Optimisation in Commerce — Maximising Profit and Minimising Cost

The techniques of maxima and minima (Sections 2–3) combine with the cost–revenue functions (Section 4) to answer the questions a business most wants answered: what output maximises profit? and what output minimises cost per unit? This is the payoff of the whole chapter.

Maximising profit. Profit is $\pi(x) = R(x) - C(x)$. To maximise it, treat $\pi$ like any function of $x$:

1. Differentiate: $\pi'(x) = R'(x) - C'(x) = \text{MR} - \text{MC}$.
2. Set $\pi'(x) = 0$. This gives

the famous rule that profit is maximised where marginal revenue equals marginal cost. It makes economic sense: as long as the next unit brings in more than it costs ($\text{MR} > \text{MC}$) profit keeps rising, and once the next unit costs more than it earns ($\text{MR} < \text{MC}$) profit falls — so the peak is exactly where they are equal.

3. Confirm it is a maximum: check $\pi''(x) < 0$ (equivalently $\text{MR}' < \text{MC}'$).

4. Substitute the optimal $x$ back to get the maximum profit, and use the demand relation to find the price that should be charged.

Minimising cost per unit. To find the output that makes each unit cheapest, minimise the average cost $\overline{C}(x) = \dfrac{C(x)}{x}$:

1. Differentiate $\overline{C}(x)$ and set $\overline{C}'(x) = 0$.
2. Confirm a minimum with $\overline{C}''(x) > 0$.
3. A useful and always-true fact drops out: at the output where average cost is least, average cost equals marginal cost, $\overline{C} = \text{MC}$. (This is why the marginal cost curve passes through the lowest point of the average cost curve.)

Maximising revenue is the special case of maximising $R(x) = p\,x$ on its own (ignoring cost): set $\text{MR} = R'(x) = 0$ and check $R''(x) < 0$.

Practical checklist for every optimisation problem:

1. Write the quantity to be optimised as a function of a single variable (using the demand or cost relations to eliminate any others). …