Worked example 1
Find a stationary point
Find stationary points of y = x² − 6x + 8.
- dy/dx = 2x − 6
- Set 2x − 6 = 0, so x = 3
- Substitute into y.
Answer: Stationary point is (3, −1), a minimum.
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Higher Mathematics
Stationary points and optimisation
Find stationary points, classify them and solve simple optimisation problems.
Before you start
- Solve simple equations after differentiating.
- Substitute x-values back into a function.
- Know what maximum and minimum mean on a graph.
Explanation
Stationary points occur where dy/dx = 0. Their nature can be classified using sign changes, a nature table or the second derivative where appropriate.
Optimisation uses the same method but asks for the maximum or minimum value in a context.
Visual support
Method and rules
- Stationary point: dy/dx = 0
- If d²y/dx² > 0, local minimum; if d²y/dx² < 0, local maximum.
- For context questions, interpret the x-value and the function value.
Worked examples
Worked example 2
Classify with second derivative
For y = −x² + 4x + 1, classify the stationary point.
- dy/dx = −2x + 4, so x = 2
- d²y/dx² = −2
- Negative second derivative means maximum.
So: Maximum at (2, 5).
Worked example 3
Optimise an area
A model area is A = −x² + 10x. Find the maximum area.
- Differentiate: dA/dx = −2x + 10
- Set equal to zero: x = 5
- Substitute into A.
Answer: Maximum area is 25 square units.
Watch out
- Stopping after finding x but not y.
- Calling every stationary point a maximum.
- Ignoring the context or valid range in optimisation.
Exam reminder
For stationary points, give the full coordinate and nature when requested. For optimisation, interpret the maximum or minimum in the context and units of the question.