Optimisation Problems
Learn optimisation problems for NSW Year 12 Mathematics Advanced (NESA outcome MAV-12-06). Businesses, designers and engineers constantly ask for the largest area, the smallest cost or the shortest time, and calculus finds these best values exactly.
You will learn to define variables, write the quantity as a function of one variable with a sensible domain, find where its derivative is zero, confirm a maximum or minimum and check endpoints, then state a clear answer in context — real optimisation in action.
Theory
Optimisation uses calculus to find the best value. This Year 12 Mathematics Advanced topic (MAV-12-06) models a quantity as a function of one variable and finds its maximum or minimum.
Write the quantity as a function of one variable (use the constraint to eliminate the other), solve \(\dfrac{dQ}{dx}=0\), confirm max or min (with \(Q''\) or a sign test), check endpoints, and answer in context.
Method
- Express the quantity as a function of one variable.
- Differentiate and solve \(\dfrac{dQ}{dx}=0\).
- Confirm the nature and answer in context.
| \(A\) | \(=\) | \(x(10-x)=10x-x^{2}\) |
| \(10-2x\) | \(=\) | \(0\) |
| \(x\) | \(=\) | \(5\) |
| \(A''\) | \(=\) | \(-2<0\ (\text{max})\) |
| \(A_{\max}\) | \(=\) | \(25\ \text{m}^{2}\) |
| \(A\) | \(=\) | \(x(12-x)\) |
| \(x\) | \(=\) | \(6,\ A=36\ \text{m}^{2}\) |
Common pitfalls
Frequently asked questions
How do you solve an optimisation problem?
Write the quantity as a function of one variable using the constraint, differentiate, set the derivative to zero, confirm it is a max or min, and answer in context.
Why must you get to one variable?
Calculus can only optimise a function of a single variable, so the constraint is used to eliminate the other.
How do you confirm a maximum?
Use the second derivative (negative means a maximum) or test the sign of the first derivative either side.
Do you need to check endpoints in optimisation?
Yes, if the variable has a restricted range, the extreme value can occur at an endpoint.