Level 3 Differentiation Walkthrough

2016 NCEA Level 3 Differentiation Question 2(e)

2016 Paper

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Question

A cone of height \(h\) and radius \(r\) is inscribed, as shown, inside a sphere of radius \(6\text{ cm}\).

The base of the cone is \(s\text{ cm}\) below the \(x\)-axis.

Cone inscribed in a sphere of radius six centimetres A sphere is shown in cross-section as a circle centred at the origin with radius six. A cone has its apex at the top of the sphere and a horizontal circular base s centimetres below the x-axis. The cone has height h and base radius r. s r h 6 6 -6 -6 x y

Find the value of \(s\) which maximises the volume of the cone.

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 volume you have found is a maximum.

First walkthrough idea

Focus to try first

Express the cone's height and radius in terms of s, build its volume function, and find the valid stationary value.

Step 1

Relate the cone dimensions to s

Use the vertical positions and the sphere's radius.

Show the first step’s working
\[h=6+s\] \[s^2+r^2=6^2=36\] \[r^2=36-s^2.\]

For the configuration shown, \(0\le s<6\).

Walkthrough overview

What this question practises

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

Method: Maximising the volume of a cone inside a sphere.

This is Question 2(e) 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 the volume of a cone inside a sphere. 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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