Maximizing Profit

Profit peaks exactly where making one more unit stops adding money. In calculus terms, that is where the derivative of profit is zero.

How it works

Optimization with derivatives

Profit $P = R - C$ is maximized where $P'(x) = 0$ , that is where marginal revenue equals marginal cost, $R'(x) = C'(x)$ , with $P''(x) < 0$ .

General formula

{\htmlData{sym=3}{P}}({\htmlData{sym=6}{x}}) = {\htmlData{sym=4}{R}}({\htmlData{sym=6}{x}}) - {\htmlData{sym=5}{C}}({\htmlData{sym=6}{x}}), \qquad {\htmlData{sym=0}{P'(x)}} = 0 \iff {\htmlData{sym=1}{R'(x)}} = {\htmlData{sym=2}{C'(x)}}

As long as one more unit brings in more than it costs, keep producing; stop where the two are equal.

$P'(x)$
Marginal profit
Extra profit from making one more unit. Zero at the peak.
$R'(x)$
Marginal revenue
Extra money brought in by selling one more unit.
$C'(x)$
Marginal cost
Extra cost of making one more unit.
$P$
Profit
Revenue minus cost.
$R$
Revenue
Money brought in from sales.
$C$
Cost
Money spent making the units.
$x$
Units
How many units are made and sold.