Level 3 Differentiation Walkthrough

2016 NCEA Level 3 Differentiation Question 3(c)

2016 Paper

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Question

A rectangle has one vertex at \((0,0)\) and the opposite vertex on the curve

\[y=(x-6)^2,\qquad 0<x<6,\]

as shown on the graph below.

Rectangle with opposite vertex on y equals x minus six squared An axes-aligned rectangle has lower-left vertex at the origin. Its upper-right vertex lies on the left branch of the upward-opening parabola y equals x minus six squared, whose vertex is at six comma zero. 6 12 (x, (x - 6)²) x y

Find the maximum possible area of the rectangle.

You must use calculus and show any derivatives that you need to find when solving this problem.

You do not need to prove that the area you have found is a maximum.

First walkthrough idea

Focus to try first

Use the opposite vertex to express the rectangle's area in one variable, then find the valid stationary point.

Step 1

Build the area function

Width times height gives an expression in x only.

Show the first step’s working
\[A=xy\] \[A(x)=x(x-6)^2.\]

Walkthrough overview

What this question practises

This 2016 walkthrough is part of AS91578 — Apply differentiation methods in solving problems.

Method: Maximising a rectangle area beneath a parabola.

This is Question 3(c) from the 2016 NCEA Level 3 Differentiation paper for AS91578 — Apply differentiation methods in solving problems. Use the guided hints to practise the method before revealing the full working.

Calc.nz is an independent learning resource. Compare questions, diagrams, and assessment information with the official NZQA resources for AS91578.

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Learning summary

Review the method, not only the answer

This walkthrough helps you practise maximising a rectangle area beneath a parabola. Use the hints to plan the method, then repeat the question without hints and check each step.

Common mistake to avoid

Finding a stationary value is only part of an optimisation argument; justify that it is the required maximum and respect the domain.

Continue practising

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