Business and Economics Applications
Textbook Reference
| Primary source | OpenStax Calculus Volume 1, Section 4.7: “Applied Optimization Problems” |
| Direct link | https://openstax.org/books/calculus-volume-1/pages/4-7-applied-optimization-problems |
| Textbook used in class | Stewart, Calculus, Section 3.7: “Optimization Problems” |
Opening Scenario
A company sells a product at a price that depends on how many units it produces: the more units produced, the lower the price must be set to sell them all. The company wants to know how many units to produce to earn the greatest profit.
This is a standard calculus problem in economics. The tools are the same as geometric optimization: set up a function of one variable (profit as a function of quantity) and find its maximum using the derivative.
Quick Reference
| Term | Definition |
|---|---|
| Demand function $p(x)$ | Price per unit when $x$ units are sold; usually decreasing in $x$ |
| Revenue $R(x)$ | Total income: $R(x) = x \cdot p(x)$ |
| Cost $C(x)$ | Total cost to produce $x$ units: fixed cost plus variable cost |
| Profit $P(x)$ | $P(x) = R(x) - C(x)$ |
| Marginal revenue $R'(x)$ | Rate of change of revenue with respect to units sold |
| Marginal cost $C'(x)$ | Rate of change of cost with respect to units produced |
Profit is maximized when $R'(x) = C'(x)$ (marginal revenue equals marginal cost), provided the second-order condition $R''(x) < C''(x)$ holds at that point.
Key Concepts
1. Why Revenue Depends on Price
If the price per unit is constant (the company is a “price taker”), $R(x) = px$ is linear. But in many markets, the more you produce, the lower the price must be to clear the market. So $p = p(x)$ is a decreasing function of $x$, called the demand function or price function.
When $p$ depends on $x$, the revenue function is: $$R(x) = x \cdot p(x).$$
This is not linear in $x$ unless $p$ is constant. Maximizing revenue requires calculus.
2. Maximizing Revenue
Example 1. A company can sell $x$ units at a price of $p(x) = 200 - 2x$ dollars per unit (for $0 \leq x \leq 100$). Find the quantity that maximizes revenue.
$R(x) = x \cdot p(x) = x(200 - 2x) = 200x - 2x^2$.
$R'(x) = 200 - 4x = 0 \Rightarrow x = 50$.
$R''(x) = -4 < 0$: local (and absolute) maximum on $[0, 100]$.
$R(50) = 50(200 - 100) = 50 \times 100 = 5000$ dollars.
Boxed answer: Revenue is maximized at $x = 50$ units, giving maximum revenue of $\$5{,}000$.
Check the boundary values: $R(0) = 0$ and $R(100) = 100(200 - 200) = 0$. The interior maximum of $5000$ is indeed the global maximum.
3. Maximizing Profit
Example 2. Using the same demand function, suppose the total cost of producing $x$ units is $C(x) = 2000 + 20x$ (dollars). Find the quantity that maximizes profit.
$P(x) = R(x) - C(x) = (200x - 2x^2) - (2000 + 20x) = -2x^2 + 180x - 2000$.
$P'(x) = -4x + 180 = 0 \Rightarrow x = 45$.
$P''(x) = -4 < 0$: maximum.
$P(45) = -2(2025) + 180(45) - 2000 = -4050 + 8100 - 2000 = 2050$ dollars.
Boxed answer: Profit is maximized at $x = 45$ units, giving maximum profit of $\$2{,}050$.
Verify via MR = MC: $R'(x) = 200 - 4x$ and $C'(x) = 20$. Set equal: $200 - 4x = 20 \Rightarrow x = 45$. Confirmed.
4. Marginal Revenue Equals Marginal Cost
The condition $R'(x) = C'(x)$ at the profit-maximizing quantity is a theorem:
$P(x) = R(x) - C(x)$ is maximized when $P'(x) = 0$, i.e., $R'(x) - C'(x) = 0$, i.e., $R'(x) = C'(x)$.
This is the MR = MC rule: at the profit-maximizing quantity, each additional unit produced adds the same amount to revenue as it adds to cost. If the extra revenue from one more unit exceeds the extra cost, it is still profitable to produce it. If the extra cost exceeds the extra revenue, the last unit was not worth producing. The optimal quantity is where these two rates are equal.
5. Average vs. Marginal
A common confusion in economics courses:
- Marginal cost $C'(x)$ is the rate of change of cost -- the approximate cost of producing one more unit.
- Average cost $\bar{C}(x) = C(x)/x$ is the cost per unit produced on average.
Minimizing average cost and maximizing profit are different objectives. At the profit-maximizing quantity, $R'(x) = C'(x)$, not $C'(x) = \bar{C}(x)$.
“The profit-maximizing quantity minimizes the average cost.” No. The profit-maximum is at $R'(x) = C'(x)$, which is a different condition from average-cost minimization. A company that minimizes its average cost is not necessarily maximizing its profit.
6. When to Set Price, Not Quantity
In some problems, the variable you optimize over is the price $p$ rather than the quantity $x$. If the demand function gives $x$ as a function of $p$ (i.e., $x = D(p)$), then $R = p \cdot D(p)$ is a function of price. The same setup-and-solve method applies with $p$ as the variable.
Example 3. Demand is $x = 1000 - 5p$ (units sold when price is $p$ dollars). Find the price that maximizes revenue.
$R(p) = p \cdot (1000 - 5p) = 1000p - 5p^2$.
$R'(p) = 1000 - 10p = 0 \Rightarrow p = 100$ dollars.
$R(100) = 100 \times 500 = 50{,}000$ dollars.
Boxed answer: Revenue is maximized at a price of $\$100$ per unit.
Common Errors Summary
| Error | Example | Correction |
|---|---|---|
| Maximizing revenue instead of profit | Finding $R'=0$ when the problem asks for profit maximization | Profit is $P = R - C$; maximize $P$, not $R$ |
| Confusing $C'(x)$ with $C(x)/x$ | Using average cost in the MR=MC condition | The MR=MC rule uses marginal cost $C'(x)$, not average cost |
| Forgetting to check that the critical number gives a maximum | Reporting $x=45$ without verifying $P''(45) < 0$ | Check the second derivative or use the FDT before claiming the critical number maximizes profit |
Leveled Practice
Level 1 -- Direct Application
Problem 1. Revenue is $R(x) = 50x - 0.5x^2$ and cost is $C(x) = 10x + 100$. Find the profit-maximizing quantity using both (a) $P'(x) = 0$ and (b) the MR = MC condition.
Show answer
(a) $P(x) = R - C = -0.5x^2 + 40x - 100$. $P'(x) = -x + 40 = 0 \Rightarrow x = 40$.
(b) $R'(x) = 50 - x$ and $C'(x) = 10$. Set equal: $50 - x = 10 \Rightarrow x = 40$.
$P''(x) = -1 < 0$: maximum. $P(40) = -800 + 1600 - 100 = 700$.
Boxed answer: Maximum profit of $\$700$ at $x = 40$ units.
Level 2 -- Multiple Steps
Problem 2. A monopolist faces demand $p = 120 - 3x$ and total cost $C(x) = x^3 - 7x^2 + 111x + 50$ (all in dollars). Find the profit-maximizing quantity and verify with the second-order condition.
Show answer
$R(x) = x \cdot p = 120x - 3x^2$.
$P(x) = R - C = 120x - 3x^2 - (x^3 - 7x^2 + 111x + 50) = -x^3 + 4x^2 + 9x - 50$.
$P'(x) = -3x^2 + 8x + 9$.
Solving $-3x^2 + 8x + 9 = 0$: use the quadratic formula with $a=-3, b=8, c=9$:
$x = \dfrac{-8 \pm \sqrt{64 + 108}}{-6} = \dfrac{-8 \pm \sqrt{172}}{-6}$.
$\sqrt{172} \approx 13.11$. The two solutions are $x \approx (-8 + 13.11)/(-6) \approx -0.85$ and $x \approx (-8 - 13.11)/(-6) \approx 3.52$.
Since $x > 0$, the relevant critical number is $x \approx 3.52$.
$P''(x) = -6x + 8$. $P''(3.52) = -6(3.52) + 8 = -21.12 + 8 = -13.12 < 0$: maximum confirmed.
Boxed answer: Profit is maximized at $x \approx 3.52$ units (interpret as $3$ or $4$ whole units in context).
Mastery Checklist
Mental Model
Business optimization problems follow the same setup-and-solve pattern as geometric ones. The vocabulary shifts -- “demand function” instead of “perimeter,” “profit” instead of “area” -- but the mathematics is identical:
- Write the objective (profit, revenue) as a function of one variable (quantity or price).
- Differentiate and solve for the critical number.
- Classify with the Second Derivative Test.
- Interpret: the critical number is the optimal quantity, and the function value there is the optimal profit.
The MR = MC rule is not a shortcut -- it is exactly $P'(x) = 0$ rewritten. Knowing both forms gives you a way to check your work.
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