot the revenue line. Revenue is $R = 3.5n$ . With no fixed part it starts at the origin (zero sales, zero income) and rises steeply, by $3.50 a sandwich. Because $3.50 is greater than $1.50, the revenue line is steeper than the cost line, so even though it starts lower it must eventually catch up and overtake.

Stage 3, find the break-even crossing. The lines cross where cost equals revenue, which is exactly the break-even point. Drop a dashed line down to the quantity axis and across to the dollar axis: the crossing is at $n = 60$ sandwiches, where both cost and revenue equal $210.

Stage 4, read off the loss and profit regions. The break-even point splits the chart in two. To the left of $n = 60$ , the cost line sits above the revenue line, so $C > R$ and the stall is making a loss (the shaded wedge). To the right, revenue overtakes cost, $R > C$ , and the stall is in profit (the accent wedge). The vertical gap between the lines is the size of the loss or profit at that quantity.

You can confirm the crossing algebraically in one line: $120 + 1.5n = 3.5n$ gives $120 = 2n$ , so $n = 60$ , and $R = 3.5(60) = 210$ , i.e. $210. Notice the contribution margin shortcut hiding in there: $n_{\text{be}} = \dfrac{120}{3.5 - 1.5} = \dfrac{120}{2} = 60$ .

Reading the gap: profit and loss as vertical distance

The most useful thing the chart teaches is that profit at any quantity is the vertical gap between the lines, with revenue on top. Take $n = 90$ . The revenue line is at $3.5(90) = 315$ , i.e. $315, and the cost line at $120 + 1.5(90) = 255$ , i.e. $255. The gap is $60, so that is $60 of profit. Now take $n = 30$ . Here the cost line ($165) is above the revenue line ($105), a gap of $60 the other way, which is a $60 loss. Each step of $30$ sandwiches past break-even adds $30 \times 2 = 60$ , i.e. $60. That is the contribution margin times the extra units, which is why the profit grows in even jumps.

How exam questions ask about break-even and linear models

The numbers change but the tasks are a fixed menu. Translate the wording:

"Find the number of units needed to break even." Solve $C = R$ , or use $n_{\text{be}} = \dfrac{F}{p - v}$ , then round up if the answer is not whole and say why.
"Find the profit if $x$ units are sold." Use $P = R - C$ at that $x$ (or the shortcut $P = (p - v)x - F$ ). Watch the sign: a negative answer is a loss, not an error.
"By how much does profit increase for each extra unit sold?" This is just the contribution margin $p - v$ , a one-line answer.
"How many units are needed to make a profit of $X?" Set $P = (p - v)n - F = X$ and solve for $n$ (then round up). Break-even is the special case $X = 0$ .
"Draw / use the graph to find the break-even point." Read the crossing of the cost and revenue lines, and remember the loss region is left of it, profit to the right.
"If the price rises to $ $p'$ (or fixed costs change), what happens to break-even?" Recompute with the new figure. A higher price or lower fixed cost raises the contribution margin or lowers $F$ , so break-even falls; the reverse raises it.

Sydney cafe break-even analysis

Break-even, profit and a price rise for a Sydney cafe

A new Sydney cafe has fixed monthly costs of $3200, a variable cost of $1.20 per coffee, and a selling price of $5.50. Find the break-even quantity, the profit at $1500$ coffees, and the effect of raising the price to $6.

Define variables and write cost and revenue functions: Naming the unknown clearly avoids confusion later. Letting $n$ be the number of coffees sold per month, we can write one function for total cost and one for total revenue.

C = 3200 + 1.20n, \qquad R = 5.50n

Find the contribution margin per coffee: The contribution margin is the amount each sale contributes towards covering fixed costs after paying its own variable cost. Subtracting the variable cost from the selling price gives this figure directly.

\text{Contribution margin} = 5.50 - 1.20 = 4.30 \text{ dollars per coffee}

Calculate the break-even quantity: At break-even, total revenue equals total cost, so profit is zero. Dividing the fixed costs by the contribution margin tells us how many coffees must be sold before the business starts to make a profit.

n_{\text{be}} = \frac{3200}{4.30} \approx 744.2, \quad \text{so } 745 \text{ coffees per month, or about } 25 \text{ a day}

Note the round-up to $745$ : at $744$ coffees the cafe is still a few cents short of covering its fixed costs.

Compute revenue, cost, and profit at 1500 coffees: Substituting $n = 1500$ into the cost and revenue functions shows the financial position when the cafe trades well above break-even. All figures are in dollars.

R = 5.50 \times 1500 = 8250

C = 3200 + 1.20 \times 1500 = 3200 + 1800 = 5000

\text{Profit} = R - C = 8250 - 5000 = 3250

Recalculate break-even after a price increase to $6: Raising the selling price increases the contribution margin, which lowers the break-even quantity. We repeat the contribution-margin and break-even steps with the new price to see by how much the target drops.

\text{New contribution margin} = 6.00 - 1.20 = 4.80 \text{ dollars per coffee}

n_{\text{be}} = \frac{3200}{4.80} \approx 667 \text{ coffees, a drop of } 78 \text{ coffees per month}

Note that the algebra cannot predict whether customers will buy fewer coffees at the higher price; that requires real sales data.

Final answer: break-even is $745$ coffees per month at $5.50; profit at $1500$ coffees is $3250; raising the price to $6 reduces the break-even quantity to approximately $667$ coffees per month.

Common traps

Mixing up cost and price: Cost is what the business pays per unit ( $v$ ); price is what the customer pays ( $p$ ). Swapping them flips the contribution margin.
Forgetting the fixed cost: Without the $F$ term, break-even is meaningless. Read the question to identify what is owed regardless of sales.
Not rounding up: The break-even formula often gives a fraction. Round up, since at the rounded-down value the business is still in loss, and say why.
Confusing revenue and profit: Revenue is total sales income; profit is revenue minus all costs. The HSC often tests whether you know the difference, especially in the "find the profit if ..." part.
Reading the gap from the wrong line: Profit is revenue minus cost, so it is the vertical gap with revenue on top. Below break-even that gap is negative, which is a loss, not a mistake.
Missing the contribution-margin shortcut: Profit at $n$ units is $(p - v)n - F$ , and break-even is $\dfrac{F}{p - v}$ . Once you have the margin, these are one-step calculations.

Exam technique

Open with "let $n$ be the number of ... sold" and write the two functions $C = F + vn$ and $R = pn$ explicitly; markers reward the set-up even before the arithmetic. Solve $C = R$ for break-even (or use the margin shortcut), and if the answer is fractional, round up and add the one-line reason. State profit with units and the right sign, calling a negative value a loss. If a graph is involved, label the fixed-cost intercept, mark the crossing, and shade or name the loss (left) and profit (right) regions. When a price or cost changes, recompute rather than guess the direction, then comment that the model assumes sales volume is unaffected by the price change.

Note

Picture running a small stall. Before you sell a thing, you already owe money for set-up things like rent or a gas bottle. That set bill is your "fixed cost". Each sale brings in cash, but part of it pays for that one item, and the rest helps pay off the fixed cost. Break-even is the moment your sales have finally paid off the fixed cost, so you have not lost money and not made any yet. Sell more than that and the extra is profit; sell less and you are still behind.

Exam-style questions

Questions in the style of NESA exam questions on this dot point, each with a worked answer. They are written by ExamExplained unless tagged "Past paper"; the year shows the paper a question is modelled on.

2023 HSC-style4 marks

A small business produces handmade soap. Fixed costs are $450 per month and variable costs are $3.50 per bar. Each bar sells for $8. Find the break-even quantity and the monthly profit if

200

bars are sold.

Show worked answer →

Let $n$ be bars sold per month.

Cost: $C = 450 + 3.50 n$ . Revenue: $R = 8 n$ .

Break-even when $C = R$ : $450 + 3.50 n = 8 n$ , so $450 = 4.50 n$ , giving $n = 100$ bars.

At $200$ bars: $R = 8 \times 200 = 1600$ , i.e. $1600. $C = 450 + 3.50 \times 200 = 450 + 700 = 1150$ , i.e. $1150.

Profit: $P = R - C = 1600 - 1150 = 450$ , i.e. $450.

Markers reward the cost and revenue equations, the break-even calculation, and the profit at $200$ bars with units.

2021 HSC-style3 marks

A market stall has a fixed daily rent of $80 and pays $5 per item bought from a wholesaler. Items sell for $12 each. How many items must be sold for the stall to break even, and what is the profit if

50

items are sold?

Show worked answer →

Let $n$ be items sold. Cost: $C = 80 + 5n$ . Revenue: $R = 12 n$ .

Break-even: $80 + 5n = 12 n$ , so $80 = 7n$ , giving $n = \frac{80}{7} \approx 11.4$ .

Round up: at least $12$ items must be sold to break even.

At $50$ items: $R = 600$ , $C = 80 + 250 = 330$ . Profit: $P = 270$ .

Markers reward the equations, the break-even rounded up because partial items are not sellable, and the profit calculation. Half marks for using $11$ or $11.4$ without explaining the round-up.

Practice questions

Original practice questions graded from foundation to exam level, each with a full worked solution. Try them before revealing the solution.

foundation2 marks

A dog-walking side business has fixed weekly costs of $60 (insurance and transport) and spends $2 per walk in treats and waste bags. Each walk is charged at $8. Let

n

be the number of walks in a week. How many walks are needed to break even?

Show worked solution →

Write the cost and revenue functions. The fixed cost is $60 and the variable cost is $2 per walk, so

C = 60 + 2n, \qquad R = 8n.

Solve for break-even. Break-even is where $C = R$ :

60 + 2n = 8n \implies 60 = 6n \implies n = 10.

Answer: the business breaks even at $10$ walks per week.

foundation2 marks

At a garage sale a student sells potted seedlings. The pots, soil and labels cost $1 per seedling, and each seedling sells for $6. (a) State the contribution margin per seedling and explain what it means. (b) If the fixed cost for the day is $50, find the break-even quantity.

Show worked solution →

Find the contribution margin. The contribution margin is the selling price minus the variable cost:

p - v = 6 - 1 = 5 \text{ dollars per seedling}.

This means that after paying the $1 cost of a seedling, $5 of its $6 price is left over to chip away at the fixed cost, and then to add to profit.

Find the break-even quantity. Divide the fixed cost by the contribution margin:

n_{\text{be}} = \frac{50}{5} = 10.

Answer: the margin is $5 per seedling and the stall breaks even at $10$ seedlings.

foundation3 marks

A student sells handmade bracelets at a market. The stall costs $90 for the day (fixed cost) and each bracelet costs $4 in materials. Bracelets sell for $10 each. Let

n

be the number of bracelets sold. (a) Write the cost function

C

and the revenue function

R

. (b) Find the break-even quantity. (c) Find the profit if

40

bracelets are sold.

Show worked solution →

Set up the functions. The fixed cost is $90 and the variable cost is $4 per bracelet, so cost is $C = 90 + 4n$ . Revenue starts at the origin at $10 each, so $R = 10n$ .

C = 90 + 4n, \qquad R = 10n

Find the break-even quantity. Break-even is where $C = R$ :

90 + 4n = 10n \implies 90 = 6n \implies n = 15.

So the stall breaks even at $15$ bracelets.

Find the profit at $40$ bracelets. Substitute $n = 40$ into revenue and cost, then subtract:

R = 10 \times 40 = 400, \qquad C = 90 + 4 \times 40 = 90 + 160 = 250

P = R - C = 400 - 250 = 150.

Check. The profit shortcut agrees: $P = (10 - 4)\times 40 - 90 = 240 - 90 = 150$ . So the profit is $150.

foundation3 marks

A home business makes jars of jam. Fixed costs are $84 per batch and each jar costs $3 in ingredients and packaging. Jars sell for $9. (a) How many jars must be sold to break even? (b) What is the profit if

30

jars are sold?

Show worked solution →

Write the cost and revenue functions. Let $n$ be the number of jars sold. The fixed cost is $84 and the variable cost is $3 per jar, so

C = 84 + 3n, \qquad R = 9n.

Find the break-even quantity. Set $C = R$ :

84 + 3n = 9n \implies 84 = 6n \implies n = 14.

So $14$ jars must be sold to break even.

Find the profit at $30$ jars. Substitute $n = 30$ :

R = 9 \times 30 = 270, \qquad C = 84 + 3 \times 30 = 84 + 90 = 174

P = R - C = 270 - 174 = 96.

Check. Using the contribution margin $9 - 3 = 6$ : $P = 6 \times 30 - 84 = 180 - 84 = 96$ . The profit is $96.

core4 marks

A small business pours scented candles. Fixed monthly costs are $200 and each candle costs $6.50 to make. Candles sell for $15 each. (a) Find the break-even quantity, giving a sensible whole-number answer and explaining your rounding. (b) Find the profit if

60

candles are sold in a month.

Show worked solution →

Set up and find the contribution margin. Let $n$ be candles sold per month, so $C = 200 + 6.5n$ and $R = 15n$ . The contribution margin is

p - v = 15 - 6.50 = 8.50 \text{ dollars per candle}.

Calculate the break-even quantity. Divide the fixed cost by the contribution margin:

n_{\text{be}} = \frac{200}{8.50} \approx 23.53.

Round up and justify. You cannot sell part of a candle, and at $23$ candles the fixed costs are not quite covered (still a loss), so round up to $24$ candles.

Find the profit at $60$ candles. Substitute $n = 60$ :

R = 15 \times 60 = 900, \qquad C = 200 + 6.50 \times 60 = 200 + 390 = 590

P = R - C = 900 - 590 = 310.

Check. Shortcut: $P = 8.50 \times 60 - 200 = 510 - 200 = 310$ . So break-even is $24$ candles and the profit at $60$ candles is $310.

core4 marks

A mobile car-wash service has fixed weekly costs of $480 (van, equipment, insurance) and spends $3 in water and supplies per wash. Each wash is charged at $15. (a) Find the break-even number of washes per week. (b) How many washes are needed to make a weekly profit of $600?

Show worked solution →

Set up the functions. Let $n$ be the number of washes per week. Then $C = 480 + 3n$ and $R = 15n$ , and the contribution margin is $15 - 3 = 12$ dollars per wash.

Find the break-even quantity. Set $C = R$ (or divide fixed cost by the margin):

480 + 3n = 15n \implies 480 = 12n \implies n = 40.

So the service breaks even at $40$ washes per week.

Find the washes needed for $600 profit. Profit is $P = (p - v)n - F$ . Set $P = 600$ :

12n - 480 = 600 \implies 12n = 1080 \implies n = 90.

Check. At $90$ washes, $R = 15 \times 90 = 1350$ and $C = 480 + 3 \times 90 = 750$ , so $P = 1350 - 750 = 600$ . The service needs $90$ washes a week.

core4 marks

A school fete lemonade stall has fixed costs of $45 (cups, ice, signage) and a variable cost of $1.25 per cup. Each cup sells for $3.50. (a) Find the break-even quantity. (b) State the contribution margin and explain what it means in this context. (c) By how much does profit increase if sales rise from

60

cups to

80

cups?

Show worked solution →

Set up the functions. Let $n$ be cups sold. Then $C = 45 + 1.25n$ and $R = 3.5n$ .

Find the break-even quantity. The contribution margin is $3.50 - 1.25 = 2.25$ dollars per cup, so

n_{\text{be}} = \frac{45}{2.25} = 20 \text{ cups}.

Explain the contribution margin. The contribution margin is $2.25 per cup: after paying the $1.25 cost of a cup, $2.25 of its $3.50 price is left over. Each cup first chips away at the $45 fixed cost, and once break-even is passed each cup adds its full $2.25 to profit.

Find the rise in profit from $60$ to $80$ cups. Using $P = 2.25n - 45$ :

P_{60} = 2.25 \times 60 - 45 = 135 - 45 = 90, \qquad P_{80} = 2.25 \times 80 - 45 = 180 - 45 = 135.

The increase is $135 - 90 = 45$ , i.e. $45.

Check. The same answer comes straight from the margin: $20$ extra cups at $2.25 each is $20 \times 2.25 = 45$ , i.e. $45.

core4 marks

A tutoring group prints revision booklets to sell at a study day. Fixed costs are $96 (printer hire and binding) and each booklet costs $2 in paper and toner. Booklets sell for $8. (a) Find the break-even number of booklets. (b) If only

10

booklets are sold, find the profit and say whether the group makes a profit or a loss.

Show worked solution →

Set up the functions. Let $n$ be the number of booklets sold. The fixed cost is $96 and the variable cost is $2 per booklet, so

C = 96 + 2n, \qquad R = 8n.

Find the break-even quantity. The contribution margin is $8 - 2 = 6$ dollars per booklet, so set $C = R$ :

96 + 2n = 8n \implies 96 = 6n \implies n = 16.

So the group breaks even at $16$ booklets.

Find the profit at $10$ booklets. Substitute $n = 10$ into revenue and cost, then subtract:

R = 8 \times 10 = 80, \qquad C = 96 + 2 \times 10 = 96 + 20 = 116

P = R - C = 80 - 116 = -36.

Since $10$ is below the break-even quantity of $16$ , the profit is negative, which is a loss.

Answer: break-even is $16$ booklets, and at $10$ booklets the group makes a loss of $36.

exam5 marks

A grower sells potted natives at a Sunday market. Fixed costs are $360 (stall fee, van and racks) and each plant costs $5 to grow and pot. Plants sell for $14. (a) Write the cost and revenue functions and find the break-even quantity. (b) Find the profit if

100

plants are sold. (c) The market operator raises the stall fee, so fixed costs rise to $450. Find the new break-even quantity and state by how many plants it changes.

Show worked solution →

Write the functions. Let $n$ be the number of plants sold. The fixed cost is $360 and the variable cost is $5 per plant, so

C = 360 + 5n, \qquad R = 14n.

Find the break-even quantity. The contribution margin is $14 - 5 = 9$ dollars per plant, so set $C = R$ :

360 + 5n = 14n \implies 360 = 9n \implies n = 40.

So the stall breaks even at $40$ plants.

Find the profit at $100$ plants. Substitute $n = 100$ :

R = 14 \times 100 = 1400, \qquad C = 360 + 5 \times 100 = 360 + 500 = 860

P = R - C = 1400 - 860 = 540.

Recalculate break-even with the higher fixed cost. The margin is unchanged at $9 per plant, but the fixed cost is now $450:

n_{\text{be}}' = \frac{450}{9} = 50.

The break-even quantity rises from $40$ to $50$ , an increase of $10$ plants.

Answer: break-even is $40$ plants, profit at $100$ plants is $540, and the higher stall fee lifts break-even to $50$ plants, a rise of $10$ .

exam6 marks

A food truck has fixed daily costs of $650 (site fee, gas, wages) and each meal costs $4.50 in ingredients. Meals sell for $12. (a) Find the break-even number of meals per day, rounding appropriately. (b) Find the profit if

150

meals are sold in a day. (c) The owner raises the price to $13. Find the new break-even quantity and state by how many meals it changes. Comment on one limitation of this model.

Show worked solution →

Set up the functions. Let $n$ be meals sold per day. Then $C = 650 + 4.5n$ and $R = 12n$ , with contribution margin $12 - 4.50 = 7.50$ dollars per meal.

Find the break-even quantity. Divide the fixed cost by the margin:

n_{\text{be}} = \frac{650}{7.50} \approx 86.67.

At $86$ meals the truck is still short of covering costs, so round up to $87$ meals.

Find the profit at $150$ meals. Substitute $n = 150$ :

R = 12 \times 150 = 1800, \qquad C = 650 + 4.50 \times 150 = 650 + 675 = 1325

P = R - C = 1800 - 1325 = 475.

Recalculate break-even at the new price. Raising the price to $12 plus $1, i.e. $13, gives a new margin of $13 - 4.50 = 8.50$ dollars per meal:

n_{\text{be}}' = \frac{650}{8.50} \approx 76.47, \quad \text{round up to } 77 \text{ meals}.

The break-even quantity falls from $87$ to $77$ , a drop of $10$ meals.

State the limitation. The model assumes the higher price does not change how many meals customers buy; in reality demand may fall at $13, which the linear model cannot predict.

Check. At the original price, profit at $150$ meals by the shortcut is $7.50 \times 150 - 650 = 1125 - 650 = 475$ , i.e. $475, matching part (b).

exam6 marks

A pop-up coffee cart trades for one weekend at a Melbourne festival. Fixed costs are $540 (site fee, urn and gas), each cup costs $2.50 in coffee, milk and a cup, and a cup sells for $6. (a) Find the break-even number of cups, rounding appropriately and justifying your rounding. (b) Find the profit if

300

cups are sold. (c) On the cost-revenue graph, state which line is higher just to the left of the break-even point and what this region represents. (d) How many cups must be sold to make a profit of $700?

Show worked solution →

Set up the functions. Let $n$ be the number of cups sold. Then $C = 540 + 2.5n$ and $R = 6n$ , with contribution margin $6 - 2.50 = 3.50$ dollars per cup.

Find the break-even quantity. Divide the fixed cost by the margin:

n_{\text{be}} = \frac{540}{3.50} \approx 154.29.

At $154$ cups the cart is still about a dollar short of covering its fixed cost, so round up to $155$ cups.

Find the profit at $300$ cups. Substitute $n = 300$ :

R = 6 \times 300 = 1800, \qquad C = 540 + 2.50 \times 300 = 540 + 750 = 1290

P = R - C = 1800 - 1290 = 510.

Read the graph to the left of break-even. Just to the left of the crossing the cost line is higher than the revenue line, so $C > R$ . This region represents a loss, because the cart has not yet sold enough cups to cover its fixed cost.

Find the cups needed for $700 profit. Use $P = (p - v)n - F$ with $P = 700$ :

3.50n - 540 = 700 \implies 3.50n = 1240 \implies n = \frac{1240}{3.50} \approx 354.29.

Round up to $355$ cups, since at $354$ the profit is $699, just short of the target.

Answer: break-even is $155$ cups, profit at $300$ cups is $510, the loss region lies to the left of break-even where cost exceeds revenue, and $355$ cups are needed for a $700 profit.

exam7 marks

A start-up prints custom T-shirts. Set-up and machine hire cost $920 for a print run (fixed cost), and each shirt costs $7.50 in blanks and ink. Shirts sell for $25. (a) Write the cost and revenue functions. (b) Find the break-even quantity, justifying the rounding. (c) Find the profit if

100

shirts are sold. (d) How many shirts must be sold to make a profit of $1000?

Show worked solution →

Write the functions. Let $n$ be the number of shirts sold. The fixed cost is $920 and the variable cost is $7.50 per shirt, so

C = 920 + 7.50n, \qquad R = 25n.

Find the break-even quantity. The contribution margin is $25 - 7.50 = 17.50$ dollars per shirt, so

n_{\text{be}} = \frac{920}{17.50} \approx 52.57.

Round up to $53$ shirts, since at $52$ shirts the $920 fixed cost is not yet fully covered.

Find the profit at $100$ shirts. Substitute $n = 100$ :

R = 25 \times 100 = 2500, \qquad C = 920 + 7.50 \times 100 = 920 + 750 = 1670

P = R - C = 2500 - 1670 = 830.

Find the shirts needed for $1000 profit. Use $P = (p - v)n - F$ with $P = 1000$ :

17.50n - 920 = 1000 \implies 17.50n = 1920 \implies n = \frac{1920}{17.50} \approx 109.71.

Round up to $110$ shirts (at $109$ the profit is just under $1000).

Check. At $110$ shirts the profit is $17.50 \times 110 - 920 = 1925 - 920 = 1005$ , i.e. $1005, which clears the $1000 target.

Practise this

Sources & how we know this

Mathematics Standard Stage 6 Syllabus (2017) — NESA
Mathematics Standard HSC exam papers — NESA