What Do Revenue, Cost, Unit Cost and Profit Mean?

Theory

Revenue

Revenue is the turnover of a business, which is the amount of money coming into its accounts. If you’re not given an expression for revenue you can use the general expression

 $I\left(x\right)=p\cdot x,$

where $x$ is the number of units sold and $p$ is the price of a unit.

Theory

Cost

Cost refers to how much money a business spends to produce $x$ units of its product.

 $C\left(x\right).$

You divide cost into fixed costs and variable costs:

Fixed costs (FC) are independent of production, like rent for production space, offices, and phone and internet bills.

Variable costs (VC) are dependent on the production—things like employee salaries and the cost of materials.

Theory

UnitCost

Unit cost is how much it costs to produce one unit. You can find the unit cost by dividing cost by the number of units produced:

Theory

Profit

Profit is how much money a business is left with after a period of time. The profit is the difference between revenue and cost during that period. It can be written as

 $\text{Profit}=P\left(x\right)=I\left(x\right)-C\left(x\right).$

Graphical Interpretation

• You look at a graph, and it shows revenue increasing from left to right.

This means that the more units you sell of a product, the more money the business receives.

• You look at another graph, and it shows cost increasing from left to right.

This means that the more units you sell, the more units you have to produce, which costs money to do.

• You view a third graph. This one shows profit in an arched shape from left to right.

This indicates low profit at the start, a maximum profit at a certain moderate level of production, and lower profit again at higher levels of production. Profit often decreases at increased levels of production due to capacity issues and wear and tear on the means of the production.

• On another graph, you see that at the point of intersection between the cost function and the revenue function, the profit is $0$. That means the profit function crosses the $x$-axis at that point.

• Where the difference between the revenue function and the cost function is the greatest, the profit is at its greatest as well, meaning $P\left(x\right)$ is at its maximum.

In other words, wherever the biggest space appears between revenue and cost on the $y-axis$ is where profit is highest. The $x$ value at this point will give you the optimal production.

• The smallest value of $x$ for the unit cost will be when

 ${C}^{\prime }\left(x\right)=U\left(x\right).$

Example 1

BMW believes that the cost, in thousands of euros, of producing $x$ units of a new car is given by

$\begin{array}{cc}C\left(x\right)=\text{}0.25\text{}{x}^{2}+100x+5000& \\ x\in \phantom{\rule{-0.17em}{0ex}}\left[0,400\right).& \end{array}$

 $C\left(x\right)=\text{}0.25\text{}{x}^{2}+100x+5000,\phantom{\rule{2em}{0ex}}x\in \phantom{\rule{-0.17em}{0ex}}\left[0,400\right).$

1.
Find an expression for the unit cost.
2.
What is the unit cost when producing 76 cars?
3.
BMW sells this new type of car for $\text{}200\phantom{\rule{0.17em}{0ex}}000\text{}$ euros per car. What’s the revenue function?
4.
What’s the revenue if they sell 132 cars?
5.
Find an expression for the profit.
6.
How many cars do BMW have to produce for the profit to be above 0?

1.
You find the unit cost by dividing the total cost $C\left(x\right)$ by the number of units $x$. $\begin{array}{llll}\hfill U\left(x\right)& =\frac{C\left(x\right)}{x}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =\frac{0.25{x}^{2}+100x+5000}{x}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =0.25x+100+\frac{5000}{x}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$

That gives you the unit cost

 $U\left(x\right)=0.25x+100+\frac{5000}{x}.$
2.
Then, the unit cost when producing $76$ cars is
$\begin{array}{llll}\hfill U\left(76\right)& =0.25\left(76\right)+100+\frac{5000}{76}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =184.789\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$

 $U\left(76\right)=0.25\left(76\right)+100+\frac{5000}{76}=184.789$

The numbers in the function represent thousands of euros, so the actual unit cost when producing $76$ cars is
 $184.789\cdot 1000=184\phantom{\rule{0.17em}{0ex}}789.$

That is, when BMW produce $76$ units of their new car, each car costs $184\phantom{\rule{0.17em}{0ex}}789$ euros to produce.

3.
The functions are in thousands of euros, so you have to divide the $200\phantom{\rule{0.17em}{0ex}}000$ euros by $1000$ for the functions to provide the right answer. That gives you
 $200\phantom{\rule{0.17em}{0ex}}000:1000=200.$

Then you can create the expression for the revenue by multiplying the price by the number of units:

 $I\left(x\right)=200x$
4.
To find the revenue when BMW sells $132$ cars, you insert 132 straight into the revenue function:
 $I\left(132\right)=200\cdot 132=26\phantom{\rule{0.17em}{0ex}}400$

You’ll have to multiply by $1000$ again, as the function is in thousands of euros. That finally gives you that when selling $132$ cars, the revenue is

 $26\phantom{\rule{0.17em}{0ex}}400\cdot 1000=26\phantom{\rule{0.17em}{0ex}}400\phantom{\rule{0.17em}{0ex}}000,$

which gives you $26\phantom{\rule{0.17em}{0ex}}400\phantom{\rule{0.17em}{0ex}}000$ euros.

5.
The profit is the difference between the revenue and the cost. That gives you
$\begin{array}{cc}P\left(x\right)=I\left(x\right)-C\left(x\right)& \\ =& \\ 200x-\phantom{\rule{-0.17em}{0ex}}\left(0.25{x}^{2}+100x+5000\right)& \\ =& \\ -0.25{x}^{2}+100x-5000.& \end{array}$

$\begin{array}{llll}\hfill P\left(x\right)& =I\left(x\right)-C\left(x\right)\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =200x-\phantom{\rule{-0.17em}{0ex}}\left(0.25{x}^{2}+100x+5000\right)\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =-0.25{x}^{2}+100x-5000.\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$

6.
Now you need to find the value of $x$ where the profit becomes positive. To find this, you can solve the inequality $\begin{array}{llll}\hfill P\left(x\right)& >0\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill -0.25{x}^{2}+100x-5000& >0.\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$

You can see that the function is a parabola with a finite maximum, because it’s a quadratic function with $a=-0.25<0$. You also know that the function is positive between its roots, which you’ll then need to find.

You solve the equation

 $-0.25{x}^{2}+100x-5000=0.$

You can solve this equation in a digital tool like `CAS` in `GeoGebra` . You’ll then get that the answer is ${x}_{1}\approx 58.6$ and ${x}_{2}=341.4$.

So the conclusion is that production of the cars will yield a profit for BMW if they produce and sell between $59$ and $341$ cars of the new type.