Optimisation problemsEdexcel International A Level Maths: Revision notes

An optimisation problem asks for the greatest or least value of a quantity such as area, volume, cost or profit. The method is always the same:

1. Define the quantity to be optimised, e.g. $A$, $S$ or $P$.
2. Use the information given to write it as a function of one variable.
3. Differentiate and solve $\frac{dy}{dx}=0$.
4. Show whether the stationary point is a maximum or minimum.
5. Substitute back to find the optimum value, and answer in the context of the question, with units.

Practical problems start with two or more variables linked by a constraint, such as a fixed volume or a fixed length of fencing. Use the constraint to eliminate one variable. Example: a farmer has 80 m of fencing for three sides of a rectangle against a wall. With perpendicular sides $x$, the parallel side is $80-2x$, so $A=x(80-2x)=80x-2x^2$. Example: an open box with square base $x$ and volume $500$ has $h=\frac{500}{x^2}$, so $S=x^2+4xh=x^2+\frac{2000}{x}$. Rewrite $\frac{2000}{x}$ as $2000x^{-1}$ before differentiating.

A full answer proves the nature of the stationary point. Use $\frac{d^2y}{dx^2}$: negative means maximum, positive means minimum. For $A=80x-2x^2$: $\frac{dA}{dx}=80-4x=0$ gives $x=20$, and $\frac{d^2A}{dx^2}=-4<0$, so the area is a maximum, $A=800$ m$^2$. If $\frac{d^2y}{dx^2}=0$ or is awkward, check the sign of $\frac{dy}{dx}$ either side. When a question says 'show that the value is a minimum' you must show this explicitly; saying 'it must be a minimum' earns nothing.

The variable usually has a practical domain: lengths must be positive, and cutting equal squares of side $x$ from a $12$ cm square card needs $0<x<6$. Check that your stationary point lies inside the domain. If the question asks for the greatest value over a closed interval, also compare the values at the end-points. Always finish by answering what was asked: the value of $x$, the maximum value of $A$, or the number of items, with units and any rounding required (e.g. 3 s.f. or money to the nearest penny).

A square card of side $12$ cm has squares of side $x$ cm cut from each corner and the sides folded up to make an open box. Find the maximum volume. Base side $12-2x$, height $x$: $V=x(12-2x)^2=4x^3-48x^2+144x$, with $0<x<6$. $\frac{dV}{dx}=12x^2-96x+144=12(x-2)(x-6)=0$, so $x=2$ (since $x<6$). $\frac{d^2V}{dx^2}=24x-96=-48<0$ at $x=2$, so $V$ is a maximum. $V=2(8)^2=128$ cm$^3$.