B. Derivatives and Their Applications · Applications of Derivatives: Motion, Rates and Optimization

Optimization in Business: Maximizing Revenue and Profit

B2.4About 70 minutes

Learning goals

I can build a revenue function from a demand formula or from a description in words.
I can find the price and quantity that maximize revenue or profit.
I can explain why marginal revenue equals marginal cost at maximum profit.
I can optimize revenue with an exponential demand model.

Words to know

demand function
A rule giving how many items sell at each price (or the price at which a given number sells).

revenue
Price × quantity sold.

profit
Revenue − cost.

marginal revenue = marginal cost
The condition R′(x) = C′(x), which holds at a profit maximum inside the domain.

Raise the fare or lower it?

A city transit system wants more money from its buses. If it raises the fare, each rider pays more, but some riders switch to cycling or driving. If it lowers the fare, more people ride, but each pays less. Somewhere is the fare that brings in the most revenue. The curriculum's own sample problem for B2.4 asks exactly this: what fare maximizes revenue, given a model of daily ridership?

In Marginal Cost, Revenue and Profit you learned the business functions. Now you will optimize them.

Building the revenue function

The key is the demand relationship: how the number sold depends on the price. Problems give it in one of two ways.

As a formula, such as riders N(p) = 21 000 − 3000p or price p(x) = 50 − 0.02x.
In words, such as "each $0.10 increase loses 300 riders". Then let n be the number of increases, and write the price and the quantity in terms of n.

Then revenue = price × quantity, and profit = revenue − cost.

At maximum profit, P′(x) = R′(x) − C′(x) = 0, so marginal revenue = marginal cost. In words: keep selling more as long as the next item brings in more than it costs.

Worked example

Example 1: The bus fare

A city's buses carry 12 000 riders a day at a fare of $3.00. A survey suggests that each $0.10 increase in the fare would lose 300 riders a day (and each $0.10 decrease would gain 300). What fare maximizes daily revenue?

Ridership changes by 300 per $0.10, which is 3000 riders per dollar. So N(p) = 12 000 − 3000(p − 3) = 21 000 − 3000p.
Revenue: R(p) = p × N(p) = 21 000p − 3000p².
Domain: N(p) ≥ 0 means 0 ≤ p ≤ 7.
R′(p) = 21 000 − 6000p = 0 at p = 3.50.
Riders: N(3.50) = 21 000 − 10 500 = 10 500. Revenue: R(3.50) = 3.50 × 10 500 = $36 750. Compare R(0) = R(7) = 0 and the current R(3) = $36 000.

Raise the fare to $3.50. Ridership falls to 10 500, but daily revenue rises to $36 750 (from $36 000). Of course, a city may care about more than revenue: fewer riders means more cars on the road.

Daily bus revenue as a function of the fare. The highest point, where the tangent is horizontal, is at a fare of $3.50.

Worked example

Example 2: Filling the arena

A junior hockey team sells 4000 tickets per game at $20. Market research shows that each $1 decrease in price would sell 250 more tickets. What price maximizes revenue?

Let n be the number of $1 decreases. Price = 20 − n; tickets = 4000 + 250n.
R(n) = (20 − n)(4000 + 250n) = 80 000 + 5000n − 4000n − 250n² = 80 000 + 1000n − 250n².
R′(n) = 1000 − 500n = 0 at n = 2.
R″ = −500 < 0, so this is a maximum.

Charge $18. The team sells 4500 tickets for revenue of $81 000 per game (up from $80 000).

Worked example

Example 3: Maximum profit

For the concert from the marginal-cost lesson, the price function is p(x) = 50 − 0.02x and the cost is C(x) = 8000 + 12x, for 0 ≤ x ≤ 2500 tickets. Find the number of tickets and the price that maximize profit.

R(x) = 50x − 0.02x². P(x) = R(x) − C(x) = −0.02x² + 38x − 8000.
P′(x) = −0.04x + 38 = 0 at x = 950. (Check: R′(950) = 50 − 38 = 12 = C′(950). Marginal revenue equals marginal cost.)
Price: p(950) = 50 − 19 = $31.
P(950) = −0.02(902 500) + 36 100 − 8000 = −18 050 + 36 100 − 8000 = $10 050. Endpoints: P(0) = −8000 and P(2500) = −133 000 + 95 000 − 8000 < 0.

Sell 950 tickets at $31 each, for a maximum profit of $10 050. Notice that the profit-maximizing quantity (950) is less than the revenue-maximizing quantity (1250): each extra ticket costs $12, so it is not worth selling tickets that bring in less than that.

Revenue (solid) and cost (dashed) for the concert. Profit is the vertical gap between them. The gap is largest where the revenue curve's tangent (dotted) is parallel to the cost line: marginal revenue = marginal cost.

Worked example

Example 4: An exponential demand model

An app developer finds that x thousand people subscribe when the yearly price is p(x) = 40e^(−0.05x) dollars. What price maximizes revenue?

R(x) = x × p(x) = 40xe^(−0.05x) thousand dollars.
Product rule: R′(x) = 40e^(−0.05x) + 40x(−0.05)e^(−0.05x) = 40e^(−0.05x)(1 − 0.05x).
e^(−0.05x) is never 0, so R′(x) = 0 only when x = 20. R′ > 0 before and R′ < 0 after, so this is the absolute maximum for x ≥ 0.
Price: p(20) = 40e^(−1) ≈ $14.72. Revenue: R(20) = 800e^(−1) ≈ 294.3 thousand dollars.

Charge about $14.72 per year. About 20 000 people subscribe, for revenue of about $294 000.

Checklist for business problems

Define your variable (price, number of increases, or quantity) and say what it means. 2. Write price and quantity in terms of it. 3. Revenue = price × quantity. 4. Profit = revenue − cost. 5. State the domain (no negative prices or quantities). 6. Differentiate, solve, check. 7. Answer with the price and the quantity, and the revenue or profit if asked.

Just for fun

The bus-fare calculus says: raise the fare by 50 cents. The riders' calculus says: d(happiness)/d(fare) < 0. Real transit planners have to balance both derivatives.

Hands-on activity

Price a fundraiser with a spreadsheet

You need: a computer with a free spreadsheet (such as LibreOffice Calc or Google Sheets), or paper and a calculator.

A school club sells pizza slices at lunch. At $3 a slice they sell 240 slices. They estimate each $0.25 increase loses 15 sales. Each slice costs the club $1.20.

Make columns: number of increases n (0 to 16), price p = 3 + 0.25n, slices q = 240 − 15n, revenue R = pq, cost C = 1.20q, profit P = R − C.
Find the rows with the largest revenue and the largest profit.
Now do it with calculus: R(n) = (3 + 0.25n)(240 − 15n) and P(n) = (1.80 + 0.25n)(240 − 15n). Differentiate, and solve R′(n) = 0 and P′(n) = 0.

What to look for: revenue is largest at n = 2 (price $3.50, 210 slices, $735). Profit is largest at n = 4.4 by calculus, so check the nearest whole steps n = 4 and n = 5: n = 4 gives $2.80 profit per slice × 180 slices = $504, and n = 5 gives $3.05 × 165 = $503.25. The club should charge $4.00 for the best profit. The profit-maximizing price is higher than the revenue-maximizing price, because each extra slice has a cost.

Your turn

A theatre sells 300 tickets at $15. Each $1 increase loses 10 sales. What price maximizes revenue?

The demand for a product is p = 80 − 0.4x. Find the quantity and price that maximize revenue.

For p = 80 − 0.4x and cost C = 1000 + 20x, find the quantity and price that maximize profit, and the maximum profit.

A bus line carries 8000 riders a day at $2.50. Each $0.25 increase loses 400 riders. Find the best fare for revenue.

Demand is p = 30e^(−0.1x) (x in thousands). Find the price that maximizes revenue.

A Niagara orchard has 50 apple trees, each producing 600 apples. Each extra tree planted reduces the yield of every tree by 10 apples. How many trees maximize the total harvest?

For Question 3, verify that marginal revenue equals marginal cost at the best quantity.

More practice: warm-up

Revenue equals…

If p = 10 − 0.01x, write R(x).

Find R′(x) for R = 10x − 0.01x².

At maximum profit (inside the domain), marginal revenue equals…

5000 riders pay $3 each. What is the revenue?

More practice: on your own

p = 50 − 0.01x and C = 0.01x² + 10x + 500. Find the quantity, price and maximum profit.

A hotel has 120 rooms, all rented at $100 a night. Each $5 increase leaves 3 more rooms empty. Each occupied room costs $20 a night to clean and supply. Find the nightly rate that maximizes profit.

The number of people who buy a concert stream at price p is 1000e^(−0.04p). Find the price that maximizes revenue.

A company sells items at $60 each. Its cost is C(x) = 0.001x³ − 0.3x² + 40x + 1000 for 0 ≤ x ≤ 400. Find the production level that maximizes profit.

An Ontario ski hill sells 2000 lift tickets a day at $60. Each $2 decrease brings 100 more skiers. Find the best price for revenue.

Challenge

For linear demand p = a − bx and cost C = c + dx (with 0 < d < a), show that profit is largest at x = (a − d)/(2b) and the price is then (a + d)/2.

Show that with linear demand p = a − bx, revenue is largest when the price is half the price at which nobody buys.

For the bus in Example 1, N(p) = 21 000 − 3000p, each rider costs the transit system $1.00 in fuel and wear. Find the fare that maximizes profit. Why is it higher than $3.50?

Check your understanding

Quiz

A store sells 200 items at $10. Each $1 increase loses 10 sales. Which function gives the revenue after n increases?

For R(p) = 21 000p − 3000p², which fare maximizes revenue?

At the quantity that maximizes profit (inside the domain), which is true?

R(x) = 40xe^(−0.05x). Where is R′(x) = 0?

Compared with the revenue-maximizing quantity, the profit-maximizing quantity (when each item has a positive cost) is usually…

Demand p = 60 − 0.5x. What is the maximum revenue?

For parents and teachers

Big idea (B2.4): to maximize revenue or profit, write it as a function of one variable (price, quantity, or the number of price changes), state the domain, and use the derivative. At maximum profit, marginal revenue equals marginal cost.

How to help: the spreadsheet activity is a good shared task. Ask your teen why the best price for profit is higher than the best price for revenue.

Common difficulties:

Setting up "each $1 increase loses 10 sales" incorrectly (the signs, or using the price instead of the number of increases).
Giving the number of increases (n = 2) instead of the actual price ($18).
Forgetting costs when the question asks about profit.
Treating a fractional answer as wrong: prices can be $22.50, but numbers of people or items may need rounding and a check of nearby whole numbers.