# Marginal Revenue, Marginal Cost and Marginal Profit

Marginal functions are growth functions. That means they tell you how much something increases or decreases per unit. Marginal functions tell you whether it is profitable to increase production, and what level of revenue, cost or profit will follow.

Theory

### MarginalFunctions

Marginal income,

 ${I}^{\prime }\left(x\right)=I\left(x+1\right)-I\left(x\right),$

refers to the change in income from increasing the production by one unit.

Marginal cost,

 ${C}^{\prime }\left(x\right)=C\left(x+1\right)-C\left(x\right),$

refers to the change in cost from increasing the production by one unit.

Marginal profit,

 ${P}^{\prime }\left(x\right)={I}^{\prime }\left(x\right)-{C}^{\prime }\left(x\right),$

refers to the change in profit from increasing the production by one unit.

Example 1

For a given purse, Valentino has the income function

 $I\left(x\right)=5{x}^{2}-150x+25000$

and the cost function

 $C\left(x\right)=15{x}^{2}-1180x+33200,$

both in thousands of euros.

1.
Find the marginal income, marginal cost and marginal profit for the purse.
2.
What is the largest obtainable profit, and how many purses do Valentino have to sell to obtain this profit?

1.
The marginal income is the derivative of the income:
 ${I}^{\prime }\left(x\right)=10x-150.$

The marginal cost is the derivative of the cost:

 ${C}^{\prime }\left(x\right)=30x-1180.$

The marginal profit can be found either by taking the derivative of the profit function or by using the formula above. In this case, the easier alternative is using the formula:

$\begin{array}{cc}{P}^{\prime }\left(x\right)={I}^{\prime }\left(x\right)-{C}^{\prime }\left(x\right)& \\ =& \\ 10x-150-\left(30x-1180\right)& \\ =& \\ -20x+1030& \end{array}$

$\begin{array}{llll}\hfill {P}^{\prime }\left(x\right)& ={I}^{\prime }\left(x\right)-{C}^{\prime }\left(x\right)\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =10x-150-\left(30x-1180\right)\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =-20x+1030\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$

2.
To find the largest profit, set ${P}^{\prime }\left(x\right)=0$ and insert the value you find for $x$ into the profit function $P\left(x\right)$: $\begin{array}{llllll}\hfill {P}^{\prime }\left(x\right)& =0\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill -20x+1030& =0\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill -20x& =-1030\phantom{\rule{2em}{0ex}}& \hfill & |÷-20\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill x& =51.5\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$

As there is no way to produce half a unit, you have to choose either $51$ or $52$. To find out whether $51$ or $52$ is the optimal amount of purses to sell, you can insert both values into the profit function and choose the $x$-value that gives you the largest profit. But first, you have to find the profit function:

$\begin{array}{llll}\hfill & P\left(x\right)=I\left(x\right)-C\left(x\right)\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =5{x}^{2}-150x+25000\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & \phantom{\rule{2em}{0ex}}-\phantom{\rule{-0.17em}{0ex}}\left(15{x}^{2}-1180x+33200\right)\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =-10{x}^{2}+1030x-8200\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$

$\begin{array}{llll}\hfill P\left(x\right)& =I\left(x\right)-C\left(x\right)\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =5{x}^{2}-150x+25000-\phantom{\rule{-0.17em}{0ex}}\left(15{x}^{2}-1180x+33200\right)\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =-10{x}^{2}+1030x-8200\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$

Now you can insert $51$ and $52$:
$\begin{array}{llll}\hfill P\left(51\right)& =-10{\left(51\right)}^{2}+1030\left(51\right)+8200\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =34720\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill P\left(52\right)& =-10{\left(52\right)}^{2}+1030\left(52\right)+8200\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =34720\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$

$\begin{array}{llll}\hfill P\left(51\right)& =-10{\left(51\right)}^{2}+1030\left(51\right)+8200\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =34720\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill P\left(52\right)& =-10{\left(52\right)}^{2}+1030\left(52\right)+8200\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =34720\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$

When the profit is equal between two or more units, you want to the choose the largest $x$-value, because the larger value looks better in your financial statement.

Example 2

You have the profit function

 $P\left(x\right)=-10{x}^{2}+1030x-8200.$

If you have a production level of 52, is it profitable to increase the production to 53 units?

To answer this question, you need to figure out whether the profit will increase or decrease after another unit is produced.

Start by finding ${P}^{\prime }\left(x\right)$ from your profit function. That means you need to take the derivative of $P\left(x\right)$, and then insert $x=52$: $\begin{array}{llll}\hfill {P}^{\prime }\left(x\right)& =-20x+1030\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill {P}^{\prime }\left(52\right)& =-20\cdot 52+1030=-10\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$

As the answer is negative, the profit will decrease by 10 if you increase the production by one unit. You won’t want to increase production this time.