Break-even is where revenue catches cost. Below it a business loses money, above it profits, and the whole analysis is one pair of simultaneous equations.
Today's hook, Every business plan contains a break-even calculation. It is a pair of straight lines and their intersection, and it is the clearest example in the course of algebra answering a real decision.
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Recall, your gut answer first
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A stall pays $\$200$ for the site and $\$3$ for each drink it sells, then sells each drink for $\$5$. Roughly how many drinks must it sell before it stops losing money? Estimate before calculating.
Before you work it out, what is your instinct? Write it down, then check it against the lesson.
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Where revenue catches cost
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Where revenue catches cost
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Cost has a fixed part and a per-unit part; revenue is usually price times quantity. The break-even point is where they are equal, so profit is zero. Below it the business runs at a loss, above it at a profit.
Setting $C = R$ and setting $P = 0$ are the same equation. Use whichever the question phrases.
Round up for break-even
If break-even is 100.4 items, selling 100 still loses money. The business must sell 101.
A quadratic model has a maximum
Revenue models where price falls as quantity rises are quadratic, and their vertex is the optimum. That is a modelling question, not calculus.
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What you'll master
Meet the destination, bring back what you already know, and gather the terms and formulas this lesson leans on.
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What you'll master
Know
Key facts
Cost is typically $C = $ fixed cost $+$ unit cost $\times$ quantity.
Revenue is typically $R = $ price $\times$ quantity.
Break-even is where $C = R$, equivalently where profit $P = R - C$ is zero.
A quadratic model reaches its optimum at its vertex, found by completing the square or $x = -\frac{b}{2a}$.
Understand
Concepts
Why break-even is an intersection of two lines rather than a single calculation.
Why a fractional break-even quantity must be rounded up rather than to nearest.
Why a revenue model becomes quadratic when the price depends on the quantity.
Can do
Skills
Build cost and revenue models and find the break-even point.
Interpret the regions either side of break-even as loss and profit.
Model a practical situation with a quadratic and find its optimum from the vertex.
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Key terms
Fixed costThe part of the cost that does not change with quantity. Like this: a $\$200$ stall fee is paid whether you sell one drink or a thousand.
Variable costThe part that rises with each additional unit. Like this: $\$3$ per drink means variable cost $3n$ for $n$ drinks.
RevenueMoney coming in, usually price times quantity. Like this: selling $n$ drinks at $\$5$ gives $R = 5n$.
Break-even pointThe quantity where revenue equals cost, so profit is zero. Like this: $5n = 200 + 3n$ gives $n = 100$ drinks.
ProfitRevenue minus cost, negative below break-even and positive above it. Like this: $P = 5n - (200 + 3n) = 2n - 200$.
OptimumThe best achievable value in a model, at the vertex when the model is quadratic. Like this: a revenue model $R = -2p^2 + 40p$ peaks at $p = 10$.
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Cost, revenue and the break-even point
Work through the core explanation before applying it.
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Cost, revenue and the break-even point
core concept
Write both models over the same variable. For the drinks stall with $n$ drinks: $C = 200 + 3n$ and $R = 5n$.
Break-even is where they meet: $5n = 200 + 3n$, so $2n = 200$ and $n = 100$ drinks. At that point both cost and revenue are $\$500$.
Equivalently, profit $P = R - C = 5n - 200 - 3n = 2n - 200$, and setting $P = 0$ gives the same $n = 100$. The two routes are the same equation rearranged.
The gradient of the profit line is the margin per unit. Here $P = 2n - 200$: each drink contributes $\$2$ towards covering the $\$200$ fixed cost, which is why 100 drinks are needed.
Quick check: a stall has $C = 200 + 3n$ and $R = 5n$. What is the break-even quantity?
Write cost and revenue over the same variable, then set them equal. Break-even is that intersection, and it is the same as setting profit $R - C$ to zero. The gradient of the profit line is the contribution each unit makes towards the fixed cost.
Pause, copy both models for the drinks stall, both routes to $n = 100$, and the margin-per-unit interpretation, into your book.
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Reading the regions, and rounding correctly
core concept
We just saw how to find the break-even quantity. That raises a question: what does the number actually tell the business, and what if it is not a whole number? This card answers it → it splits the range into a loss region and a profit region, and a fractional answer must be rounded up.
Below break-even, cost exceeds revenue and the business runs at a loss. Above it, revenue exceeds cost and it profits. The break-even quantity is the boundary.
If break-even comes out at 100.4 items, selling 100 still leaves a loss, because 100 is below the boundary. The business must sell **101**. Round up, not to nearest.
A question asking "how many must be sold to make a profit" wants the first whole number strictly above break-even, which is 101 here, not 100.4 and not 100.
Round to nearest is wrong here. 100.4 rounds to 100 by the usual rule, but 100 items still loses money. Context beats convention: the answer is the first quantity that actually achieves the goal.
Fill the blank: if break-even is 100.4 items, the business must sell at least items to profit.
Break-even is the boundary between a loss region below and a profit region above. A fractional break-even must be rounded up, because the whole number below it still loses money. Round to nearest is wrong in this context.
Pause, copy the loss and profit regions, and the round-up rule with the 100.4 example, into your book.
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Quadratic models and their optimum
core concept
We just saw break-even with straight-line models. That raises a question: what happens when the price itself depends on how much you sell, so revenue is no longer linear? This card answers it → revenue becomes quadratic, and the vertex is the optimum.
If raising the price reduces the number sold, revenue is price times quantity where both depend on each other, which produces a quadratic.
Suppose at price $p$ dollars a stall sells $40 - 2p$ items. Revenue is $R = p(40 - 2p) = 40p - 2p^2$, a concave-down parabola.
Its maximum is at the vertex: $p = -\frac{40}{2(-2)} = 10$, giving $R = 40(10) - 2(100) = 200$. So the revenue-maximising price is $\$10$ and the maximum revenue is $\$200$.
No calculus needed. The syllabus places this in MAV-11-01, so the optimum comes from the vertex, using $x = -\frac{b}{2a}$ or completing the square. Differentiation for optimisation is Year 12.
Which is NOT a valid way to find the optimum of a quadratic model at this stage?
When price and quantity depend on each other, revenue is quadratic and concave down, so its vertex is the maximum. Find it with $x = -\frac{b}{2a}$ or by completing the square, then substitute back for the optimal value. Calculus is Year 12.
Pause, copy the worked $R = p(40-2p)$ giving an optimum price of $\$10$, and the note that the vertex, not differentiation, is the Year 11 method, into your book.
Worked examples · 3 in a row, reveal as you go
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Work examples end to end
Follow the reasoning through complete worked solutions.
PROBLEM 1 · BREAK-EVEN
A stall pays $\$200$ for the site and $\$3$ per drink, selling each at $\$5$. Find the break-even quantity and the profit on 250 drinks.
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$C = 200 + 3n$ and $R = 5n$, where $n$ is the number of drinks
Both models over the same variable.
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Break-even: $5n = 200 + 3n$, so $2n = 200$ and $n = 100$
Profit is the margin per drink times quantity, less the fixed cost.
PROBLEM 2 · ROUNDING AT BREAK-EVEN
A workshop has fixed costs of $\$340$ and makes $\$8$ profit per item. How many items must it sell to make a profit?
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Profit $P = 8n - 340$
Margin per item times quantity, less fixed costs.
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Break-even: $8n = 340$, so $n = 42.5$
Not a whole number.
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$42$ items still loses money, so it must sell at least $43$
Round up, since the goal is to be strictly above break-even.
PROBLEM 3 · QUADRATIC MODEL
At a price of $p$ dollars, a stall sells $40 - 2p$ items. Find the price that maximises revenue and the maximum revenue.
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$R = p(40 - 2p) = 40p - 2p^2$
Revenue is price times quantity.
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$a = -2$, $b = 40$, so the vertex is at $p = -\dfrac{40}{2(-2)} = 10$
Concave down, so the vertex is a maximum.
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$R = 40(10) - 2(10)^2 = 400 - 200 = 200$ dollars
Substitute back for the maximum revenue.
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Quick-fire practice
Work through the core explanation before applying it.
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Quick-fire practice
+10 XP
If $C = 150 + 4n$ and $R = 10n$, find the break-even quantity.
A business has $P = 6n - 400$. How many units must it sell to profit?
Write the revenue model if $n$ items sell at $\$7$ each.
Find the vertex of $R = 30p - 3p^2$.
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Revisit your estimate
Run the quick drill and copy the summary into your book.
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Revisit your estimate
At the start you estimated how many drinks the stall must sell. Compare your estimate with the worked answer of 100, and explain in one sentence why the margin per drink, rather than the selling price, is what determines it.
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Multiple choice
Answer the drill bank and rate your confidence.
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Multiple choice
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Short answer
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Short answer
ApplyBand 44 marks
Q1. A club pays $\$480$ in fixed costs and $\$6$ per shirt, selling each shirt for $\$22$. Find the break-even quantity and the profit if 90 shirts are sold. (4 marks)
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ApplyBand 54 marks
Q2. At a price of $p$ dollars a stall sells $60 - 3p$ items. Find the price that maximises revenue, and the maximum revenue. (4 marks)
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UnderstandBand 42 marks
Q3. A break-even calculation gives 87.2 items. Explain why the answer to "how many must be sold to make a profit" is 88 and not 87. (2 marks)
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📖 Comprehensive answers (click to reveal)
Practice 1: $10n = 150 + 4n$ gives $6n = 150$, so $n = 25$. Practice 2: $6n = 400$ gives $n \approx 66.7$, so 67 units. Practice 3: $R = 7n$. Practice 4: $p = -\frac{30}{2(-3)} = 5$, $R = 150 - 75 = 75$.
Q1 (4 marks): $C = 480 + 6s$ and $R = 22s$, where $s$ is the number of shirts [1]. $22s = 480 + 6s$, so $16s = 480$ and $s = 30$ [1]. Profit $P = 16s - 480$ [1]. At $s = 90$: $P = 1440 - 480 = 960$ dollars [1].
Q2 (4 marks): $R = p(60 - 3p) = 60p - 3p^2$ [1]. $a = -3$, $b = 60$, so $p = -\frac{60}{2(-3)} = 10$ [1]. The parabola is concave down, so this is a maximum [1]. $R = 60(10) - 3(100) = 600 - 300 = 300$ dollars [1].
Q3 (2 marks): Break-even at 87.2 means 87.2 items is where revenue exactly equals cost, so any quantity below it runs at a loss [1]. Selling 87 items is below 87.2 and therefore still loses money, so the first whole number that produces a profit is 88 [1].
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Review and finish
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Boss battle · Break-even Blitz
earn bronze · silver · gold
Build cost and revenue models, find break-even points and optimise quadratic models. Beat the boss to bank a tier, gold (90% + speed), silver (75%), or bronze (50%). Replays welcome.