IGCSE · Practice questions · Stationary points

Flat for a moment.

Extended. Six original Cambridge-style questions on stationary points: finding them, deciding their nature and using them to solve maximum and minimum problems.

Original questions All questions on this page are original work, written in the Cambridge IGCSE style. They are not from past papers. They test the same concepts and skills the syllabus rewards.

The method

Differentiate, set to zero, find \(y\), classify.

Differentiate: expand first if needed, then find \(\dfrac{dy}{dx}\).
Solve \(\dfrac{dy}{dx} = 0\); a cubic gives a quadratic, so look for two answers.
Coordinates: substitute each \(x\) into the original equation.
Classify: \(\dfrac{d^2y}{dx^2} > 0\) minimum, \(< 0\) maximum; or check the gradient either side.
In context: build a one-variable formula, reject impossible values, and answer the actual question.

Recall

[2 marks]

Find the coordinates of the stationary point of \(y = x^2 + 6x - 1\).

\(\dfrac{dy}{dx} = 2x + 6 = 0\), so \(x = -3\). ✓
\(y = 9 - 18 - 1 = -10\): the point is \((-3, -10)\). ✓

Calculation

[4 marks]

Find the coordinates of the two stationary points of \(y = x^3 - 6x^2 + 9x\).

\(\dfrac{dy}{dx} = 3x^2 - 12x + 9 = 3(x^2 - 4x + 3) = 3(x - 1)(x - 3)\). ✓
\(\dfrac{dy}{dx} = 0\) when \(x = 1\) or \(x = 3\). ✓
\(x = 1\): \(y = 1 - 6 + 9 = 4\). \(x = 3\): \(y = 27 - 54 + 27 = 0\). ✓
The points are \((1, 4)\) and \((3, 0)\). ✓

Analysis

[2 marks]

Use the second derivative to determine the nature of each stationary point of \(y = x^3 - 6x^2 + 9x\).

\(\dfrac{d^2y}{dx^2} = 6x - 12\). ✓
At \(x = 1\): \(-6 < 0\), so \((1, 4)\) is a maximum. At \(x = 3\): \(6 > 0\), so \((3, 0)\) is a minimum. ✓

Calculation

[3 marks]

Find the stationary points of \(y = 12x - x^3\) and state which is the maximum.

\(\dfrac{dy}{dx} = 12 - 3x^2 = 0\), so \(x^2 = 4\) and \(x = \pm 2\). ✓
\(x = 2\): \(y = 24 - 8 = 16\). \(x = -2\): \(y = -24 + 8 = -16\). ✓
\(\dfrac{d^2y}{dx^2} = -6x\) is \(-12\) at \(x = 2\), so \((2, 16)\) is the maximum (and \((-2, -16)\) the minimum). ✓

Analysis

[3 marks]

The curve \(y = x^4 - 8x^2\) has a stationary point at the origin.

By finding the gradient at \(x = -1\) and \(x = 1\), decide whether the origin is a maximum or a minimum.

\(\dfrac{dy}{dx} = 4x^3 - 16x\). ✓
At \(x = -1\): \(-4 + 16 = 12\), positive. At \(x = 1\): \(4 - 16 = -12\), negative. ✓
The gradient goes \(+, 0, -\), so the origin is a maximum. ✓

Analysis

[5 marks]

An open box is made from a square sheet of card 18 cm by 18 cm by cutting a square of side \(x\) cm from each corner and folding up the sides.

(a) Show that the volume is \(V = 4x^3 - 72x^2 + 324x\).

(b) Find the value of \(x\) that gives the greatest volume, and the greatest volume.

(a) Base \((18 - 2x)\) by \((18 - 2x)\), height \(x\): \(V = x(324 - 72x + 4x^2) = 4x^3 - 72x^2 + 324x\). ✓
(b) \(\dfrac{dV}{dx} = 12x^2 - 144x + 324 = 12(x^2 - 12x + 27) = 12(x - 3)(x - 9)\). ✓
\(x = 9\) uses the whole sheet (no base), so \(x = 3\). ✓
\(V = 3 \times 12^2 = 3 \times 144 = 432\ \text{cm}^3\). ✓
(c) \(\dfrac{d^2V}{dx^2} = 24x - 144 = 72 - 144 = -72 < 0\) at \(x = 3\), so it is a maximum. ✓

Mark this once you have attempted all six and checked your working. It records a Practiced badge on the topic and adds a one-time bonus. Revealing the solutions alone does not count.