Business and Economics Optimization

6 min read

Reference: Stewart §3.7

Maximizing Profit, Minimizing Cost

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.

Prerequisite Map

Quick Reference

Property	Value
Concept	Optimization
Chapter	3.7
Difficulty	Intermediate
Time	~20 minutes

Key Economic Functions

The Cost Function

The cost function $C(x)$ gives the total cost of producing $x$ units.

$$C(x) = \text{fixed costs} + \text{variable costs}$$

Example: $C(x) = 5000 + 8x + 0.01x^2$

Fixed costs: \$5000 (rent, equipment)
Variable costs: $8x + 0.01x^2$ (materials, labor)

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

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.

The Revenue Function

Revenue is what you earn from selling:

$$R(x) = x \cdot p(x) = \text{(quantity)} \times \text{(price per unit)}$$

The marginal revenue is:

$$\text{Marginal Revenue} = R'(x)$$

The Profit Function

Profit is revenue minus cost:

$$P(x) = R(x) - C(x)$$

The marginal profit is:

$$\text{Marginal Profit} = P'(x) = R'(x) - C'(x)$$

Summary Table

Function	Symbol	Definition	Derivative (Marginal)
Cost	$C(x)$	Total production cost	$C'(x)$ = marginal cost
Price/Demand	$p(x)$	Price per unit	(not separately defined)
Revenue	$R(x)$	$x \cdot p(x)$	$R'(x)$ = marginal revenue
Profit	$P(x)$	$R(x) - C(x)$	$P'(x)$ = marginal profit

The Fundamental Principle

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.

Worked Example: Maximizing Revenue with Rebates

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

Let $x$ = number of laptops sold per week
Starting point: 500 laptops at \$800 each

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.

Practice Problems

Level 1 Identifying Economic Functions

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 $P(x)$.

Thought Process

The cost function is already given in standard form: constant + (linear term).

Fixed costs are the constant term (costs that don’t depend on production level)
Variable costs are everything else

Revenue = quantity × price = $x \cdot p(x)$

Profit = Revenue - Cost

Show Answer

(a) Fixed costs: \$2000 (the constant term in $C(x)$)

(b) Variable cost per widget: \$15 (coefficient of $x$)

(d) Profit function: $$P(x) = R(x) - C(x) = (50x - 0.02x^2) - (2000 + 15x)$$ $$P(x) = -0.02x^2 + 35x - 2000$$

Level 2 Finding the Demand Function

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.

Thought Process

We need to relate price to quantity sold.

Starting point: 300 loaves at \$4.50

If we sell fewer loaves (below 300), price is higher
If we sell more loaves (above 300), price is lower

Change in quantity: $(x - 300)$ For each decrease of 15 loaves, price increases by \$0.25. So for each unit decrease in sales, price increases by $\frac{0.25}{15}$.

Equivalently: for each unit increase in sales, price decreases by $\frac{0.25}{15}$.

Show Answer

Starting point: $(x_0, p_0) = (300, 4.50)$

Rate of change: A decrease of 15 loaves corresponds to a \$0.25 increase.

So the slope of $p$ vs $x$ is: $$\frac{\Delta p}{\Delta x} = \frac{0.25}{-15} = -\frac{1}{60}$$

Using point-slope form: $$p - 4.50 = -\frac{1}{60}(x - 300)$$

$$p = 4.50 - \frac{x - 300}{60} = 4.50 - \frac{x}{60} + 5 = 9.50 - \frac{x}{60}$$

Demand function: $p(x) = 9.50 - \frac{x}{60}$ dollars per loaf

Check: When $x = 300$: $p(300) = 9.50 - 5 = 4.50$ ✓

Level 3 Maximizing Revenue

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?

Thought Process

Find the demand function from the given information
Write Revenue = quantity × price
Take the derivative and set equal to zero
Solve for the optimal quantity
Find the corresponding price

The theater is decreasing price to increase attendance, so the demand function will have a negative slope.

Show Answer

Step 1: Find demand function

Starting: 800 tickets at \$12 Slope: $\frac{-0.50}{25} = -0.02$ dollars per additional ticket

$$p(x) = 12 - 0.02(x - 800) = 12 - 0.02x + 16 = 28 - 0.02x$$

Step 2: Revenue function

$$R(x) = x \cdot p(x) = x(28 - 0.02x) = 28x - 0.02x^2$$

Step 3: Find critical point

$$R'(x) = 28 - 0.04x = 0$$ $$x = 700$$

This gives fewer than 800 tickets, which would mean raising prices. Recheck:

Actually, the slope calculation: a \$0.50 decrease causes a 25 increase in attendance. So if $x$ increases by 25, $p$ decreases by 0.50. Slope = $\frac{-0.50}{25} = -0.02$ ✓

Hmm, but $x = 700 < 800$ means selling fewer tickets. Let me verify by checking the endpoints and the critical point makes sense.

$R'(x) = 28 - 0.04x$ At $x = 800$: $R'(800) = 28 - 32 = -4 < 0$

So at the current level, revenue is decreasing as we sell more (lower prices). We should actually raise prices (sell fewer tickets) to increase revenue.

$x = 700$ gives: $p(700) = 28 - 14 = \$14$

Verify it’s a maximum: $R''(x) = -0.04 < 0$ ✓

Answer: Ticket price of \$14 maximizes revenue.

This means raising the price from $12 to $14, accepting fewer customers (700 instead of 800), but earning more total revenue.

Level 4 Maximizing Profit with Cost Constraints

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.

Thought Process

Profit = Revenue - Cost

Find $R(x) = x \cdot p(x)$
Find $P(x) = R(x) - C(x)$
Set $P'(x) = 0$ and solve
Calculate maximum profit
Verify $R'(x) = C'(x)$ at the critical point

Show Answer

(a) Find optimal production level

Revenue: $R(x) = x(450 - 0.003x) = 450x - 0.003x^2$

Profit: $$P(x) = R(x) - C(x) = (450x - 0.003x^2) - (25000 + 180x + 0.002x^2)$$ $$P(x) = -0.005x^2 + 270x - 25000$$

$$P'(x) = -0.01x + 270 = 0$$ $$x = 27000 \text{ units}$$

(b) Maximum profit

$$P(27000) = -0.005(27000)^2 + 270(27000) - 25000$$ $$= -0.005(729000000) + 7290000 - 25000$$ $$= -3645000 + 7290000 - 25000$$ $$= \$3{,}620{,}000$$

(c) Verify MR = MC

Marginal revenue: $R'(x) = 450 - 0.006x$ At $x = 27000$: $R'(27000) = 450 - 162 = 288$

Marginal cost: $C'(x) = 180 + 0.004x$ At $x = 27000$: $C'(27000) = 180 + 108 = 288$

$R'(27000) = C'(27000) = 288$ ✓

At optimal production, the cost of making one more phone equals the revenue it would bring in.

Level 5 Average Cost Minimization

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.

(d) Explain intuitively why “marginal = average at minimum average” makes sense.

Thought Process

For part (a): Use the quotient rule to find $\bar{C}'(x)$ and set it equal to zero.

For part (b): Apply the result from (a), or directly minimize $\bar{C}(x)$.

For part (d): Think about what happens when you add something to an average. If you add a value equal to the average, the average stays the same.

Show Answer

(a) Proof that MC = AC at minimum average cost

$$\bar{C}(x) = \frac{C(x)}{x}$$

Using the quotient rule: $$\bar{C}'(x) = \frac{C'(x) \cdot x - C(x) \cdot 1}{x^2} = \frac{C'(x) \cdot x - C(x)}{x^2}$$

At a minimum, $\bar{C}'(x_0) = 0$: $$\frac{C'(x_0) \cdot x_0 - C(x_0)}{x_0^2} = 0$$

$$C'(x_0) \cdot x_0 = C(x_0)$$

$$C'(x_0) = \frac{C(x_0)}{x_0} = \bar{C}(x_0)$$

$\square$

(b) Find minimum average cost

$$\bar{C}(x) = \frac{10000 + 50x + 0.01x^2}{x} = \frac{10000}{x} + 50 + 0.01x$$

$$\bar{C}'(x) = -\frac{10000}{x^2} + 0.01 = 0$$

$$\frac{10000}{x^2} = 0.01$$

$$x^2 = 1000000$$

$$x = 1000 \text{ units}$$

(c) Verify MC = AC at x = 1000

Marginal cost: $C'(x) = 50 + 0.02x$ $C'(1000) = 50 + 20 = 70$

Average cost: $\bar{C}(1000) = \frac{10000}{1000} + 50 + 0.01(1000) = 10 + 50 + 10 = 70$

$C'(1000) = \bar{C}(1000) = 70$ ✓

(d) Intuitive explanation

Think of average cost like your GPA. If you take a course and get exactly your current GPA, your GPA stays the same.

Similarly:

If marginal cost < average cost, producing one more unit brings down the average (you’re adding a “below average” data point)
If marginal cost > average cost, producing one more unit raises the average
Only when marginal = average does the average stay constant, which is exactly what happens at a minimum (or maximum)

At the minimum average cost, adding one more unit costs exactly the average. Below that point, each unit costs less than average (pulling average down). Above that point, each unit costs more than average (pushing average up).

CCI-Style Conceptual Questions

Level 2 Conceptual: When to Stop Producing

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

Thought Process

Marginal revenue = what you earn from one more unit Marginal cost = what it costs to make one more unit

If MC > MR, then making one more unit costs more than it earns: not profitable.

What about making one fewer? Then you save MC but lose MR. Since MC > MR, you save more than you lose.

Show Answer

(B) is correct.

Since marginal cost (\$52) exceeds marginal revenue (\$45), producing the last unit actually decreased profit by \$7. The company should produce fewer units.

At maximum profit, MR = MC. Here MR < MC, so the company has overproduced.

Level 3 Conceptual: Interpreting Marginal Cost

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

Thought Process

$C'(x)$ is the marginal cost: the rate of change of cost with respect to quantity.

$C'(500) = 13$ means the derivative at $x = 500$ is 13.

The derivative gives the instantaneous rate of change, which approximates the change from producing one additional unit.

Show Answer

(D) is correct.

$C'(500) = 13$ means that at $x = 500$, cost is changing at a rate of \$13 per item.

This approximates the cost of producing the 501st item (the marginal unit): $$C(501) - C(500) \approx C'(500) \cdot 1 = \$13$$

Note: (B) is close, but “increasing at \$13 per item” is awkward phrasing. (C) is a common misconception: the derivative gives an approximation, not the exact cost of any single item.

Mastery Checklist

I can identify cost, revenue, and profit functions from a problem description
I can construct a demand function from market data
I understand that marginal = derivative (marginal cost = $C'$, marginal revenue = $R'$)
I can find the production level that maximizes profit
I understand why maximum profit occurs when $MR = MC$
I can interpret marginal quantities in practical terms
I can find the production level that minimizes average cost

Mental Model

The Balancing Point:

Imagine you’re filling a bathtub with the drain open.

Water flowing in = Revenue from selling
Water flowing out = Cost of producing

At first, water flows in faster than it drains out (MR > MC): 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.

Connections

Looking back:

Optimization Problem Setup teaches the translation from words to functions
Optimization Solution Methods provides the calculus techniques

Looking ahead:

Integral calculus can recover total cost from marginal cost
Multivariable calculus extends these ideas to multiple products

Real-world connections:

Every business uses these principles, whether explicitly or intuitively
Pricing algorithms (airlines, rideshare) optimize revenue continuously
The MR = MC rule is a cornerstone of microeconomics

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Last updated: 2026-01-22