Finding Maxima and Minima with Calculus Revision Notes | GCSE AQA Higher Further Maths

1. Why Maxima and Minima Matter

Imagine you are designing a box to hold the maximum possible volume, or a company trying to minimise its costs. Both problems share the same mathematical structure: find the highest or lowest value a quantity can reach. Calculus gives you a precise, reliable method to do exactly that.

You already know how to differentiate polynomial functions to find the gradient at any point. The key insight here is that at a peak or a trough on a curve, the gradient is zero — the curve is momentarily flat. Setting $\frac{dy}{dx} = 0$ and solving is how you locate those special points.

2. Stationary Points: The Core Idea

There are two types of stationary point you need to know at GCSE: a maximum, where the curve peaks and then falls away, and a minimum, where the curve dips and then rises again. Both have gradient zero, so you find them the same way — but you classify them differently.

The curve above has a local maximum on the left (where the curve peaks) and a local minimum on the right (where it troughs). Both points have a horizontal tangent — that is the visual signature of $\frac{dy}{dx} = 0$.

3. The Four-Step Method

Every question asking you to find and classify stationary points follows the same four steps. Practise this sequence until it becomes automatic, because the method earns marks even if arithmetic slips occur later.

• Step 1 — Differentiate: find $\frac{dy}{dx}$.
• Step 2 — Set equal to zero: solve $\frac{dy}{dx} = 0$ to find the $x$-coordinate(s).
• Step 3 — Find $y$: substitute each $x$ back into the original equation.
• Step 4 — Classify: differentiate again to get $\frac{d^2y}{dx^2}$, then substitute the $x$-value and check the sign.

Worked example: Finding and Classifying Stationary Points

Find the coordinates of the stationary points of $y = x^3 - 3x^2 - 9x + 2$ and determine whether each is a maximum or a minimum.

1. $\frac{dy}{dx} = 3x^2 - 6x - 9$ — Bring down the power and reduce it by one for each term. The constant term disappears. [M1 — correct differentiation]
2. $3x^2 - 6x - 9 = 0$ $x^2 - 2x - 3 = 0$ $(x - 3)(x + 1) = 0$ $x = 3$ or $x = -1$ — Divide through by 3 first to simplify, then factorise. Both values of $x$ are stationary points. [M1 — setting to zero and solving]
3. When $x = 3$: $y = (3)^3 - 3(3)^2 - 9(3) + 2 = 27 - 27 - 27 + 2 = -25$ When $x = -1$: $y = (-1)^3 - 3(-1)^2 - 9(-1) + 2 = -1 - 3 + 9 + 2 = 7$ — Substitute each $x$ back into the original equation. Stationary points are $(3, -25)$ and $(-1, 7)$. [A1 — correct coordinates]
4. $\frac{d^2y}{dx^2} = 6x - 6$ — Differentiate $\frac{dy}{dx} = 3x^2 - 6x - 9$ again. [M1 — correct second derivative]
5. At $x = 3$: $\frac{d^2y}{dx^2} = 6(3) - 6 = 18 - 6 = 12 > 0$ → minimum At $x = -1$: $\frac{d^2y}{dx^2} = 6(-1) - 6 = -6 - 6 = -12 < 0$ → maximum — Positive second derivative means minimum; negative means maximum. [A1 — correct classification]

Answer: Maximum at $(-1, 7)$; minimum at $(3, -25)$.

4. Classifying with the Second Derivative

The second derivative $\frac{d^2y}{dx^2}$ tells you how the gradient itself is changing. At a maximum, the gradient is decreasing (going from positive to zero to negative), so $\frac{d^2y}{dx^2}$ is negative. At a minimum, the gradient is increasing (negative to zero to positive), so $\frac{d^2y}{dx^2}$ is positive.

A useful memory anchor: think of a hill. At the top (maximum), the slope is tilting downward — the gradient is getting more negative, so the second derivative is negative. At the bottom of a valley (minimum), the slope is tilting upward — the gradient is getting more positive, so the second derivative is positive.

5. A Quadratic Example — Simpler Case First

Before tackling cubics in exam questions, it helps to see the method on a quadratic, where there is only one stationary point. Quadratics always produce exactly one stationary point — the vertex of the parabola.

Worked example: Stationary Point of a Quadratic

Find the coordinates of the stationary point of $y = 2x^2 - 8x + 5$ and state whether it is a maximum or minimum.

1. $\frac{dy}{dx} = 4x - 8$ — Differentiate term by term: $2x^2 \to 4x$, $-8x \to -8$, constant $5 \to 0$. [M1]
2. $4x - 8 = 0$ $4x = 8$ $x = 2$ — Solve for $x$ to find the $x$-coordinate of the stationary point. [M1]
3. $y = 2(2)^2 - 8(2) + 5$ $= 2 \times 4 - 16 + 5$ $= 8 - 16 + 5$ $= -3$ — Substitute $x = 2$ into the original equation. Note: $2^2 = 2 \times 2 = 4$. [A1]
4. $\frac{d^2y}{dx^2} = 4$ Since $4 > 0$, the stationary point is a minimum. — The second derivative of $4x - 8$ is $4$, a positive constant. This confirms a minimum for all $x$. [A1]

Answer: Minimum at $(2, -3)$.

Notice that for a quadratic $y = ax^2 + bx + c$, the second derivative is always $2a$. When $a > 0$ (a U-shaped parabola), the second derivative is positive, confirming the single stationary point is always a minimum. When $a < 0$ (an inverted parabola), it is always a maximum.

6. Applying Calculus to Contextual Problems

Exam questions often embed the optimisation in a real-world context — a farmer fencing a field, a manufacturer minimising material costs, or a projectile reaching its peak height. The maths is identical; the extra skill is setting up the function correctly before differentiating.

Worked example: Contextual Optimisation Problem

A ball is thrown upward. Its height $h$ metres after $t$ seconds is given by $h = 20t - 5t^2$. Find the maximum height reached and the time at which it occurs.

1. $\frac{dh}{dt} = 20 - 10t$ — Treat $h$ as the dependent variable and $t$ as the independent variable. Differentiate term by term. [M1]
2. $20 - 10t = 0$ $10t = 20$ $t = 2$ — The ball reaches its maximum height when the rate of change of height is zero — it is momentarily stationary in the vertical direction. [M1]
3. $h = 20(2) - 5(2)^2$ $= 40 - 5 \times 4$ $= 40 - 20$ $= 20$ — Substitute $t = 2$ into the original equation. Note: $2^2 = 2 \times 2 = 4$. [A1]
4. $\frac{d^2h}{dt^2} = -10$ Since $-10 < 0$, this is a maximum. — The second derivative is negative, confirming the stationary point is a maximum — consistent with the ball reaching a peak. [A1]

Answer: Maximum height of $20$ metres, reached at $t = 2$ seconds.

The structure here is exactly the same as the pure maths examples. The only difference is that you interpret the answer in context: $t = 2$ seconds and $h = 20$ metres carry physical meaning, so always state units in your final answer.

7. Practice Problems

The following problems cover the full range of question types you will meet. Work through each one using the four-step method before checking the solution.

8. Exam Technique and Common Mistakes

This is where marks are most commonly dropped. The errors below are predictable — which means they are avoidable once you know what to watch for.

Finding maxima and minima with calculus comes down to one core skill: recognising that gradient zero means stationary, and that the second derivative's sign tells you which type. Master the four-step method, practise substituting carefully, and always state full coordinates with units where the context demands them.