Cost, Revenue, Profit, and Break-Even | Malone Math

The Central Idea

Let $x$ represent the number of units produced or sold.

A simple linear cost model has the form

\[ \boxed{C(x)=F+vx}, \]

where:

$F$ is fixed cost;
$v$ is variable cost per unit;
$x$ is the number of units.

If the selling price per unit is $p$, revenue is

\[ \boxed{R(x)=px}. \]

Profit is

\[ \boxed{P(x)=R(x)-C(x)}. \]

Fixed Cost

A fixed cost does not depend on the number of units produced within the model.

Examples may include:

rent;
equipment leases;
licenses;
certain salaried costs;
setup costs.

\[ C(x)=F+vx, \]

the fixed cost is the intercept

\[ \boxed{C(0)=F}. \]

Variable Cost

A variable cost changes with the number of units produced.

If each unit costs $v$ dollars to produce, then producing $x$ units creates total variable cost

\[ \boxed{vx}. \]

Therefore,

\[ \boxed{C(x)=F+vx}. \]

Example: Build a Cost Function

A company has fixed costs of $\$2{,}000$ and spends $\$12$ to produce each unit.

The cost function is

\[ \boxed{C(x)=2000+12x}. \]

For $100$ units,

\[ C(100)=2000+12(100). \] \[ C(100)=3200. \]

Therefore,

\[ \boxed{\$3{,}200}. \]

Revenue

Revenue is the money received from sales.

If each unit sells for constant price $p$, then

\[ \boxed{R(x)=px}. \]

This model assumes all $x$ units are sold at the same price.

Example: Build a Revenue Function

A product sells for $\$30$ per unit.

Revenue from selling $x$ units is

\[ \boxed{R(x)=30x}. \]

At $100$ units,

\[ R(100)=30(100)=3000. \]

Therefore,

\[ \boxed{\$3{,}000}. \]

Profit

Profit is revenue minus cost:

\[ \boxed{P(x)=R(x)-C(x)}. \]

If profit is positive, revenue exceeds cost.

If profit is negative, the business has a loss under the model.

If profit is zero, the business is at break-even.

Deriving the Linear Profit Function

Suppose

\[ C(x)=F+vx \]

and

\[ R(x)=px. \]

Then

\[ P(x)=px-(F+vx). \]

Simplify:

\[ P(x)=px-F-vx. \] \[ \boxed{ P(x)=(p-v)x-F }. \]

Contribution Margin per Unit

The difference

\[ p-v \]

is called the contribution margin per unit in this simple model.

It measures how much each additional unit contributes toward covering fixed cost and then generating profit.

\[ \boxed{m=p-v}. \]

Profit Using Contribution Margin

\[ m=p-v, \]

then the profit function can be written

\[ \boxed{P(x)=mx-F}. \]

This form makes the role of each additional unit especially clear.

Example: Profit Function

Fixed cost is $\$4{,}000$, variable cost is $\$18$ per unit, and the selling price is $\$50$ per unit.

Cost:

\[ C(x)=4000+18x. \]

Revenue:

\[ R(x)=50x. \]

Profit:

\[ P(x)=50x-(4000+18x). \] \[ \boxed{P(x)=32x-4000}. \]

Evaluate Profit at a Given Sales Level

Continue with

\[ P(x)=32x-4000. \]

At $150$ units,

\[ P(150)=32(150)-4000. \] \[ P(150)=4800-4000. \] \[ P(150)=800. \]

Therefore,

\[ \boxed{\$800\text{ profit}}. \]

Loss as Negative Profit

\[ P(x)<0, \]

then the model indicates a loss.

Using

\[ P(x)=32x-4000, \]

at $100$ units,

\[ P(100)=3200-4000=-800. \]

Therefore,

\[ \boxed{\$800\text{ loss}}. \]

Break-Even

The break-even point occurs when revenue equals cost:

\[ \boxed{R(x)=C(x)}. \]

Equivalently,

\[ \boxed{P(x)=0}. \]

Deriving the Break-Even Quantity

Start with

\[ px=F+vx. \]

Subtract $vx$:

\[ (p-v)x=F. \]

\[ p-v>0, \]

then

\[ \boxed{ x=\frac{F}{p-v} }. \]

Example: Find the Break-Even Quantity

Suppose fixed cost is $\$4{,}000$, variable cost is $\$18$ per unit, and price is $\$50$ per unit.

Contribution margin:

\[ 50-18=32. \]

Break-even quantity:

\[ x=\frac{4000}{32}=125. \]

Therefore,

\[ \boxed{125\text{ units}}. \]

Check the Break-Even Example

Revenue at $125$ units:

\[ R(125)=50(125)=6250. \]

Cost:

\[ C(125)=4000+18(125). \] \[ C(125)=4000+2250=6250. \]

Therefore,

\[ R(125)=C(125), \]

so profit is zero.

Break-Even Revenue

Once the break-even quantity $x_b$ is known, break-even revenue is

\[ \boxed{ R(x_b)=px_b }. \]

In the previous example,

\[ R(125)=6250. \]

Therefore, break-even revenue is

\[ \boxed{\$6{,}250}. \]

Graphical Meaning of Break-Even

In a graph of cost and revenue against quantity:

the cost graph begins at fixed cost $F$;
its slope is variable cost $v$;
the revenue graph begins at $0$;
its slope is selling price $p$.

Break-even occurs where the two lines intersect.

Profit as Vertical Difference

At any quantity $x$,

\[ P(x)=R(x)-C(x). \]

Graphically, profit is the vertical difference between the revenue and cost lines.

Above break-even, revenue lies above cost.

Below break-even, cost lies above revenue.

Target Profit

A business may want to know how many units must be sold to earn a desired profit $P_t$.

Start with

\[ P(x)=(p-v)x-F. \]

Set

\[ P(x)=P_t. \]

Then

\[ (p-v)x-F=P_t. \]

Therefore,

\[ \boxed{ x=\frac{F+P_t}{p-v} }. \]

Example: Units Needed for a Target Profit

A product has fixed cost $\$5{,}000$, variable cost $\$20$ per unit, and selling price $\$45$ per unit. How many units must be sold for a $\$2{,}500$ profit?

Contribution margin:

\[ 45-20=25. \]

Use

\[ x=\frac{5000+2500}{25}. \] \[ x=300. \]

Therefore,

\[ \boxed{300\text{ units}}. \]

Target Loss

A negative target profit represents a specified loss.

For example, if

\[ P_t=-1000, \]

then

\[ x=\frac{F-1000}{p-v} \]

gives the quantity at which the loss has narrowed to $\$1{,}000$, assuming the model remains valid.

Unknown Selling Price

Suppose fixed cost, variable cost, quantity, and target profit are known, but selling price $p$ is unknown.

Use

\[ px-(F+vx)=P_t. \]

Solve for $p$:

\[ px=F+vx+P_t. \] \[ \boxed{ p=\frac{F+vx+P_t}{x} }. \]

Example: Required Selling Price

Fixed cost is $\$3{,}000$, variable cost is $\$15$ per unit, and the business expects to sell $200$ units. What price per unit produces a $\$2{,}000$ profit?

\[ p= \frac{3000+15(200)+2000}{200}. \] \[ p= \frac{3000+3000+2000}{200}. \] \[ p=\frac{8000}{200}=40. \]

Therefore,

\[ \boxed{\$40\text{ per unit}}. \]

Unknown Variable Cost

If $v$ is unknown, start with

\[ P_t=px-F-vx. \]

Rearrange:

\[ vx=px-F-P_t. \]

Therefore,

\[ \boxed{ v=\frac{px-F-P_t}{x} }. \]

Example: Required Variable Cost

A product sells for $\$60$ per unit. Fixed cost is $\$6{,}000$. At $300$ units, the desired profit is $\$3{,}000$. What variable cost per unit is allowed?

\[ v= \frac{60(300)-6000-3000}{300}. \] \[ v= \frac{18000-9000}{300}. \] \[ v=30. \]

Therefore,

\[ \boxed{\$30\text{ per unit}}. \]

Unknown Fixed Cost

From

\[ P_t=(p-v)x-F, \]

solve for fixed cost:

\[ \boxed{ F=(p-v)x-P_t }. \]

Example: Infer Fixed Cost

A product sells for $\$35$, costs $\$20$ per unit to produce, and generates $\$1{,}500$ profit at $200$ units.

Then

\[ F=(35-20)(200)-1500. \] \[ F=3000-1500. \] \[ F=1500. \]

Therefore,

\[ \boxed{\$1{,}500}. \]

Contribution Margin Ratio

The contribution margin per unit is

\[ m=p-v. \]

Relative to the selling price, the contribution margin ratio is

\[ \boxed{ \frac{p-v}{p} }. \]

This describes the fraction of each sales dollar available to cover fixed costs and profit in the simple model.

Example: Contribution Margin Ratio

A product sells for $\$50$ and has variable cost $\$30$.

Contribution margin:

\[ 50-30=20. \]

Contribution margin ratio:

\[ \frac{20}{50}=0.40. \]

Therefore,

\[ \boxed{40\%}. \]

Break-Even in Sales Dollars

If the contribution margin ratio is

\[ \frac{p-v}{p}, \]

then break-even sales revenue can be written as

\[ \boxed{ \text{break-even sales} = \frac{F}{(p-v)/p} }. \]

This is equivalent to finding break-even units first and multiplying by the selling price.

Example: Break-Even Sales Dollars

Fixed cost is $\$12{,}000$, selling price is $\$80$, and variable cost is $\$50$ per unit.

Contribution margin ratio:

\[ \frac{80-50}{80} = \frac{30}{80} = 0.375. \]

Therefore,

\[ \frac{12000}{0.375}=32000. \]

Break-even sales revenue is

\[ \boxed{\$32{,}000}. \]

Whole-Unit Constraints

Many products cannot be sold in fractional units.

If algebra gives a break-even quantity of

\[ 142.4, \]

then selling $142$ units is still below the calculated threshold.

To avoid a loss, the business must sell at least

\[ \boxed{143\text{ units}}. \]

Example: Round Up for Break-Even

Fixed cost is $\$2{,}500$, price is $\$28$, and variable cost is $\$11$ per unit.

\[ x_b= \frac{2500}{28-11} = \frac{2500}{17} \approx147.06. \]

Therefore, at least

\[ \boxed{148\text{ whole units}} \]

must be sold to reach or exceed break-even.

When Break-Even Is Impossible

\[ p-v\le0, \]

then each additional unit contributes no positive amount toward covering fixed cost.

If $p=v$, profit remains

\[ P(x)=-F. \]

If $p<v$, each additional unit makes profit more negative.

Therefore, with positive fixed cost, ordinary break-even is impossible under this model.

Example: Price Below Variable Cost

Fixed cost is $\$1{,}000$, selling price is $\$15$, and variable cost is $\$18$ per unit.

Profit is

\[ P(x)=(15-18)x-1000. \] \[ P(x)=-3x-1000. \]

This is negative for every

\[ x\ge0. \]

Therefore,

\[ \boxed{\text{no break-even quantity exists in the model}}. \]

Cost and Revenue from Data

Sometimes the equations are not given directly.

If cost is linear and two cost observations are known, use them to find the cost slope and intercept.

For points

\[ (x_1,C_1) \]

and

\[ (x_2,C_2), \]

variable cost per unit is

\[ \boxed{ v=\frac{C_2-C_1}{x_2-x_1} }. \]

Example: Build Cost from Two Data Points

Total cost is $\$2{,}600$ at $100$ units and $\$3{,}800$ at $200$ units.

Variable cost:

\[ v= \frac{3800-2600}{200-100} = \frac{1200}{100} = 12. \]

Use

\[ C(x)=F+12x. \]

Substitute $x=100$:

\[ 2600=F+1200. \] \[ F=1400. \]

Therefore,

\[ \boxed{C(x)=1400+12x}. \]

Revenue from Price Data

If every unit sells at the same price, revenue is direct variation:

\[ R(x)=px. \]

If revenue is $\$4{,}500$ from $150$ units, then

\[ p=\frac{4500}{150}=30. \]

Therefore,

\[ \boxed{R(x)=30x}. \]

Break-Even from Two Functions

If cost and revenue are already given as functions, solve

\[ C(x)=R(x). \]

For example, if

\[ C(x)=2500+14x \]

and

\[ R(x)=39x, \]

then

\[ 2500+14x=39x. \] \[ 2500=25x. \] \[ x=100. \]

Therefore,

\[ \boxed{100\text{ units}}. \]

Profit Inequalities

Sometimes the goal is not an exact profit but a minimum.

To require profit of at least $P_t$, write

\[ \boxed{ P(x)\ge P_t }. \]

Then solve the inequality.

Example: At Least a Target Profit

Suppose

\[ P(x)=25x-5000. \]

Require at least $\$3{,}000$ profit:

\[ 25x-5000\ge3000. \] \[ 25x\ge8000. \] \[ x\ge320. \]

Therefore,

\[ \boxed{x\ge320}. \]

Maximum Acceptable Loss

If a business wants its loss to be no more than $\$L$, then profit must satisfy

\[ \boxed{ P(x)\ge-L }. \]

The negative sign translates loss into the profit-function framework.

Domain Restrictions

In most quantity models,

\[ x\ge0. \]

If units are indivisible,

\[ x \]

should also be a whole number.

Capacity may impose an upper bound such as

\[ 0\le x\le500. \]

Capacity Can Make a Target Impossible

Suppose a target profit requires

\[ x\ge600 \]

units, but production capacity is only $500$ units.

The algebraic target quantity exists, but it is infeasible under the stated capacity constraint.

Applied algebra must interpret solutions within real constraints.

Break-Even and Intersections

Break-even is one example of solving an intersection problem.

We seek the quantity $x$ at which

\[ R(x)=C(x). \]

This is the same algebraic structure used whenever two models are set equal to find when they represent the same value.

Profit and the Difference of Functions

Profit is not a separate unrelated model.

It is the difference

\[ \boxed{P=R-C}. \]

Thus:

break-even means $P=0$;
profit means $P>0$;
loss means $P<0$.

Interpreting Slopes

In the simple linear models,

\[ C(x)=F+vx \]

has slope

\[ v, \]

while

\[ R(x)=px \]

has slope

\[ p. \]

The profit slope is

\[ p-v. \]

This tells us how much profit changes for each additional unit.

Example: Interpret the Profit Slope

\[ P(x)=18x-2700, \]

then each additional unit increases profit by

\[ \boxed{\$18}. \]

The intercept

\[ -2700 \]

represents the loss at zero sales caused by fixed cost.

Average Cost

Average cost per unit is total cost divided by quantity:

\[ \boxed{ A_C(x)=\frac{C(x)}{x} }. \]

For

\[ C(x)=F+vx, \]

we have

\[ A_C(x) = \frac{F}{x}+v, \qquad x>0. \]

Example: Average Cost

Suppose

\[ C(x)=3000+20x. \]

At $100$ units,

\[ A_C(100) = \frac{3000+2000}{100} = 50. \]

Therefore,

\[ \boxed{\$50\text{ per unit}}. \]

Why Average Cost Can Fall

\[ A_C(x)=\frac{F}{x}+v, \]

the fixed-cost portion per unit,

\[ \frac{F}{x}, \]

decreases as $x$ increases.

This simple model illustrates how fixed cost can be spread across more units.

Linear Models Are Simplifications

Real cost and revenue relationships may not remain linear across all quantities.

Variable cost can change, price can depend on demand, capacity can create step costs, and discounts can alter revenue.

The linear model is useful when the stated problem assumes approximately constant unit cost and unit price over the relevant range.

Checking with Units

\[ vx, \]

if $v$ is dollars per unit and $x$ is units, then

\[ \frac{\$}{\text{unit}}(\text{unit}) = \$. \]

Cost, revenue, and profit should all have monetary units.

Checking Break-Even by Substitution

After solving for break-even quantity $x_b$, verify

\[ C(x_b)=R(x_b). \]

Alternatively verify

\[ P(x_b)=0. \]

Both checks should agree.

Checking with Direction

\[ p>v, \]

then every additional unit should improve profit by

\[ p-v. \]

Profit should therefore increase with quantity in the simple model.

A result showing decreasing profit would indicate an error unless the model includes additional effects.

Common Error: Confusing Revenue with Profit

Revenue is money received from sales:

\[ R(x)=px. \]

Profit subtracts cost:

\[ \boxed{P(x)=R(x)-C(x)}. \]

Large revenue does not automatically imply large profit.

Common Error: Forgetting Fixed Cost

If cost is

\[ C(x)=F+vx, \]

using only

\[ vx \]

ignores expenses that exist even at zero production.

Common Error: Subtracting in the Wrong Order

Profit is

\[ \boxed{R-C}, \]

not

\[ C-R. \]

Reversing the order changes profit to its negative.

Common Error: Adding Fixed Cost to Revenue

Fixed cost belongs to the cost function.

Revenue under a constant-price model is

\[ R(x)=px, \]

not

\[ F+px. \]

Common Error: Dividing Fixed Cost by Price Instead of Margin

Break-even quantity is not generally

\[ \frac{F}{p}. \]

Each unit must also cover its variable cost.

Therefore,

\[ \boxed{ x_b=\frac{F}{p-v} }. \]

Common Error: Rounding Break-Even Down

If $147.06$ units are required to break even and units are indivisible, selling $147$ units is not enough.

Round up:

\[ \boxed{148\text{ units}}. \]

Common Error: Ignoring Feasibility

A solution may require more units than the market or production system can support.

Always compare the algebraic result with any stated domain, capacity, or demand restrictions.

Common Error: Assuming Linear Models Always Hold

The formulas in this lesson depend on constant unit price and constant variable cost.

If a problem gives quantity discounts, nonlinear demand, tiered costs, or changing unit cost, the model must be adjusted.

A Reliable Procedure for Cost-Revenue-Profit Problems

Define the quantity variable $x$.
Identify fixed cost $F$.
Identify variable cost per unit $v$.
Write $C(x)=F+vx$.
Identify selling price $p$.
Write $R(x)=px$.
Write $P(x)=R(x)-C(x)$.
Evaluate or solve the function requested by the problem.

A Reliable Procedure for Break-Even Problems

Write the cost function.
Write the revenue function.
Set $C(x)=R(x)$, or set $P(x)=0$.
Solve for $x$.
Round up when whole units are required.
Substitute back into cost and revenue.
Verify that both values are equal.
Check feasibility against capacity or domain constraints.

A Reliable Procedure for Target-Profit Problems

Write $P(x)=(p-v)x-F$.
Set $P(x)$ equal to the desired target profit.
Solve for the unknown quantity, price, cost, or other parameter.
Apply whole-unit restrictions when necessary.
Substitute the result into the original profit model.
Verify that the target profit is achieved.

Practice 1: Cost Function

A business has fixed cost $\$2{,}400$ and variable cost $\$16$ per unit. Write the cost function.

Practice 2: Revenue Function

A product sells for $\$42$ per unit. Write the revenue function.

Practice 3: Profit Function

Using Practices 1 and 2, write the profit function.

Practice 4: Evaluate Profit

Using the same model, find profit at $120$ units.

Practice 5: Break-Even Quantity

Using the same model, find the break-even quantity.

Practice 6: Target Profit

Fixed cost is $\$6{,}000$, variable cost is $\$25$ per unit, and price is $\$55$ per unit. How many units produce a $\$3{,}000$ profit?

Practice 7: Required Selling Price

Fixed cost is $\$4{,}000$, variable cost is $\$18$ per unit, and expected sales are $250$ units. What selling price produces a $\$2{,}500$ profit?

Practice 8: Unknown Variable Cost

A product sells for $\$70$ per unit. Fixed cost is $\$5{,}000$. At $200$ units, profit is $\$3{,}000$. Find variable cost per unit.

Practice 9: Infer Fixed Cost

Price is $\$50$, variable cost is $\$32$, and profit at $300$ units is $\$2{,}400$. Find fixed cost.

Practice 10: Round Up Break-Even

Fixed cost is $\$3{,}200$, price is $\$35$, and variable cost is $\$14$. Find the minimum whole number of units needed to reach break-even.

Practice 11: Impossible Break-Even

Fixed cost is $\$2{,}000$, selling price is $\$20$, and variable cost is $\$22$ per unit. Explain whether break-even is possible.

Practice 12: Build Cost from Data

Total cost is $\$3{,}000$ at $100$ units and $\$4{,}400$ at $200$ units. Assuming a linear cost model, find $C(x)$.

Practice 13: Profit Inequality

\[ P(x)=28x-5600, \]

how many units are needed for profit of at least $\$2{,}800$?

Practice 14: Average Cost

\[ C(x)=5000+25x, \]

find the average cost per unit at $250$ units.

Practice 15: Identify the Error

A product has fixed cost $\$4{,}000$, selling price $\$50$, and variable cost $\$30$ per unit. A student says break-even occurs at

\[ \frac{4000}{50}=80 \]

units. Explain the error and find the correct break-even quantity.

Practice Answers

Practice 1

\[ \boxed{C(x)=2400+16x}. \]

Practice 2

\[ \boxed{R(x)=42x}. \]

Practice 3

\[ P(x)=42x-(2400+16x). \] \[ \boxed{P(x)=26x-2400}. \]

Practice 4

\[ P(120)=26(120)-2400. \] \[ P(120)=3120-2400. \] \[ \boxed{\$720}. \]

Practice 5

\[ 26x-2400=0. \] \[ 26x=2400. \] \[ x=\frac{2400}{26} = \frac{1200}{13} \approx92.31. \]

If whole units are required, the minimum non-loss quantity is

\[ \boxed{93\text{ units}}. \]

Practice 6

Contribution margin:

\[ 55-25=30. \]

Target-profit equation:

\[ 30x-6000=3000. \] \[ 30x=9000. \] \[ \boxed{x=300\text{ units}}. \]

Practice 7

\[ p= \frac{4000+18(250)+2500}{250}. \] \[ p= \frac{4000+4500+2500}{250} = \frac{11000}{250}. \] \[ \boxed{\$44\text{ per unit}}. \]

Practice 8

\[ v= \frac{70(200)-5000-3000}{200}. \] \[ v= \frac{14000-8000}{200} = 30. \] \[ \boxed{\$30\text{ per unit}}. \]

Practice 9

\[ F=(50-32)(300)-2400. \] \[ F=18(300)-2400. \] \[ F=5400-2400. \] \[ \boxed{\$3{,}000}. \]

Practice 10

\[ x_b= \frac{3200}{35-14} = \frac{3200}{21} \approx152.38. \]

Therefore,

\[ \boxed{153\text{ units}}. \]

Practice 11

Contribution margin:

\[ 20-22=-2. \]

Therefore,

\[ P(x)=-2x-2000. \]

Profit is negative for every

\[ x\ge0. \]

Thus,

\[ \boxed{\text{break-even is impossible under the model}}. \]

Practice 12

Variable cost:

\[ v= \frac{4400-3000}{200-100} = \frac{1400}{100} = 14. \]

Use the first point:

\[ 3000=F+14(100). \] \[ 3000=F+1400. \] \[ F=1600. \]

Therefore,

\[ \boxed{C(x)=1600+14x}. \]

Practice 13

\[ 28x-5600\ge2800. \] \[ 28x\ge8400. \] \[ x\ge300. \]

Therefore,

\[ \boxed{x\ge300}. \]

Practice 14

\[ A_C(250) = \frac{5000+25(250)}{250}. \] \[ A_C(250) = \frac{5000+6250}{250}. \] \[ A_C(250) = \frac{11250}{250} = 45. \]

Therefore,

\[ \boxed{\$45\text{ per unit}}. \]

Practice 15

The student divided fixed cost by selling price and ignored the variable cost required to produce each unit.

Contribution margin:

\[ 50-30=20. \]

Therefore,

\[ x_b= \frac{4000}{20}=200. \]

Thus,

\[ \boxed{200\text{ units}}. \]

What to Remember

A simple linear cost function is

\[ \boxed{C(x)=F+vx}. \]

Revenue at constant unit price is

\[ \boxed{R(x)=px}. \]

Profit is

\[ \boxed{P(x)=R(x)-C(x)}. \]

Therefore,

\[ \boxed{P(x)=(p-v)x-F}. \]

Break-even occurs when

\[ \boxed{R(x)=C(x)} \]

or equivalently

\[ \boxed{P(x)=0}. \]

When $p>v$, the break-even quantity is

\[ \boxed{ x_b=\frac{F}{p-v} }. \]

The most important habit is to distinguish revenue from profit and to use contribution margin $p-v$, not selling price alone, when determining how each unit helps cover fixed cost.

Where We Go Next

Cost, revenue, and profit models describe money generated by production and sales. Another major family of financial models describes money changing over time through interest, growth, and repeated compounding.

The next lesson, Simple and Compound Interest, develops those time-based financial relationships.