A. Characteristics of Functions · Quadratic Functions

Real-World Problems: Revenue and Profit

A2.3 A2.2About 75 minutes

Learning goals

I can build a revenue or profit function from a description.
I can find the maximum revenue or profit and the input that produces it.
I can find break-even points and solve "at least" questions using zeros and a graph.
I can interpret answers with correct units and sensible restrictions.

Words to know

revenue
The money taken in from sales: price × number sold.

cost
The money spent to make and sell a product, often fixed costs plus a cost per item.

profit
Revenue − cost.

break-even point
A value where profit is 0 (revenue equals cost).

quadratic inequality
A statement such as P(x) ≥ 4000 that is true for an interval of x-values.

Where do business functions come from?

A ski hill, a bakery or a video company has to choose prices and budgets. Raise the price too much and fewer people buy; spend too little on advertising and nobody hears about you. Many of these trade-offs give quadratic functions, and the best choice is the vertex.

The key relationships are:

Revenue = (price per item) × (number of items sold)
Profit = Revenue − Cost

When the price goes up and the number sold goes down in a straight-line way, the product of the two is quadratic.

Worked example

Example 1: A profit problem based on the curriculum's sample

The profit of a video company, P(x), in thousands of dollars, is given by P(x) = −5x² + 550x − 5000, where x is the amount spent on advertising, in thousands of dollars. (a) Find the maximum profit. (b) For what amounts of advertising does the company make a profit? (c) For what amounts is the profit at least $4 000 000?

(a) x = −b/(2a) = −550/(−10) = 55. P(55) = −5(3025) + 550(55) − 5000 = −15 125 + 30 250 − 5000 = 10 125.
So the maximum profit is $10 125 000, when $55 000 is spent on advertising.
(b) Find where P(x) = 0: −5x² + 550x − 5000 = 0. Divide by −5: x² − 110x + 1000 = 0, so (x − 10)(x − 100) = 0 and x = 10 or x = 100.
The parabola opens down, so P(x) > 0 between the zeros: 10 < x < 100, that is, between $10 000 and $100 000 of advertising.
(c) $4 000 000 is 4000 thousand. Solve P(x) = 4000: −5x² + 550x − 9000 = 0, so x² − 110x + 1800 = 0, and (x − 20)(x − 90) = 0.
The graph is at or above P = 4000 between these values: 20 ≤ x ≤ 90, that is, from $20 000 to $90 000 of advertising.

Maximum profit $10 125 000 at $55 000 of advertising; a profit for $10 000 < advertising < $100 000; at least $4 000 000 for $20 000 to $90 000.

The profit function P(x) = −5x² + 550x − 5000. The shaded band shows the advertising amounts (20 ≤ x ≤ 90) where the graph is on or above the dashed line P = 4000.

Solving "at least" and "at most" questions

To find where f(x) ≥ m, solve f(x) = m to find the boundary values, then use the graph. If the parabola opens down, f(x) ≥ m between the boundary values. If it opens up, f(x) ≥ m outside them. Always sketch the graph.

Worked example

Example 2: Building a revenue function

A ski hill rents snowshoes for $8 a day and rents 300 pairs. A survey suggests that for every $0.50 increase in price, 10 fewer pairs are rented. What price gives the maximum revenue?

Let n be the number of $0.50 increases. Price: 8 + 0.5n. Pairs rented: 300 − 10n.
R(n) = (8 + 0.5n)(300 − 10n). This is in factored form!
Zeros: 8 + 0.5n = 0 gives n = −16; 300 − 10n = 0 gives n = 30.
Axis: n = (−16 + 30)/2 = 7.
R(7) = (8 + 3.5)(300 − 70) = 11.50 × 230 = 2645.
Check in standard form: R(n) = −5n² + 70n + 2400, and −70/(2 × (−5)) = 7 ✓.

A price of $11.50 (7 increases) gives the maximum daily revenue of $2645 (230 pairs).

Revenue R(n) = (8 + 0.5n)(300 − 10n) for n increases of $0.50. Raising the price helps at first, but past n = 7 the lost rentals cost more than the higher price earns. (The vertical axis starts at $2000.)

Worked example

Example 3: Profit with costs

A sugar bush sells bottles of maple syrup. At a price of p dollars, it sells 400 − 10p bottles a season. Each bottle costs $8 to produce, and fixed costs are $1000. Find the price that maximizes profit, the maximum profit, and the break-even prices.

Profit per bottle is p − 8, so P(p) = (p − 8)(400 − 10p) − 1000.
The part (p − 8)(400 − 10p) is 0 at p = 8 and p = 40, so the axis is p = (8 + 40)/2 = 24. (Subtracting 1000 does not move the axis.)
P(24) = (16)(160) − 1000 = 2560 − 1000 = 1560.
Break-even: P(p) = 0. Expand: P(p) = −10p² + 480p − 4200 = 0, so p² − 48p + 420 = 0.
p = (48 ± √(2304 − 1680))/2 = (48 ± √624)/2 ≈ (48 ± 24.98)/2, so p ≈ 11.51 or p ≈ 36.49.

Charge $24 for a maximum profit of $1560. The business breaks even at about $11.51 and $36.49, and makes a profit for prices between them.

Hands-on activity

Price survey at home

You need: paper, a pencil and at least three people (family, friends or neighbours).

Pick a homemade item, such as a cookie at a school bake sale.
Ask each person how many cookies they would buy at $0.50, $1.00, $1.50 and $2.00. Add up the totals for each price.
For each price, compute revenue = price × total sold. Which price gives the most?
Check whether the number sold drops by roughly the same amount for each $0.50 increase. If so, write a revenue function R(n) = (0.50 + 0.5n)(q − dn), where q is the number sold at $0.50 and d is the drop per increase, and find its vertex.

What to look for: a sample result is 24, 18, 12 and 6 cookies, so revenues of $12, $18, $18 and $12. The drop is 6 cookies per $0.50, so R(n) = (0.5 + 0.5n)(24 − 6n), with zeros n = −1 and n = 4, axis n = 1.5 and maximum R(1.5) = 1.25 × 15 = $18.75 at a price of $1.25.

Just for fun

Funny but true: in the snowshoe example, raising the price from $8.00 to $11.50 earns more money even though 70 fewer pairs are rented. Then again, the snowshoes probably appreciate the day off.

Your turn

For P(x) = −5x² + 550x − 5000 (Example 1), find and interpret P(30).

A hoodie sells for $20 and 150 are sold each month. For each $1 increase, 5 fewer are sold. Write a revenue function R(x) for x increases.

Find the price that gives the maximum revenue for the hoodies, and the maximum revenue.

Write the hoodie revenue function in standard form and check your answer with x = −b/(2a).

A profit function is P(x) = −2x² + 120x − 1000 dollars, where x is the price in dollars. Find the maximum profit and the break-even prices.

For the same P(x), find the prices that give a profit of at least $600.

A theatre sells 400 tickets at $15. For each $1 decrease in price, 40 more tickets sell. What price maximizes revenue?

More practice: warm-up

A store sells 85 items at $12 each. What is the revenue?

Revenue is $5000 and costs are $3200. What is the profit?

What does "break even" mean?

P(x) = −(x − 40)² + 900. What is the maximum profit, and for what x?

P(x) = −3(x − 10)(x − 50). Find the break-even values and the maximum profit.

More practice: on your own

A lemonade stand sells 120 cups a day at $2.00. Each $0.25 increase loses 8 sales. Find the best price if prices must be whole multiples of $0.25.

For the video company in Example 1, find the advertising amounts that give a profit of at least $8 000 000. Round to the nearest $100.

A bus tour costs $80 per person for 40 people. For each extra person, everyone's fare drops by $1. The bus has 70 seats. How many people give the maximum revenue?

A company's profit for selling x items is P(x) = −0.1x² + 60x − 2000 dollars. Find the maximum profit and the whole numbers of items that give a profit.

Explain why the highest possible price usually does not give the highest revenue.

Challenge

A new tax costs the video company in Example 1 a fixed $500 000 a year. Does its best advertising amount change? What is its new maximum profit?

At price p dollars, a company sells 1200 − 40p items. Each item costs $4 to make and fixed costs are $2000. Find the best price and the maximum profit.

A store sells 200 toques a week at $24. Each $1 decrease sells 20 more. Can weekly revenue reach $6000? Use the discriminant.

Check your understanding

Quiz

What is the revenue from selling 150 items at $8 each?

For P(x) = −5x² + 550x − 5000, the maximum profit occurs when x =

A profit function opens down and has zeros 10 and 100. When is the profit positive?

R(x) = (10 + x)(50 − 2x). For what x is revenue greatest?

P(x) = −(x − 30)² + 400. Which statement is correct?

P(x) = −2x² + 120x − 1000. For which prices x is the profit at least $600?

For parents and teachers

Big idea: Expectation A2.3: solve problems involving quadratic functions from real-world applications, represented with function notation. Example 1 is based on the curriculum's sample problem (a video company's profit as a function of advertising, and the amounts that give a profit of at least $4 000 000); part (b) is added here.

How to help: Ask your teen to state the meaning of every variable and its units before solving, and to answer in a sentence with units (dollars, thousands of dollars, price per item).

Common difficulties:

Units: in Example 1 both x and P are in thousands of dollars.
Answering with the number of increases instead of the price.
Choosing the wrong side of the zeros for an inequality. A quick sketch prevents this.
Forgetting that some variables must be whole numbers (tickets, people, $0.25 steps).