Level 3 Differentiation Walkthrough

2017 NCEA Level 3 Differentiation Question 2(b)

2017 Paper

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Question

The percentage of seeds germinating depends on the amount of water applied to the seedbed that the seeds are sown in, and may be modelled by the function

\[P(w)=96\ln(w+1.25)-16w-12,\]

where \(P\) is the percentage of seeds that germinate and \(w\) is the daily amount of water applied (litres per square metre of seedbed), with \(0\le w\le15\).

Find the amount of water that should be applied daily to maximise the percentage of seeds germinating.

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

First walkthrough idea

Focus to try first

Differentiate the percentage model, solve for an interior stationary point, then use the derivative or concavity to justify that it gives the maximum in the allowed domain.

Step 1

Differentiate the model

An interior maximum of a differentiable model occurs at a stationary point, so begin by finding the derivative.

Show the first step’s working
\[P'(w)=96\left(\frac{1}{w+1.25}\right)-16\] \[P'(w)=\frac{96}{w+1.25}-16\]

The logarithm uses the chain rule; the derivative of \(w+1.25\) is \(1\).

Walkthrough overview

What this question practises

This 2017 walkthrough is part of AS91578 — Apply differentiation methods in solving problems.

Method: Maximising a logarithmic germination model over its domain.

This is Question 2(b) from the 2017 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 logarithmic germination model over its domain. 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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