Level 3 Differentiation Walkthrough

2023 NCEA Level 3 Differentiation Question 1(e)

2023 Paper

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Question

The graph of

\[ y=x(x-2m)^2,\qquad m>0 \]

is shown, and the total shaded area between the curve and the \(x\)-axis from \(x=0\) to \(x=2m\) is

\[ A=\frac{4m^4}{3}. \]

A right-angled triangle is constructed with one vertex at \((0,0)\) and another on the curve.

Show that the maximum area of such a triangle is \(\frac{3}{8}\) of the total shaded area.

You must use calculus and show any derivatives that you need to find when solving this problem. You do not have to prove that the area you found is a maximum.

First walkthrough idea

Hint to try first

If the point on the curve is \((x,y)\), then the triangle area is \(\frac{1}{2}xy\).

Step 1

Write the triangle area in terms of \(x\)

Start from \(\frac{1}{2}xy\), then substitute \(y=x(x-2m)^2\).

Show the first step’s working

Start from \(\frac{1}{2}xy\), then substitute \(y=x(x-2m)^2\).

Key result

\[ A=\frac{x^2}{2}(x-2m)^2 \]

Walkthrough overview

What this question practises

This 2023 walkthrough is part of AS91578 — Apply differentiation methods in solving problems.

Method: Maximising a triangle area on the curve \(y=x(x-2m)^2\).

This is Question 1(e) from the 2023 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 triangle area on the curve \(y=x(x-2m)^2\). 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

← Question 1(d) Question 2(a) →