Business Applications

Main idea

In business, "marginal" means derivative: the extra cost, revenue or profit from making one more unit. Keep making more while the next unit brings in more than it costs, so profit peaks where marginal revenue equals marginal cost.

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

What the symbols mean

$R(x)$
Revenue
Money taken in from selling x units: price times quantity.
$C(x)$
Cost
Fixed costs plus the cost of making x units.
$R'(x) = C'(x)$
Marginal revenue = marginal cost
One more unit brings in exactly what it costs: the profit-maximizing point.

Average cost

{\htmlData{sym=0}{\overline{C}(x)}} = \frac{C(x)}{x}

What the symbols mean

$\overline{C}(x)$
Average cost
Cost per unit. It's smallest where it equals the marginal cost.

Explanation

Business applications of derivatives are simply questions like "how many should we make to earn the most?" answered with the tools from this unit. All we're really saying is that profit is what you take in minus what you spend. The formula is just that: $R(x)$ (revenue, the money coming in from selling $x$ units), $C(x)$ (cost, the money going out to make them) and $P(x) = R(x) - C(x)$ (profit, what's left). Economists say "marginal" for "the extra from one more unit", and that's a derivative. The vocabulary is new, but the math is the optimization you already know.

Marginal means derivative. $C'(x)$ approximates $C(x + 1) - C(x)$ , the cost of the next unit. That's the linear approximation from a few lessons ago with a step of $dx = 1$ .

Demand and revenue. If the price people will pay is $p(x)$ when $x$ units are sold, then $R(x) = x\,p(x)$ .

Why $R' = C'$ maximizes profit. Think about the next unit. If it brings in more than it costs ( $R' > C'$ ), making it adds to your profit, so make it. If it costs more than it brings in, making it eats into profit, so don't. You keep going until the two balance, and that's the best output. It's the same as setting $P'(x) = R'(x) - C'(x) = 0$ .

A common mistake. It's tempting to think the best output is where revenue is largest, because more money coming in sounds better. It isn't: past a point, the extra sales cost more to make than they bring in. Profit peaks earlier, where $R' = C'$ , not where $R' = 0$ .

Average cost is minimized where $\overline{C}'(x) = 0$ , which works out to $\overline{C}(x) = C'(x)$ : when the next unit costs exactly the current average.

See it

This is a company's revenue, R(x) = 50x − 0.04x², and its cost, C(x) = 1500 + 14x. Profit is the vertical gap between them. Notice that the gap is widest at x = 450 units, a profit of $6600, and that's exactly where the revenue curve's slope drops to 14, the cost line's slope. Revenue keeps climbing until 625 units, but the extra sales past 450 cost more than they bring in.

Worked examples

Example 1

A company has $C(x) = 1000 + 10x + 0.01x^2$ and demand $p(x) = 50 - 0.01x$ . Maximize profit.

1
Revenue is price times quantity, $R(x) = x(50 - 0.01x) = 50x - 0.01x^2$ , so $P(x) = R(x) - C(x) = 40x - 0.02x^2 - 1000$ .

Practice

Type your answer and check it. Solve them all to complete the lesson.

Problem 1+5 XP

Find the marginal cost of $C(x) = 500 + 4x + 0.002x^2$ at $x = 1000$ .

Problem 2+5 XP

For $R(x) = 60x - 0.05x^2$ , what output maximizes revenue?

Problem 3+5 XP

With that revenue and $C(x) = 1000 + 20x$ , what output maximizes profit?

Problem 4+5 XP

For $C(x) = 800 + 2x + 0.02x^2$ , what output minimizes the average cost?

Problem 5+5 XP

What is the minimum average cost there?

Problem 6+5 XP

A coffee cart sells $400$ cups a day at $\$ 3 $each. Market research says every$ 10 $¢ price increase loses$ 20 $customers a day, and every$ 10 $¢ cut gains$ 20$. What price maximizes daily revenue?

Challenge problem+25 XP

The hardest one here. Get it, and you've got this lesson.

A company faces demand $p = 100 - 2x$ (price $p$ in dollars at $x$ units) and costs $C(x) = 20x$ . The government adds a tax of $\tau$ dollars per unit, and the company then picks the output that maximizes its profit. What tax rate $\tau$ brings in the most tax revenue?

Where it's used

Pricing, production planning and the decision to run one more shift are all made at the margin, with exactly these derivatives.

Maximizing ProfitCalc I · Economics & Finance

Newton's Method Indefinite Integrals

Business Applications

Main idea

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

What the symbols mean

$R(x)$
Revenue
Money taken in from selling x units: price times quantity.
$C(x)$
Cost
Fixed costs plus the cost of making x units.
$R'(x) = C'(x)$
Marginal revenue = marginal cost
One more unit brings in exactly what it costs: the profit-maximizing point.

Average cost

{\htmlData{sym=0}{\overline{C}(x)}} = \frac{C(x)}{x}

What the symbols mean

$\overline{C}(x)$
Average cost
Cost per unit. It's smallest where it equals the marginal cost.

Explanation

Marginal means derivative. $C'(x)$ approximates $C(x + 1) - C(x)$ , the cost of the next unit. That's the linear approximation from a few lessons ago with a step of $dx = 1$ .

Demand and revenue. If the price people will pay is $p(x)$ when $x$ units are sold, then $R(x) = x\,p(x)$ .

Average cost is minimized where $\overline{C}'(x) = 0$ , which works out to $\overline{C}(x) = C'(x)$ : when the next unit costs exactly the current average.

See it

Worked examples

Example 1

A company has $C(x) = 1000 + 10x + 0.01x^2$ and demand $p(x) = 50 - 0.01x$ . Maximize profit.

1
Revenue is price times quantity, $R(x) = x(50 - 0.01x) = 50x - 0.01x^2$ , so $P(x) = R(x) - C(x) = 40x - 0.02x^2 - 1000$ .

Practice

Type your answer and check it. Solve them all to complete the lesson.

Problem 1+5 XP

Find the marginal cost of $C(x) = 500 + 4x + 0.002x^2$ at $x = 1000$ .

Problem 2+5 XP

For $R(x) = 60x - 0.05x^2$ , what output maximizes revenue?

Problem 3+5 XP

With that revenue and $C(x) = 1000 + 20x$ , what output maximizes profit?

Problem 4+5 XP

For $C(x) = 800 + 2x + 0.02x^2$ , what output minimizes the average cost?

Problem 5+5 XP

What is the minimum average cost there?

Problem 6+5 XP

Challenge problem+25 XP

The hardest one here. Get it, and you've got this lesson.

Where it's used

Pricing, production planning and the decision to run one more shift are all made at the margin, with exactly these derivatives.

Maximizing ProfitCalc I · Economics & Finance