12.11PureStretch

Modelling with Differentiation

Differentiation finds rates of change and optimum values in real problems — the largest volume, the minimum cost, the fastest speed. The technique is always the same: differentiate, set to zero, solve, then check it is the max or min you want.

30 min Video by Zeeshan Zamurred Differentiation

Edexcel AS Level Maths: 12.11 Modelling with DifferentiationWatch the full walkthrough before the notes below.

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What you'll be able to do

Interpret the derivative as a rate of change
Set up an optimisation problem
Find a maximum or minimum value in context
Verify the nature of the optimum

Derivative as a rate

In context, $\frac{d y}{d x}$ is a rate of change — for example $\frac{d V}{d r}$ is how fast volume changes with radius. Differentiating a model gives the rate directly.

Optimisation

To find a maximum or minimum of a quantity, differentiate it, set the derivative to zero, and solve. This locates the optimum — then confirm with the second derivative.

\frac{d V}{d x} = 0 \Rightarrow optimum, check \frac{d ^{2} V}{d x ^{2}}

Differentiate, solve, then verify max or min.

\frac{d V}{d x} = 12 - 3 x^{2} = 0 \Rightarrow x^{2} = 4 \Rightarrow x = 2

(positive).

\frac{d ^{2} V}{d x ^{2}} = - 6 x = - 12 < 0

, so it is a maximum.

Answermaximum at

x = 2

Tip — Always state which solution is physically valid and confirm it is the max/min the question asks for.

Setting up the model

Often you must first build the quantity to optimise from the problem, using a constraint to reduce it to one variable. Then optimise as above. Reject impossible solutions (negative lengths, etc.).

Formula recap

\frac{d y}{d x} = rate of change

Interpretation in context.

\frac{d V}{d x} = 0

Condition for an optimum.

\frac{d ^{2} V}{d x ^{2}} ≶ 0 \Rightarrow max / min

Verify the nature.

Common mistakes to avoid

Finding the optimum x but not the actual maximum/minimum value.

Substitute back into the original model to get the optimal value.

Keeping a physically impossible solution (e.g. negative length).

Reject solutions that make no sense in context.

Key takeaways

dy/dx is a rate of change in applied problems.
Optimise by setting the derivative to zero and solving.
Confirm max/min with the second derivative; reject impossible solutions.

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