A. Quadratic Functions · Quadratic Functions

Solving Problems: Area, Revenue and Profit

A3.3 A2.4About 70 minutes

Learning goals

I can write a quadratic function for an area, revenue or profit situation.
I can find a maximum or minimum value and the input that produces it.
I can find break-even points.
I can check that my answer makes sense in the context.

Words to know

optimization
Finding the best (maximum or minimum) value of a quantity.

revenue
Money taken in: price × number sold.

profit
Revenue minus cost.

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

uniform width
The same width all the way around, as in a border or path.

Making the most of it

A farmer wants the biggest field for a fixed amount of fence. A ski club wants the ticket price that brings in the most money. A small business wants to know how many items to make to earn the most profit, and when it would just break even. These are optimization problems: find the best value. Very often the function involved is quadratic, and the best value is at the vertex.

A plan for optimization problems

Draw a diagram or make a table. Name the variable (with units) and write what it means.
Write the function: area = length × width; revenue = price × number sold; profit = revenue − cost.
Find the vertex, using the zeros (factored form), completing the square, or h = −b ÷ (2a).
Answer the question asked: the best value of the variable, the maximum (or minimum) value, or a related quantity such as the price.
Check that the answer is in the sensible domain (positive lengths, enough seats, whole numbers if needed).

Worked example

Example 1: Fencing beside a river

A farmer has 120 m of fence to enclose a rectangular field beside a straight river. No fence is needed along the river. What dimensions give the largest area?

Let x be the length (m) of each side perpendicular to the river. The side parallel to the river is 120 − 2x.
Area: A(x) = x(120 − 2x). Domain: x > 0 and 120 − 2x > 0, so 0 < x < 60.
Zeros: x = 0 and x = 60. The vertex is halfway: x = 30.
A(30) = 30(120 − 60) = 30 × 60 = 1800 m².
The field is 30 m by 60 m, with the 60 m side along the river.

30 m by 60 m (60 m along the river); maximum area 1800 m².

Only three sides need fence: two sides of length x and one of length 120 − 2x. The area is a quadratic function of x.

Worked example

Example 2: Ticket prices

A ski club charges $12 for a movie night and 400 people come. A survey shows that for every $1 increase in price, 20 fewer people will come. What price gives the greatest revenue?

Let x be the number of $1 increases. Price: 12 + x. Attendance: 400 − 20x.
Revenue: R(x) = (12 + x)(400 − 20x). It is already in factored form.
Zeros: 12 + x = 0 gives x = −12; 400 − 20x = 0 gives x = 20.
Axis halfway: x = (−12 + 20) ÷ 2 = 4.
R(4) = (16)(320) = $5120. The best price is 12 + 4 = $16, with 320 people.
Check: at $12 the revenue is 12 × 400 = $4800, and at $17 it is 17 × 300 = $5100. Both are less. ✓

Charge $16; the maximum revenue is $5120.

Revenue R(x) = (12 + x)(400 − 20x), where x is the number of $1 price increases. Raising the price helps at first, but past 4 increases the lost customers cost more than the higher price earns.

Worked example

Example 3: Profit and break-even

A small workshop makes hockey sticks. Its daily profit in dollars from making x sticks is P(x) = −2x² + 80x − 600. Find the break-even points and the maximum profit.

Break-even means profit = 0. Solve −2x² + 80x − 600 = 0. Divide by −2: x² − 40x + 300 = 0, so (x − 10)(x − 30) = 0.
Break-even at 10 sticks and 30 sticks. Between them the workshop makes a profit; outside, it loses money.
Maximum: halfway, x = 20. P(20) = −800 + 1600 − 600 = $200.
Since x counts sticks, the domain is whole numbers, and x = 20 is a whole number. ✓

Break-even at 10 and 30 sticks; maximum profit $200 at 20 sticks.

Worked example

Example 4: A picture frame (solving an equation)

A photo 20 cm by 30 cm is surrounded by a frame of uniform width x cm. The total area of photo and frame is 1000 cm². How wide is the frame?

Outer dimensions: (20 + 2x) by (30 + 2x).
(20 + 2x)(30 + 2x) = 1000, so 600 + 100x + 4x² = 1000, and 4x² + 100x − 400 = 0. Divide by 4: x² + 25x − 100 = 0.
Quadratic formula: x = (−25 ± √(625 + 400)) ÷ 2 = (−25 ± √1025) ÷ 2 ≈ (−25 ± 32.02) ÷ 2.
x ≈ 3.51 or x ≈ −28.5 (reject: a width can't be negative).
The frame is about 3.5 cm wide. Check: 27.0 × 37.0 ≈ 1000 ✓.

The frame is about 3.5 cm wide.

Just for fun

A shop owner kept raising prices to earn more. Revenue went up, up, up… and then down. The owner had walked right past the vertex. Every parabola has a peak, and business is no exception.

Hands-on activity

The best field beside a wall

You need: a 30 cm piece of string or a strip of paper 30 cm long, a hardcover book, a sheet of graph paper (1 cm squares if you have them), a ruler, a pencil and a calculator.

Lay the book on the graph paper. Its straight edge is the "river": no fence needed there.
Use the 30 cm string as the fence. Bend it into three sides of a rectangle against the book: two ends of length x and a front of length 30 − 2x.
Try x = 3, 5, 7, 7.5, 9, 11 and 13 cm. For each, trace the rectangle and find its area by counting squares or by A(x) = x(30 − 2x).
Record the results in a table and plot A against x.
Use the zeros of A(x) to find the vertex. Does it match the best shape in your table?
Write two sentences comparing your best shape with the river field in Example 1.

What to look for: zeros 0 and 15, vertex x = 7.5 cm, maximum area 7.5 × 15 = 112.5 cm². As in Example 1, the best shape has the front side twice as long as each end.

Your turn

Two numbers add to 30. What is the largest possible product?

A rectangle has a perimeter of 60 m. What is its largest possible area?

80 m of fence encloses a rectangle against a barn (three sides). Find the maximum area and the dimensions.

R(x) = (10 + x)(200 − 10x) is the revenue in dollars when the price $10 is raised by $x. Find the best price and the maximum revenue.

P(x) = −x² + 14x − 40. Find the break-even points and the maximum profit.

A rectangle's length is 3 m more than its width, and its area is 70 m². Find its dimensions.

More practice: warm-up

A(w) = w(10 − w). Find A(4).

State the maximum value of R(x) = −2(x − 5)² + 300 and where it occurs.

Revenue = ______ × ______.

What does "break-even" mean for a profit function?

Two numbers add to 10 and are equal. What is their product?

More practice: on your own

A charter bus costs $50 per person for 30 people. For every extra person, the price drops by $1 for everyone. The bus holds 50. What number of passengers gives the most revenue?

A 10 m by 6 m garden has a path of uniform width x around it. The garden and path cover 96 m². Find x.

600 m of fence encloses a rectangle and divides it in two with a fence parallel to the width (3 widths and 2 lengths). Find the maximum total area.

P(x) = −5x² + 200x − 1500 is the profit in dollars when an item sells for $x. Find the best price, the maximum profit and the break-even prices.

Two numbers differ by 8. What is the smallest possible product?

Challenge

A theatre with 500 seats charges $20 and sells 300 tickets. For each $1 decrease, it sells 25 more. Find the price that maximizes revenue.

A rectangular lot beside a road needs fence on all four sides. The fence along the road costs $10/m; the other three sides cost $5/m. The budget is $1200. Find the largest area.

Prove that among all rectangles with perimeter P, a square has the greatest area.

Check your understanding

Quiz

Two numbers add to 16. What is their largest possible product?

100 m of fence encloses a rectangle (all four sides). What is the largest area?

R(x) = (8 + x)(100 − 5x). How many $1 increases give the maximum revenue?

P(x) = −(x − 2)(x − 12). What are the break-even points?

P(x) = −(x − 2)(x − 12). What is the maximum profit?

Why must the domain of a revenue model R(x) = (price)(number sold) be restricted?

For parents and teachers

Big idea: expectation A3.3 asks students to solve problems arising from real-world applications, given the algebraic representation of a quadratic function. This lesson covers area, revenue, profit and number problems, many of which are optimization problems solved at the vertex.

How to help: start every problem with a sketch or a small table (try x = 0, 1, 2…). Ask "What does x stand for? What are its units? What does the question really want: x, the maximum, or something else like the price?"

Common difficulties:

Giving the vertex x-value when the question asks for the price, the dimensions, or the maximum value.
Setting up revenue as price + number sold instead of price × number sold.
Forgetting that a fence along a river or wall covers only three sides.
Ignoring domain limits (seats on a bus, negative widths).