Maximum & Minimum Problems

What this method is for

Optimisation questions ask for the largest or smallest possible value of something, the maximum area a fence can enclose, the minimum surface area of a container, the greatest volume of a box. This method uses differentiation to find that best value exactly, rather than by guessing.

The idea rests on turning points. At a maximum or minimum the curve momentarily flattens, so its gradient is zero: $\frac{dy}{dx}=0$ .

By expressing the quantity we care about as a single-variable function and solving $\frac{dy}{dx}=0$ , we locate the turning point, and the second derivative tells us whether it is a peak or a trough. In Add Math these questions carry good marks in the Differentiation chapter and often come dressed as a real-world scenario, so recognising the underlying structure is half the battle.

When to reach for it

Reach for this method whenever a question asks for a maximum or minimum value, the largest or smallest, the greatest area or volume, or the value that makes a cost 'as small as possible'. These words signal a turning point, and turning points come from $\frac{dy}{dx}=0$ .

A reliable clue is that the problem gives a constraint, a fixed length of fencing, a fixed perimeter, a set amount of material, alongside the quantity to optimise. That constraint is what lets you reduce two variables down to one.

If you can write the target quantity as a function of a single variable and the question wants its best value, this is exactly the method to use.

The method, step by step

Stationary point and its natureMust memorise

\frac{dy}{dx}=0 \;\text{ gives a turning point};\quad \frac{d^{2}y}{dx^{2}}<0 \Rightarrow \text{max},\;\; \frac{d^{2}y}{dx^{2}}>0 \Rightarrow \text{min}

1
Name the quantity
Write an expression for the thing to be maximised or minimised, such as area $A$ or volume $V$ .
2
Use the constraint
Use the given fixed condition to eliminate one variable, so the quantity depends on one variable only.
3
Differentiate
Find the first derivative of the single-variable expression.
4
Solve the derivative equals zero
Set the first derivative to $0$ and solve to find the value at the turning point.
5
Confirm the type
Find the second derivative; a negative value means a maximum, a positive value means a minimum.
6
State the answer
Substitute back to find the required maximum or minimum value, with units.

Worked example

Q1[6 marks]

A rectangular garden is to be fenced on three sides, with an existing wall forming the fourth side. The total length of fencing available is $12$ m.

Find the maximum area of the garden.

Show worked solution

Let the two sides perpendicular to the wall each have length $x$ m, and let the side parallel to the wall have length $y$ m.

The fencing covers the two $x$ -sides and the one $y$ -side, so the constraint is $2x+y=12$ , giving $y=12-2x$ .

The area is $A=xy$ . Substitute the constraint:

A=x(12-2x)=12x-2x^{2}

Differentiate and set equal to zero:

\frac{dA}{dx}=12-4x=0\;\Rightarrow\; x=3

Confirm it is a maximum using the second derivative:

\frac{d^{2}A}{dx^{2}}=-4<0 \;\Rightarrow\; \text{maximum}

When $x=3$ , $y=12-2(3)=6$ . The maximum area is $A=3\times 6=18$ m $^{2}$ .

Common mistakes to avoid

Trying to differentiate a two-variable expression. You must use the constraint to get down to one variable first.
Forgetting the second-derivative check, so a minimum is reported when a maximum was wanted, or vice versa.
Solving $\frac{dy}{dx}=0$ for $x$ and stopping, you still need to find the actual maximum or minimum value.
Setting up the constraint wrongly, for example counting all four sides when the wall replaces one of them.
Dropping units in the final answer, or giving a length when an area or volume was requested.

How one-to-one teaching helps

The step that decides these questions is the setup: translating the words into a quantity and a constraint, then reducing to one variable. Students who rush this end up differentiating the wrong expression and cannot recover.

In a one-to-one lesson our teachers slow down that first stage with you, labelling the diagram and writing the constraint before any calculus begins, so the differentiation is straightforward. Because the paper awards method marks, a clear $\frac{dy}{dx}=0$ line and a second-derivative check earn credit even under time pressure.

Teachers at spmaddmath.com.my are experienced, and lessons are online and taught in English. To practise optimisation with guided setups, message us on WhatsApp to arrange a one-hour paid class from RM50/hr at the teacher's rate.

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Frequently asked questions

Why does

\frac{dy}{dx}=0

give a maximum or minimum?

At a maximum or minimum the curve levels off for an instant, so its gradient is zero. Setting $\frac{dy}{dx}=0$ finds exactly those flat points, called stationary points, where a maximum or minimum can occur.

How do I tell a maximum from a minimum?

Use the second derivative. If $\frac{d^{2}y}{dx^{2}}<0$ at the stationary point, the curve bends downward, so it is a maximum.

If $\frac{d^{2}y}{dx^{2}}>0$ , the curve bends upward, so it is a minimum.

What is the role of the constraint in the problem?

The constraint is the fixed condition, such as a set length of fencing, that links your variables. It lets you replace one variable in terms of another so the quantity you want to optimise becomes a function of a single variable, which is what you can differentiate.

Do I always need the second derivative test?

It is the clearest way to confirm the nature of a turning point and is expected working. You can also test the sign of $\frac{dy}{dx}$ just before and after the point, but the second-derivative test is usually quicker and tidier for these questions.

Source:SRC-DSKP-EN

Written by the spmaddmath.com.my editorial team.· Last updated 5 September 2026

Solving maximum and minimum problems