Quadratic applications and model checks | Year 10

01 / LEARN

Understand the idea.

Define the variable before forming a rule

A quadratic model begins with a defined variable and its unit. Translate each relationship separately, then use the relevant quantity: rectangle area is width times length, and revenue is price times attendance.

State the feasible inputs before solving. Lengths must be positive, counts may be whole numbers, and a stated model interval limits what can be inferred even when the algebra accepts other inputs.

Interpret and test the result

Choose a method for the question being asked. Roots locate a target output such as zero; vertex form reveals a greatest or least value. A vertex gives an attainable optimum only if its input is allowed.

Check the result against the original relationships, units and restrictions. If an input must be a whole number, compare feasible nearby values when needed. Predictions also depend on the model’s assumptions continuing to hold.

02 / WORKED EXAMPLES

See the method in action.

EXAMPLE 01

An area fixes the dimensions

A rectangle is 5 m longer than it is wide. Its area is 84 m². Find its dimensions.

Worked solution

1
Define the positive width.
$\begin{array}{c} width = x m, x > 0 \\ length = (x + 5) m \end{array}$
The length exceeds the width by five metres.
2
Use the area relationship.
$\begin{aligned} x (x + 5) & = 84 \\ x^{2} + 5 x - 84 & = 0 \end{aligned}$
Multiplying the two side lengths gives the area in square metres.
3
Factor and solve.
$\begin{array}{c} (x + 12) (x - 7) = 0 \\ x = - 12 or x = 7 \end{array}$
The factor constants multiply to $- 84$ and add to $5$ .
4
Apply the domain and check both dimensions.
$\begin{array}{c} x = 7 m, x + 5 = 12 m \\ 7 \times 12 = 84, 12 - 7 = 5 \end{array}$
The negative root cannot be a width; the positive pair satisfies both the area and length difference.

Answer

Width $7$ m and length $12$ m.

EXAMPLE 02

Revenue with changing demand

A model predicts that increasing a 20-dollar ticket by x dollars changes attendance from 100 to $(100 - 2 x)$ . Find the revenue-maximising whole-dollar increase for $0 \leq x \leq 50$ .

Worked solution

1
Write price, attendance and allowed increases.
$\begin{array}{c} price = 20 + x \\ attendance = 100 - 2 x \\ x \in {0, 1, \dots, 50} \end{array}$
The model increases price and decreases attendance, with a whole-dollar increase in the stated interval.
2
Form and expand revenue.
$\begin{aligned} R & = (20 + x) (100 - 2 x) \\ = - 2 x^{2} + 60 x + 2000 \end{aligned}$
Revenue is the ticket price multiplied by the predicted number of tickets sold.
3
Complete the square.
$\begin{aligned} R & = - 2 (x^{2} - 30 x) + 2000 \\ = - 2 [{(x - 15)}^{2} - 225] + 2000 \\ = - 2 {(x - 15)}^{2} + 2450 \end{aligned}$
The negative squared term cannot increase revenue above the constant term.
4
Check that the maximum is feasible.
$\begin{array}{c} x = 15 \in {0, 1, \dots, 50} \\ R_{max} = 2450 \end{array}$
The square is zero at an allowed whole-dollar increase, so the continuous maximum is also attainable here.
5
Interpret and verify the prediction.
$\begin{array}{c} price = 20 + 15 = 35 dollars \\ attendance = 100 - 2 (15) = 70 \\ 35 \times 70 = 2450 dollars \end{array}$
The predicted maximum revenue is 2450 dollars from a 15-dollar increase, using the given attendance model.

Answer

Increase $15$ ; predicted maximum revenue $2450$ .

03 / YOUR TURN

Try it yourself.

Check your understanding

A width x satisfies $(x - 2) (x + 7) = 0$ and must be positive. What is the width?

Show answer & reasoning

$x = 2$

The factors give $x = 2$ or $x = - 7$ . Only $x = 2$ satisfies the positive-width condition, so that is the permitted width.

Keep practising

6 worksheet questions · 19 marks

Open worksheet Worked solutions

Read the questions online

A rectangle is 3 cm longer than it is wide and has area 54 cm². Find its dimensions.
3 marks
A photo is 6 cm by 10 cm. A uniform border makes the total area 96 cm². Find the border width.
4 marks
A ball’s height is $h = - 5 t^{2} + 20 t + 1$ metres after $t$ seconds. Find its maximum height and when it occurs.
3 marks
A model gives ticket revenue $R = p (80 - 2 p)$ dollars when price $p$ dollars satisfies $0 \leq p \leq 40$ . Find the price giving greatest predicted revenue.
3 marks
A design width $x$ metres must satisfy $x^{2} - 5 x + 6 = 0$ and $x > 2$ . Which root is allowed?
2 marks
A proposed height model is $h = - t^{2} + 6 t + 2$ . At $t = 4$ , a measurement is 9.5 m. Find the model’s error there and explain whether one check proves the model valid at all times.
4 marks

Read the worked solutions

Equivalent correct methods earn the corresponding marks. Include units where asked.

A rectangle is 3 cm longer than it is wide and has area 54 cm². Find its dimensions.
3 marks
$x (x + 3) = 54$ , so $(x + 9) (x - 6) = 0$ .
The positive width is $6$ .
$6$ cm by $9$ cm.
A photo is 6 cm by 10 cm. A uniform border makes the total area 96 cm². Find the border width.
4 marks
$(6 + 2 x) (10 + 2 x) = 96$ .
$4 x^{2} + 32 x - 36 = 0$ , or $(x + 9) (x - 1) = 0$ .
The positive border width is $1$ .
$1$ cm.
A ball’s height is $h = - 5 t^{2} + 20 t + 1$ metres after $t$ seconds. Find its maximum height and when it occurs.
3 marks
$h = - 5 {(t - 2)}^{2} + 21$ .
Maximum $21$ m at $t = 2$ s.
A model gives ticket revenue $R = p (80 - 2 p)$ dollars when price $p$ dollars satisfies $0 \leq p \leq 40$ . Find the price giving greatest predicted revenue.
3 marks
$R = - 2 {(p - 20)}^{2} + 800$ .
The allowed price $p = 20$ gives 40 tickets.
Price $20$ ; revenue $800$ .
A design width $x$ metres must satisfy $x^{2} - 5 x + 6 = 0$ and $x > 2$ . Which root is allowed?
2 marks
$(x - 2) (x - 3) = 0$ gives $2, 3$ .
The strict condition excludes $2$ .
$x = 3$ m.
A proposed height model is $h = - t^{2} + 6 t + 2$ . At $t = 4$ , a measurement is 9.5 m. Find the model’s error there and explain whether one check proves the model valid at all times.
4 marks
$h (4) = - 16 + 24 + 2 = 10$ .
$10 - 9.5 = 0.5$ .
Test more values in the intended interval and consider the physical assumptions.
Prediction $10$ m; overestimate $0.5$ m. One check does not establish validity for all times.

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Curriculum reference & lesson scope

Australian Curriculum v9.0 and South Australian Curriculum

Year 10 quadratics, including the fuller school-course algebra requested for Ascent: non-monic factorisation, completing the square and the quadratic formula. These lessons consolidate Year 9 monic quadratics and extend the available solving methods.

Official curriculum source ↗

In this lesson

Understand the idea.

Define the variable before forming a rule

Interpret and test the result

See the method in action.

An area fixes the dimensions

Define the positive width.

Use the area relationship.

Factor and solve.

Apply the domain and check both dimensions.

Revenue with changing demand

Write price, attendance and allowed increases.

Form and expand revenue.

Complete the square.

Check that the maximum is feasible.

Interpret and verify the prediction.

Try it yourself.

Check your understanding

Keep practising