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MATH1202 Chap.8 Curve Analysis and Optimisation

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Chapter 8 of 10 · MATH1202

Curve Analysis and Optimisation

This chapter turns derivatives into shape and decisions. First-derivative signs identify intervals of increase and decrease and classify turning points. Second derivatives describe concavity and provide a quick nature test when conclusive. Optimisation adds an objective, a constraint and a feasible domain, so stationary points must compete with endpoints and contextual restrictions.

Algebraic candidates are therefore only the beginning of the argument. Sign charts, endpoint values and the meaning of the variables determine which candidate is locally or absolutely optimal. A complete conclusion reports the decision, the resulting objective value and appropriate units.

In this chapter

What this chapter covers

  • 01

    Critical and stationary inputs

  • 02

    First-derivative sign charts

  • 03

    Local maximum and minimum

  • 04

    Second derivative and concavity

  • 05

    Constraint reduction

  • 06

    Endpoint and feasibility checks

Worked example · free

Maximise area under a perimeter constraint

Q [4 marks]. A rectangle has 40 metres of edging. Find the dimensions that maximise its area. The mark allocation shown here is not an official university marking scheme.
First derivative sign chartThe derivative changes from positive to negative at x equals one and from negative to positive at x equals five. x=1x=5 f′ > 0 ↗f′ < 0 ↘f′ > 0 ↗ local maximumlocal minimum
Nature comes from a sign change, not from a zero alone. Positive-to-negative derivative gives a local maximum; negative-to-positive gives a local minimum.
  • +1Define dimensions and reduce the perimeter constraint.
  • +1Write area as a one-variable function with a feasible domain.
  • +1Differentiate and solve for the stationary candidate.
  • +1Classify and state dimensions plus maximum area.
The maximum occurs for a 10 metre by 10 metre square, with area 100 square metres.
Sia tip — A stationary point is only a candidate until the domain and nature test approve it.
Glossary

Key terms

Stationary Point
A curve point where the first derivative is zero.
Local Maximum
A point whose nearby function values are not larger.
Concavity
The bending direction indicated by the second derivative sign.
Feasible Domain
The set of inputs permitted by the model and context.
FAQ

Curve Analysis and Optimisation FAQ

Does every stationary point give a turning point?

No. A derivative can equal zero without changing sign, producing a stationary inflection or another flat non-turning point. Use a sign chart or a conclusive second-derivative test to determine the nature.

Why must endpoints be checked?

On a closed feasible interval, an absolute maximum or minimum can occur at an endpoint even though the ordinary interior derivative test does not identify it. Evaluate endpoints alongside all feasible stationary candidates.

What should an optimisation conclusion include?

Name the decision input, the optimum objective value and the units. If the decision must be a whole number, compare nearby feasible integers and explain the implemented choice.

How does a first-derivative sign chart classify a point?

Factor the derivative, locate every critical input in the function's domain and test one value in each interval. A change from positive to negative gives a local maximum, while negative to positive gives a local minimum. No sign change means the stationary point is not a turning point, even though its derivative equals zero.

When is the second-derivative test inconclusive?

At a stationary point, a positive second derivative gives a local minimum and a negative value gives a local maximum. If the second derivative is zero or undefined, the test does not decide the nature. Return to a first-derivative sign chart or compare nearby function values rather than declaring an inflection automatically.

What is the role of a constraint in optimisation?

A constraint links the decision variables and limits feasible choices. Use it to rewrite the objective in one variable, then transfer all practical restrictions into that variable's domain. Candidates obtained after substitution must still satisfy the original constraint. Checking this relationship catches impossible dimensions, quantities or resource allocations.

How do local and absolute extrema differ?

A local extremum is best only among nearby inputs, while an absolute extremum is best over the entire stated domain. Stationary-point tests establish local behaviour. To establish an absolute result on a closed interval, evaluate every feasible stationary point and both endpoints, then compare their actual objective values.

Why should an optimisation variable be defined with units?

The definition determines what can vary, which values are feasible and how the objective should be interpreted. It prevents confusing a dimension with an area, a quantity with profit, or a continuous input with a whole-number decision. Carry the units through the constraint and objective, then report both the chosen input and resulting optimum in the original terms of the problem.

Study strategy

Exam move

Build sign charts from factored derivatives and test one simple point in each interval. Sketch the resulting rise and fall before looking at the original graph, then reconcile any disagreement. Use the second derivative only when its sign gives a conclusive classification, and practise one stationary inflection so zero does not become synonymous with turning point.

For applications, define the decision variable, objective and constraint separately, reduce to one variable and state the feasible domain before differentiating. List stationary points, endpoints and discrete neighbours in a candidate table with objective values and units. Check each candidate against the original constraint and practical restrictions.

Finish with a decision sentence containing both the chosen input and the optimum output, plus any required rounding rationale.

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