Interpreting intersections and break-even points | Year 12

01 / LEARN

Understand the idea.

Equal values for the same input

An intersection compares two outputs at the same input. For two prices, it gives an equal-cost quantity and the shared cost. For revenue and cost, it gives the quantity at which the business neither gains nor loses money.

To compare either side, subtract the models or test suitable inputs. A low fixed fee can be outweighed by a high per-use charge, so the cheapest choice may change as usage increases.

Break-even is not maximum profit

Profit is revenue minus the entire cost, including fixed charges. Break-even means $P = 0$ ; positive profit requires $P > 0$ . Keep this distinction when choosing the first profitable whole-number quantity.

A break-even calculation does not identify maximum profit. To find a practical maximum, the model also needs limits such as capacity, stock or demand; otherwise increasing sales may keep increasing the calculated profit.

P = R - C

02 / WORKED EXAMPLES

See the method in action.

EXAMPLE 01

Choosing a plan

Plan A costs AUD 25 plus AUD 3 per use. Plan B costs AUD 61 plus AUD 1 per use. Compare them.

Worked solution

1
Set the plan costs equal.
$25 + 3 n = 61 + n$
Equal costs occur where the two pricing rules intersect.
2
Find the equal-use threshold and cost.
$\begin{aligned} 2 n & = 36 \\ n & = 18 \\ 25 + 3 (18) & = 61 + 18 = 79 \end{aligned}$
Both plans cost AUD 79 at 18 uses.
3
Compare the difference for all allowed uses.
$A - B = (25 + 3 n) - (61 + n) = 2 n - 36$
A negative difference means A costs less; a positive difference means B costs less.
4
State the whole-number comparison.
$\begin{aligned} 0 \leq n < 18 & : A < B \\ n = 18 & : A = B \\ n > 18 & : B < A \end{aligned}$
Use counts are whole numbers, and the sign changes at the equal-cost threshold.

Answer

A is cheaper below 18 uses, B above 18; equal at 18.

EXAMPLE 02

An event

Tickets sell for AUD 15. Costs are AUD 120 plus AUD 7 per attendee. Find break-even and first profitable attendance.

Worked solution

1
Write revenue and total cost.
$R = 15 n, C = 120 + 7 n$
Each attendee pays AUD 15 and adds AUD 7 to the fixed AUD 120 cost.
2
Form the profit rule.
$\begin{aligned} P & = R - C \\ = 15 n - (120 + 7 n) \\ = 8 n - 120 \end{aligned}$
Each ticket contributes AUD 8 towards the fixed cost before a profit is made.
3
Solve for break-even.
$\begin{aligned} 8 n - 120 & = 0 \\ n & = 15 \end{aligned}$
Break-even is zero profit, so 15 attendees exactly cover the costs.
4
Find the first profitable attendance.
$\begin{aligned} 8 n - 120 & > 0 \\ n & > 15 \\ P (16) & = 8 \end{aligned}$
The first allowed integer above 15 is 16, giving AUD 8 profit.

Answer

Break-even at 15; first profit at 16 attendees.

03 / YOUR TURN

Try it yourself.

Check your understanding

Revenue is $9 n$ and cost is $52 + 5 n$ . Find break-even.

Show answer & reasoning

13 items; both are AUD 117.

Solve $9 n = 52 + 5 n$ to obtain $4 n = 52$ and $n = 13$ . Revenue is then $9 (13) = 117$ and cost is $52 + 5 (13) = 117$ , both in AUD.

Keep practising

6 worksheet questions · 12 marks

Open worksheet Worked solutions

Read the questions online

Find equal cost for $A = 40 + 2 n$ and $B = 16 + 5 n$ .
2 marks
For the same plans, which is cheaper at 12 uses?
2 marks
A product sells for AUD 28 and costs AUD 210 fixed plus AUD 13 per unit. Find break-even.
2 marks
Find profit on 20 products in Question 3.
2 marks
Profit is $P = 12 n - 95$ . Find the smallest profitable whole-number attendance.
2 marks
Why do $A = 30 + 4 n$ and $B = 80 + 4 n$ never have equal cost?
2 marks

Read the worked solutions

Equivalent correct methods earn the corresponding marks. Include units where asked.

Find equal cost for $A = 40 + 2 n$ and $B = 16 + 5 n$ .
2 marks
$40 + 2 n = 16 + 5 n$ gives $n = 8$ .
8 uses, AUD 56.
For the same plans, which is cheaper at 12 uses?
2 marks
$A = 64$ and $B = 76$ .
A, by AUD 12.
A product sells for AUD 28 and costs AUD 210 fixed plus AUD 13 per unit. Find break-even.
2 marks
$28 n = 210 + 13 n$ gives $15 n = 210$ .
14 products.
Find profit on 20 products in Question 3.
2 marks
$P = 15 n - 210$ ; at 20 this is $300 - 210 = 90$ .
AUD 90.
Profit is $P = 12 n - 95$ . Find the smallest profitable whole-number attendance.
2 marks
$n > \frac{95}{12}$ , about 7.92; the first permitted integer is 8.
8 people.
Why do $A = 30 + 4 n$ and $B = 80 + 4 n$ never have equal cost?
2 marks
The gradients match and $B - A = 50$ for all $n$ .
B always costs AUD 50 more.

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Curriculum reference & lesson scope

SACE Board of South Australia

Stage 2 General Mathematics: Modelling with Linear Relationships, a compulsory school-assessed topic. The external examination scope differs from the full course.

Official curriculum source ↗

In this lesson

Understand the idea.

Equal values for the same input

Break-even is not maximum profit

See the method in action.

Choosing a plan

Set the plan costs equal.

Find the equal-use threshold and cost.

Compare the difference for all allowed uses.

State the whole-number comparison.

An event

Write revenue and total cost.

Form the profit rule.

Solve for break-even.

Find the first profitable attendance.

Try it yourself.

Check your understanding

Keep practising