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Every business faces the same fundamental questions: How many units should we produce? What price should we charge? How do we minimize costs while meeting demand?
These questions translate directly into optimization problems. The quantity to maximize or minimize is profit, revenue, or cost—and the constraint comes from the relationship between price and demand. Calculus gives businesses a precise tool for finding optimal production levels and pricing strategies.
The key insight: in economics, rates of change have names. The derivative of cost is marginal cost. The derivative of revenue is marginal revenue. At the optimal production level, these quantities have a special relationship.
| Property | Value |
|---|---|
| Concept | Optimization |
| Chapter | 3.7 |
| Difficulty | Intermediate |
| Time | ~20 minutes |
The cost function $C(x)$ gives the total cost of producing $x$ units.
$$C(x) = \text{fixed costs} + \text{variable costs}$$
Example: $C(x) = 5000 + 8x + 0.01x^2$
The marginal cost is the rate of change of cost with respect to quantity:
$$\text{Marginal Cost} = C'(x)$$
This approximates the cost of producing one additional unit.
The demand function (or price function) $p(x)$ gives the price per unit when $x$ units are sold.
Typically, $p(x)$ is decreasing: to sell more units, you must lower the price.
Example: $p(x) = 100 - 0.5x$
When 80 units are sold, the price is $p(80) = 100 - 40 = \$60$ per unit.
Revenue is what you earn from selling:
$$R(x) = x \cdot p(x) = \text{(quantity)} \times \text{(price per unit)}$$
The marginal revenue is:
$$\text{Marginal Revenue} = R'(x)$$
Profit is revenue minus cost:
$$P(x) = R(x) - C(x)$$
The marginal profit is:
$$\text{Marginal Profit} = P'(x) = R'(x) - C'(x)$$
| Function | Symbol | Definition | Derivative (Marginal) |
|---|---|---|---|
| Cost | $C(x)$ | Total production cost | $C'(x)$ = marginal cost |
| Price/Demand | $p(x)$ | Price per unit | — |
| Revenue | $R(x)$ | $x \cdot p(x)$ | $R'(x)$ = marginal revenue |
| Profit | $P(x)$ | $R(x) - C(x)$ | $P'(x)$ = marginal profit |
At maximum profit:
$$\boxed{R'(x) = C'(x)}$$
Marginal revenue equals marginal cost.
Why? Profit is $P(x) = R(x) - C(x)$, so: $$P'(x) = R'(x) - C'(x)$$
Setting $P'(x) = 0$ gives $R'(x) = C'(x)$.
Intuition: If producing one more unit brings in more revenue than it costs ($R' > C'$), you should produce it. If it costs more than it brings in ($R' < C'$), you shouldn't. At the optimal level, they're equal.
Problem: A store sells 500 laptops per week at $800 each. Market research shows that for each $20 rebate, sales increase by 40 units per week. What rebate maximizes weekly revenue?
Step 1: Define variables
Step 2: Find the demand function
Sales increase from 500 by $(x - 500)$ units. Each increase of 40 units requires a $20 price decrease. So each additional unit requires a $\frac{20}{40} = \$0.50$ price decrease.
$$p(x) = 800 - 0.50(x - 500) = 800 - 0.5x + 250 = 1050 - 0.5x$$
Step 3: Write the revenue function
$$R(x) = x \cdot p(x) = x(1050 - 0.5x) = 1050x - 0.5x^2$$
Step 4: Find the critical point
$$R'(x) = 1050 - x = 0 \Rightarrow x = 1050$$
Step 5: Verify maximum
$R''(x) = -1 < 0$, so the parabola opens downward. This is indeed a maximum.
Step 6: Find the optimal price and rebate
$$p(1050) = 1050 - 0.5(1050) = 1050 - 525 = \$525$$
Original price was $800, so the rebate is $800 - 525 = \$275$.
Answer: A rebate of $275 maximizes weekly revenue.
A company's weekly cost to produce $x$ widgets is $C(x) = 2000 + 15x$ dollars, and the demand function is $p(x) = 50 - 0.02x$ dollars per widget.
(a) What are the fixed costs? (b) What is the variable cost per widget? (c) Write the revenue function $R(x)$. (d) Write the profit function $P(x)$.
A bakery sells 300 loaves of bread per day at $4.50 each. The owner estimates that for each $0.25 increase in price, 15 fewer loaves will be sold. Find the demand function $p(x)$ where $x$ is the number of loaves sold.
A movie theater charges $12 per ticket and sells an average of 800 tickets per showing. Research indicates that for every $0.50 decrease in ticket price, attendance increases by 25 people. What ticket price maximizes revenue?
A manufacturer can produce smartphones at a cost of $C(x) = 25000 + 180x + 0.002x^2$ dollars for $x$ units. The demand function is $p(x) = 450 - 0.003x$ dollars per phone.
(a) Find the production level that maximizes profit. (b) What is the maximum profit? (c) Verify that marginal revenue equals marginal cost at this production level.
The average cost per unit is defined as $\bar{C}(x) = \frac{C(x)}{x}$.
(a) Show that if the average cost is minimized at $x = x_0$, then the marginal cost equals the average cost at that point: $C'(x_0) = \bar{C}(x_0)$.
(b) For $C(x) = 10000 + 50x + 0.01x^2$, find the production level that minimizes average cost.
(c) Verify that marginal cost equals average cost at your answer from part (b).
(d) Explain intuitively why "marginal = average at minimum average" makes sense.
A company is currently producing at a level where marginal revenue is $45 per unit and marginal cost is $52 per unit. To increase profit, they should:
(A) Produce more units, because revenue exceeds cost (B) Produce fewer units, because the next unit costs more than it brings in (C) Stay at the current level, because they're at maximum profit (D) Cannot determine without knowing the price
If $C(x) = 5000 + 12x + 0.001x^2$ gives the cost in dollars of producing $x$ items, what does $C'(500) = 13$ mean in practical terms?
(A) It costs $13 to produce 500 items (B) At a production level of 500 items, the cost is increasing at $13 per item (C) The 500th item costs exactly $13 to produce (D) Producing 501 items instead of 500 increases cost by approximately $13
The Balancing Point:
Imagine you're filling a bathtub with the drain open.
At first, water flows in faster than it drains out (MR > MC)—so you should "produce more."
Eventually, the flow rates balance (MR = MC)—that's maximum profit.
If you keep going, water drains out faster than it flows in (MR < MC)—you're losing money on each additional unit.
The optimal production level is where the rates balance.
Looking back:
Looking ahead:
Real-world connections:
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| Optimization Solution Methods | Skills Index | Newton's Method |
Last updated: 2026-01-22