Optimisation problems ask for the largest or smallest possible value of a quantity under stated conditions. The calculus is often short. The difficult part is building the correct function, respecting its domain and interpreting the stationary point in the original problem.
1. The optimisation workflow
- Define the quantity to optimise.
- Use the constraints to express it in one variable.
- State the meaningful domain.
- Differentiate and find stationary candidates.
- Classify or compare candidates and endpoints.
- Answer in the original context with units.
2. Worked rectangle problem
A rectangle has perimeter 40 units. Find its maximum area.
Let length be x and width be y. Then 2x+2y=40, so y=20−x.
Area A=xy=x(20−x)=20x−x², with 0<x<20 for a nondegenerate rectangle.
dA/dx=20−2x. Set to zero: x=10. Then y=10 and A=100.
d²A/dx²=−2<0, so this is a maximum. The maximum area is 100 square units.
3. Why the square appears
The calculus result agrees with a broader geometric principle: among rectangles with fixed perimeter, the square has maximum area. Agreement with known structure is a useful check, not a substitute for the requested method.
4. Restricted intervals require endpoint checks
Suppose f(x)=x³−3x on −2≤x≤2. Stationary points satisfy 3x²−3=0, so x=±1.
Evaluate f at endpoints and stationary points:
- f(−2)=−2
- f(−1)=2
- f(1)=−2
- f(2)=2
On this closed interval, the absolute maximum is 2 and absolute minimum is −2. They occur at more than one allowed point.
5. Local versus absolute extrema
A local maximum is larger than nearby values. An absolute maximum is largest over the entire stated domain. A stationary point can be a local maximum without being the absolute maximum if the domain extends elsewhere.
6. Optimisation from a product constraint
Suppose positive x and y satisfy x+y=12. Maximise P=xy.
Write y=12−x. Then P=x(12−x)=12x−x².
P′=12−2x=0 gives x=6, hence y=6 and maximum product 36.
7. A minimum example
Find the minimum of f(x)=x+9/x for x>0.
f′(x)=1−9/x². Set to zero: x²=9, and x>0 gives x=3.
f″(x)=18/x³>0 for x>0, so x=3 gives a minimum. The minimum value is 3+3=6.
8. Domain is part of the model
In the previous example, x>0 is essential. Without it, the expression also has negative inputs and very different behaviour. In geometry, lengths must usually be positive; in time models, negative time may be excluded; in a closed interval, endpoints may be candidates.
9. Use exact values when possible
If a stationary point occurs at x=√3, keep √3 exact through substitution unless the question requests a decimal. Premature rounding can shift a maximum or minimum value and make later checking harder.
10. Geometry-to-function translation
Many optimisation problems begin with a diagram. Label variable lengths, write the constraint, eliminate extra variables and only then differentiate.
The derivative cannot repair an incorrect area, volume or perimeter model. Model construction is therefore the first mathematical checkpoint.
11. Classification choices
You can classify an interior stationary point using a derivative sign change or an appropriate second-derivative test. On a restricted closed interval, direct comparison of all candidates is often the clearest way to identify absolute extrema.
12. Check whether the answer is physically possible
If a calculated “maximum length” is negative, or an area exceeds a simple bounding rectangle, revisit the model or algebra. Context provides constraints that pure differentiation may not display automatically.
13. Common mistakes
- Differentiating before reducing the model to one independent variable.
- Finding a stationary point and declaring it a maximum without classification.
- Ignoring endpoints of a restricted domain.
- Reporting the variable value when the question asks for maximum area or minimum cost.
- Forgetting units.
- Keeping a stationary candidate outside the meaningful domain.
14. Independent practice
- A rectangle has perimeter 24. Find its maximum area.
- Positive x and y satisfy x+y=20. Maximise xy.
- Find the minimum of x+16/x for x>0.
- Find the absolute maximum and minimum of f(x)=x²−4x+1 on 0≤x≤5.
- Explain why endpoints must be checked in a closed-interval optimisation problem.
- A stationary candidate for a length problem is x=−3. What should you check before accepting it?
15. Answers
- Let sides x and 12−x. A=12x−x²; A′=12−2x=0 gives x=6. Maximum area 36.
- y=20−x; P=20x−x²; maximum at x=10, y=10, product 100.
- f′=1−16/x²=0 gives x=4; minimum value 8.
- f′=2x−4=0 gives x=2. Values: f(0)=1, f(2)=−3, f(5)=6. Absolute minimum −3; absolute maximum 6.
- An endpoint can be an absolute maximum or minimum even when the derivative is not zero there.
- Whether the domain permits a negative length; normally it does not.
16. What mastery looks like
Optimisation mastery is the ability to move through context → constraint → one-variable function → derivative → classification → interpretation without losing the original meaning of the quantities.
Continue with Stationary Points, Tangents and Normals, or return to the Additional Mathematics Hub.