Grade by Grade
Quadratic Relations of the Form y = ax² + bx + c · Quadratic Expressions and Equations

Solving Problems: Area, Revenue and Number Puzzles

4.2 4.1 3.8About 75 minutes
Learning goals
  • I can write a quadratic equation or relation to model an area, revenue or number problem.
  • I can solve the equation or find the vertex, and interpret the answer in context.
  • I can reject solutions that do not make sense in the situation.
  • I can follow a problem-solving plan and answer in a sentence with units.

Words to know

revenue
The total money taken in from sales: price × number sold.
uniform width
The same width all the way around, as in a path or frame.
parabolic arch
An arch whose shape can be modelled by a parabola that opens down.
model
An equation or relation that represents a real situation.

Quadratics at work

Quadratic relations are not only about thrown balls. They appear whenever two quantities that both depend on x are multiplied:

  • area = length × width, when both sides change with x;
  • revenue = price × number sold, when raising the price lowers sales;
  • number puzzles where a number is multiplied by another related number.

The plan is always the same: define a variable, write an equation or relation, solve or find the vertex, and check that the answer makes sense.

A problem-solving plan

  1. Understand: What is asked? What is given? Draw a diagram.
  2. Define a variable with units ("let x be the width of the path in metres").
  3. Write an equation (if a value is given) or a relation (if you need a max or min).
  4. Solve it: factor, use the formula, complete the square, or use technology.
  5. Check each answer in the original situation and reject impossible ones.
  6. Answer in a sentence with units.
garden10 m by 6 mpath, width x10 + 2x6 + 2xTotal area = 96 m²(10 + 2x)(6 + 2x) = 964x² + 32x + 60 = 96x² + 8x − 9 = 0(x + 9)(x − 1) = 0x = 1 (reject x = −9)(drawn with x = 1, to scale)
A garden with a path of equal width x all the way around. Each outside dimension grows by 2x, because the path is on both sides.
Worked example

Example 1: A path around a garden

A community garden is 10 m by 6 m. A path of uniform width will surround it, and the garden and path together must cover 96 m². How wide is the path?

  1. Let x be the width of the path in metres. The outer rectangle is (10 + 2x) by (6 + 2x).
  2. Equation: (10 + 2x)(6 + 2x) = 96.
  3. Expand: 60 + 20x + 12x + 4x² = 96, so 4x² + 32x − 36 = 0.
  4. Divide by 4: x² + 8x − 9 = 0. Factor: (x + 9)(x − 1) = 0.
  5. x = −9 (impossible for a width) or x = 1.
  6. Check: 12 m × 8 m = 96 m² ✓.

The path is 1 m wide.

Worked example

Example 2: Maximum revenue

A school club sells hoodies for $30 and sells 200. A survey suggests that for every $1 increase in price, 5 fewer hoodies will sell. What price gives the maximum revenue, and what is that revenue?

  1. Let x be the number of $1 increases. Price: 30 + x. Number sold: 200 − 5x.
  2. Revenue: R = (30 + x)(200 − 5x).
  3. Zeros: 30 + x = 0 gives x = −30; 200 − 5x = 0 gives x = 40.
  4. Axis: x = (−30 + 40) ÷ 2 = 5.
  5. R = (35)(175) = 6125.
  6. Compare: at the current price, R = 30 × 200 = 6000. The increase earns $125 more.

Charge $35; they should sell 175 hoodies for a maximum revenue of $6125.

051015202530354045010002000300040005000600070008000xrevenue ($)max (5, 6125)(0, 6000)(40, 0)
Revenue for the hoodie problem, R = (30 + x)(200 − 5x), where x is the number of $1 increases, drawn for 0 ≤ x ≤ 40. The maximum is at x = 5 (a $35 price). At x = 40 (a $70 price), the model predicts no sales.
Worked example

Example 3: Will the truck fit?

A parabolic arch spans 40 m and is 10 m high at the centre. A truck 4 m tall drives with its side 5 m from one edge of the arch. Will it fit?

  1. Put the origin at the left edge. Zeros: x = 0 and x = 40. So y = ax(x − 40).
  2. The vertex is at x = 20 with y = 10: 10 = a(20)(−20) = −400a, so a = −0.025.
  3. Model: y = −0.025x(x − 40).
  4. At x = 5: y = −0.025(5)(−35) = 4.375.
  5. The arch is 4.375 m high there, and 4.375 > 4.

Yes, with about 0.375 m (37.5 cm) to spare at the truck's outside edge.

4 m10 m highspan 40 m5 m0y = −0.025x(x − 40)At x = 5:y = −0.025(5)(−35)= 4.375 m4.375 > 4, so thetruck fits (barely!)
The arch y = −0.025x(x − 40) and a 4 m tall truck whose side is 5 m from the left edge (drawn to scale). At x = 5 the arch is 4.375 m high, just above the truck's 4 m roof.
Worked example

Example 4: A number problem

Two numbers differ by 4, and their product is 96. Find the numbers.

  1. Let the smaller number be x. The larger is x + 4.
  2. x(x + 4) = 96, so x² + 4x − 96 = 0.
  3. Factor: (x + 12)(x − 8) = 0, so x = −12 or x = 8.
  4. If x = 8, the numbers are 8 and 12. If x = −12, they are −12 and −8.
  5. Both pairs check: 8 × 12 = 96 and (−12)(−8) = 96.

8 and 12, or −12 and −8. (In a number problem, negative answers are allowed unless the question says otherwise.)

Hands-on activity

Frame a photo

You need: a postcard or photo that is about 10 cm by 15 cm (or draw a 10 cm by 15 cm rectangle), a large sheet of paper, a ruler, a pencil, and scissors with a grown-up's help.

The goal: make a paper frame of equal width x all the way around, so that the frame alone has the same area as the photo.

  1. The photo's area is 10 × 15 = 150 cm². The photo plus frame must be 300 cm².
  2. Write the equation: (10 + 2x)(15 + 2x) = 300.
  3. Expand, rearrange and solve. (Hint: you should get 4x² + 50x − 150 = 0.)
  4. Draw the outer rectangle and cut out the frame. Measure to check.

Model answer: 4x² + 50x − 150 = 0, or 2x² + 25x − 75 = 0. Factoring gives (2x − 5)(x + 15) = 0, so x = 2.5 cm. The outer rectangle is 15 cm by 20 cm = 300 cm² ✓, and the frame area is 300 − 150 = 150 cm² ✓.

Just for fun

Why was the student council treasurer so happy to see the revenue parabola?

Because for once, reaching the top meant the club could finally afford the pizza party.

Your turn

1

Two numbers have a sum of 15 and a product of 56. Find them.

2

A rectangle's length is 5 m more than its width, and its area is 84 m². Find its dimensions.

3

In Example 1, what would the path width be if the total area had to be 140 m²?

4

In Example 2, at what price would the model predict no hoodies sold?

5

In Example 2, what is the revenue if the price is $40? Is that more or less than at $35?

6

For the arch y = −0.025x(x − 40), how high is the arch 10 m from the left edge?

More practice: warm-up

1

A square with side (x + 3) cm has area 49 cm². Find x.

2

The product of two consecutive even numbers is 48. Find them.

3

For R = (20 + x)(100 − 2x), find R when x = 0.

4

Find the zeros of R = (20 + x)(100 − 2x).

5

Find the maximum of R = (20 + x)(100 − 2x).

More practice: on your own

1

A rectangular pool is 8 m by 4 m. A deck of uniform width surrounds it, and the pool and deck cover 96 m². Find the deck's width.

2

A bus tour charges $40 and has 60 riders. For each $1 decrease, 2 more people ride. What price maximizes revenue?

3

For the arch in Example 3, a 4 m truck drives with its side 4 m from the edge instead. Does it fit?

4

Solve the photo frame equation (10 + 2x)(15 + 2x) = 300.

5

The sum of a number and its square is 72. Find the number.

Challenge

1

A farmer has 100 m of fence to make a rectangular pen divided into two parts by one extra fence parallel to the width. What dimensions give the largest total area?

2

A theatre sells 400 tickets at $12. Each $0.50 increase loses 10 sales. Find the price for maximum revenue.

3

The two legs of a right triangle differ by 7 cm, and the hypotenuse is 13 cm. Find the legs.

Check your understanding

Quiz

1.

Two numbers differ by 3 and their product is 40. Which could they be?

2.

A 10 m by 6 m garden has a path of width x around it, with total area 96 m². What is x?

3.

Revenue is R = (30 + x)(200 − 5x), where x is the number of $1 increases on a $30 price. What price gives the maximum revenue?

4.

To find the maximum revenue from a quadratic revenue model, you should find:

5.

An arch is modelled by y = −0.025x(x − 40). What is its maximum height?

6.

The sum of a number and its square is 30. What could the number be?

For parents and teachers

Big idea: This lesson continues expectation 4.2: solving problems from realistic situations represented by a quadratic relation. It uses area, revenue and number contexts, and the problem-solving process (understand, model, solve, check, answer). These problems are good preparation for Grade 11 functions courses, where quadratic models return in more depth.

How to help: Ask your teen to draw a diagram and write "Let x be ..." with units before writing any equation. The photo-frame activity shows a real answer you can measure. When checking, ask whether each root makes sense.

Common difficulties:

  • Adding only x (not 2x) to each dimension for a border.
  • Mixing up the number of increases (x) with the price (30 + x) in the final answer.
  • Forgetting to answer the actual question (for example, giving x = 5 instead of "charge $35").