Optimisation Problems
Key Terms
- Identify variables
- define what you are maximising or minimising (the objective function) and any constraint linking the variables
- Use the constraint
- to eliminate one variable, expressing the objective function in terms of one variable only
- Determine the domain
- — realistic values of the variable (e.g. lengths must be positive)
- Find critical points
- set f′(x) = 0 and solve within the domain
- Confirm max or min
- using f″(x): if f″ < 0 at the critical point it is a maximum; if f″ > 0 it is a minimum
- Interpret the answer
- in context — state the optimal value with correct units
1. Define variables and write the objective function Q
2. Apply the constraint to write Q in terms of one variable
3. State the domain (e.g. 0 < x < 10)
4. Differentiate: find Q′(x), set Q′(x) = 0
5. Find Q″(x): if Q″ < 0 ⇒ maximum; if Q″ > 0 ⇒ minimum
6. Calculate the optimal value and interpret
Check endpoints: for a closed domain [a, b], also evaluate Q at x = a and x = b
Step 1 — Define variables:
Let width = x m, length = y m. Objective: maximise A = xy.
Step 2 — Constraint:
Perimeter: 2x + 2y = 40 ⇒ y = 20 − x
So A(x) = x(20 − x) = 20x − x²
Step 3 — Domain: x > 0 and 20 − x > 0 ⇒ 0 < x < 20
Step 4 — Differentiate:
A′(x) = 20 − 2x = 0 ⇒ x = 10
Step 5 — Confirm maximum:
A″(x) = −2 < 0 ⇒ maximum confirmed
Step 6 — Answer:
x = 10, y = 20 − 10 = 10. Max area = 10 × 10 = 100 m²
The rectangle is a square with side 10 m.
Step 1 — Set up:
After cutting, base = (30 − 2x) × (30 − 2x), height = x
V(x) = x(30 − 2x)²
Step 2 — Domain: x > 0 and 30 − 2x > 0 ⇒ 0 < x < 15
Step 3 — Expand and differentiate:
V(x) = x(900 − 120x + 4x²) = 900x − 120x² + 4x³
V′(x) = 900 − 240x + 12x² = 12(x² − 20x + 75) = 12(x − 5)(x − 15)
V′(x) = 0 ⇒ x = 5 or x = 15
Only x = 5 is in the domain (0, 15)
Step 4 — Confirm maximum:
V″(x) = 24x − 240; V″(5) = 120 − 240 = −120 < 0 ⇒ maximum
Answer: Cut squares of side x = 5 cm. Max volume = 5(20)² = 2000 cm³
What Is Optimisation?
Optimisation means finding the best possible value of some quantity — the maximum profit, the minimum cost, the greatest area, the shortest distance. These problems arise constantly in engineering, business, science and everyday life. Calculus gives us a systematic method: use derivatives to locate where the function reaches its maximum or minimum value.
At its heart, optimisation is about translating a real-world situation into a mathematical function, then finding the input value that produces the best output. The function we are maximising or minimising is called the objective function. Any equation relating the variables is called a constraint.
The Optimisation Strategy
Step 1: Identify and define variables. Read the problem carefully. Decide what quantity you are optimising (your objective function Q) and what variable controls it. Give each quantity a clear symbol and, if geometric, draw a diagram.
Step 2: Write the objective function. Express Q in terms of your variables. If there are two variables, you will need a constraint to eliminate one — this gives Q as a function of a single variable.
Step 3: Determine the domain. What values of the variable are physically meaningful? Lengths must be positive; percentages must be between 0 and 100; if cutting squares from a corner, the cut cannot exceed half the side length. Stating the domain explicitly is important for exam marks and for making sure you only consider valid solutions.
Step 4: Differentiate and find critical points. Solve Q′(x) = 0. Make sure every solution lies within the domain.
Step 5: Confirm maximum or minimum. Use the second derivative test: Q″(x) < 0 at a critical point confirms a local maximum; Q″(x) > 0 confirms a local minimum. Alternatively, use a sign table for Q′: if Q′ changes from positive to negative, you have a maximum; negative to positive gives a minimum.
Step 6: Calculate the optimal value and answer the question. Substitute the optimal x back to find Q. Don’t forget to state the answer in context with units.
Closed Domains and Endpoint Checking
When the domain is a closed interval [a, b], the global maximum or minimum might occur at an endpoint rather than at a critical point inside the interval. In such cases, evaluate Q at x = a, at x = b, and at each critical point within (a, b), then compare all values. The largest is the global maximum; the smallest is the global minimum.
For open domains (0, b) or (0, +∞), the only candidates are interior critical points. If the function has exactly one critical point and Q″ confirms it is a maximum (or minimum), that is the global optimum for the domain.
Common Problem Types
Area problems: maximise the area of a shape subject to a perimeter or material constraint. Set up area as the objective and use the perimeter equation to eliminate one dimension.
Volume problems: typically maximise the volume of a box, cylinder, or cone subject to a surface area constraint. Express volume in terms of one variable and differentiate.
Profit and revenue: a common economic application. A profit function P(x) = R(x) − C(x) is maximised where P′(x) = 0, i.e. marginal revenue equals marginal cost. These functions are usually quadratic or cubic.
Distance and time: find the shortest distance between a point and a curve, or the fastest route. Minimise the distance or time function, which often involves Pythagoras.
Number problems: find two numbers satisfying a sum or product condition that optimise some expression. These provide clean algebraic practice.
Interpreting Results Realistically
After finding the mathematical answer, ask: does this make sense? A negative length is impossible. An x-value outside the domain must be discarded. A maximum profit at x = −5 units sold is meaningless. Checking your answer against the context is both good mathematical practice and something examiners look for in problem-solving questions.
It is also worth noting that the second derivative test can fail if Q″ = 0 at the critical point. In that case, fall back on the sign table for Q′ to classify the stationary point.
Mastery Practice
- Fluency Find the maximum value of f(x) = −x² + 6x + 1 on the interval [0, 5].
- Fluency A farmer has 100 m of fencing for three sides of a rectangular pen; the fourth side is a wall. Find the dimensions that give the maximum area.
- Fluency Find two positive numbers with sum 20 whose product is a maximum.
- Fluency Find the minimum value of S = x + 36/x for x > 0.
- Understanding An open rectangular box with a square base has volume 32 m³. Find the dimensions that minimise the total surface area.
- Understanding A cylinder with no lid has volume 250π cm³. Find the radius that minimises the total surface area.
- Understanding A profit function is P(x) = −2x² + 120x − 800 dollars, where x is the number of units sold. Find the quantity that maximises profit and the maximum profit.
- Understanding A particle moves along a line with position x(t) = t³ − 6t² + 9t (metres) for 0 ≤ t ≤ 4. Find the maximum displacement from the origin in this interval.
- Problem Solving A 10 cm wire is cut into two pieces. One piece is bent into a circle, the other into a square. Where should the wire be cut to minimise the total area enclosed?
- Problem Solving Find the dimensions of the largest rectangle that can be inscribed in a semicircle of radius 5.