LEARN · EXPLAIN · REVISE

A rectangular display has perimeter 28 m. Let one side be x metres. Form its area function and find the greatest possible area in square metres.

Read the idea, work independently, then explain what changed.

TOPIC 01

Quadratic equations and functions

The downward-opening area graph has vertex (7, 49). Checking the domain matters: a negative or zero side length would not describe the display.

PREDICT → MOVE → EXPLAIN

What does each coefficient change?

Before moving a slider, predict the opening, vertex and intercepts. Then compare the graph with your prediction.

Try it. Leave your reasoning visible.

Use one hint at a time. A correction explains what changed, not just the final answer.

01 / Transfer#Worked example

A rectangular display has perimeter 28 m. Let one side be x metres. Form its area function and find the greatest possible area in square metres.

2(x+\ell)=28,\qquad 0<x<14

Working and explanation

BUILD THE REASONING

Hint 1

The sum of the two different side lengths is half the perimeter.

Hint 2

Write the area as 14x − x², then complete the square.

Worked solution

Express the other side in terms of x.
$\ell=14-x$
Multiply the side lengths and complete the square.
$A(x)=x(14-x)=49-(x-7)^2$
The squared term is nonnegative, and x = 7 is in the allowed domain.
$A(x)\le49,\qquad A(7)=49$

The greatest area is 49 m², attained by a 7 m × 7 m square.

Checks and common pitfalls: The downward-opening area graph has vertex (7, 49). Checking the domain matters: a negative or zero side length would not describe the display.

Think first. Reveal a hint when the class is ready.